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R.D. Sharma Solutions Class 10th: Ch 4 Triangles Exercise 4.6

February 4, 2019, 10:08 pm
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Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math Exercise 4.6

Exercise 4.6

1. Triangles ABC and DEF are similar
(i) If area (△ABC) = 16cm2, area (△DEF) = 25 cm2 and BC = 2.3 cm, find EF.
(ii) If area (△ABC) = 9cm2, area (△DEF) = 64 cm2 and DE = 5.1 cm, find AB.
(iii)If AC = 19cm and DF = 8 cm, find the ratio of the area of two triangles.
(iv)If area (△ABC) = 36cm2, area (△DEF) = 64 cm2 and DE = 6.2 cm, find AB.
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of △ABC and △DEF.

Solution

(i)

(ii) 

(iii) 

(iv) 

(v)

2. In fig. below ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB): area (∆APQ).
Solution


3. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians ?

Solution



4. The areas of two similar triangles are 169 cm2 and 121 cm2 respectively. If the longest side
of the larger triangle is 26 cm, find the longest side of the smaller triangle.

Solution


5. The areas of two similar triangles are 25 cm2 and 36 cm2 respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.

Solution



6. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. 

Solution


7. ABC is a triangle in which ∠A =90°, AN⊥ BC, BC = 12 cm and AC = 5cm. Find the ratio of the areas of ∆ANC and ∆ABC. 

Solution



8. In Fig. 4.178, DE || BC 
(i) If DE = 4 cm, BC = 6 cm and Area (∆ADE) = 16 cm2 , find the area of ∆ABC.
(ii) If DE = 4cm, BC = 8 cm and Area (∆ADE) = 25 cm2 , find the area of ∆ABC.
(iii)If DE : BC = 3 : 5. Calculate the ratio of the areas of ∆ADE and the trapezium BCED .

Solution


9.In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC . 

Solution

10. The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude the
bigger triangle is 5 cm, find the corresponding altitude of the other . 

Solution


11. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the
first triangle is 12.1 cm, find the corresponding median of the other.  

Solution


12. If ∆ABC ~ ∆DEF such that AB = 5 cm, area (∆ABC) = 20 cm2 and area (∆DEF) = 45 cm2 , determine DE.  

Solution


13. In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find BP/AB . 

Solution


14. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If  BC = 4.5 cm, find the length of QR. 

Solution


15. ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that area of ∆APQ is one- sixteenth of the area of ABC. 

Solution

16. If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a Point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE.

Solution




17. If ∆ABC and ∆BDE are equilateral triangles, where D is the mid-point of BC, find the ratio of areas of ∆ABC and ∆BDE. 

Solution


18. Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.

Solution


19. In Fig., 4.180, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect at O, prove that Area(∆ABC)/Area(∆DBC) = AO/DO . 

Solution


20. ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove
that: (i) ∆AOB and ∆COD (ii) If OA = 6 cm, OC = 8 cm,
Find :
(a) area(∆AOB)area(∆COD)
(b) area(∆AOD)area(∆COD) 

Solution


21. In ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.

Solution



22. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE): Area (∆ABC) = 3: 4.

Solution

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R.D. Sharma Solutions Class 10th: Ch 4 Triangles Exercise 4.7

February 5, 2019, 12:12 am
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Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math Exercise 4.7

Exercise 4.7

1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle. 

Solution

Side of traingles are:
AB = 3cm
BC = 4 cm
AC = 6 cm
Now, we know that the triangle can be right angled if it follows the pythagoras theorem.
∴ AB2 = 32 = 9
BC2 = 42 = 16
AC2 = 62 = 36
Since, AB2 + BC2≠ AC2
Then, by converse of Pythagoras theorem, Triangle is not a right angled.

2. The sides of certain triangles are given below. Determine which of them right triangles are.
(i) a = 7 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = l6 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm

Solution



3. A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point ?


Solution

4. A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building .  

Solution



5. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops . 

Solution

6. In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC. 

Solution


7. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach ?

Solution



8. Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. 

Solution

9. Using Pythagoras theorem determine the length of AD in terms of b and c shown in Fig. 4.221.

Solution


10. A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. 

Solution



11. ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ∆FBE =
108 cm2 , find the length of AC. 

Solution



12. In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC.

Solution

13 . In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that 
(i) AD = a√3
(ii) Area (∆ABC) =√3a2 

Solution



14. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. 

Solution

15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.  

Solution


16. Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

Solution


17.  In Fig., 4.221, ∠B < 90° and segment AD ⊥ BC, show that 
(i) b2 = h2 + a2 + x2 - 2ax 
(ii) b2 = a2 + c2 - 2ax  

Solution


18. In an equilateral ∆ABC, AD ⊥ BC, prove that AD2 = 3BD2 .  

Solution




19. △ABD is a right triangle right angled at A and AC ⟂ BD. Show that:
(i) AB2 = CB × BD 
(ii) AC2 = DC × BC
(iii) AD2 = BD × CD
(iv) AB2/AC2 = BD/DC  

Solution


(i)
 

(ii) 

(iii) 

(iv) 

20. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ? 

Solution


21. Determine whether the triangle having sides (a − 1) cm, 2√a cm and (a + 1) cm is a right - angled triangle.

Solution



22. In an acute-angled triangle, express a median in terms of its sides.  

Solution


23. In right-angled triangle ABC in which ∠C = 90°, if D is the mid-point of BC, prove that AB2 = 4 AD2 - 3 AC2 . 

Solution


24. In Fig. 4.223, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that: 
(i) b2 = p2+ax+a4
(ii) c2 = p2 - ax+a2/4
(iii) b2+c2 = 2p2 + a2/2 

Solution 

(i) 
(ii) 


(iii)

25. In ∆ABC, ∠A is obtuse, PB ⊥AC and QC ⊥ AB. Prove that:
(i) AB × AQ = AC × AP
(ii) BC2 = (AC × CP + AB × BQ)

Solution


26. In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that CD2 = 4(AD2 - AC2) .

Solution

27. In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2 , prove that ∠ACD = 90°.  

Solution


28. An aeroplane leaves an airport and flies due north at a speed of 1000km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1 hours ? 

Solution

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R.D. Sharma Solutions Class 10th: Ch 4 Triangles MCQ

February 5, 2019, 10:48 pm
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Chapter 4 triangles R.D. Sharma Solutions for Class 10th Math MCQ's

Multiple Choice Questions

1. Sides of two similar triangles are in the ratio 4:9 . Areas of these triangles are in the ratio. 
(a) 2:3 
(b) 4:9 
(c) 81:16 
(d) 16:81

Solution

Given: Sides of two similar triangles are in the ration 4:9
To find: Ratio of these triangles
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of corresponding sides.
ar(triangle 1)/ar(triangle 2) = (side 1/side 2)2 = (4/9)2
ar(triangle 1)/ar(triangle 2) = 16/81
Hence the correct answer is option d.

2. The areas of two similar triangles are in respectively 9 cm2 and 16 cm2. The ratio of their corresponding sides is
(a) 3:4
(b) 4:3
(c) 2:3
(d) 4:5

Solution

Given: Areas of two similar triangles are 9 cm2 and 16 cm2.
To find : Ratio of their corresponding sides .
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
ar(triangle1)/ar(triangle2) = (side 1/side 2)2 
9/16 = (side1/side2)2
Taking square root on both sides, we get
side1/side2 = 3/4
So, the ratio of their corresponding sides is 3:4
Hence the correct answer is a.

3. The areas of two similar triangles △ABC and △DEF are 144 cm2 and 81 cm2 respectively . If longest sides of larger △ABC be 36 cm, then the longest side of the smallest triangle △DEF is
(a) 20 cm
(b) 26 cm
(c) 27 cm
(d) 30 cm

Solution

Given : Areas of two similar triangles △ABC and △DEF are 144 cm2 and 81 cm2 .
If the longest side of larger △ABC is 36 cm .
To find : the longest side of the smaller triangle △DEF
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of theirs corresponding sides.
ar(△ABC)ar(△DEF) = (longest side of larger △ABC/ longest side of smaller △DEF)2
144/81 = (36/ longest side of smaller △DEF)2
Taking square root on both sides, we get
12/9 = 36/longest side of smaller △DEF
Longest side of smaller △DEF = 36 × 9/12 = 27 cm
Hence the correct answer is c .

4. △ABC and △BDE are two equilateral triangles such that D is the mid – point of BC , The ratio of the areas of triangle ABC and BDE is
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:4

Solution

Given △ABC and △BDE are two equilateral triangles such that D is the midpoint of BC.
To find : ratio of areas of △ABC and △BDE . 

△ABC and △BDE are equilateral triangles ; hence they are similar triangles .
Since D is the midpoint of BC, BD = DC
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides.
ar(△ABC)ar(△BDE) = (BC/BD)2
ar(△ABC)ar(△BDE) = (BD + DC/BD)2 [D is the midpoint of BC]
ar(△ABC)ar(△BDE) = (BD + DC/BD)2
ar(△ABC)ar(△BDE) = (2BD/BD)2
ar(△ABC)ar(△BDE) = 4/1
Hence the correct answer is c .

5. If △ABC and △DEF are similar such that 2AB = DE and BC = 8 cm, then DF =
(a) 16 cm
(b) 12 cm
(c) 8 cm
(d) 4 cm

Solution

Given : △ABC and △DEF are similar such that 2AB = DE and BC = 8 cm .
To find : EF
We know that if two triangles are similar then there sides are proportional .
Hence, for similar triangles △ABC and △DEF
AB/DE = BC/EF = CA/FD
AB/DE = BC/EF
1/2 = 8/EF
EF = 16 cm
Hence the correct answer is a .
   
6. In △ABC and △DEF are two triangles such that AB/DE = BC/EF = CA/FD = 2/5, then Area (△ABC) : Area (△DEF) =
(a) 2:5
(b) 4: 25
(c) 4:15
(d) 8:125

Solution

Given : △ABC and △DEF are two triangles such that AB/DE = BC/EF = CA/FD = 2/5 .
To find : ar(△ABC) : ar(△DEF)
We know that if the sides of two triangles are proportional, then the two triangles are similar.
Since AB/DE = BC/EF = CA/FD = 2/5, therefore, △ABC and △DEF are similar .
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
ar (△ABC)/ ar(△DEF) = AB2/DE2
ar (△ABC)/ar(△DEF) = 22/52
ar (△ABC)/ar(△DEF) = 4/25
Hence the correct answer is b .

7. XY is drawn parallel to the base BC of a △ABC Cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =
(a) 2 cm
(b) 4 cm
(c ) 6 cm
(d) 8 cm

Solution

Given : XY is drawn parallel to the base BC of a △ABC Cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm.
To find : AY    
In △AXY and △ABC ,
∠AXY = ∠B (Corresponding angles)
∠A = ∠A (common)
∴ △AXY ~ △ABC (AA similarity)
We know that if two triangles are similar , then their sides are proportional .
It is given that AB = 4BX = X.
Then, AX = 3X
AX/BX = AY/ YC
3x/1x = AY/2
AY = 3x × 2/1x
AY = 6 cm
Hence the correct answer is c .

8. Two poles of height 6 m and 11 m stand vertically upright on a plane ground . If the distance between their foot is 12 m, the distance between their tops is
(a) 12 m
(b) 14 m
(c) 13 m
(d) 11 m

Solution

Given : Two poles of heights 6m and 11m stand vertically upright on a plane ground . Distance between their foot is 2 m.
To find : Distance between their tops.
Let CD be the pole with height 6m .
AB is the pole with the height 11 m, distance between their foot i.e. DB is 12 m .
Let us assume a point EE on the pole AB which is 6 m from the base of AB.
Hence
AE = AB – 6 = 11 – 6 = 5 m
Now in right triangle AEC, Applying Pythagoras theorem
AC2 = AE2 + EC2
AC2 = 52 + 122 (Since CDEB forms a rectangle and opposite sides of rectangle are equal )
AC2 = 25 + 144
AC2 = 169
AC = 13 m
Thus, the distance between their tops is 13m.
Hence correct answer is c.

9. In △ABC, D and E are points on side AB and AC respectively such that DE | | BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =
(a) 1.1 cm
(b) 4 cm
(c) 4.4 cm
(d) 5.5 cm

Solution

Given : In △ABC , D and E are points on the side AB and AC respectively such that DE | | BC and AD : DB = 3:1. Also , EA = 3.3 cm
To find : AC
In △ABC, DE | | BC,
Using corollary of basis proportionality theorem, we have
AD/AB = EA/AC
AD/AD+BD = 3.3/AC
AD/AD + 1/3 AD = 3.3/AC
EC = 4.4 cm
Hence the correct answer is C .

10. In triangle ABC and DEF , ∠A = ∠E = 40° , AB : ED = AC : EF and ∠F = 65° , then ∠B =
(a) 35°
(b) 65°
(c) 75°
(d) 85°

Solution

Given : In △ABC and △DEF
∠A = ∠E = 40°
AB : ED = AC : EF
∠F = 65°
To find : Measure of angle B .
In △ABC and △DEF
∠A = ∠E = 40°
AB : ED = AC : EF
△ABC ~ △DEF (S.A.S Similarity crieteria)
Hence in similar triangles △ABC and △DEF 
∠A = ∠E = 40°
∠C = ∠F = 65°
∠B = ∠D
We know that sum of all angles of a triangle is equal to 180° .
∠A + ∠B + ∠C = 180°
40° + ∠B + 65° = 180°
∠B + 115° = 180°
∠B = 180° - 115°
∠B = 75°
Hence the correct answer is c .

11.If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 83° , then ∠C =
(a)50°
(b)60°
(c)70°
(d)80°

Solution 

Given: If  △ABC  and △DEF are similar triangles such that
∠A = 47°
 ∠E = 83°
To find : Measure of angle C
In similar △ABC  and △DEF ,
∠A = ∠D = 47° 
∠B = ∠E = 83°
∠C = ∠F
We know that sum of all the angles of a triangle is equal 180° .
∠A + ∠B + ∠C = 180°
47° + 83° + ∠C = 180°
∠C + 130° = 180°
∠C = 180° - 130°
∠C = 50°
Hence the correct answer is b .

12. If D,E,F are the mid-points of sides BC,CA and AB respectively △ABC, then the ratio of triangles DEF and ABC is
(a) 1:4
(b) 1:2
(c) 2:3
(d) 4:5

Solution

Given : In △ABC, D, E and F are the midpoints of BC, CA, and AB respectively .
To find : Ratio of the areas of △DEF and △ABC
Since it is given that D and, E are the midpoints of BC, and AC respectively.
Therefore DE | | AB, DE | | FA …(1)
Again it is given that D and , F are the midpoints of BC, and, AB respectively .
Therefore, DF | | CA, DF | | AE...(2)
From (1) and (2) we get AFDE is a parallelogram.
Similarly we can prove that BDEF is a parallelogram
Now, in △ADE and △ABC
∠FDE = ∠A (Opposite angles of | |gm AFDE)
∠DEF = ∠B (Opposite angles of | |gm BDEF)
⇒ △ABC ~ △DEF (AA similarity criterion) 
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
ar(△DEF)/ar(△ABC) = (DE/AB)2
ar(△DEF)/ar(△ABC) = [(1/2(AB))/AB] (Since DE = 1/2 AB)
ar(△DEF)/ar(△ABC) = (1/4)
Hence the correct option is a .

13. In a △ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm . If AD ⊥ BC, then AD =
(a) 13/2 cm
(b) 60/13 cm
(c) 13/60 cm
(d) 2√15/13 cm

Solution

Given : In △ABC ∠A = 90°, AD ⊥ BC, AC = 12 cm, and AB = 5cm
To find : AD
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
In ∠ACB and ∠ADC ,
∠C = ∠C  (Common)
∠A = ∠ADC =  90° 
∴ △ACB ~ △ADC (AA Similarity)
AD/AB = AC/BC 
AD = AB × AC/BC 
AD = 12×5/13
AD = 60/13 
We got the result as b .  

14. If △ABC is an equilateral triangle such that AD ⊥ BC, then AD2 = 
(a) (3/2) DC2
(b) 2 DC2 
(c) 3 CD2 
(d) 4 DC2

Solution

Given : In an equilateral △ABC, AD ⊥ BC .
Since AD ⊥ BC, BD = CD = BC/2

In △ADC
AC2 = AD2 + DC2
BC2 = AD2 + DC2 (Since AC = BC)
(2DC)2 = AD2 + DC2 (Since BC = 2DC)
4DC2 = AD2 + DC2
3DC2 = AD2
3CD2 = AD2
We got the result as c .

15. In a △ABC , AD is the bisect of ∠BAC . If AB = 6 cm , AC = 5 cm and BD = 3 cm, then DC =
(a) 11.3 cm
(b) 2.5 cm
(c) 3:5 cm
(d) None of these

Solution

Given : In a △ABC, AD is the bisector of ∠BAC . AB = 6 cm and AC = 5 cm and  BD = 3 cm.
To find : DC 
We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle , 
Hence , 
AB/AC = BD/DC
6/5 = 3/DC
DC = 5×3/6
DC = 2.5 cm
Hence we got the result b.

16. In △ABC, AD is the bisector of ∠BAC . If AB = 8 cm, BD = 6 cm and DC = 3 cm . Find AC
(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 8 cm

Solution

Given: In a △ABC, AD is the bisector of ∠BAC . AB = 8 cm, and DC =3 cm and BD = 6 cm .
To find : AC
We know that the internal bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle .
AB/AC = BD/DC
8/AC = 6/3
AC = 8×3/6
AC = 4 cm
Hence we got the result a.

17. ABCD is a trapezium such that BC | | AD and AD = 4 cm . If the diagonals AC and BD intersect at O such that AO/OC = DO/OB = 1/2 , then BC =
(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 6 cm  

Solution

Given : ABCD is a trapezium in which BC | | AD and AD = 4 cm
The diagonals AC and BD intersect at O such that AO/OC = DO/OB = 1/2
To find : DC  

In △AOD and △COB
∠OAD = ∠OCB (Alternate angles)
∠ODA = ∠OBC (Alternate angles)
∠AOD = ∠BOC (Vertically opposite angles)
So, △AOD ~ △COB (AAA similarity)
Now, corresponding sides of similar △’s are proportional .
AO/CO = DO/BO = AD/BC
⇒ 1/2 = AD/BC
⇒ 1/2 = 4/BC
⇒ BC = 8 cm
Hence the correct answer is b .

18. If ABC is a right – angled at B and M, N are the mid – points of AB and BC respectively, then 4(AN2 + CM2) = 
(a) 4 AC2
(b) 5 AC2
(c ) 5/4 AC2
(d) AC2

Solution


M is the mid – point of AB.
∴ BM = AB/2
N is the mid – point of BC .
∴ BN = BC/2
Now,
AN2 + CM2 = (AB2 + (1/2 BC)2) + ((1/2 AB)2 + BC2
= AB2 + 1/4 BC2 + 1/4 AB2 + BC2
= 5/4 (AB2 + BC2)
⇒ 4(AN2 + CM2) = 5 AC2
Hence option b is correct .

19. If in △ABC and △DEF , AB/DE = BC/FD , then △ABC ~ △DEF when
(a) ∠A = ∠F
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠B = ∠E 


Solution


Given : In △ABC and △DEF , AB/DE = BC/FD .
We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal , then the two triangles are similar .
Then ,
∠B = ∠D

Hence, △ABC is similar to △DEF , we should have ∠B = ∠D
Hence the correct answer is c.

20. If in two triangles ABC and DEF , AB/DE = BC/FE = CA/FD , then
(a) △FDE ~ △CAB
(b) △FDE ~ △ABC
(c) △CBA ~ △FDE
(d) △BCA ~ △FDE

Solution

We know that if two triangles are similar if their corresponding sides are proportional .

It is given that △ABC and △DEF are two triangles such that AB/DE = BC/EF = CA/FD .
∠A = ∠D
∠B = ∠E
∠C = ∠F
∴ △CAB ~ △FDE
Hence the correct answer is a .

21. △ABC ~ △DEF, ar(△ABC) = 9 cm2 . If BC = 2.1 cm, then the measure of EF is
(a) 2.8 cm
(b) 4.2 cm
(c) 2.5 cm
(d) 4.1 cm

Solution

Given : Ar(△ABC) = 9 cm2 , Ar(△DEF) = 16 cm2 and BC = 2.1 cm
To find : measure of EF
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
Ar(△ABC)/Ar(△DEF) = BC2/EF2
9//16 = 2.12
3/4 = 2.1/EF
EF = 2.8 cm
Hence the correct answer is a .

22. The length of the hypotenuse of an isosceles right triangle whose one side is 4√2 cm is
(a) 12 cm
(b) 8 cm
(c) 8√2 cm
(d) 12√2 cm

Solution

Given : One side of isosceles right triangle is 4√2 cm
To find : Length of the hypotenuse .
We know that in isosceles triangle two sides are equal . 

In isosceles right triangle ABC , let AB and AC be the two equal sides of measures 4√2 cm .
Applying Pythagoras theorem, we get
BC2 = AB2 + AC2
BC2 = (4√2)2 + (4√2)2
BC2 = 32 + 32
BC2 = 64
BC = 8 cm
Hence correct answer is b .

23. A man goes 24 m due west and then 7 m due north. How far is he from the starting point ?
(a) 31 m
(b) 17 m
(c) 25 m
(d) 26 m

Solution

A man goes 24 m due to west and then 7 m due north.
Let the man starts from point B and goes 24 m due to west and reaches point A, then walked 7 m north and reaches point C.
Now we have to find the distance between the starting point and the end point i.e. BC.

In right triangle ABC , applying Pythagoras theorem we, get
BC2 ,= AB2 + AC2
BC2 = (24)2 + (7)2
BC2 = 576 + 49
BC2 = 625
BC = 25 m
Hence correct answer is c .

24. △ABC ~ △DEF . If BC = 3 cm , EF = 4 cm and ar(△ABC) = 54 cm2 , then ar(△DEF) = (a) 108 cm2
(b) 96 cm2
(c) 48 cm2
(d) 100 cm2

Solution

Given : In △ABC and △DEF
△ABC ~△DEF
Ar(△ABC ) = 54 cm2 BC = 3 cm and EF = 4 cm
To find : Ar(△DEF)
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
Ar(△ABC )/Ar(△DEF) = BC2/EF2
54/Ar(△DEF) = 32/42
54/ Ar(△DEF) = 9/16
Ar(△DEF) = 16×54/9
Ar(△DEF) = 96 cm2
Hence the correct answer is b .

25. △ABC ~△PQR such that ar(△ABC) = 4 ar(△PQR). If BC = 12 cm, then QR =
(a) 9 cm
(b) 10 cm
(c) 6 cm
(d) 8 cm 

Solution

Given : In △ABC and △PQR
△ABC ~△PQR
Ar(△ABC ) = 4Ar (△PQR)
BC = 12 cm
To find : Measure of QR
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .
Ar(△ABC )/Ar(△PQR) = BC2/QR2
4 Ar(△PQR)/ Ar(△PQR) = 122/QR2 (Given Ar(△ABC ) = 4 Ar(△PQR))
4/1 = 122/QR2
2/1 = 12/QR
QR = 6 cm
Hence the correct answer is c .

26. The areas of two similar triangles are 121 cm2 and 64cm2 respectively .If the median of the first triangle is 12.1 cm, then the corresponding median of the other triangle is
(a) 11 cm
(b) 8.8 cm
(c) 11.1 cm
(d) 8.1 cm

Solution

Given : The area of two similar triangles is 121 cm2 and 64 cm2 respectively . The median of the first triangle is 2.1 cm . To find : Corresponding medians of the other triangle
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians .
ar(triangle1)/ar(triangle2) = (median1/median2)2
121/64 = (12.1/median2)2
Taking square root on both side, we get
11/8 = 12.1 cm/median2
⇒ median2 = 8.8 cm
Hence the correct answer is b .

27. In an equilateral triangle ABC if AD ⊥ BC, then AD2 = 
(a) CD2
(b)2CD2
(c) 3CD2
(d) 4CD2

Solution

In an equilateral △ABC , AD ⊥ BC .

In △ADC, applying Pythagoras theorem , we get
AC2 = AD2 + DC2
BC2 = AD2 + DC2 (∵BC = 2DC)
4DC2 = AD2 + DC2
3DC2 = AD2
Hence the correct option is c.

28. In an equilateral triangle ABC if AD ⊥ BC, then 
(a) 5AB2 = 4AD2
(b) 3AB2 = 4AD2
(c) 4AB2 = 3AD2 
(d) 2AB2 = 3AD2

Solution

△ABC is an equilateral triangle and AD ⊥ BC .
In △ABD, applying Pythagoras theorem, we get
AB2 = AD2 + BD2
AB2 = AD2 + (1/2BC)2 (∵ BD = 1/2 BC)
AB2 = AD2 + (1/2 AB)2 (∵ AB = BC)
AB2 = AD2 + 1/4 AB2
3AB2 = 4AD2
We got the result as b .   

29. △ABC is an isosceles triangle in which ∠C = 90 . If AC = 6 cm, then AB =
(a) 6 √2 cm
(b) 6 cm
(c) 2√6 cm 
(d) 4√2 cm

Solution

Given : In an isosceles △ABC , ∠C = 90° . If AC = 6 cm .
To find : AB
In an isosceles △ABC, ∠C = 90° .
Therefore , BC = AC = 6 cm 
 Applying Pythagoras theorem in △ABC , we get
AB2 = AC2 + BC2
AB2 = 62 + 62 (AC = BC, sides of isosceles triangle)
AB2 = 36 + 36
AB2 = 72
AB = 6√2 cm
We got the result as a.

30. If in two triangle ABC and DEF , ∠A = ∠E, ∠B = ∠F, then which of the following is not true?
(a) BC/DF = AC/DE
(b) AB/DE = BC/DF
(c) AB/EF = AC/DE
(d) BC/DF = AB/EF 

Solution


In △ABC and △DEF
∠A = ∠E
∠B = ∠F
∴ △ABC and △DEF are similar triangles.
Hence AB/EF = BC/FD = CA/DE
Hence the correct answer is b .

31 . In the given figure the measure of ∠A and ∠F are respectively
(a) 50°, 40°
(b) 20°, 30°
(c) 40°, 50°
(d) 30°, 20°

Solution

△ABC and △DEF , 
AB/AC = EF/ED
∠A = ∠E = 130°
△ABC ~ △EFD (SAS Similarity)
∴ ∠F = ∠B = 30°
∠D = ∠C = 20°
Hence the correct answer is b . 

32. In the given figure , the value of x for which DE | | AB is 
(a) 4
(b) 1
(c) 3
(d) 2

Solution 

Given : In △ABC , DE | | AB .
To find : the value x
According to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In △ABC , DE | | AB .
AD/DB = AE/EC
x+3/3x +19 = x/3x+4
(x+3)(3x+4) = (x)(3x+19)
3x2 + 4x + 9x + 12 = 3x2 + 19x
19x – 13x = 12
6x = 12
x = 2
Hence we got the result d .

33. In the given figure, if ∠ADE = ∠ABC, then CE = 
(a) 2
(b) 5
(c) 9/2
(d) 3 

Solution

Given: ∠ADE = ∠ABC
To find: The value of CE
Since ∠ADE = ∠ABC
∴ DE | | BC (Two lines are parallel if the corresponding angles formed are equal)
According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
AD/DB = AE/EC
2/3 = 3/EC
EC = 3×3/2
EC = 9/2
Hence we got the result c .

34. In the given figure, RS | | DB | | PQ. If CP = PD = 11 cm and DR = RA = 3 cm . Then the values of x and y are respectively . 
(a) 12,10 
(b) 14,6
(c) 10,7
(d) 16,8


Solution

Given : RS | | DB | | PQ . CP = PD = 11 cm and DR = RA = 3 cm
To find : the value of x and y respectively .
In △ASR and △ABD ,
∠ASR = ∠ABQ (Corresponding angles)
∠A = ∠A (Common)
∴ △ASR ~ △ABD (AA Similarity)
AR/AD = AS/AB = RS/DB
3/6 = RS/DB
1/2 = x/y
x = 2y
This relation is satisfied by option d.
Hence, x = 16 cm and y =8 cm
Hence the result is d .

35. In the given figure, if PB | | CF and DP | | EF, then AD/ DE =
(a) 3/4
(b) 1/3
(c) 1/4
(d) 2/3 

Solution

Given : PB | | CF and DP | | EF . AB = 2 cm and AC = 8 cm .
To find : AD:DE
According to basic proportionality theorem, if a line is drawn parallel to one side of a triangle intersecting the other two sides . then it divides the two sides in the same ratio .
AB/BC = AP/PF
AP/PF = 2/8-2
AP/PF = 2/6
AP/PF = 1/3 …(1)
Again, DP | | EF .
AD/DE = AP/PF
AD/DE = 1/3
Hence we got the result option b .

36. A chord of a circle of radius 10 cm subtends a right at the centre. The length of the chord (in cm) is
(a) 5 √2
(b) 10√2
(c) 5 /√2
(d) 105√3

Solution

In right ΔOAB,
AB2 = OA2 + OB2 (Pythagoras theorem)
⇒ AB2 = (10)2 + (10)2 (OA = OB = 10 cm)
⇒ AB2 = 100 + 100 = 200
⇒ AB = √200 = 10√2 cm
Thus, the length of the chord is 10√2 cm
Hence the correct answer is option B.

37. A vertical stick 20 m long casts a shadow 10 m long on the ground . At the same time , a tower casts a shadow 50 m long on the ground . The height of the tower is
(a) 100 m
(b) 120 m
(c) 25 m
(d) 200 m 

Solution

Given : Vertical stick 20 m long casts a shadow 10 m long on the ground . At the same time a tower casts the shadow 50 m long on the ground.
To determine : Height of the tower
Let AB be the vertical stick and AC be its shadow . Also , let DE be the vertical tower and DF be its shadow.
Join BC and EF.

In ΔABC and ΔDEF , we have
∠A = ∠D = 90°
∠C = ∠F
ΔABC ~ ΔDEF
We know that in any two similar triangles , the corresponding sides are proportional . Hence ,
AB/DE = AC/DF
20/DE = 10/50
DE = 100 m
Hence the correct answer is option a .

38. Two isosceles triangles have equal angles and their areas are in the ratio 16:25. The ratio of their corresponding heights is
(a) 4:5
(b) 5:4
(c) 3:2
(d) 5:7 

Solution

Given two isosceles triangle have equal vertical angles and their areas are in the ratio of 16:25 .
To find : ratio of their corresponding heights .


Let ΔABC and ΔPQR be two isosceles triangles such that ∠A = ∠P . Suppose AD ⊥ BC and PS ⊥ QR .
In ΔABC and ΔPQR ,
AB/PQ = AC/PR
∠A = ∠P
∴ ΔABC ~ ΔPQR (SAS similarity)
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes .
Hence,
Ar(ΔABC)/Ar(ΔPQR) = (AD/PS)2
⇒ 16/25 = (AD/PS)2
⇒ AD/PS = 4/5
Hence we got the result as a .

39. ΔABC is such that AB = 3 cm , BC = 2 cm and CA = 2.5 cm . If ΔDEF ~ ΔABC and EF = 4 cm , then perimeter of ΔDEF is 
(a) 7.5 cm
(b) 15 cm
(c) 22.5 cm
(d) 30 cm 

Solution

Given : In ΔABC , AB = 3 cm, BC = 2 cm , CA = 2.5 cm . ΔDEF ~ ΔABC and EF = 4 cm 
To find : Perimeter of ΔDEF .
We know that if two triangles are similar , then their sides are proportional
Since ΔABC and ΔDEF are similar ,
AB/DE = BC/EF = CA/FD
3/DE = 2/4 = 2.5/FD
3/DE = 2/4
DE = 6 cm …(1)
2/4 = 2.5/FD
FD = 5 cm …(2)
From (1) and (2) , we get
Perimeter of ΔDEF = DE + EF + FD = 6 + 4 + 5 = 15 cm
Hence the correct answer is b .

40. In ΔABC , a line XY parallel to BC cuts AB at X AC at Y . If BY bisects ∠XYC , then
(a) BC = CY
(b) BC = BY
(c) BC ≠ CY
(d) BC ≠ BY

Solution

Given XY | | BC and BY is bisector of ∠XYC .

Since XY | | BC
So ∠YBC = ∠BYC (Alternate angles)
Now, in triangles BYC two angles are equal . Therefore , the two corresponding sides will be equal
Hence, BC = CY
Hence option a is correct.

41. In a ΔABC , ∠A = 90°, AB = 5 cm and AC = 12 cm . If AD ⊥ BC , then AD =
(a) 13/2 cm
(b) 60/13 cm
(c) 13/60 cm
(d) 2√15/13 cm

Solution

Given : In ΔABC ∠A = 90°, AD ⊥ BC , AC = 12 cm, and AB = 5 cm .
To find : AD
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides .

In ΔACB and ΔADC ,
∠C = ∠C (Common)
∠A = ∠ADC = 90°
∴ ΔACB ~ ΔADC (AA Similarity)
AD/AB = AC/BC
AD = AB × AC/BC
AD = 12×5/13
AD = 60/13
We got the result as b .

42. In a ΔABC, perpendicular AD from A and BC meets BC at D . If BD = 8 cm, DC = 2 cm and AD = 4 cm, then
(a) ΔABC is isosceles
(b) ΔABC is equilateral
(c) AC = 2AB
(d) ΔABC is right – angled at A

Solution

Given : In ΔABC, AD ⊥ BC , BD = 8 cm, DC = 2 cm and AD = 4 cm .

In ΔADC ,
AC2 = AD2 + DC2
AC2 = 42 + 22
AC2 = 20 …..(1)
Similarly , in ΔADB
AB2 = AD2 + DC2
AC2 = 42 + 82
AB2 = 80 …..(2)
Now, in ΔABC
BC2 = (CD + DB)2 = (2+8)2 = (10)2 = 100
and
AB2 + CA2 = 80 + 20 = 100
∴ AB2 + CA2 = BC2
Hence, triangle ABC is right angled at A .
We got the result as d .

43 . In a ΔABC, point D is on side AB and point E is on side AC , such that BCED is a trapezium . If DE : BC = 3:5, then Area (ΔADE) : Area(ΔBCED) =
(a) 3:4
(b) 9:16
(c) 3:5
(d) 9:25 

Solution

Given : In ΔABC,D is on side AB and point E is on side AC, such that BCED is a trapezium . DE : BC = 3:5.
To find : Calculate the ratio of the areas of ΔADE and the trapezium BCED .
In ΔADE and ΔABC ,
∠ADE = ∠B (Corresponding angles)
∠A = ∠A (Common)
∴ ΔADE ~ ΔABC (AA Similarity) .

We know that
Ar(ΔADE)/Ar(ΔABC) = DE2/BC2
Ar(ΔADE)/Ar(ΔABC) = 32/52
Ar(ΔADE)/Ar(ΔABC) = 9/25
Let Area of ΔADE = 9x sq. units and Area of ΔABC = 25x sq. units
Ar[trap BCED] = Ar(ΔABC) – Ar(ΔADE)
= 25x – 9x
= 16x sq units
Now,
Ar(ΔADE)/Ar(trap BCED) = 9x/16x
Ar(ΔADE)/Ar(trap BCED) = 9/16
Hence the correct answer is b .

44. If ABC is an isosceles triangle and D is a point of BC such that AD ⊥ BC , then
(a) AB2 - AD2 = BD.DC
(b) AB2 – AD2 = BD2 – DC2
(c) AB2 + AD2 = BD.DC
(d) AB2 + AD2 = BD2 – DC2

Solution

Given : ΔABC is an isosceles triangle , D is a point on BC such that AD ⊥ BC
We know that in an isosceles triangle the perpendicular from the vertex bisects the Base .
∴ BD = DC

Applying Pythagoras theorem in ΔABD
AB2 = AD2 + BD2
⇒ AB2 – AD2 = BD × BD
Since BD = DC
⇒ AB2 – AD2 = BD × BD
Since BD = DC
⇒ AB2 – AD2 = BD × BD
Hence correct answer is a .

45 . ΔABC is a right triangle right – angled at A and AD ⊥ BC . Then , BD/DC =
(a) (AB/AC)2
(b) AB/AC
(c) (AB/AD)2
(d) AB/AD

Solution

Given : In ΔABC, ∠A = 90° and AD ⊥ BC .
To find : BD : DC
 

∠CAD + ∠BAD = 90° …(1)
∠BAD + ∠ABD = 90° …(2) (∠ADB = 90°)
From (1) and (2),
∠CAD = ∠ABD
In ΔADB and ΔADC,
∠ADB = ∠ADC (90° each)
∠ABD = ∠CAD (Proved)
∴ ΔADB ~ ΔADC (AA Similarity)
⇒ CD/AD = AC/AB = AD/BD (Corresponding sides are proportional)
Disclaimer : The question is not correct . The given ratio cannot be evaluated using he given conditions in the questions .

 46. If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA, then AB2 + BC2 + CA2 =
(a)2 BE2
(b) 3 BE2
(c) 4 BE2
(d) 6 BE2

Solution

In triangle ABC , E is a point on AC such that BE ⊥ AC .
We need to find AB2 + BC2 + AC2.

Since BE ⊥ AC , CE = AE = AC/2 (In a equilateral triangle, the perpendicular from the vertex bisects the base.
In triangle ABE, we haveAB2 = BE2 + AE2
Since AB = BC = AC
Therefore , AB2 = BC2 = AC2 = BE2 + AE2
⇒ AB2 + BC2 + AC2 = 3BE2 + 3AE2
Since in triangle BE is an altitude, so BE = √3/2 AB
BE = √3/2 AB = √3/2 × AC
= √3/2 × 2AE = √3AE
⇒ AB2 + BC2 + AC2 = 3BE2 + 3(BE/√3)2
= 3BE2 + BE2 = 4BE2
Hence option c is correct . 

47. In a right triangle ABC right – angled at B, if P and Q are points on the sides AB and AC respectively , then
(a) AQ2 + CP2 = 2(AC2 + PQ2)
(b) 2(AQ2 + CP2) = AC2 + PQ2
(c) AQ2 + CP2 = AC2 + PQ2
(d) AQ + CP = 1/2 (AC + PQ)

Solution  

Disclaimer : There is mistake in the problem . The question should be ‘’ In a right triangle ABC right – angled at B, if P and Q are points on the sides AB and BC respectively , then ‘’
Given : In the right △ABC , right angled at B. P and Q are points on the sides AB and BC respectively .

Applying Pythagoras theorem,
In △AQB,
AQ2 = AB2 + BQ2 …(1)
In △PBC
CP2 = PB2 + BC2 …(2)
Adding (1) and (2), we get
AQ2 + CP2 = AB2 + BQ2 + PB2 + BC2
…(3)
In △ABC ,
AC2 = AB2 + BC2 …(4)
In △PBQ ,
PQ2 = PB2 + BQ2 …(5)
From (3), (4) and (5) , we get
AQ2 + CP2 = AC2 + PQ2
We got the result as c .

48. If △ABC ~ △DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of △ABC is
(a) 18 cm
(b) 20 cm
(c) 12 cm
(d) 15 cm

Solution

Given : △ABC and △DEF are similar triangles such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm and BC = 4 cm .
To find : Perimeter of △ABC .
We know that if two triangles are similar then their corresponding sides are proportional .
Hence , AB/DE = BC/EF = CA/FD
Substituting the values we get
AB/BC = DE/EF
AB/4 = 3/2
AB = 6 cm …(1)
Similarly,
CA/BC = DF/EF
CA/4 = 2.5/2
CA = 5 cm …(2)
Perimeter of △ABC = AB + BC + CA
= 6+4+5
= 15 cm
Hence the correct option is d .

49. In △ABC ~ △DEF such that AB = 9.1 cm and DE = 6.5 cm . If the perimeter of △DEF is 25 cm , then the perimeter of △ABC is
(a) 36 cm
(b) 30 cm
(c) 34 cm
(d) 35 cm

Solution

Given △ABC is similar to △DEF such that AB = 9.1 cm , DE = 6.5 cm . Perimeter of △DEF is 25 cm .To find : Perimeter of △ABC . 
We know that the ratio of corresponding sides of similar triangles is equal to the ratio of their perimeters .
AB/DE = BC/EF = AC/DE = P1/P2
AB/DE = P(△ABC )/P(△DEF)
9.1/6.5 = P(△ABC )/25
P(△ABC ) = 9.1×25/6.5
P(△ABC ) = 35 cm
Hence the correct answer is d .

50. In an isosceles triangle ABC if AC = BC and AB2 = 2AC2, then ∠C =
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Solution

Given : In Isosceles △ABC , AC = BC and AB2 = 2AC2 .
To find : Measure of angle C
In isosceles △ABC ,
AC = BC
∠B = ∠A (Equal sides have equal angles opposite to them )
AB2 = 2AC2
AB2 = AC2 + AC2
AB2 = AC2 + BC2 (AC = BC)
⇒ △ABC is right angle triangle , with ∠C = 90°
Hence the correct answer is c.
↧

R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios Exercise 5.1

February 6, 2019, 3:52 am
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Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math Exercise 5.1

Exercise 5.1

1. In each of the following one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios. 

(i) sinA = 2/3 

Solution

 

(ii) cosA = 4/5

Solution


(iii) tanθ = 11 

Solution 


(iv)  Sinθ = 11/5

Solution


(v) tanα = 5/12 

Solution


(vi) Sinθ = √3/2 

Solution


(vii) Cosθ = 7/25

Solution


(viii) tanθ = 8/15  

Solution


(ix) cotθ = 12/5  

Solution            


 (x) secθ = 13/5 

Solution


(xi) cosecθ = √10 

Solution


(xii) cosecθ = 12/5

Solution


2. In a ∆ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) Sin A, Cos A
(ii) Sin C, cos C 

Solution


3. In Fig below, Find tan P and cot R. Is tan P = cot R ?

Solution


4. If sinA = 9/41, compute cosA and tanA.

Solution

   

5. Given 15 cot A = 8, find SinA and SecA . 

Solution


6. In ∆PQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R. 

Solution



7. If cot θ = 7/8 , evaluate :
(i) (1+sinθ)(1-sinθ)/(1+cosθ)(1-cosθ)
(ii) cot2θ  

Solution
(i)

(ii)

8. If 3 cot A = 4, check whether 1-tan2 A/1+tan2 A = cos2 A - sin2 A or not.

Solution


9. If tan θ = a/b , find the value of cosθ + sinθ/cosθ - sinθ 

Solution


10. If 3 tan θ = 4 , find the value of 4cosθ - sinθ/2cosθ + sinθ 

Solution


11. If 3 cot θ = 2 , find the value of 4 sinθ - 3 cosθ/2 sinθ + 6 cosθ  

Solution


12. If tan θ = a/b , prove that a sin θ - b cos θ/a sin θ + b cos θ = a2 - b2/a2 + b2 . 

Solution



13. If sec θ = 13/5, show that 2cos θ - 3cos θ/4sin θ - 9cos θ = 3. 

Solution  


14. If cos θ = 12/13, show that sin θ (1 - tan θ) = 35/156 . 

Solution


15. If cot θ = 1/√3 , show that  1- cos2  θ/2 - sin2 θ = 3/5 .  

Solution


16. If tan θ = 1/√7 cosec2 θ - sec2  θ/cosec2 θ + sec2  θ = 3/4 .  

Solution 


17. If Sec θ = 5/4 , find the value of sin θ - 2 cos θ/ tan θ - cot θ 

Solution


18. If tan θ = 12/13, find the value of 2 sin θ cosθ/cos2 θ - sin2 θ

Solution



19. If cos θ = 3/5 , find the value of sinθ - (1/tanθ)/2 tan θ 

Solution


20. If sin θ = 3/5 , evaluate cosθ - (1/tanθ)/2 cot θ 

Solution



21. If tan θ = 24/7 , find that sin θ + cos  θ 

Solution



22. If sin θ = a/b , find sec θ + tan θ in terms of a and b . 

Solution


23. If 8 tan A = 15, find sin A - cos A .  

Solution



24. If tan θ = 20/21, show that 1-sin θ + cos θ/1+ sin θ + cos θ = 3/7  .

Solution


25 . If cosec A = 2 find 1/Tan A + sin A/1+cos A  

Solution


26. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Solution


27. In a ∆ABC, right angled at A, if tan C = √3, find the value of sin B cos C + cos B sin C.

Solution


28. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) Sec A = 12/5 for some value of angle A.
(iii) Cos A is the abbreviation used for the cosecant of angle A.
(iv) Sin θ = 4/3 for some angle θ . 

Solution



29. If sin θ = 12/13 find sin2θ - cos2θ/2 sinθ cosθ × 1 tan2θ  .

Solution


30. If cos θ = 5/13 find sin2θ - cos2θ/2 sinθ cosθ × 1 tan2θ .

Solution



31.  If sec A = 5/4, verify that 3 sinA - 4 sin3A/4 cos3 A - 3 cosA = 3 tan A - tan3 A /1 - 3 tan2A  .

Solution



32. 

Solution


33. If sec A = 17/8 , verify that 3-4 sin2 A/4 cos2 A-3 = 3-tan2 A/1-3 tan2 A.  

Solution 


34. 

Solution


35. If  3 cos θ - 4 sin θ = 2 cos θ + sin θ, find tan θ.  

Solution


36. If ∠A and ∠P are acute angles such that tan A = P, then show that ∠A = ∠P. 

Solution

↧

R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios Exercise 5.2

February 6, 2019, 4:10 am
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Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math Exercise 5.2

Exercise 5.2

Evaluate each of the following (1 – 19):

1. sin45°sin30° + cos45°cos30° 

Solution


2. Sin 60° cos 30° + cos 60° sin 30° 

Solution



3. Cos 60° cos 45° - sin 60° ∙ sin 45°

Solution


4. Sin2 30° + sin2 45° + sin2 60° + sin2 60° + sin2 90°   

Solution 

 
5. cos2 30° + cos2 45° + cos2 60° + cos2 90°   

Solution


6. tan2 30° + tan2 60° + tan2 45°

Solution


7. 2 sin2 30° - 3 cos2 45° + tan2 60°  

Solution 


8. sin2 30° cos2 45° + 4tan2 30° + 1/2 sin2 90° - 2 cos2 90° + 1/24 cos2 0° .

Solution


9. 4(sin4 60° + cos4 30°) + 3(tan2 60° - tan2 45°) + 5 cos2 45°  

Solution


10. (cosec2 45° sec2 30°)(sin2 30° + 4 cot2 45° - sec2 60°) 

Solution


11. cosec3 30° cos 60° tan3 45° + sin2 90° sec2 45° cot 30°   

Solution


12. cot2 30° - 2cos2 60° - 3/4 sec2 45° - 4 sec2 30°     

Solution


13. ( cos 0° + sin 45° + sin 30°)(sin 90° - cos 45° + cos 60°)  

Solution


14. sin 30° - sin 90° + 2 cos 0°/tan 30° tan 60° 

Solution

 

15. 4/cot2 30° + 1/ sin2 60° - cos2 45° 

Solution


16. 4(sin4 30° + cos2 60°) - 3(cos2 45° - sin2 90°) - sin2 60°  

Solution


17.  tan2 60° + 4 cos2 45° + 3 sec2 30° + 5 cos2 90°/cosec 30° + sec 60° - cot2 30° 

Solution


18. sin 30°/sin 45° + tan 45°/sec 60° - sin 60°/cot 45° - cos 30° sin 90° 

Solution 


19. tan 45°/sec 60° - sin 60°/cot 45° - 5 sin 90°/ 2 cos 0°

Solution


20. 2sin 3x = √3/2 = ? 

Solution

sin 3x = √3/2
sin 3x = sin60°
Equating angles we get,
3x = 60°
x = 20°

21. 2 sin x/2 = 1 x = ? 

Solution 

sin x/2 = 1/2
sin x/2 = sin 30°
x/2 = 30°
x = 60°

22. √3 sin x = cos x

Solution 

√3 tan x = 1
tan x = 1/√3
∴ tan x = tan 30°
x = 30°

23. tan x = sin 45° cos 45° + sin 30°

Solution


24. √3 tan 2x = cos 60° + sin 45° cos 45°

Solution


25.Cos 2x = cos 60° cos 30° + sin 60° sin 30° .

Solution


26. If θ = 30° verify 
(i) Tan 2 θ = 2 tan θ/1-tan2 θ 
(ii) Sin θ = 2 tan θ/1-tan2 θ
(iii) Cos 2 θ = 1-tan2θ/1+tan2θ 
(iv) Cos 3 θ = 4 cos3 θ - 3 cos θ 

Solution

(i)
 

(ii) 

(iii) 

(iv) 


27. If A = B = 60° verify 
(i) cos (A-B) = Cos A cos B + sin A sin B
(ii) sin (A-B) = sin A cos B - cos A sin B 
(iii) tan (A-B) = tan A - tan B/1 + tan A tan B 

Solution 

(i)

(ii)

(iii) 

28. If A = 30° B = 60° verify 
(i) Sin(A+B) = Sin A Cos B + cos A sin B
(ii) Cos(A+B) = cos A cos B - sin A sin B  

Solution

(i)

(ii)

29. If sin (A+B) = 1 and cos (A-B) = 1,0°< A+B ≤ 90°, find A and B.   

Solution

Given :
Sin(A+B) = 1…(1)
Cos(A-B) = 1…(2)
We know that ,
Sin 90° = 1…(3)
Cos 0° = 1 …(4)
Now by comparing equation (1) and (3)
We get,
A+B = 90 …(5)
Now by comparing equation (2) and (4)
We get,
A-B=0 …(6)
Now we get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get ,
A+B = 90
+A-B = 0
2A + 0 = 90
Therefore,
2A = 90
⇒ A = 90/2
⇒ A = 45°
Hence A = 45°
Now by subtracting equation (6) from equation (5)
We get,
A+B = 90
-A-B = 0
(-)(+)(-)
0+2B = 90
Therefore,
2B = 90
⇒ B = 90/2
⇒ B = 45°
Hence B = 45°
Therefore the values of A and B are as follows
A = 45° and B = 45°

30. If tan (A-B) 1/√3 and tan (A+B) = √3, 0° < A+B ≤ 90°, A≥B, Find A&B.

Solution 


31. If sin (A-B) = 1/2 and cos (A+B) = 1/2 , 0°< A + B ≤ 90°, A < B find A and B .

Solution

Given :
Sin(A-B) = 1/2 …(1)
Cos(A+B) = 1/2 …(2)
We know that ,
sin 30° = 1/2 …(3)
cos 60° = 1/2 …(4)
Now by comparing equation (1) and (3)
We get,
A-B = 30 …(5)
Now by comparing equation (2) and (4)
We get,
A+B = 60 …(6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
A-B = 30
+ A+B = 60
2A + 0 = 90
Therefore,
2A = 90
⇒ A = 90/2
⇒ A = 45°
Hence A = 45°
Now by subtracting equation (5) from equation (6)
We get,
A+B = 60
-A-B = 30
(-)(+)(-)
0 + 2B = 30
Therefore ,
2B = 30
⇒ B = 30/2
⇒ B = 15°
Hence B = 15°
Therefore the values of A and B are as follows
A = 45° and B = 15°

32. In a △ABC right angled at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B

Solution


33. Find the acute angles A&B, if sin (A+2B) = √3/2 cos(A+4B) = 0, A>B.

Solution


34. In ∆PQR, right angled at Q, PQ = 3cm PR = 6cm. Determine ∠P = ? ∠R = ?

Solution



35. If sin (A-B) = sin A - cos A sin B and cos (A-B) = cos A cos B + sin A sin B , find the values of sin 15° and cos 15° .

Solution


36. In a right triangle ABC, right angled at C , if ∠B = 60° and AB = 15 units . Find the remaining angles and sides .

Solution


37 . If △ABC is a right triangle such that ∠C = 90° ∠A = 45°, BC = 7 units find ∠B , AB and AC .

Solution


38. In rectangle ABCD AB = 20 cm ∠BAC = 60° BC, calculate side BC and diagonals AC and BD. 

Solution


39. If A and B are acute angles such that tan A = 1/2 , tan B = 1/3 and tan (A+B) = tan A + tan B/1-tan A tan B , find A + B .

Solution


40. Prove that (√3+1) (3-cot 30°) = tan3 60° - 2 sin 60°.

Solution

↧
↧

Notes of Ch 10 Financial Markets| Class 12th Business Studies

February 6, 2019, 10:53 pm
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Summary and Notes of Ch 10 Financial Markets| Class 12th Business Studies 

Meaning of Financial Markets

Financial Market means the market that creates and exchanges financial assets.

Functions of Financial Market

• Mobilisation of savings and channeling them into the most Productive Uses- It helps to transfer of savings from savers to investors.

• Facilitating Price Discovery- The forces of demand and supply established a price of any commodity or services in the market.

• Providing Liquidity to financial Assets- Financial Markets provides easy sale and purchase of financial assets.

• Reducing the cost of transactions- Provides information related to financial securities.

Financial market is divided into two categories

Money Market

It is a market for a short term funds which deals in monetary assets whose period of maturity is up to one year. The major participants in the market are the Reserve Bank of India, Commercial Banks, Non- Banking Finance Companies, State Governments, Large Corporate Houses and Mutual Funds.

Money Market Instruments

• Treasury Bill: It is a short-term instrument issued by the Reserve Bank of India on behalf of the central government to meets its short- term requirement of funds. Its maturing period cannot be more than one year. Treasury Bills are also known as Zero Coupon Bond.

• Commercial Paper: Commercial Papers are short- term unsecured promissory notes which are issued by large and creditworthy companies to raise short-term funds at lower rates of interests than market rates. It is sold at discount and redeemed at par. It usually has a maturity period of 15 days to one year.

• Call Money: Call Money is short term finance repayable on demand. The maturity period of the loan is between one and 15 days.

• Certificate of Deposit: Certificates of deposit are unsecured, negotiable, short-term instruments in bearer form issued by commercial banks and development financial institutions.

• Commercial Bill: Commercial Bill is a short-term negotiable, self- liquidating instrument, bill of exchange used to finance the working capital requirements of business firms.

Capital Market

It refers to that market where long term funds, both debt and equity are raised and invested. Share and debenture are main securities in this market. An ideal capital market is one where finance is available at reasonable cost.

The capital market can be divided into two parts:

Primary market: Primary market refers to that market in which securities are sold for the first time for collecting long-term capital. Primary market is also known as the New Issues Market and also deals with new securities being issued for the first time. A company can raise capital through the primary markets in the form of equity shares, preference shares, debentures, loans and deposits.

Secondary market: The secondary market is a market for purchase and sale of existing securities in open market without intervention of the issuing company. The chief purpose of the secondary market is to create liquidity in securities.

Distinction between Capital Market and Money Market

Basis for comparisonMoney MarketCapital Market
Meaning Deals with short-term securities.Deals with long-term securities.
InstrumentsThe main instruments are T-bills, trade bills reports, commercial paper and certificate.The main instruments are equity shares, debentures, bonds, preference shares etc.
ParticipantsThe participants in the Money market are large undertaken by institutional participants such as the RBI, Banks, financial institutions and finance companies.The participants in the capital market are financial institutions, banks, corporate entities, foreign investors and ordinary retail investors.
DurationWithin a year.More than a year.
LiquidityHighLow
SafetyLow riskyComparatively high 
Expected returnLessComparatively High

Stock Exchange

It is institution which provides a platform for buying and selling of existing securities. It helps companies to raise finance, provide liquidity and safety of investment to the investors and enhance the credit worthiness of individual companies.

Definition of stock exchange according to securities contracts (Regulation) Act 1956

Stock exchange means anybody of individuals, whether incorporated or not, constituted for the purpose of assisting, regulating or controlling the business of buying and selling or dealing in securities.

Functions of a Stock Exchange

•Providing Liquidity and Marketability to Existing Securities.

•Share prices are determined by the forces of demand and supply.

•The investing public gets a safe and fair deal because its dealings are well defined according to the existing legal framework.

•A stock exchange provides liquidity to securities. This gives the investor a double benefit –first, the benefit of the change in the market price of securities can be taken advantage of, and secondly, in case of need for money they can be sold at the existing market price at any time.

•The stock exchange ensuring a wider ownership by regulating new issues, better trading practices and taking effective steps in educating the public about investment.

•When securities are purchase with a view to getting profit as a result of change in their market price, it is called speculation.

Trading

All buying and selling of shares and debentures are done through a computer terminal. Trading in securities is done through brokers who are member of stock exchange. Every broker has to have access to a computer terminal that is connected to the main stock exchange.

Advantage of electronic trading systems

It ensures transparency, increases efficiency of information, increases the efficiency of operations.
Steps in the Trading and settlement Procedure

•If an investors wishes to buy or sell any security he has to firs registered as a broker or sub-broker, then enter in to an agreement. This required information is mandatory for him

PAN number, Date of birth and address, Educational qualification and occupation, residential status, Bank account details, Depository account details, name of any other broker with whom registered, Client code number in the client registration form.

•The investors has to open a ‘demat’ account or ‘beneficial owner’ account.

•The investor places an order with the broker.

•The broker will go online and match the share and best prices available.

•When final deal is done, the broker will issue a trade confirmation slip to the investor and contract note.

•After receiving a contract note, the investor has to deliver the shares sold or pay cash for the shares bought.

Dematerialisation and Depositories

Dematerialisation: Dematerialisation is the process of converting a share certificate from its physical form to electronic form.

Depository: The investor has to open a Demat Account with an organisation called a depository.

Rematerialisation: Physical shares can be converted into electronic form or electronic holdings can be reconverted into physical certificates called rematerialisation.

NSE (National Stock Exchange of India)

NSE was incorporated in 1992 and was recognised as a stock exchange in April 1993 but started operations in 1994. The NSE was set up by leading financial institutions, banks, insurance companies and other financial intermediaries.

Objective of NSE

•Establishing a nationwide trading facility for all types of securities.

•Ensuring equal access to investors all over the country through an appropriate communication network.

•Providing a fair, efficient and transparent securities market using electronic trading system.
•Enabling shorter settlement cycles and book entry settlements.
•Meeting international benchmarks and standards.

Market Segments of NSE

The Exchange provides trading in the following two segments:

• Whole sale Debt Market Segment: This segment provides a trading platform for a wide range of fixed income securities that include central government securities, treasury bills, and State development loans, bonds issued by public sector undertakings, floating rate bonds, zero coupon bonds, index bonds, commercial paper, certificate of deposit, corporate debentures and mutual funds.

•Capital Market Segment: The capital Market segment of NSE provides an efficient and transparent platform for trading in equity, preference, debentures, exchange traded funds as well as retail Government Securities.

BSE (Bombay Stock Exchange Ltd)

BSE Ltd was established in 1875 and was Asia’s first stock Exchange. It was granted permanent recognition under the securities contract Act, 1956. It has contributed to the growth of the corporate sector by providing a platform for raising capital.

Objective of BSE

•To provide fair and transparent market for trading in equity, debt instruments, derivatives, and
mutual funds.

•Trading platform for equities of small and medium enterprises.

•To conform to international standards.

•Instead of trading BSE also provides risk management, clearing, settlement, market data and education.

SEBI (Securities and Exchanging Board of India)

SEBI was established in 1988 and was given statutory status through an Act in 1992. The SEBI was set-up to protect the interests of investors, development and regulation of securities market

Purpose and role of SEBI

•The main purpose of SEBI is to create an environment to facilitate efficient mobilisation and allocation of resources through the securities markets.

•It also aims to stimulate competition and encourage innovation.

The environment aims at meeting the needs of the three groups which basically constitute the market:-
→ To the issuers- It aims to provide a market place in which they can confidently look forward to raising finances they need in an easy, fair and efficient manner.
→ To the investors- It should provide protection of their rights and interest.
→ To the intermediaries-It should offer a competitive, professionalised and expanding market with adequate and efficient infrastructure so that they are able to render better service to the investors and issuers.

Objective of SEBI

•The objective of SEBI is to regulate stock exchanges and securities industry to promote their orderly functioning.

•To protect the rights and interest of investors means protecting investors from wrong information, particularly individual investor and educate them regarding rules and regulations.

• To prevents investors from insider trading and malpractices.

•To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers etc.

Functions of SEBI

The function of the SEBI can be divided into three parts:

Regulatory Functions

•To register brokers, sub-brokers, transfer agents and other players in the market.

•To register collective investment schemes and mutual Funds.

•Regulation of stock brokers, portfolio exchanges, underwriters and merchant bankers.

•Regulation of takeover bids by companies.

Development functions

•Providing training of intermediaries of the securities market and conducting research and publishing information useful to all market participants.

•Help in developing capital markets by adapting a flexible approach.

Protective Functions

•Taking steps for investor protection and controlling insider trading.

•Promotion of fair practices and code of conduct in securities market and prohibition of fraudulent and unfair trade practices like making misleading statements, manipulations, price rigging etc.

The organisation Structure of SEBI

SEBI has decided its activity into five operational departments. Each department is headed by an executive director. The office of SEBI is situated at Mumbai with its regional offices at Kolkata, Delhi and Chennai. The SEBI also formed two advisory committee primary market advisory committee and Secondary market advisory. These committees consist of the market players, the investors associations recognised by the SEBI.

The objectives of the two committees are as follows:

•To advise SEBI on matters relating to the regulation of intermediaries for ensuring investors protection.

•To advise SEBI related to primary market development.

•To advise SEBI on disclosure requirements for companies.

•To advise the board in matters relating to the development and regulation of the secondary market in the country.

•To advise for changes in legal framework to introduce simplification and transparency in the primary market.

NCERT Solutions of Chapter 10 Financial Markets

↧

R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios Exercise 5.3

February 7, 2019, 12:15 am
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Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math Exercise 5.3

Exercise 5.3

1. Evaluate the following .
(i) sin 20° / cos70° 

Solution



(ii) cos19°/sin71° 

Solution


(iii) sin21° / cos 69° 

Solution


(iv) tan10° / cot 80°  

Solution


(v) sec 11° / cosec 79°    

Solution


2. Evaluate the following : 

(i) (sin 49° / cos 41°)2  + (cos 41° / sin 49°)2 

Solution


(ii) cos 48° - sin 42°

Solution


(iii) cot 40°/ tan 50° - 1/2 (cos 35°/sin 55°)

Solution



(iv) (sin 27°/cos 63°)2 - (cos 63°/sin 27°)2
Solution


(v) (tan 35°/ cot 55°) + (cot 78° / tan 12°) - 1

Solution

 

(vi) sec 70°/cosec 20° + sin 59°/cos 3°

Solution


(vii) cosec 31° + sec 59°

Solution

 

(viii) (sin72° + cos 18°)(sin 72° - cos 18° )

Solution

We have to find : (sin 72° + cos 18°)( sin 72° - cos 18° )
Since sin(90°- θ) = cosθ . So
(sin72° + cos 18°)(sin 72° - cos 18°) = (sin 72°)2 – (cos 18°)2
= [sin(90° - 18°)]2 - (cos 18°)2
= (cos 18°)2 - (cos 18°)2
= cos2 18° - cos2 18° 
So value of (sin 72° + cos 18°) (sin 72° + cos 18°) is 0 .

(ix) sin 35° sin 55°- cos 35° cos 55°

Solution

We find : sin 35° sin 55°- cos 35° cos 55°
Since sin (90° - θ) = cos θ and cos (90°- θ) = sin θ
sin 35° sin 55° - cos 35° cos55° = sin (90°- 55°)sin 55° - cos(90°- 55°) cos 55°
= cos 55° sin 55° - sin 55° cos 55°
= 1 – 1
= 0
So value of sin 35° sin 55° - cos 35° cos 55° is 0.

(x) tan 48° tan 23° tan 42° tan 67°

Solution

We have to find tan 48° tan 23° tan × 42° tan 67°
Since tan (90° - θ) = cot θ . So
tan 48° tan 23° tan 42° tan 67° = tan(90° - 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (tan 67°cot 67°) (tan 42° cot 42°)
= 1×1
= 1
So value of tan 48° tan 23° tan 42° tan 67° is 1 .

(xi) sec 50° sin 40° + cos 40° cosec 50°  

Solution

We find to find sec 50° sin 40° + cos 40° cosec 50° 
Since cos (90°- θ) = sin θ, sec(90°- θ) = cosec θ and sin θ cosec θ = 1. So
sec 50° sin 40° + cos 40° cosec 50° = sec(90° - 40°) sin 40° + cos(90° - 50°) cosec 50°
= cosec 40° sin 40° + sin 50° cosec 50°
= 1 + 1
= 2
So value of sec 50° sin 40° + cos 40° cosec 50° is 2.

3. Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) Sin 59° + cos 56°

Solution

sin 59° = sin(90°-31°) = cos 31°
cos 56° = cos (90° -34°) = sin 34°
⇒ cos 31° + sin 34°

(ii) Tan 65° + cot 49°  

Solution

tan 65° = tan (90°-25°) cot 25°
cot 49° = cot(90°-41°) tan 41°
⇒ cot 25° + tan 41°

(iii) Sec 76° + cosec 52°

Solution

sec 76° = sec(90°-14°) = cosec 14°
cosec 52° = cosec(90°-88°) =  sec 38°
⇒ cosec 14° + sec 38°

(iv) Cos 78° + sec 78°

Solution

cos 78° = cos (90°-12°) = sin 12°
sec 78° = sec(90°-12°) = cosec 12°
⇒ sin 12° + cosec 12°

(v)  Cosec 54° + sin 72°

Solution

cosec 54° = cosec (90°-36°) = sec 36°
sin 72° = sin (90°-18°) cos 18°
⇒ sec 36° + cos 18°

(vi) Cot 85° + cos 75° 

Solution

cot 85° = cot (90°-5°) = tan 5°
cos 75° = cos (90°-15°) = sin 15°
tan 5° + sin 15°

(vii) Sin 67° + cos 75°

Solution

sin 67° = sin (90°-23°) = cos 23°
cos 75° = cos(90° - 15°) = sin 15°
= cos 23° + sin 15°

4. Express Cos 75° + cot 75° in terms of angles between 0° and 30°.

Solution

cot 75° = cos (90°-15°) = sin 15°
cot 75° = cot (90° 15°) = tan 15°
= sin 15° + tan 15°

5. If Sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A = ?

Solution


6. If A, B, C are interior angles of a triangle ABC, prove that

(i) tan [c+A/2] = cot B/2

Solution


(ii) sin [B+C/2] = cos A/2

Solution


7. Prove that

(i) tan 20° tan 35° tan 45° tan 55° Tan 70° = 1

Solution


(ii) sin 48° sec 42° + cosec 42° = 2

Solution


(iii) sin 70°/ cos 20° +  cosec 20°/ sec 70° - 2 cos 70° cosec 20° = 0 

Solution


(iv) cos 80°/sin 10° + cos 59° cosec 31° = 2 

Solution


8. Prove the following .

(i) sin θ sin (90 - θ) - cos θ cos (90 - θ) = 0  

Solution


(ii) cos (90°-θ) sec (90°-θ) tan θ/cosec (90°-θ)sin(90°-θ)cot(90°-θ) + tan(90°-θ)/cot θ = 2  

Solution


(iii) (tan(90-A)cot A /cosec2 A) - cos2 A = 0 

Solution


(iv) cos(90°-A) sin(90°-A)/tan(90°-A) - sin2 A = 0 

Solution  


(v) sin(50°+ θ) - cos(40° - θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1Sol.

Solution


9. Evaluate : 
(i) 2/3 (cos430°-sin445°)-3(sin2 60°-sec245°)+1/4cot230° 

Solution


(ii) 4 (sin430° + cos460°) - 2/3(sin260° - cos245°) + 1/2 tan2 60° 

Solution


(iii) (sin 50°/cos40° + cosec 40°/sec 50°) - 4 cos 50° cosec 40°   

Solution


(iv) ) Tan 35° tan 40° tan 50° tan 55°

Solution


(v) Cosec (65°  + θ) – sec (25°  – θ) – tan (55°  – θ) + cot (35° + θ)    

Solution


(vi) tan 7° tan 23° tan 60° tan 67° tan 83° .

Solution


(vii) (2 sin 68°/ cos 22° ) - (2 cot 15°/5 tan 75°) - 3 tan 45° tan 20° tan 40° tan 50° tan 70°/ 5 .

Solution


(viii) 3 cos 55° / 7 sin 35° - 4(cos 70° cosec 20°) /7 (tan 5° tan 25° tan 45° tan 65° tan 85°) 

Solution



(ix) sin 18°/cos 72° + √3 { tan 10° tan 30°  tan 40° tan 50°  tan 80°}

Solution


(x) (cos 58°/sin 32°) + (sin 22°/cos 68°)- cos38°cosec52°/tan18°tan35°tan60°tan72°tan65°

Solution

10. If Sin θ = cos (θ – 45°), where θ – 45° are acute angles, find the degree measure of θ.

Solution


11. If A, B, C are the interior angles of a ∆ABC,  show that: 
(i) sin(B+C/2) = cos A/2
(ii) cos [B+C/2] = sin A/2

Solution


12. If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying Sin (20 + 45°) = cos (30 - θ°) 

Solution


13. If θ is a positive acute angle such that sec θ = cosec 60°, find 2 cos2 θ – 1 . 

Solution


14. If cos 2θ = sin 4 θ where 2θ , 4θ  are acute angles, find the value of θ . 

Solution


15. If Sin 3θ  = cos (θ  – 6°) where 3 θ  and θ − 6° are acute angles, find the value of θ . 

Solution


16. If Sec 4A = cosec (A – 20°) where 4A is acute angle, find the value of A. 

Solution


17. If Sec 2A = cosec (A – 42°) where 2A is acute angle. Find the value of A .

Solution

↧

NCERT Solutions for Class 12th: Ch 9 Financial Management (MCQ and Short Questions)

February 7, 2019, 9:47 pm
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NCERT Solutions for Class 12th: Ch 9 Financial Management (MCQ and Short Questions)

Exercises

Page No. 263

Objective–type questions

1. The cheapest source of finance is:
a. debenture 
b. equity share capital
c. preference share
d. retained earning
► c. preference share

2. A decision to acquire a new and modern plant to upgrade an old one is a:
a. financing decision
b. working capital decision
c. investment decision
d. None of the above
► b. working capital decision

3. Other things remaining the same, an increase in the tax rate on corporate profits will:
a. make the debt relatively cheaper
b. make the debt relatively the dearer
c. have no impact on the cost of debt
d. we can’t say
► a. make the debt relatively cheaper 

Page No. 264

4. Companies with a higher growth pattern are likely to:
a. pay lower dividends
b. pay higher dividends
c. dividends are not affected by growth considerations
d. none of the above
► a. pay lower dividends

5. Financial leverage is called favourable if:
a. Return on Investment is lower than the cost of debt
b. ROI is higher than the cost of debt
c. Debt is easily available
d. If the degree of existing financial leverage is low
► b. ROI is higher than the cost of debt

6. Higher debt-equity ratio results in:
a. lower financial risk
b. higher degree of operating risk
c. higher degree of financial risk
d. higher EPS
► c. higher degree of financial risk

7. Higher working capital usually results in:
a. higher current ratio, higher risk and higher profits
b. lower current ratio, higher risk and profits
c. higher equity, lower risk and lower profits
d. lower equity, lower risk and higher profits
► a. higher current ratio, higher risk and higher profits

8. Current assets are those assets which get converted into cash:
a. within six months
b. within one year
c. between one and three years 
d. between three and five years
► a. within six months

9. Financial planning arrives at:
a. minimising the external borrowing by resorting to equity issues
b. entering that the firm always have significantly more fund than required so that there is no paucity of funds
c. ensuring that the firm faces neither a shortage nor a glut of unusable funds
d. doing only what is possible with the funds that the firms has at its disposal
► c. ensuring that the firm faces neither a shortage nor a glut of unusable funds

Page No. 265 

10. Higher dividend per share is associated with:
a. high earnings, high cash flows, unstable earnings and higher growth opportunities
b. high earnings, high cash flows, stable earnings and high growth opportunities
c. high earnings, high cash flows, stable earnings and lower growth opportunities
d. high earnings, low cash flows, stable earnings and lower growth opportunities
► c. high earnings, high cash flows, stable earnings and lower growth opportunities

11. A fixed asset should be financed through:
a. a long-term liability
b. a short-term liability
c. a mix of long and short-term liabilities
► c. a mix of long and short-term liabilities

12. Current assets of a business firm should be financed through:
a. current liability only
b. long-term liability only
c. both types (i.e. long and short term liabilities)
► a. current liability only

Short Answer Questions

1. What is meant by capital structure?

Answer

Capital structure refers to the mix between owners and borrowed funds. It can be calculated as debt/equity ratio.

2. Discuss the two objectives of financial planning.

Answer

Two objective of financial planning 
• To ensure availability of funds whenever required.
• To see that the firm does not raise resources unnecessarily.

3. What is financial risk? Why does it arise?

Answer

Financial risk refers to a position when a company is unable to meet its fixed financial charges for example interest payment, preference dividend, and repayment obligations.
Financial risk arise when higher fixed coast arise. And it is obligatory for the company to pay the interest charges on debt along with the principle amount.

4. Define ‘Current Assets’. Give four examples of such assets?

Answer

A ‘current assets’ is a assets which are expected to generate economic benefit or which is help in running our business and within one year we can converted in to cash.
For example-Cash, gold, investment, foreign currency.

5. Financial management is based on three broad financial decisions. What are these?

Answer 

• The size and the composition of the fixed assets of the business.
• All items in the profit and Loss Account , e.g.. Interest, Expense, Depreciation etc.
• Break- up of long–term financing in to debt, equity etc.

6. What are the main objective of financial management? Briefly explain.

Answer

The main objective of the financial management is to maximize shareholders’ wealth. This is because company funds belong to the shareholders and the manner in which they are invested and the return earned by them determines their market value and price. All financial decisions aim at ensuring that each decisions is efficient and add some value.

7. How does working capital affect both the liquidity as well as profitability of a business?

Answer

The working capital should neither be more nor less than, if working capital is more than required, it will increase liquidity but decrease profitability  For instance, if large amount of cash is kept as working capital,  then this excessive cash will remain idle and cause the profitability to fall.
On the contrary, if the amount of cash and other current assets are very little then lot of difficulties will have to be faced in meeting daily expenses of the organisation
and making payment to the creditors. So, working capital affect both the liquidity as well as profitability of a business.

NCERT Solutions of Chapter 9 Financial Mangement (Long Answer Questions)

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NCERT Solutions for Class 12th: Ch 9 Financial Management (Long Answer Questions)

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NCERT Solutions for Class 12th: Ch 9 Financial Management (Long Answer Questions)

Exercises

Page No. 263

Long Answer Type Questions

1. What is meant by working capital? How is it calculated? Discuss five important determinants of working capital requirements.

Answer

Working capital is that part of total capital which is required to meet day to day expenses and to pay wages and other activities of the organisation and also help to run our business smooth.

It is calculated by

Working capital = Current assets – Current liabilities.

Five important determinants of working capital requirements-

1. Nature of business: The nature of business influences the amount of working capital required. For example trading organisation usually needs a smaller amount of working capital as compared to manufacturing organisation. because in trading organisation there is no need of processing, no need of any raw material, and no long procedure for sales and marketing where as in manufacturing business raw material converted in to finished goods before any sales become possible.

2. scales of operations : Those organisations which operate on higher scale of operation , needs a more working capital because the quantum of inventory and debtors required is generally high ,where as those organisation which operate on a lower scale needs less working capital.

3. Business cycle: Working capital is depends on phase of a business cycle. In case of a boom, the sales as a well as production are likely to be larger so, large amount of working capital required. And the period of depression, the sales as well as production will be small, so lower amount of working capital required.

4. Seasonal factors: working capital depends on a seasonal factor because in peak season higher level of activity, larger amount of working capital is required and in during the lean season requirement of working capital and activities are low.

5. Production cycle: some businesses have a longer production cycle and some have a shorter production cycle. Duration of production affect the working capital because length of production cycle, affect the amount of funds required for raw material and expenses. So, working capital requirement is higher in firm with longer processing cycle and lower in firms with a shorter processing cycle.

6. Credit allowed: Different firms allow different credit terms to their customers. A liberal credit policy results in higher amount of debtors, increasing the requirements of working capital.

7. Credit Availed: Just as a firm allows credit to its customers it also may get credit from its suppliers. The more credit a firm avails on its purchases, the working capital requirement is reduced.

2."Capital structure decision is essentially optimisation of risk- return relationship." Comment.

Answer

Capital Structure refers to the combination of different financial sources used by a company for raising funds. The sources of raising funds can be classified on the basis of ownership into two categories as borrowed funds and owners’ fund. Borrowed funds are in the form of loans, debentures, borrowings from banks, public deposits, etc. On the other hand, owners’ funds are in the form of reserves, preference share capital, equity share capital, retained earnings, etc. Thus, capital structure refers to the combination of borrowed funds and owners’ fund. For simplicity, all borrowed funds are referred as debt and all owners’ funds are referred as equity. Thus, capital structure refers to the combination of debt and equity to be used by the company. The capital structure used by the company depends on the risks and returns of the various alternative sources.

Both debt and equity involve their respective risk and profitability considerations. While on one hand, debt is a cheaper source of finance but involves greater risk, on the other hand, although equity is comparatively expensive, they are relatively safe.

The cost of debt is less because it involves low risk for lenders as they earn an assured amount of return. Thereby, they require a low rate of return which lowers the costs to the firm. In addition to this, the interest on debt is deductible from the taxable income (i.e. interest that is to be paid to the debt security holders is deducted from the total income before paying the tax). Thus, higher return can be achieved through debt at a lower cost. In contrast, raising funds through equity is expensive as it involves certain floatation cost as well. Also, the dividends are paid to the shareholder out of after tax profits.

Though debt is cheaper, higher debt raises financial risk. This is due to the fact that debt involves obligatory payments to lenders. Any default in payment of the interest can lead to the liquidation of the firm. As against this, there is no such compulsion in case of dividend payment to shareholders. Thus, high debt is related to high risk.

Another factor that affects the choice of capital structure is the return offered by various sources. The return offered by each source determines the value of earning per share. High use of debt increases the earning per share of a company (this situation is called Trading on Equity). This is because as debt increases the difference between Return on Investment and the cost of debt increases and so does the EPS. Thus, there is a high return on debt. However, even though higher debt leads to higher returns but it also increases the risk to the company.

Therefore, the decision regarding the capital structure should be taken very carefully, taking into consideration the return and risk involved.

3. A capital budgeting decision is capable of changing the financial fortune of a business. Do you agree? Why or why not?

Answer

Yes, I agree a capital budgeting is capable of changing the financial fortune of a business because a long term investment decision is called a capital budgeting decision and long term investment affect our business.

There are certain factors which affect our capital budgeting decisions –

Cash flows of the project- when a company takes an investment decision involving huge amount to generate some cash flows over a period. These cash flows are in the form of a series of a cash receipt and payment over the life of an investment. The amount of these cash flows analysed before considering capital budgeting decisions.

The rate of return: The most important criterion is the rate of return of the project. These calculations are based on the expected returns from each proposal and assessment of the risk involved.

The investment criteria involved: the decision to invest in a particular project involves a number of calculations regarding the amount of investment, interest rate, cash flow and rate of return.

Above discussed decision are very important for any business. They affect its earning capacity over the long-run, assets of a firm, profitability and competitiveness, are all affected by the capital budgeting decisions. Moreover, these decisions normally involve huge amounts of investment and are irreversible except at a huge cost. Therefore, once made, it is almost impossible for a business to wriggle out of such decisions. Therefore, they need to be taken with utmost care. These decisions must be taken by those who understand them comprehensively. A bad capital budgeting decision normally has the capacity to severely damage the financial fortune of a business.

4. Explain the factors affecting the dividend decision.

Answer

Dividend decision of a company deals with what portion of the profit is to be distributed as dividend between the shareholder and what portion is to be kept as retained earnings.

The factors affecting the dividend decision

• Amount of earnings: dividends are paid out of current and past earning. So earning s of amount will affect our dividend decision because a company having higher earnings will be in a position to pay a higher amount of dividend to its shareholders. In contrast to this, a company having low or limited earnings would distribute low dividends among shareholder.

• Stability earning: when a company having a stable earning is in a better position to declare higher dividends and when a company having unstable earning position they declare smaller dividend.

• Stability of dividend: Companies generally follow the practice of stabilising their dividend per share. They try to avoid frequent fluctuations in dividend per share and opt for increasing (or decreasing) the value only when there is a consistent rise (or fall) in the earnings of the company.

• Growth opportunities: companies having good growth opportunities retain more money out of their earnings so as to finance the required investment.so, dividend in growth companies is smaller than non-growth companies.

• Cash flow position: if a company is low on cash then the dividend will be lower as compared to the company which has more liquidity. So dividend totally depends on cash flow even a company has high profit, it will not be able to distribute high dividends if it does not have enough cash.

• Shareholders’ preference: while declaring a dividend, companies must keep in mind the preference of the shareholders. If the shareholder in general desires that at least a certain amount should be paid as dividend because there are always some shareholders who depend upon a regular income from their investment.

• Taxation policy: if tax on dividend is higher than company pay less dividend and when the company pay lower tax than they pays a higher dividend to the shareholders.

• Stock Market Reaction: if an increase in dividend and stock prices is good and taken positively. Similarly, a decrease in dividend may have negative impact on the share price in the stock market. So, it is important factors considered by the management while taking a decision about a dividend decision.

• Access to Capital Market: Large and reputed companies generally have easy access to the capital market and therefore, depend less on retained earnings to finance their growth. These companies tend to pay higher dividends than the smaller companies which have relatively low access to the market.

• Legal constraints: some rules and regulation of the company act which restrict to pay as dividend. Such provisions must be kept in mind while declaring the dividend.

• Contractual constraints: while granting a loan to a company. Sometimes, the lender may impose certain restrictions on the payment of dividend in the future. On that time companies may ensure that dividend does not violate the terms of the loan agreement.

5. Explain the term ‘Trading on Equity’. Why, when and how it can be used by a company?

Answer

Trading on equity refers to a practice of raising the proportion of debt in the capital structure Such that the earning per share increases. A company resorts to trading on equity when the rate of return on investment is greater than the rate of interest on the borrowed fund. Trading on equity occurs when a corporation uses bonds and other debt.


Company 'X'Company 'Y'
Share Capital₹10 lakhs₹ 6 lakhs
Loan@15%pa
-
₹ 4 lakhs
₹ 10 lakhs₹ 10 lakhs
Profit before Interest + Tax₹ 3 lakhs₹ 3 lakhs
Interest
Nil
₹ 0.09 lakhs
Profit before tax₹ 3 lakhs₹ 2.01 lakhs
Tax@50%₹ 1.5 lakhs₹ 1.05 lakhs
Profit after tax₹ 1.5 lakhs₹ 1.05 lakhs
No. Of equity shares₹ 10 lakhs₹ 4 lakhs
Rate of return on share
15%
26.25%

It should be clear from the above example that shareholders of the company ‘X’ have a higher rate of return than company ‘Y’ due to loan component in the total capital of the company.

Case problem

‘S’ limited is manufacturing steel at its plant in India. It is enjoying a buoyant demand for its product as economic growth is about 7%-8% and the demand for steel is growing. It is planning to set up a new steel plant to cash on the increased demand.it is estimated that it will require about Rs. 5000 crores to set up and about Rs 500 crores of working capital to start the new plant.

Question

1. Describe the role and objective of financial management for this company.

Answer

The role and objective of the financial management for this company is:

(i) Managerial decisions relating to procurement of long term and short term funds
(ii) Keeping the risk associated with respect to procured funds under control.
(iii) Utilisation of funds in the most productive and effective manner
(iv) Fixed debt equity ratio in capital.

The main objective of the financial management is to maximize shareholders’ wealth which is
referred to as the wealth maximisation concept, and management also ensure that all the decision is efficient and adds some value.  The investment decision, financial decision and dividend decision help an organisation to achieve this objective. In the given situation, S limited envisages growth prospects of steel industry due to the growing demand. To expand the production capacity, the company needs to invest. However, investment decision will depend on the availability of funds, the financing decision and the dividend decision. However, the company will take those financial decisions which result in value addition examples the benefits are more than the cost. This leads to an increase in the market value of the shares of the company.

2. Explain the importance of having a financial plan for this company. Give an imaginary plan to support your answer.

Answer 

Financial planning is an important part of overall planning of any business organisation.

The importance of financial planning

• It helps in forecasting what may happen in future under different business situation.
• It helps in avoiding business shocks and surprises and helps the company in preparing for the future.
• It helps in co-ordinating various business functions, by providing clear policies and procedure.
• Detailed plans of action prepared under financial planning reduce waste, duplication of efforts, and gaps in planning.
• It tries to link the present with the future.
• It provides a link between investment and financing decisions on a continuous basis.

3. What are the factors which will affect the capital structure of this company?

Answer

Capital structure refers to the mix between owners fund and borrowed fund. Deciding about the capital structure of a firm involves determining the relative proportion of various types of fund. This depends on various factors which are cash flow position, interest coverage ratio, debt service coverage ratio.
The factor affecting the choice of capital structure:
1. cash flow position.
2. Interest coverage ratio.
3. Debt service coverage Ratio.
4. Return on investment.
5. Cost of debt.
6. Tax rate.
7. Cost of equity.
8. Floatation costs.
9. Risk consideration.
10. Flexibility.
11. Control.

4. Keeping in mind that it is a highly capital intensive sector what factors will affect the fixed and working capital. Give reasons with regard to both in support of your answer.

Answer

The working and fixed capital requirement of ‘S’ Limited will be high due to the following reasons
(i) Working capital depends on nature of business.
(ii) Higher scale of operation need higher level of working capital whereas operation is lower level need low working capital.
(iii) In case of steel industry, the major input is iron ore and coal. The ratio of cost of raw material to total cost is very high. Hence, higher will be the need for working capital.
(iv) The longer the operating cycle, the larger is the amount of working capital required as the funds get locked up in the production process for a long period of time.
(v) Terms of credit for buying and selling goods, discount allowed by suppliers and to the customers also determines the quantum of working capital.

NCERT Soutions of Chapter 9 Financial Management (MCQ and Short Questions)

Notes of Chapter 9 Financial Management

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R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios MCQ

February 8, 2019, 11:04 pm
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Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math MCQ's

Multiple Choice Questions

Mark the correct alternative in each of the following :

1. If θ is an acute angle such that cos θ = 3/5, then sin θ tan θ - 1/2 tan2θ = 
(i) 16/625
(ii) 1/36
(iii) 3/160
(iv) 160/3

Solution


2. If tan θ = a/b, then a sin θ + b cos θ/a sin θ - b cos θ is equal to
(a) a2+b2/a2-b2
(b) a2-b2/a2+b2
(c) a+b/a-b
(d) a-b/a+b    

Solution


3. If 5 tan θ - 4 = 0, then the value of 5 sin θ - 4 cos θ/5 sin θ + 4 cos θ is 
(a) 5/3
(b) 5/6
(c) 0
(d) 1/6 

Solution


4.If 16 cot x = 12, then sin x - cos x/sin x + cos x equals 
(a) 1/7
(b) 3/7
(c) 2/7
(d) 0  

Solution


5. If 8 tan x = 15, then sin x - cos x is equal to
(a) 8/17
(b) 17/7
(c) 1/17
(d) 7/17

Solution



6. If tan θ = 1/√7 , then cosec2 θ - sec2 θ/cosec2 θ - sec2 θ =   
(a) 5/7
(b) 3/7
(c) 1/12
(d) 3/4

Solution


7. If tan θ = 3/4, then cos2 θ - sin2 θ =
(a) 7/25 
(b) 1
(c) -7/25
(d) 4/25

Solution


8. If θ is an acute angle such that tan2 θ = 8/7, then the value of (1+sin θ)(1-sin θ)/(1+cos θ)(1-cos θ) is
(a) 7/8
(b) 8/7
(c) 7/4
(d) 64/49   

Solution


9. If 3 cos θ = 5 sin θ, then the value of 5 sin θ - 2 sec3 θ+2 cos θ/5 sin θ + 2 sec3 θ-2 cos θ 
(a) 271/979
(b) 316/2937
(c) 542/2937
(d) None of these 

Solution


10. If tan2 45° - cos2 30° = x sin 45° cos 45°, then x =
(a) 2
(b) -2
(c) -(1/2)
(d) 1/2

Solution


11. The value of cos2 17° - sin2 73° is 
(a) 1
(b) 1/3
(c) 0
(d) -1

Solution


12. The value of cos3° 20° - cos3 70°/ sin3 70° - sin3 20° is
(i) 1/2
(ii) 1/√2
(iii) 1
(iv) 2 

Solution


13. If x cosec2 30° sec2 45°/ 8 cos2 45° sin2 60°
(a) 1
(b) -1
(c) 2
(d) 0 

Solution


14. If A and B are complementary angles , then 
(a) sin A = sin B 
(b) cos A = cos B
(c) tan A = tan B 
(d) sec A = cosec B 

Solution


15. If x sin (90° - θ) cot (90° - θ) , then x = 
(a) 0 
(b) 1
(c) -1
(d) 2

Solution


16. If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
(a) 1
(b) √3
(c) 1/2
(d) 1/√2 

Solution



17. If angles A,B,C to a △ABC from an increasing AP, then sin B =
(a) 1/2
(b) √3/2
(c) 1
(d) 1/√2 

Solution


18. If θ is an acute angle such that sec2 θ = 3, then the value of tan2 θ-cosec2 θ/tan2 θ + cosec2 θ is
(a) 4/7
(b) 3/7
(c) 2/7
(d) 1/7

Solution


19. The value of tan 1° tan 2° tan 3° ... tan 89° is 
(a) 1
(b) -1
(c) 0
(d) None of these   

Solution


20. The value of cos 1° cos 2° cos 3° ... cos 180° is 
(a) 1
(b) -1
(c) 0
(d) None of these 

Solution

Here we have to find: cos 1° cos 2° cos 3° ...cos 180°
cos 1° cos 2° cos 3° ...cos 180°
= cos 1° cos 2° cos 3°...cos 89° cos 90° cos 91° ...cos 180° [since cos 90° = 0]
= cos 1° cos 2° cos 3°...0 × cos 90° ...cos 180°
= 0

21.The value of tan 10° tan 15° tan 75° tan 80° is 
(a) -1
(b) 0
(c) 1
(d) None of these 

Solution

Here we have to find : tan 10° tan 15° tan 75° tan 80°
Now
tan 10° tan 15° tan 75° tan 80°
= tan (90° - 80°) tan (90° - 75°)tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= (cot 80° tan 80°)(cot 75° tan 75°)
= 1×1 [since cotθ tanθ = 1]
= 1

22. The value of cos(90°-θ) sec(90°-θ) tan θ/cosec (90°-θ) sin (90°-θ) cot (90°-θ)  + tan (90°-θ)/cot θ is 
(a) 1
(b) -1
(c) 2
(d) -2 

Solution

We have to find : cos(90°-θ)sec(90°-θ)tanθ/cosec(90°-θ)sin(90°-θ)cot(90°-θ) + tan(90°-θ)/cot θ
so cos(90°-θ)sec(90°-θ)tanθ/cosec(90°-θ) sin(90°-θ) cot(90°-θ) + tan(90°-θ)/cot θ
= sinθ cosecθ tanθ/secθ cosθ tanθ + cotθ/cotθ
= 1×tanθ/1×tanθ + cotθ/cotθ
= 1+1
= 2

23. If θ and 2 θ - 45° are acute angles such that sin θ = cos (2θ - 45°), then tan θ is equal to
(a) 1
(b) -1
(c) √3
(d) 1/√3  

Solution

Given that : sin θ = cos (2θ - 45°) and θ and 2θ - 45 are acute angles
we have to find tan θ
⇒ sin θ = cos(2θ - 45°)
⇒ cos (90°- θ) = cos(2θ - 45°)
⇒ 90°- θ = 2θ - 45°
⇒ 3θ = 135°
Where θ and 2θ - 45° are acute angles
since θ = 45°
Now
tan θ
= tan 45° put θ = 45°
= 1 

24. If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ - √3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1 
(d) 1+√3  

Solution

We are given that 5θ and 4θ are acute angles satisfying the following condition
sin 5θ = cos 4θ . we are asked to find 2 sin 3θ - √3tan 3θ
⇒ sin 5θ = cos 4θ
⇒ cos(90°-5θ) = cos 4θ
⇒ 90° = 5θ = 4θ
⇒ 9θ = 90°
Where 5θ and 4θ are acute angles
⇒ θ = 10°
Now we have to find :
2 sin 3θ - √3 tan 3θ
= 2 sin 30° - √3 tan 30°
= 2 × 1/2 - √3 × 1/√3
= 1-1
= 0

25. If A + B = 90°, then tan A tan B + tan A cot B/sin A sec B - sin2 B/cos2 A is equal to
(a) cot2A
(b) cot2B
(c) -tan2A
(d) -cot2A   

Solution 


26. 2 tan 30°/1+tan230° is equal to
(a) sin 60° 
(b) cos 60° 
(c) tan 60° 
(d) sin 30°

Solution


27. 1-tan2 45°/1+tan2 45° is equal to
(a) tan 90°
(b) 1
(c) sin 45° 
(d) sin 0° 

Solution


28.  Sin 2A = 2 sin A is true when A = 
(a) 0°
(b) 30°
(c) 45°
(d) 60°

Solution
We are given, sin2A = 2sinA.cosA
so
⇒ sin 2A = 2sinA
⇒ 2 sinA.cosA = 2sinA
⇒ cos A = 1
As A = 0°

29. 2 tan 30°/1-tan2 30° is equal to
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°

Solution


30. If A,B and C are interior angles of a triangle ABC, then sign (B+C/2) =
(a) sin A/2
(b) cos A/2
(c) - sin A/2
(d) - cos A/2

Solution

We know that in triangle ABC
A+B+C =
⇒ B+C = 180° - A
⇒ B+C/2 = 90°/2 - A/2
⇒ sin (B+C/2) = sin(90°-A/2)
Since sin(90°-A) = cos A
So
sin (B+C/2) = cos A/2

31. If cos θ = 2/3, then 2 sec2 θ + 2 tan2 θ - 7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4

Solution


32. tan 5° × tan 30° × 4 tan 85° is equal to 
(a) 4/√3
(b) 4√3
(c) 1
(d) 4 

Solution

We have to find tan5° × tan30° × 4 tan85°
We know that
tan(90°-θ)= cotθ
tanθ cotθ = 1
tan 30° = 1/√3
so
tan5° × tan30° × 4 tan85°
= tan(90°- 85°) × tan30° × 4 tan85°
= cot 85° × tan30° × 4 tan85°
= 4 cot85° × tan85° tan30°
= 4 × 1 × 1/√3
= 4/√3   

33. The value of tan 55°/cot35° + cot 1° cot 2° cot 3°... cot 90°, is
(a) -2
(b) 2
(C) 1
(d) 0

Solution 

We have to find the value of the following expression
tan 55°/cot 35° + cos 1° cot 2° cot 3° ...cot 90° 
= tan 55°/cot 35° + cos 1° cot 2° cot 3° ...cot 90°  
= tan(90°-35°)/cot 35° + cot(90°-89°)cot(90°-88°)cot(90°-87°)...cot87 cot88°cot89°...cot90°
= cot35°/cot35° + tan 89° tan 88° tan 87°...cot87 cot 88° cot 89°...cot 90° 
= 1+1×1×1...× 0
= 1
As cot 90° = 0 

34.  In Fig. 5.47, the value of cos φ  is 
(a) 5/4
(b) 5/3
(c) 3/5
(d) 4/5

Solution

We should proceed with the fact that sum of angles on one side of a straight line is 180° .
so from the given figure ,
θ + φ + 90° = 180°
so, θ = 90° - ...(1)
Now from the triangle ?ABC,
sin θ  = 4/5
Now we will use equation (1) in the above,
sin(90° - φ ) = 4/5
Therefore, cos φ = 4/5

35. In fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot  θ
(a) 12/5
(b) 5/12
(c) 13/12 
(d) 12/13

Solution 

We have the following given data in the figure, AD = 4 cm, BD = 3 cm, CB = 12 cm
Now we will use pythagoras theorem in △ABD,
AB = √32 + 42
5 cm
Therefore
cot θ = CB/AB
= 12/5 
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NCERT Solutions for Class 12th: Ch 10 Financial Markets (MCQ and Short Questions)

February 12, 2019, 4:07 am
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NCERT Solutions for Class 12th: Ch 10 Financial Markets (MCQ and Short Questions)

Exercises

Page No. 289

Multiple Choice questions

 1. Primary and secondary markets:
a. Compete with each other
b. Complement each other
c. Function independently
d. Control each other
► b. Complement each other

Page No. 290

2. The total number of Stock Exchanges in India is:
a. 20
b. 21
c. 22
d. 23
► d. 23

3. The settlement cycle in NSE is:
a. T + 5
b. T + 3
c. T + 2
d. T+1
► b. T + 3

4. The National Stock Exchange of India was recognized as stock exchange in the year:
a. 1992
b. 1993
c. 1994
d. 1995
► a. 1992

5. NSE commenced futures trading in the year:
a. 1999
b. 2000
c. 2001
d. 2002
► b. 2000

6. Clearing and settlement operations of NSE are carried out by:
a. NSDL
b. NSCCL
c. SBI
d. CDSL
► a. NSDL

7. A Treasury Bill is basically:
a. An instrument to borrow short-term funds
b. An instrument to borrow long-term funds
c. An instrument of capital market
d. None of the above
► a. An instrument to borrow short-term funds

Short answer questions

1. What are the functions of a financial market?

Answer

• Mobilisation of savings and Channeling them in to the most productive uses.
•Facilitating price discovery.
•Providing liquidity to financial Assests.
•Reducing the cost of transactions.

2. "Money market is essentially a market for short term funds." Discuss.

Answer

Money market is a market for short- term funds because it deals in monetary assets whose period of maturity is less than one year. The instrument of money market includes Treasury bill, commercial bill, call money, certificate of deposit, Commercial bills.

3. What is a Treasury Bill?

Answer

They are also known as zero coupon bonds issued by the Reserve bank of India on behalf of the central government to meet its short-term requirement of funds.

4. Distinguish between capital market and money market.

Answer

Basis of differenceCapital market Money market
ParticipantsThe participants in the capital market are financial institutions, banks, corporate entities, foreign investors and ordinary retail investors from members of the public.The participation in the money market is by large undertaken by institutional participants such as the RBI, banks, financial institutions and finance companies.
InstrumentsEquity shares, debentures, bonds, preference shares etc are the main instruments traded in capital market.Short term debt instruments such as T- bills, trade bills reports, commercial paper and certificate of deposit.
Time span of securities Maturity period is more than one year.The maturity period can vary from one day to a maximum of one year.
RiskCapital market securities involve greater risk in terms of repayment of the principal amount.Money market securities are less risky due to short time period and sound financial position of the issuers.
Returns expectedExpected returns are higher due to the possibility of capital gains in long-term and regular dividends or bonus.Expected returns are lower due to shorter duration.

5. What are the functions of a stock exchange?

Answer

The function of the stock exchange is:

• Providing liquidity and marketability to existing securities: The main function of the stock exchange is the creation of a continuous market where securities are bought and sold and also it gives investors the chance to disinvest and reinvest.

• Pricing of securities: Price of the stock exchange is determined by the forces of demand and supply.

• Safety of transaction: The membership of a stock exchange is well- regulated and its dealings are well defined according to the legal framework. This ensures that the investing public gets a safe and fair deal.

• Contribution to economic growth: The process of disinvestment and reinvestment saving get channelised into productive investment avenues. This leads to capital formation and economic growth.

• Spreading of equity cult: The stock exchange can play a vital role in ensuring wider share ownership by regulating new issues. Better trading practices and taking effective steps in educating the public about investments.

• Providing scope for speculation: The stock exchange provides sufficient scope within the provisions of law for speculative activity in a restricted and controlled manner.

6. What are the objectives of the SEBI?

Answer

The overall objective of SEBI is to protect the interests of investors and regulate the securities market. This may be elaborate as:

• To regulate stock exchanges and the securities industry to promote their orderly functioning.

•  To protect the rights and interests of investors.

• To prevent trading malpractices.

• To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers etc.

7. State the objectives of the NSE.

Answer

The objectives of the NSE –

• Establishing a nationwide trading facility for all types of securities.

• Through an appropriate communication network, ensuring equal access to investors.

• Through electronic trading system, provides a fair, efficient and transparent securities market.

• It enables shorter settlement cycles and book entry settlements.

• Meeting international benchmarks and standards.

8. What is OTCEI?

Answer

The OTCEI is a company incorporated under the Companies Act, 1956. It was set up to provide small and medium companies an access to the capital market for raising finance in a cost effective manner. It is fully computerised, transparent, single window exchange which commenced trading in 1992. This exchange is established on the lines of NASDAG the OTC exchange in USA. If has been promoted by UTI, ICICI, IDBI, IFCI, LIC, GIC, SBI capital markets and can bank financial services.
It is a negotiated market place that exists anywhere as opposed to the auction market place, represented by the activity on securities exchange. Thus, in the OTC exchange, trading takes place when a buyer or seller walks up to an OTCEI counter, taps on the computer screen, finds quotes and effects a purchase or sale depending on whether the prices meet their target.

NCERT Solutions of Chapter 10 Financial Markets (Long Answer Questions)


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NCERT Solutions for Class 12th: Ch 10 Financial Markets (Long Answer Questions)

February 12, 2019, 9:05 am
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NCERT Solutions for Class 12th: Ch 10 Financial Markets (Long Answer Questions)

Exercises

Page No. 289

Long Answer Type Questions

1. Explain the various money Market Instruments.

Answer

Money Market Instruments are-

1. Treasury Bill: it is a short-term instrument and issued by the reserve bank of India on behalf of the central government to meet its short-term requirement of funds. Treasury bills are issued in the form of a promissory note. They are also known as zero coupon bonds. They are highly liquid and issued at a price which is lower than their face value and repaid at par. Treasury bills are available for a minimum amount of Rs 25,000.

2. Commercial paper: commercial paper is a short term unsecured promissory note.it is issued by large and creditworthy companies to raise short term funds at lower rates of interest than market rates. It usually has a maturity period of 15 days to one year. It is sold at a discount and redeemed at par. The main purpose of commercial paper was to provide short – term funds for seasonal and working capital needs.

3. Call money: Call money is short- term finance repayable on demand, with a maturity period of one day to fifteen days. Call money is a method by which banks borrow from each other to be able to maintain the cash reserve ratio. The reserve banks of India changes the cash reserve ratio from time to time. Which affect the amount of funds available to be given as loans by Commercial Banks. The interest rate paid on call money loans is known as the call rate. There is an inverse relationship between call rates and other short term money market, a rise in call rates money makes other sources of finance such as commercial paper and certificate of deposit cheaper in comparison for banks raise funds from these sources.

4. Certificate of Deposit: Certificates of deposit are unsecured, negotiable, short-term instruments in bearer form, issued by commercial banks and development financial institutions. They can be issued to individuals, corporations and companies. They help to mobilise a large amount of money for short periods.

5. Commercial Bill: A commercial bill is a bill of exchange used to finance the working capital requirements of business firms. It is a short term negotiable, self-liquidating instrument which is used to finance the credit sales of firms, when goods are sold on credit, the buyer becomes liable to make payment on a specific date in future.

2. What are the methods of floatation in Primary Market?

Answer

Methods of floatation in Primary market:

1. Offer through prospectus: This involves inviting subscription from the public through issue of prospectus. A prospectus makes a direct appeal to investors to raise capital, through an advertisement in newspapers and magazines. The issues may be underwritten and also required to be listed on at least one stock exchange. The contents of the prospectus have to be in accordance with the provisions of the Companies Act and SEBI disclosure and investor protection guidelines.

2. Offer for sale: Under this method, securities are not issued directly to the public but are offered for sale through intermediaries like issuing houses or stock brokers. In this case, a company sells securities enbloc at an agreed price to brokers who, in turn, resell them to the investing public.

3. Private placement: it is the allotment of securities by a company to institutional investors and individual. It helps companies to raise capitals more quickly than a public issue. It is expensive on account of various mandatory and non-mandatory expenses. Some companies cannot afford a public issue so they choose to use private placement.

4. Rights issue: under this method shareholder are offered the ‘ right’ to buy new shares in proportion to the number of shares. This is a privilege only given to the existing shareholder to subscribe new issues according to the terms and condition.

5. e-IPOs: A company proposing to issue capital to the public through the on-line system of the stock exchange has to enter into an agreement with the stock exchange. This is called an initial public offer (IPO). Only SEBI registered brokers have to select for the accepting applications and placing orders with the company. The lead manager coordinates all the activities amongst intermediaries connected with the issue.

3. Explain the recent capital market reforms in India.

Answer

The National Stock Exchange is the latest, most modern and technology driven exchange. NSE has setup a nationwide fully automated screen based trading system. The NSE was setup by leading financial institutions, banks, insurance companies and others financial intermediaries. It is managed by professionals, who do not directly or indirectly trade on the exchange. The trading rights are with the trading members who offer their services to the investors. The Board of NSE comprises senior executives from promoter institutions and eminent professionals, without having any representation from trading members.

Objectives of NSE

• Establishing a nationwide trading facility for all types of securities.

• Ensuring equal access to investors all over the country through an appropriate communication network.

• Providing a fair, efficient and transparent securities market using electronic trading system.

• Enabling shorter settlement cycles and book entry settlements.

• Meeting international bench marks and standards.

Within a span of 10 year, NST was able to achieve its objectives for which it was set up. It has been playing a leading role as a change agent in transforming the Indian capital market.

4. Explain the objective and functions of the SEBI?

Answer

The objective of SEBI:

• To regulate stock exchanges and securities industry to promote their orderly functioning.

• To protect the rights and interests of investors, particularly individual investors and to guide and educate them.

• To prevent trading malpractices and achieve a balance between self – regulation by the securities industry and its regulation.

• To regulate and develop a code of conduct and fair practices by intermediaries like brokers, merchant bankers etc. with a view to making them competitive and professional.

Function of SEBI

SEBI has generally doing twin task both regulation and development:

Regulatory functions

• Registration of brokers and sub-brokers and other players in the market.

• Registration of collective investment schemes and mutual funds.

• Regulation of stock brokers, portfolio exchange, underwriters and merchant bankers and the business in stock exchange and any other securities market.

• Regulation of takeover bids by companies.

• Calling for information by under- taking inspection, conducting enquiries and audits of stock exchange and intermediaries.

• Levying fee or other charges for carrying out the purposes of the act.

Development function

• Training of intermediaries of the securities market.

• Conducting research and publishing information useful to all market participants.

• Undertaking measures to develop the capital markets by adapting a flexible approach.

Protective functions

• Prohibition of fraudulent and unfair trade practices.

• Controlling insider trading and imposing penalties for such practices.

• Undertaking steps for investor protection.

• Promotion of fair practices and code of conduct in securities.

5. Explain the various segment of the NSE.

Answer

The exchange provides trading in the following two segments

• Whole Sale Debt Market Segments: This segments provides a trading platform for a wide range of fixed income securities that include central government securities, treasury bill, state development loans, bond issued by public sector undertakings, floating rate bond, commercial paper, certificate of deposit, corporate debentures and mutual funds.

• Capital Market Segment: this segment of NSE provides an efficient and transparent platform for trading in equity, preference, debentures, exchange traded funds as well as retail government securities.

Case Problems

1. ‘R’ Limited is a real estate company which was formed in 1950. In about 56 years of its existence the company has managed to carve out a niche for itself in this sector. Lately, this sector is witnessing a boom due to the fact that the Indian economy is on the rise. The incomes of middle class are rising. More people can afford to buy homes for themselves due lo easy availability of loans and accompanying tax concessions.

To expand its business in India and abroad the company is weight various options to raise money through equity offerings in India. Whether to tap equity or debt, market whether to raise money from domestic market or international market or combination of both? When their to raise the necessary finance from money market or capital market. It is also planning to list itself in New York Stock Exchange to raise money through ADR’s. To make its offerings attractive it is planning to offer host of financial plans products to its stakeholders and investors and also expand it’s listing at NSE after complying with the regulations of SEBI.

1. What benefits will the company derive from listing at NSE?

Answer

• NSE provides nationwide trading facility for all types of securities.

• Providing a fair, efficient and transparent securities market using electronic trading system.

• The NSE network is used to disseminate information and company announcements across the country.

• Meeting international benchmarks and standards.

2. What are the regulations of SEBI that the company must comply with?

Answer

Following are the regulations of the SEBI for new issue that the company must comply with:

• Prospectus has to be attached with every application.

• Objective of the issue and cost of project should be mentioned in the prospectus.

(iii) Company’s management, past history and present business of the firm should be highlighted in the prospectus.

(iv) Subscription list for public issue should be kept open for a minimum of 3 days and maximum of 10 days.

(v) Collections agents are not allowed to collect application money in cash.

(vi) Issue should make adequate disclosure regarding the terms and conditions of redemption, security conversion and other relevant features of the new instrument so that an investor can make reasonable determination of risks, returns, safety and liquidity of the instrument. The disclosure shall be vetted by SEBI in this regard.

3. How does the SEBI exercise control over ‘R’ limited in in the interest of investors?

Answer

SEBI should provide protection of their rights and interests through adequate, accurate information and disclosure of information on a continuous basis. Prohibition of fraudulent and unfair trade practices like making misleading statements, manipulations, price rigging etc.

Case Problem II

World Markets
NSE Indices

IndexCurrentPrev.%CHGIndexCurrentPrev.% change
S&P CNX Nifty3641.137770.55-3.43%NYSE Composite8926.889120.93-2.13%
CNX Nifty junior6458.556634.85-2.66%NASDAQ Composite2350.572402.29-2.15%
CNX IT5100.55314.05-4.02%DOW Jones I. A.1207612318.6-1.97%
Bank Nifty5039.055251.55-4.05%S&P 5001377.951406.6-2.04%
CNX1003519.353640.35-3.32%NIKK16676.917178.8-2.92%

The above figures are taken from the website of national stock exchange of India. They illustrate the movement of NSE stock indices as well as world stock indices?

1. What do you mean by stock index? How is it calculated?

Answer

A stock market index is a barometer of market behaviour. It measures overall market sentiments through a set of stocks that are representative of the market. Most of the stock market uses the following 3 methods of calculating index.
• Price Weighted Index An index reflecting the sum of the prices of the sample share in a certain year/month/week/day with reference to a base year.
• Equal Weighted Index An index reflecting the simple arithmetic average of the price relatives of a sample of shares in a certain period with reference to base year.
• Value Weighted Index It is an index reflecting the aggregate market capitalisation of the sample shares in certain period in relation to base year.

2. What conclusions can you draw from the various movements of NSE stock indices?

Answer

When comparing the previous and current index of NSE it is clear that all market Index is downward for all sectors so, we can say that there is a depression in the market.

3. What factors affect the movement of stock indices? Elaborate on the nature of these factors.

Answer

In the above table, there is a fall in the domestic market. We can see the world Indices and NSE indices are moving in the same direction.

4. What relationship do you see between the movement of indices in world markets and NSE indices?

Answer

There is a direct relation between NSE indices and world market indices. As there is a negative trend in the world index, it brings a negative trend in NSE index also.

5. Give details of all the indices mentioned above, you can find information on the web or business magazines.

Answer

The above figure shows that the movement of NSE stock indices, as well as world stock indices on the date, indicated.



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R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.1

February 14, 2019, 8:35 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.1

1. Calculate the mean for the following distribution:
x: 5 6 7 8 9
f: 4 8 14 11 3  

Solution


2. Find the mean of the following data:
x: 19 21 23 25 27 29 31
f.  13 15 16 18 16 15 1

Solution


3. If the mean of the following data is 20.6. Find the value of p.
x: 10 15 p 25 35
f:  3 10 25 7 5

Solution


4. If the mean of the following data is 15, find p.
x: 5 10 15 20 25
f: 6 p 6 10 5

Solution


5. Find the value of p for the following distribution whose mean is 16.6
x: 8  12  15  p  20  25 30
f. 12 16 20 24 16 8 4

Solution


6. Find the missing value of p for the following distribution whose mean is 12.58 .
x: 5 8 10 12  p 20 25
f. 2 5 8 22 7 4 2

Solution


7. Find the missing frequency (p) for the following distribution whose mean is 7.68.
x: 3 5 7 9 11 13
f: 6 8 15 p 8 4 

Solution


8. The following table gives the number of boys of a particular age in a class of 40 students.
Calculate the mean age of the students 
Age (in years): 15 16 17 18 19 20
No. of students: 3  8   10 10  5  4

Solution 


9. Candidates of four schools appear in a mathematics test. The data were as follows:
 
If the average score of the candidates of all the four schools is 66, find the number of
candidates that appeared from school III.

Solution


10. Five coins were simultaneously tossed 1000 times and at each toss the number of heads
were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained
are shown in the table below. Find the mean number of heads per toss.

Solution


11. The arithmetic mean of the following data is 14. Find the value of k
xi : 5 10 15 20 25
fi : 7 k 8 4 5

Solution


12. The arithmetic mean of the following data is 25, find the value of k.
xi : 5 15 25 35 45
fi : 3 k 3 6 2 

Solution


13.  If the mean of the following data is 18.75. Find the value of p.
xi : 10 15 p 25 30
fi : 5 10 7 8 2

Solution


14. Find the value of p, if the mean of the following distribution is 20.
x: 15 17 19 20+p 23
f: 2 3 4 5p 6 

Solution


15. Find the missing frequencies in the following frequency distribution if it is known that the
mean of the distribution is 50.
x: 10 30 50 70 90
f:  17 f1 32 f2 19

Solution

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R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.1

February 14, 2019, 9:13 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.1

1. Calculate the mean for the following distribution:
x: 5 6 7 8 9
f: 4 8 14 11 3  

Solution


2. Find the mean of the following data:
x: 19 21 23 25 27 29 31
f.  13 15 16 18 16 15 1

Solution


3. If the mean of the following data is 20.6. Find the value of p.
x: 10 15 p 25 35
f:  3 10 25 7 5

Solution


4. If the mean of the following data is 15, find p.
x: 5 10 15 20 25
f: 6 p 6 10 5

Solution


5. Find the value of p for the following distribution whose mean is 16.6
x: 8  12  15  p  20  25 30
f. 12 16 20 24 16 8 4

Solution


6. Find the missing value of p for the following distribution whose mean is 12.58 .
x: 5 8 10 12  p 20 25
f. 2 5 8 22 7 4 2

Solution


7. Find the missing frequency (p) for the following distribution whose mean is 7.68.
x: 3 5 7 9 11 13
f: 6 8 15 p 8 4 

Solution


8. The following table gives the number of boys of a particular age in a class of 40 students.
Calculate the mean age of the students 
Age (in years): 15 16 17 18 19 20
No. of students: 3  8   10 10  5  4

Solution 


9. Candidates of four schools appear in a mathematics test. The data were as follows:
 
If the average score of the candidates of all the four schools is 66, find the number of
candidates that appeared from school III.

Solution


10. Five coins were simultaneously tossed 1000 times and at each toss the number of heads
were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained
are shown in the table below. Find the mean number of heads per toss.

Solution


11. The arithmetic mean of the following data is 14. Find the value of k
xi : 5 10 15 20 25
fi : 7 k 8 4 5

Solution


12. The arithmetic mean of the following data is 25, find the value of k.
xi : 5 15 25 35 45
fi : 3 k 3 6 2 

Solution


13.  If the mean of the following data is 18.75. Find the value of p.
xi : 10 15 p 25 30
fi : 5 10 7 8 2

Solution


14. Find the value of p, if the mean of the following distribution is 20.
x: 15 17 19 20+p 23
f: 2 3 4 5p 6 

Solution


15. Find the missing frequencies in the following frequency distribution if it is known that the
mean of the distribution is 50.
x: 10 30 50 70 90
f:  17 f1 32 f2 19

Solution

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R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.2

February 14, 2019, 10:09 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.2

1. The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:
No. of calls(x): 0 1 2 3 4 5 6
No. of intervals (f): 15 24 29 46 54 43 39
Compute the mean number of calls per interval.  

Solution 


2. Five coins were simultaneously tossed 1000 times, and at each toss the number of heads was observed. The number of tosses during which 0,1,2,3,4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss
No. of heads per toss (x): 0     1    2    3      4     5
No. of tosses (f):              38 144 342 287 164 25

Solution


3. The following table gives the number of branches and number of plants in the garden of a
school.
No. of branches (x): 2   3   4   5   6
No. of plants     (f): 49 43 57 38 13
Calculate the average number of branches per plant. 

Solution 


4. The following table gives the number of children of 150 families in a village
No. of children (x): 0 1 2 3 4 5
No. of families (f): 10 21 55 42 15 7
Find the average number of children per family.

Solution 


5. The marks obtained out of 50, by 102 students in a Physics test are given in the frequency

table below:
Marks(x):       15 20 22 24 25 30 33 38 45
Frequency (f): 5  8   11 20 23 18 13  3  1 
Find the average number of marks. 

Solution  


6. The number of students absent in a class were recorded every day for 120 days and the information is given in the following frequency table:
No. of students absent (x): 0 1   2   3   4   5 6  7
No. of days (f):                   1 4 10 50 34 15 4 2
Find the mean number of students absent per day.

Solution 


7. In the first proof reading of a book containing 300 pages the following distribution of
misprints was obtained:
No. of misprints per page (x): 0     1     2    3   4   5
No. of pages (f):                      154 95  36   9   5   1
Find the average number of misprints per page.  

Solution  


8. The following distribution gives the number of accidents met by 160 workers in a factory
during a month.
No. of accidents (x): 0 1 2 3 4
No. of workers (f): 70 52 34 3 1
Find the average number of accidents per worker.

Solution  


9. Find the mean from the following frequency distribution of marks at a test in statistics:
Marks(x): 5 10 15 20 25 30 35 40 45 50
No. of students (f): 15 50 80 76 72 45 39 9 8 6

Solution 

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Search

NCERT Solutions for Class 12th: Ch 11 Marketing (VSQs and SAQs)

February 14, 2019, 12:07 pm
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NCERT Solutions for Class 12th: Ch 11 Marketing (VSQs and SAQs)

Exercises

Page No. 289

Very Short Answer Questions:

 1. Explain the advantage of branding to marketers of goods and services.

Answer

There are following factors helps to branding to marketers of goods and services:

• Brand name helps in creating product differentiation.
• Helps in advertising and Display programmes.
• Branding enables a firm to charge different price for its products than that charged by its competitors.
• Ease in introduction of new product.

2. List the characteristics of a good brand name?

Answer

• The brand name should be short, easy to pronounce, spell, recognise and remember.
• A brand name should be distinctive.
• Must be good packing and labelling.

3. What is the societal concept of marketing?

Answer

The societal concept of marketing is not pays attention towards customer benefit and their interest they also pay attention to the social, ethical and ecological aspects of marketing.

4. List the characteristics of convenience products?

Answer

Convenience products are purchased frequently, immediately with least time and efforts and it is regular and continuous in demand. Example: Ice- cream, medicines, stationary items.

5. Enlist the advantages of packaging of a consumer products?

Answer

• Rising standards of health and sanitation.

• Self service outlets.

• Innovational opportunity e.g., new types of packaging availability have made it easier to market the product.

• Product Differentiation Packaging is one of the very important means creating product differentiation.

6. What are the limitations of a advertising as a promotional tool? Enlist.

Answer

Following are the limitations of advertising:

• Less forceful.

• Lack of feedback.

• Inflexibility.

• Low effectiveness.

7. List five shopping products purchased by you or your family during the last few months.

Answer 

• Curtains
• Shirts
• Cosmetics
• Travelling bags
• Hand bags

Short Answer Type Questions

1. What is marketing? What functions does it play with process of exchange of goods and services? Explain.

Answer

Marketing has been referred to as performance of business activities that direct the flow of goods and services from producers to consumers which involves many activities.

The functions does it play with process of exchange of goods and services:

• Gathering and analysing market information: collecting and analysing market information helps to take a decision for the successful marketing of the products and services and needs of customers and know market trends.

• Marketing planning: Another important activity or area of work of a marketer is to develop appropriate marketing plans, so that the marketing objective of the organisation can be achieved.

• Product Designing and Development: Designing of a product makes it attractive so, customer can easily attract. A good design can improve performance of a product and give it a competitive advantage in the market.

• Standardisation and Grading: Standardisation refers to producing goods of predetermined specification which helps in achieving uniformity and consistency in the output which reduces the need for inspection, testing and evaluation of the products.

Grading is the process of classification of products into different groups, on the basis of its features such as quality, size etc. It ensures that goods belong to a particular quality helps in realising higher prices for high quality output.

• Packaging and Labelling: Packaging refers to designing the package for the products. Labelling refers to designing the label to be put on the package. Packaging provides protection to the product and also helps in its promotion. Labelling helps in self -service. Through labelling we know the content of the product, price.

• Branding: Brand names help in creating product differentiations i.e., how the product can be distinguished from its competitors.

• Customer Support Service: Marketing management relates to developing customer support service such as after sales services, handling customer complaints. All these aim at provides customer satisfaction which is a key to marketing success.

• Pricing of Product: Price is an important factor affecting the success or failure of a product in the market. The marketers have to analyse properly the factors determining the price of a product.

• Promotion: Promotion of products and services involves informing the customers about the firm’s product, its features etc and persuading them to purchase these products. It includes four method advertising, sales promotion, personal selling and publicity

• Physical Distribution: The important decisions areas under physical distribution include managing inventory, storage, warehousing and transportation of goods from one place to the other.

• Transportation: Transportation involves physical movement of goods from one place to another. A marketer has to perform this function very efficiently keeping in mind the nature of product, cost, location of target market etc.

• Storage or Warehousing: In order to maintain smooth flow of products in the market, there is a need for proper storage of the products. Further, there is a need for storage of adequate stock of goods to protect against unavoidable delays in delivery or to meet ‘out contingencies in the demand. Wholesalers and retailers are playing an important role.

2. Distinguish between the product concept and production concept of marketing.

Answer

Basis of Difference
Production Concept
Product Concept
Starting Point
FactoryFactory
Main FocusQuantity of ProductQuality Performance, Lectures of product
MeansAvailability and Affordability of ProductProduct Improvements
EndsProfit through volume of productProfit through product quality

3. Product is a bundle of utilities? Do you agree? Comment.

Answer

Yes, I agree anything that can be of value to the buyer can be termed as a product, satisfied consumer needs.

4. What are industrial products? How are they different from consumer products? Explain.

Answer

Industrial products are those products, which are used as inputs in producing other products.
Example of industrial products is raw materials, engines, lubricants, etc. it is used for non- personal and business uses. The market for industrial product consists of manufacturers, transport agencies, banks etc.
Whereas Example of consumer products is soap, edible oils, eatables. It is used it is used for personal and non- business.

5. Distinguish between convenience product and shopping product.

Answer

Convenience Product
Shopping Product
These products are purchased at convenient locations, with least effort and time.The shopping products are generally of durable nature.
These products have small unit of purchased and low price.High price, customer pre- planned before making purchased.

6. ‘ Product is a mixture of tangible and intangible attributes’. Discuss
Answer

When we think about a product, it refers only to the tangible attributes of a product example we say have bought a bike. But while we take a buying decision we along with tangible attributes also consider the intangible attributes of a product example brand name, image of the company, guarantee, warranty, packing. So, product is a mixture of tangible and intangible attributes which are capable of being exchanged for a consideration and satisfy the customer needs.

7. Describe the functions of labelling in the marketing of products.
Answer

The functions of labelling in the marketing of product:

• Describe the product and specify its contents.

• Identification of the product or brand.

• Grading of product.

• Help in promotion of products.

• Providing information required by law.

8. Discuss the role of intermediaries in the distribution of consumer non-durable products.
Answer
The term channels of distribution refers the facilitate to the movement of goods and services and their title between the point of production and point of consumption, by performing a variety of marketing activities. Following are the functions performed by the channels of distribution

• Accumulation: It aims at holding the stock to match between the consumer demand and supply condition, warehousing helps in maintaining continuous flow of goods and services.

• Promotion: The marketing channels also help in promoting the demand for the product by displaying demonstrating and participating in various promotional activities organised by the producers.

• Negotiating: The marketing channels are the intermediaries between the producers and the consumers. They attempt to reach final agreement on price and other terms of the offer, so that transfer of ownership is properly affected.

• Risk Taking: Risk taking is the basic responsibility of the intermediaries. It may arise out of physical deteriorations, changes in price levels, natural calamities, change in fashion etc. These are unavoidable as they hold sufficiently large and variety of inventories till the sale of stock.

• Grading/Sorting: Grading is the process whereby they sort the products on the basis of different sizes, qualities, moisture contents and so on. It helps us realising the time value for the product and at the same time ultimate consumer feels satisfied with the uniform quality of the product.

• Packaging: The products are packed in the small tradable lots for the convenience of the consumer.

• Assembling/Assortment: Marketing channels aim at satisfying the needs of the customers. The products desired by the consumer may not be available in the market. They procure such goods from different sources, assemble or assort them as per the requirements of the consumers.

9. Explain the factors determining choice of channels of distribution.
Answer
The choice of channels depend on various factors, which are discussed as follows

• Consideration related to product: When the manufacturer selects some channel of distribution he/she should take care of following factors like unit value of the product, standardised or customised product, perishability, technical nature.

• Consideration related to market: the important market characteristics affecting the choice of channel distribution like number of buyers, types of buyers, their buying habit, buying quantity and size of market also depend.

• Consideration related to the company: The important company characteristics affecting the choice of channel distribution include goodwill, financial strength and desire to control the channel of distribution.

• Consideration related to government also affect the selection of channel because only licence holder can carry.

10. Explain briefly the components of physical distribution.
Answer
There are mainly four components of physical distribution

• Transportation: transportation is that activity through which products are moved from one place to another. By making the products reach a desirable place can increase the importance and value of those products. while choosing the means of transportation, the following elements kept in mind: cost, speed, dependability, frequency, power, safety.

• Inventory: Inventory Control Inventories ensure the availability of the product as and when consumer demand arises. There are various factors which influence a firm decision regarding the level of inventory e.g., degree of accuracy of sales forecast, cost of blocking of the working capital etc.

• Ware housing: under ware housing activity the following decision regarding the inventory of material are taken:

• Which is better option? (to own or to rent a warehouse)
• Which is the right location for a warehouse? (nearer the factory or nearer the market)
• Which decision is more appropriate?
Warehousing requires investment, so after analysing its advantages and usefulness desirable decision should be taken .

11. Define advertising. What are its main features? Explain.
Answer
Advertising means providing adequate knowledge about some special product/ service/ idea to potential consumers so that they are stimulated to buy it.

Features of Advertising

• Paid form: only that attempt of the seller on which he/ she has spent some money to communicate information to the consumer is called advertising. It means any information about a product communicated free of cost cannot be termed as advertising it may be called propaganda or publicity.

• Impersonal presentation: advertising is non- personal presentation of information. It means the advertiser and the consumer do not come into personal contact.

• Speedy and mass communication: advertising is a speedy medium of communication in other words, it reaches millions of people simultaneously.

• Identified sponsor: sponsor name should be identified when the sponsor name not be identified it will be called publicity, not advertisement.

12. Discuss the role of ‘sales promotion’ as an element of promotion mix?
Answer
Sales promotion includes those marketing activities other than personal selling, advertising and publicity that stimulate consumer purchasing and dealer effectiveness, such as display, shows and exhibitions, demonstrations and various non-current selling efforts not in the ordinary routine, The main objectives of sales promotion activities are

• Creation of demand for the product.

• Educating the consumers about new products or new uses of the old product.

• Building brand loyalty for the product among the consumers.

NCERT Solutions of Chapter 11 Marketing (LAQs)

Notes of Chapter 11 Marketing

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NCERT Solutions for Class 12th: Ch 11 Marketing (LAQs)

February 15, 2019, 6:07 am
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NCERT Solutions for Class 12th: Ch 12 (VSQs and SAQs)

Exercises

Page No. 289

Very Short Answer Questions:

 1) Define marketing. How is it different from selling? Discuss.

Answer

It refers to that process under which valuable goods/ services are created, offered and by doing transaction independently, the needs are satisfied. These activities include the following:

Gathering and analysing market information, Marketing planning, product designing, standardisation and grading, packaging and labelling, branding etc.

Basis
Selling
Marketing
Starting point
The selling starts after production has taken place.It starts before goods are produced.
MeansIt is achieved through advertisement personal selling and sales promotion.Various marketing activities.
Main focusAttracting consumers.Consumer satisfaction.
Ends/profit maximisation  Profit maximisation is achieved through volume of sales.Profit maximisation is achieved through customer satisfaction.

2. What is marketing concept? How does it help in the effective marketing of goods and services?

Answer

Orientation of marketing implies that focus on the satisfaction of customers need, is the key to the success of any organisation in the market. All the decisions in the firm are taken from the point of view of the customers. e.g., What product will be produced, with what features and at what price shall it be sold or where shall it be made available for sale will depend on what do the customer wants.
Marketing concept helps in effective marketing of goods and services by using the following
(i) Identification of market or customer who are chosen e.s the target of marketing effort.
(ii) Understanding needs and wants of customers in the target market.
(iii) Development of products or services for satisfying needs of the target market.
(iv) Satisfying needs of target market better than the competitors.
(v) Doing all this at a profit.

3. What is marketing mix? What are its main elements? Explain.

Answer

Marketing mix refers to the aggregate of policies formulated with a view of successfully completing the different marketing activities.

Elements of marketing mix:

• Product mix – Product mix refers to all the decisions relating to the product. These decisions are mainly with regard to branding, packaging, labelling, colour, design, quality, size and weight of the product.

• Price mix- It refers to all those decisions which are concerned with the price fixation of any product or service. Under price mix, besides fixing the price of the product or service, decisions regarding credit sale, discount, etc are also included. Price mix includes the following decisions:
→ Determining of price of product or service.
→ Determining of trade and cash discount.
→ Credit policy.
→ Period of credit and procedure to be adopted in case of non-receipt of payment.

• Promotion mix- promotion mix refers to a combination of promotional tools used by the business to inform and persuade the customers about the products. Promotion mix include advertising, personal selling, sales promotion, publicity.

• Place mix- Place mix refers to the combination of all the decisions relating to make products available to consumers. If the product is not available on the right time, in the right quantity and at the right place then the consumer will not be able to buy it.

4. How does branding help in creating product differentiation? Does it help in marketing of goods and services? Explain.

Answer

Branding helps a firm in distinguishing its products from that of its competitors. This helps the firm to secure and control the market for its products. If products were sold by generic names, it would be very difficult for marketers to distinguish their products from its competitors. Thus, most marketers give a name to their product, which helps in identifying and distinguishing their products from their competitor product. This process of giving a name or a sign or a symbol etc to a product is called Branding.

5. What are the factors affecting determination of the price of a product or service? Explain.

Answer

The factors affecting determination of the price of a product or service:

• Cost of production: Before price fixation, it is necessary to compile data relating to the cost of production and keep that in mind. Two types of cost
(i) Fixed cost.
(ii) Variables cost.
At least the price should be able to recover the variables cost.

• Demand for product: If demand is relatively more than supply, higher price can be fixed.

• Price of competing firms: The price is also affected by the nature and degree of competition. The price will tend to reach the upper limit in case there is less degree of competition while under free competition, the price will tend to be set at the lowest level.

• Purchasing power of customers: what is the purchasing power of the customers and at what price and how much they can purchase?

• Government regulation: sometimes price of the commodity is also fixed by the government.
Marketing methods used: Price is also affected by other elements of marketing such as distribution system, quality of salesmen employed, quality and amount of advertising, sales promotion efforts, the type of packaging, product differentiation, credit facility and customer service provided.

6. What do you mean by ‘Channels of distribution’? What functions do they play in the distribution of goods and services? Explain.

Answer

Channels of distribution Smoothen the flow of goods by creating possession, place and time utilities. They facilitate movement of goods by overcoming various time, place and possession barriers that exist between the manufacturers and consumers.
The important functions they play in the distribution of goods and services.

• Sorting: Middleman procure supplies of goods from a variety of sources which is often not of the same quality, nature and size.

• Accumulation: This function involves accumulation of goods into larger homogeneous stock, which help in maintaining continuous flow of supply.

• Allocation: Allocation involves breaking homogenous stock in to smaller, marketable lots.

• Assorting: Middlemen build assortment of products for resale. There is usually a difference between the product lines made by manufacturers and the assortment or combinations desired by the users. Middleman produces variety of goods from different sources and delivers them in combinations, desired by customers.

• Product promotion: Advertising and other sales promotion activities are organised by manufacturers. Middlemen also participate in certain activities such as demonstrations, special displays, contest, etc. to increase the sale of products.

• Negotiation: Channels operate with manufacturers on the one hand and customer on the other. They negotiate the price, quality, guarantee and other related matters with customers, so that transfer of ownership is properly affected.

• Risk Taking: Risk taking in the sense of the price, demand fluctuations, spoilage, destructions.

7. Explain the major activities involved in the physical distribution of the product?

Answer

The major activities involved in the physical distribution of the product are transportation, inventory, warehousing, order processing.

• Transportation: Transportation is that activity through which products are moved from one place to another. Transportation involves cost. While choosing the means of transportation, the following elements should be kept in consideration: Cost, speed, dependability, frequency, power, safety.

• Inventory: Inventory means the stock of raw materials, semi-finished goods and finished goods held in anticipation of sales or use. The main reason as to why this decision is important is that if the inventory is either more or less than required.
Warehousing: warehousing requires investment so after analysing its advantages and usefulness desirable decision should be taken.
→ Which is better option? (To own or to rent a warehouse)
→ Which is the right location for a warehouse? (Nearer the factory or nearer the market)
Order processing: order processing means the process which is followed to fulfil the materials order of the customer. Customer satisfaction and speed of shipment are directly interrelated.

8. ‘Expenditure on advertising is a social waste’ Do you agree? Discuss.

Answer

Advertising is regarded as the life-blood of modern business but on the other hand, some scholars are of view that money spent on advertising is a waste.

Following are the points of criticism:

• Adds to costs:  An organisation has to spend large amount on advertising, It increases the cost of the products to meet this expenditure, price of the product is raised because of this reason advertising costs are passed on to the consumers in the form of high prices.

Undermines Social Values: Advertising undermines social values and promotes materialism. It breeds discontentment among people as they come to know about new products and feel dissatisfied with their present state of affairs. This criticism is not entirely time. Advertisement in fact helps buyer by informing them about the new products which may be improvement over the existing products.

Confuses the Buyers: Through advertisement so many products are being advertised which makes similar claims that the buyers gets confused as to which one is true and which one should be relied upon.

Encourage sale of Inferior products: Advertisement does not distinguish between superior and inferior products and persuade people to purchase even the inferior products.

Some Advertisement are in Bad taste: Many times, foul language and objectionable picture are used in advertising in order to attract a particular class and may be hurt their feelings.

9. Distinguish between advertising and personal Selling.

Answer

Basis of difference 
Advertising
Personal selling
FormThis is impersonal.This is personal.
Message There is uniformity means message is same for all the customers.The message has no uniformity which means it can be changed keeping in view the behaviour of the customers.
FlexibilityIt lacks flexibility.It is completely flexible.
ReachMessage can be carried to numerous of people at same time.Through this medium the message is conveyed to a single people or a group of person at a time.
CostLess costly method.More costly method.
Time It takes a little time in conveying any information to the customers.It takes a more time in conveying information to the customers.
Media T.V, radio, newspaper, magazines.Through salesman.
FeedbackThis gives no information about the reactions of the customers.The reaction of the customers becomes immediately known or clear.

Application Type Questions

1. As marketing manager of a big hotel located at an important tourist destination, what societal concerns would be facted by you and what steps would you plan to take care of these concerns? Discuss.

Answer

The societal marketing concept holds that the task of any organisation is to identify the needs and wants of the target market and deliver the desired satisfaction in an effective and efficient manner, so that the long term well-being of the consumers and the society is taken care of. In case any business activity encourages pollution, deforestation, storage of resources, population explosion, then its benefits can not be justified. As marketing manager of a big hotel located at an important tourist destination certain care need to be taken regarding environment.
(i) Proper drainage facility
(ii) In-built plant for re-cycling the waste
(iii) Solar geysers to be used
(iv) Rain water harvesting
(v) proper maintainance of greenery-lawns, parks, gardens
(vi) CNG based vehicles to be used to provide transport facility to the guests.

2. Suppose you are the marketing Vice President of an insurance company, planning to design a new mediclaim policy for senior citizens. What information would you like lo collect to perform this task and how will, you collect such information? Discuss.

Answer

The following information about the senior citizens should be collected
(i) Age of the senior citizen
(ii) Their source of income
(iii) Medical background
(iv) Present working states
There are various sources of collecting the above information – personal visits, questionnaires, medical reports from nursing homes, doctor clinics etc.

3. What shopping products have been purchased by you/your family in the last six months? Make a list and specify what factors influenced the purchase of each of these products.

Answer

The following shopping goods were purchased by my family
• Clothes Price, fashion, occasion for which they were bought.
• Refrigerator Brand, price, features, books, durability.
• Shoes Size, price, brand , material used.
• Furniture (study table) Design, quality, finishing, comfort level, cost.

4. What information is generally placed on the package of a food product? Design a label for one of the food product of your choice.

Answer

The following information is normally placed on the package of good product
• Name of the product
• Brand name
• Veg/Non-veg sign (green/red dot)
• Price
• Manufacturing date and date of expiry
• Ingredients
• Net weight
• Directions for use
• FPO mark
• Preservatives used

5. For buyers of consumer durable products, what ‘customer care services’ would you plan as a manager of a firm marketing new brand of motorcycle. Discuss.

Answer

For marketing motorcycles, the following customer care services can be planned
• Specified period warranties
• Easy monthly instalments
• Exchange offer
• 0% finance scheme
• Free servicing

Case Problem

Nokia takes four-lane road to consumers.

NEW DELHI After having grabbed a king size 79% share of the 15,000 crore mobile handset market in India, Nokia India has found a new way of connecting people.
The mobile handset manufacturer has embarked upon a brand new retail strategy that is based on a classification of its consumers into four major groups that separates people in terms of usage, income level and lifestyle.

The classification is based on an extensive survey— the Nokia Segmentation Study —that was carried over two years involving 42,000 consumers from 16 countries. It studied the impact lifestyle choices and attitudes have on the mobile devices consumers buy and how they use them.
The strategy, which was announced globally in June last year, is being unfolded in India now. While the nitty-gritty of the new strategy is still being worked out, it is likely that the company would follow separate marketing strategies for the four different segments. The advertising campaigns could be different for the segments.

Nokia’s entire product portfolio has now been re-aligned towards these four groups to address the specific needs of each. The first of these segments Live, aimed a first time users whose basic need is to stay in touch with voice as the main driver, would have basic handsets low on features and price.
“These may be functional phones but the target group for these phones range from SEC C (low socio-economic class) to SEC A1+ (very high socio-economic class) markets”, says Nokia India marketing head Devinder Kishore. The second segment Connect looks at more evolved users who look for more functionality and features and connectivity. Accordingly, phones in this segment would have GPRS, camera and music capabilities.

The next two categories, Achieve and Explore, are aimed at high-end users and have Nokia’s top-end handsets, e.g., Achieve segment looks at enterprise users who need to have business functionalities in their phones. Nokia’s new E-series has been put under this segment with handsets having QWERTY keyboards and full Internet capabilities.

Aimed at high-end lifestyle users, Explore would be the most prominent segment for the company in the coming years. Says Nokia India multimedia business director Vineet Taneja, “This segment would see the most vibrant growth in the coming year. It will look at five different areas—applications, imaging, mobile, TV, music and gaming. We are fast developing (the ecosystem to support these areas.”

Nokia acquired music solution and content provider LoudEye and GPS solution provider Gate5. It is all slated to launch its most high-profile handset, which boasts of having a 5 mega pixel camera and GPS capabilities apart from iPod quality music, in February.

Says Taneja, “There is increasing demand for convergence and multiple functionalities in high-end handsets. The N-series will try to address that”. Nokia feels that the new platform strategy wherein different handsets are launched under a platform, like the N-Series, will become a status and style statement and drive numbers.

1. Identify the four market segments that Nokia plans to address as per the news report above.

Answer

Live, Connect, Achieve and Explore.

2. What is the basis of classification of the market used by the company?

Answer

SEC Socio-Economic Class, usage and lifestyle

3. What do you mean by realignment of product portfolio? Illustrate this from the case above.

Answer

It means that whatever product Nokia is planning to develop now, it will be according to the needs of the consumers. The four different handsets are Live, Connect, Achieve and Explore, being planned keeping the needs of four different types of users.

4. Identify the points that can be highlighted in marketing campaigns for each segment.

Answer

The points that can be highlighted in marketing campaigns for each segment can be
• Latest model
• Reasonable price
• Better performance
• Advanced technology
• Consumer friendly

5. What are the different considerations in the mind of consumers of each segment while purchasing mobile phones as given in the above case?

Answer

Different considerations in the mind of each segment while purchasing mobile phones are:

• The first of these segments Live, aimed at first time users whose basic need is to stay in touch with voice as the main driver. So, here price of the phone is the main consideration.

• The second segment connect looks at more evolved users who look for more functionality and features and connectivity. So, here the features of the phone as well as an economic price tag are considered.

• The next two categories, Achieve and Explore, are aimed at high-end users. So, here the uniqueness of the handset and its business functionalities are the main points considered by the consumer.

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R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.3

February 15, 2019, 10:40 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.3

1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.  
Find the average expenditure (in rupees) per household. 

Solution


2. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house .
Number of plants: 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses: 1  2  1 5 6  2  3
Which method did you use for finding the mean, and why?

Solution


3. Consider the following distribution of daily wages of 50 workers of a factory
Daily wages (in Rs). 100-120 120-140 140-160 160-180 180-200
Number of workers: 12 14  8  6  10
Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution


4. Thirty women were examined in a hospital by a doctor and the number of heart beats per
minute recorded and summarized as follows. Find the mean heart beats per minute for these
women, choosing a suitable method.
Number of heat beats per minute: 65 - 68 68 - 71 71 - 74 74 - 77 77 - 80 80 - 83 83 -86
Number of women: 2  4   3  8  7   4  2

Solution


Find the mean of each of the following frequency distributions: (5 − 14)
5. Class interval: 0-6 6-12 12-18 18-24 24-30
Frequency:  6  8  10  9  7

Solution 

 

6. Class interval: 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 150 – 170
Frequency: 18 12 13 27 8 22

Solution


7. Class interval: 0-8 8- 16 16- 24 24-32 32-40
Frequency: 6 7 10 8 9

Solution


8. Class interval: 0-6 6-12 12-18 18-24 24-30
Frequency: 7 5 10 12 6

Solution


9. Class interval: 0- 10 10- 20 20-30 30-40 40-50
Frequency: 9 12 15 10 14

Solution 


10. Class interval: 0-8 8- 16 16-24 24-32 32-40
Frequency: 5 9 10 8 8

Solution

 

11. Class interval: 0-8 8- 16 16- 24 24-32 32-40
Frequency: 5 6 4 3 2

Solution


12. Class interval: 10-30 30-50 50-70 70-90 90-110 110- 130
Frequency: 5  8 12 20 3 2

Solution


13. Class interval: 25-35 35-45 45-55 55 - 65 65 -75
Frequency: 6 10 8 12 4

Solution


14. Classes: 25 -29 30-34 35-39 40-44 45-49 50-54 55-59
Frequency: 14 22 16 6 5 3 4

Solution


15. For the following distribution, calculate mean using all suitable methods:
Size of item: 1-4 4-9 9- 16 16-27
Frequency: 6 12 26 20

Solution


16. The weekly observations on cost of living index in a certain city for the year 2004 - 2005 are given below. Compute the weekly cost of living index.

Solution


17. The following table shows the marks scored by 140 students in an examination of a certain paper:
Marks: 0-10 10-20 20-30 30-40 40-50
Number of students: 20 24 40 36 20
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.


Solution


18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequency f1 and f2 .
Class: 0-20 20-40 40-60 60-80 80-100 100–120
Frequency: 5 f1 10 f2 7 8

Solution


19. The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
Class interval: 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Frequency: 7 6 9 13 - 5 4

Solution


20. If the mean of the following distribution is 27, find the value of p.
Class: 0-10 10-20 20-30 30-40 40-50
Frequency: 8  p 12 13 10

Solution


21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes
contained varying number of mangoes. The following was the distribution of mangoes
according to the number of boxes.
Number of mangoes: 50-52 53-55 56-58 59-61 62-64
Number of boxes: 15 110 135 115 25
Find the mean number of mangoes kept in a packing box. Which method of finding the
mean did you choose ?

Solution


22. The table below shows the daily expenditure on food of 25 households in a locality
Daily expenditure (in Rs): 100-150 150-200 200-250 250-300 300-350
Number of households: 4 5 1 2  2 2
Find the mean daily expenditure on food by a suitable method.

Solution


23. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was
collected for 30 localities in a certain city and is presented below . 
Find the mean concentration of SO2 in the air. 

Solution


24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days:  0-6 6-10 10-14 14-20 20-28 28-38 38-40
Number of students: 11 10  7  4  4  3 1  

Solution


25. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %): 45-55 55-65 65-75 75-85 85-95
Number of cities: 3 10 11  8 3

Solution


26 . The following is the cummlative frequency distribution (of less than type) of 100 persons each of age 20 years and above . Determine the mean age .
Age below (in years) : 30  40    50   60  70    80
Number of persons :  100 220 350 760  950 1000

Solution

We know that mean , x̅ = A + h(1/N Σfiui)
Now, we have N = Σfi = 1000, h = 10, A = 55, Σfiui = - 370
x̅ = 55 + 10 [1/100 × (-370)]
= 55 - 3.7
= 51.3 years

27. If the mean of the following frequency distribution is 18, find the missing frequency .
Class intervals : 11-13 13 -15 15 -17 17-19 19-21 21-23 23-25
Frequency : 3 6  9 13  f  5 4 

Solution


We know that mean , x̅ = A + h(1/N Σfiui)
Now, we have N = Σfi = 40 + f, h = 2, A = 18, Σfiui = -8 + f
= 18 = 18 + 2 [1/(40+f) × (-8 + f)
⇒ 0 = -16+2f/40+f
⇒ -16 + 2f = 0
⇒ f = 8

28. Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.
Class: 0-20 20-40 40-60 60-80 80-100
Frequency :  17 fi  32 f2 19 

Solution

We know that mean , x̅ = A + h(1/N Σfiui)
Now, we have N = 68 + f1 + f2, h = 20, A = 50, Σfiui = 4 - f1 + f2.
50 = 50 + 20 (4-f1+f2/68+f1+f
2)
⇒ f1 - f2 = 4
⇒ f1 = 4 + f2 .....(1)
Now, N = 68 + f1 + f2 = 120
f1 + f2 = 120 - 68 = 52
⇒ 4 + f2 + f2 = 52 [Using (1)]
⇒ f2 = 24
so,
f1 = 4 + 24 = 28

29. The daily incomes of a sample of 50 employees are tabulated as follows : 
Income: 1-200 201-400 401-600 601-800
No.of employees : 14  15  14  7
Find the mean daily income of employees . 

Solution


We know that mean , x̅ = A + h(1/N Σfiui)
Now, we have N = Σfi = 50, h = 200, A = 500.5, Σfiui = -36
x̅ = 500.5 + 200 [1/50 × (-36)]
= 500.5 - 144
= 356.5  
↧

R.D. Sharma Solutions Class 10th: Ch 7 Statistics Exercise 7.4

February 16, 2019, 12:21 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.4

1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median:
715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution


2. The following is the distribution of height of students of a certain class in a certain city:
Height (in cm): 160-162 163-165 166-168 169-171 172-174
No. of students:  15  118 142 127 18
Find the median height. 

Solution


3. Following is the distribution of I.Q. of loo students. Find the median I.Q.
I.Q.:   55-64  65-74  75-84  85-94  95-104  105-114  115-124  125-134  135-144
No of Students: 1  2  9  22  33  22  8  2  1

Solution


4. Calculate the median from the following data:
Rent (in Rs.): 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95
No. of Houses: 8  10  15  25  40  20  15  7 

Solution


5. Calculate the median from the following data:
Marks below: 10 20 30 40 50  60   70 80
No. of students: 15 35 60 84 96 127 198 250

Solution


6. Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Age in years:  0-10 10-20 20-30 30-40 40-50
No. of persons: 5  25  ?  18  7

Solution


7. The following table gives the frequency distribution of married women by age at marriage:
Calculate the median and interpret the results

Solution


8. The following table gives the distribution of the life time of 400 neon lamps:  
Find the median life .

Solution


9. The distribution below gives the weight of 30 students in a class. Find the median weight of students:
Weight (in kg): 40-45 45-50 50-55 55-60 60-65 65-70 70-75
No. of students:  2  3  8  6  6  3  2

Solution


10. Find the missing frequencies and the median for the following distribution if the mean is
1.46.
No. of accidents:  0  1  2  3   4  5  Total
Frequency (No. of days): 46  ?  ?  25  10  5  200

Solution


11. An incomplete distribution is given below:
Variable: 10-20  20-30  30-40  40-50  50-60  60-70  70-80
Frequency: 12  30  -  6  5 -  25   18
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.

Solution

(i)

(ii) 

12. If the median of the following frequency distribution is 28.5 find the missing frequencies:
Class interval: 0-10 10-20 20-30 30-40 40-50 50-60 Total
Frequency:         5        f 1     20       15     f 2      5      60 

Sol.


13. The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data: 

Solution 


14. If the median of the following data is 32.5, find the missing frequencies.
Class interval: 0- 10  10-20  20-30  30-40  40-50  50-60  60-70  Total
Frequency: f1  5  9  12  f2  3  2  40

Solution


15. Compute the median for each of the following data:  
(i)
(ii)

Solution 

(i)

(ii)

16. A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and
the following data was obtained:
Height in cm  Number of Girls
Less than 140   4
Less than 145  11
Less than 150  29
Less than 155  40
Less than 160  46
Less than l65   51
Find the median height.

Solution

To calculate the median height, we need to find the class intervals and their corresponding frequencies .
The given distribution being of thee less than type 140, 145, 150,…..,165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 145, 150,….., 160, 165 observe that from the given distribution, we find that there are 4-girls with height less than 140 is 4. Now there are 4 girls with heights less than 140. Therefore, the number of girls with height in the interval 140, 145 is 11- 4=7, similarly. The frequencies of 145 150 is 29-11=18, for 150-155 it is 40-29=11, and so on so our frequencies distribution becomes.

17. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
Age in years  Number of policy holders
Below 20   2
Below 25   6
Below 30   24
Below 35   45
Below 40   78
Below 45   89
Below 50   92
Below 55   98
Below 60   100

Solution



18. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length (in mm): 118-126  127-135  136-144  145-153  154-162  163-171  172-180
No. of leaves:  3   5   9  12   5  4  2
Find the mean length of life. 

Solution


19. An incomplete distribution is given as follows:
Variable: 0-10 10-20  20-30  30-40  40–50  50-60  60-70
Frequency: 10  20  ?  40  ?  25 15
You are given that the median value is 35 and the sum of all the frequencies is 170. Using
the median formula, fill up the missing frequencies. 

Solution


20. The median of the distribution given below is 14.4 . Find the values of x and y , if the total frequency is 20. 
Class interval : 0-6  6-12 12-18 18-24 24-30
Frequency :        4       x       5        y       1

Solution


21.  The median of the following data is 50. Find the value of p and q , if the sum of all the frequencies is 90. 

Solution


↧

R.D. Sharma Solutions Class 10th: Ch 7 Statistics MCQ's

February 16, 2019, 3:28 am
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Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math MCQ's

Multiple Choice Questions

1. Which of the following is not a measure of central tendency ? 
(a) Median
(b) Median
(c) Mode
(d) Standard deviation 

Solution

Standard deviation is not a measure of central tendency
Hence , the correct option is c

2. The algebraic sum of the deviations of a frequency distribution from its mean is
(a) Always positive
(b) Always Negative
(c) 0
(d) A non – zero number

Solution

The algebraic sum of the deviation of a frequency distribution from its mean is zero .
Hence , the correct option is c. 

3. The arithmetic mean of 1,2,3,…,n is
(a) n+1/2
(b) n-1/2
(c) n/2
(d) n/2 + 1 

Solution

Arithmetic mean of 1,2,3,…,n
= 1+2+3+…+n/n
= n(n+1)/2/n
= n+1/2
Hence , the correct option is a .

4. For a frequency distribution, mean, median and mode are connected by the relation
(a) Mode = 3 Mean – 2 Median
(b) Mode = 2 median – 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean

Solution

The relation between mean, median and mode is
Mode = 3 Median – 2 Mean 
Hence, the correct option is c .

5. Which of the following cannot be determined graphically ?
(a) Mean
(b) Median
(c) Mode
(d) None of these

Solution

‘Mean’ cannot be determined by graphically.
Hence, the correct option is a .

6. The median of a given frequency distribution is found graphically with the help of 
(a) Histogram
(b) Frequency curve
(c) Frequency Polygon
(d) Ogive

Solution

The median of a given frequency distribution is found graphically with the help of ‘ogive’ .
Hence , the correct option is d .

7. The mode of a frequency distribution can be determined graphically from
(a) Histogram
(b) Frequency polygon
(c) Ogive
(d) Frequency curve 

Solution

The mode of a frequency distribution can be determined from ‘’Histogram’’ .
Hence, the correct option is a.      

8. Mode is
(a) Least frequency value
(b) Middle most value
(c) Most frequent value
(d) None of these

Solution

Mode is ‘’ Most frequent value’’ .
Hence , the correct option is c .

9. The mean of n observation is x̅ . If the first item is increased by 1, second by 2 and so on, then the new mean is 
(a) x̅ + n
(b) x̅ + n/2
(c) x̅ + n+1/2
(d) None of these

Solution

Let x1 , x2 , x3, …..,xn be the n observations .
Mean = x̅ = x1+x2+…+xn/n
⇒ x1 + x2 +x3 + …+ xn = n x̅
If the first item is increased by 1, second by 2 and so on .
Then, the new observations are x1 + x2 + 2 , x3 + 3, …, xn + n.
New mean = (x1 + 1) + (x2+2) + (x3 + 3) + … +(xn+n)/n
= x1+ x2 + x3 + … + xn + (1+2+3+…+n)
= n x̅ + (n(n+1)/2)/n
= x̅ + n+1/2
Hence , the correct answer is c .

10. One of the methods of determining mode is
(a) Mode = 2 median – 3 mean
(b) Mode = 2 median + 3 mean
(c) Mode = 3 median – 2 mean
(d) Mode = 3 median + 2 mean

Solution

We have ,
Mode = 3 median – 2 mean
Hence , the correct option is c .

11. If the mean of the following distribution is 2.6 , then the value of y is
Variable (x) : 1 2 3 4 5
Frequency     4 5 y 1 2

Solution

Now,
Mean = Σfx/ Σf
2.6/1 = 28 + 3y/12 + y
31.2 + 2.6y = 28 + 3y
0.4y = 3.2
y = 3.2/0.4
y = 8
Hence, the correct option is b .

12. The relationship between mean, median and mode for a moderately skewed distribution is
(a) Mode = 2 median – 3 mean
(b) Mode = median – 2 mean
(c) Mode = 2 median – mean
(d) Mode = 3 median – 2 mean

Solution

Mode = 3 median - 2 mean 
Hence, the correct option is d . 

13. The mean of a discrete frequency distribution xi / fi, i = 1,2,…,n is given by

Solution

The mean of discrete frequency distribution is
x̅ = Σfixi/Σfi
Hence, the correct option is a .

14. If the arithmetic mean of x, x+3, x+6, x+9, and x+12 is 10, the x =
(a) 1
(b) 2
(c) 6
(d) 4

Solution

The given observations are x, x+3, x+6, x+9, and x +12.
∴ Σ x = 5x + 30, n = 5, x̅ = 10
Now,
x̅ = Σx/n
10 = 5x + 30/5
50 = 5x + 30
5x = 20
x = 4
Hence, the correct option is d.

15. If the median of the data : 24,25,26,x+2x+3,30,31,34 is 27.5 , then x =
(a) 27
(b) 25
(c) 28
(d) 30

Solution

The given observations are 24, 25, 26, x + 2, x + 3, 30, 31, 34
Median = 27.5
Here, n = 8
Median = (n/2)th term + (n/2 + 1)th term / 2
27.5 = 4th term + 5 th term/2
27.5 = (x+2) + (x+3) / 2
27.5 = 2x + 5 / 2
2x + 5 = 55
2x = 50
x = 25
Hence , the correct option is b .

16. If the median of the data : 6,7,x-2, x,17,20 written ascending order is, 16 . Then x =
(a) 15
(b) 16
(c) 17
(d) 18

Solution

The given observations arranged in ascending order are
6,7,x-2,x,17,20
n = 6(even), median = 16
Median = (n/2)th term + (n/2 +1)th term/2
= 3rd term + 4th term/2
= x – 2 + x /2
= 2x – 2 / 2
= 16 = 2x – 2 / 2
= 2x – 2 = 32
x = 17
Hence the correct option is c.

17. The median of first 10 prime numbers is
(a) 11
(b) 12
(c) 13
(d) 14

Solution

First 10 prime numbers are 2 ,3 , 5 ,7 , 11 , 13 , 17, 19 , 23, 29 .
n = 10 (even)
Median = (n/2)th term + (n/2+1)th term
= 5th term + 6th term / 2
= 11 + 13 / 2
= 24/2
= 12
Hence, the correct option is b.

18. If the mode of the data : 64,60,48, x, 43, 34 is 43, then x + 3 
(a) 44
(b) 45
(c) 46
(d) 48

Solution


It is given that the mode of the given date is 43 . So, it is the value with the maximum frequency .
Now, this is possible only when x = 43. In this case, the frequency of the observation 43 would be 3.
Hence,
x + 3 = 46
Hence, the correct option is c .

19. If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
(a) 15
(b) 16
(c) 17
(d) 19

Solution


It is given that the mode of the data is 15. So, it is the observation with the maximum frequency .
This is possible only when x = 15. In this case, the frequency of 15 would be 3.
Hence , the correct answer is a .

20. The mean of 1,3,4,5,7,4 is m. The numbers 3,2,2,4,3,3,p have mean m – 1 and median q. Then, p+q = 
(a) 4
(b) 5
(c) 6
(d) 7

Solution 

1,3,4,5,7,4
Mean = 1+3+4+5+7+4/6
= 24/6
= 4
Consider the numbers 3,2,2,4,3,3,p.
Mean = 3+2+2+4+3+3+p / 7.
⇒ 7× (4 – 1) = 17 + p
⇒ 21 = 17 + p
⇒ p = 4
Arranging the numbers 3,2,2,4,3,3,4 in ascending order, we have
2,2,3,3,3,4,4
Median = (n+1/2)th term
q = (7+1/2)th term
= (8/2)th term
= 4th term
q = 3
so,
p+q = 4 + 3
= 7
Hence , the correct option is d.

21. If the mean of frequency distribution is 8.1 and Σfixi = 132 + 5k, Σfi = 20, then k =
(a) 3
(b) 4
(c) 5
(d) 6

Solution

Given:
Σfixi = 132 + 5k, Σfi = 20 and mean = 8.1
Then,
Mean = Σfixi / Σfi
8.1 = 132 + 5k / 20
162 = 132 + 5k
5k = 30
k = 6
Hence , the correct option is d .

22. If the mean of 6,7,x,8,y,14 is 9, then
(a) x+y = 21
(b) x+y = 19
(c) x-y = 19
(d) x-y = 21

Solution

The given observation are 6,7,x,8,y,14.
Mean = 9 (Given)
⇒ 6+7+x+8+y+14 / 6 = 9
⇒ 35 + x + y = 54
⇒ x + y = 54 – 35 = 19
Hence , the correct option is b . 

23. 

Solution



24. If the mean of first n natural numbers is 5n/9 then n =
(a) 5
(b) 4
(c) 9
(d) 10

Solution



25. The arithmetic mean and mode of a data are 24 and 12 respectively , then its median is 
(a) 25
(b) 18
(c) 20
(d) 22

Solution

Given :
Mean = 24 and Mode = 12
We know that
Mode = 3 Median – 2 Mean
⇒ 12 = 3 median – 2 × 24
⇒ 3 median = 12 + 48 = 60
⇒ Median = 20

26. The mean of first n odd natural number is
(a) n+1/2
(b) n/2
(c) n
(d) n2

Solution



27. The mean of first n odd natural numbers is n2/81, then n =
(a) 9
(b) 81
(c) 27
(d) 18

Solution



28. If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36

Solution

Given :
Mode – Median = 24
We know that
Mode = 3 median – 2 mean
Now,
Mode – median = 2(Median – Mean)
⇒ 24 = 2(Median – Mean)
⇒ Median – Mean = 12
 29. If the arithmetic mean , 7,8,x,11,14 is x, then x = 
(a) 9
(b) 9.5
(c) 10
(d) 10.5

Solution

The given observations are 7,8,x,11,14 .
Mean = x (Given)
Now,
Mean = 7+8+x+11+14 / 5
⇒ x = 40+x / 5
⇒ 5x = 40 + x
⇒ 4x = 40
⇒ x = 10

30. If the mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10

Solution

Given: Mode – Mean = 12
We know that
Mode = 3 median – 2 Mean
∴ Mode – Mean = 3(Median – Mean)
⇒ 12 = 3(Median – Mean)
⇒ Median – Mean = 4 …(1)
Again ,
Mode = 3 Median – 2 Mean
⇒ 2 Mode = 6 Median – 4 Mean
⇒ Mode – Mean + Mode = 6 Median – 5 Mean
⇒ 12 + (Mode – Median) = 5(Median – Mean)
⇒ 12 + (Mode – Median) = 20 [Using (1)]
⇒ Mode – Median = 20 – 12 = 8

31. If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29

Solution



32.

Solution


33. 

Solution




34. 

(a) 23 
(b) 24
(c) 27
(d) 25

Solution



35. If 35 is removed from the data: 30,34,35,36,37,38,39,40 then the median increased by
(a) 2
(b) 1.5
(c) 1
(d) 0.5

Solution


36. While computing mean of grouped data, we assume that the frequencies are
(a) Evenly distributed over all the classes.
(b) Centred at the class marks of the classes.
(c) Centred at the upper limit of the classes .
(d) Centred at the lower limit of the classes .

Solution

We know that while computing the mean of a grouped data, the frequencies are centered at the class marks of the classes . 

37.

Solution


38. For the following distributions :
Class :  0-5  5-10  10-15  15-20  20-25
Frequency : 10 15  12  20   9
The sum of the lower limits of the median and modal class is
(a) 15
(b) 25
(c) 30
(d) 35

Solution
 

Here, N = 66.
∴ N/2 = 33, which lies in the interval 10-15. 
So, the lower limit of the median class is 10.
The highest frequency is 20, which lies in the interval 15-20.
Therefore, the lower limit of modal class is 15.
So, the required sum is 10 +15 = 25 .

39. For the following distribution :
Below: 10   20   30   40   50   60
Number of students : 3  12  27  57   75   80
the modal class is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60

Solution 

Here, N = 80 
∴ N/2 = 40, which lies in the interval 30-40.
Therefore, the modal class is 30-40.

40. Consider the following frequency distribution :
Class: 65-85  85-105  105-125  125-145  145-165  165-185  185-205
Frequency : 4 5 13 20 14 7 4
The difference of the upper limit of the median class and the lower limit of the modal class is :

Solution

Here, N = 67
∴ N/2 = 33.5, which lies in the interval 125-145.
Therefore, the lower limit of the median class is 125.
The highest frequency is 20, which lies in the interval 125-145.
Therefore, the upper limit of modal class is 145.
So, the required difference is 145-125 = 20.

41.  

Solution

The given formula represents the formula to find the mean by assumed mean method . 
Here, di = xi – a where xi is the ith observation and a is assumed mean .
So, di’s are the deviation from a of mid – points of the classes .

42. The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its
(a) Mean
(b) Median
(c) Mode
(d) All the three above

Solution

The less than ogive and more than ogive when drawn on the same graph intersect at a point . From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median.
 Thus, the abscissa of the point if intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.

43. Consider the following frequency distribution :
Class:  0-5  6-11  12-17  18-23   24-29
Frequency: 13  10  15   8   11
The upper limit of the median class is
(a) 17
(b) 17.5
(c) 18
(d) 18.5 

Solution

The given classes in the table are non-continuous. So , we first make the classes continuous by adding 0.5 to the upper limit and substracting 0.5 from the lower limit in each class .    
Now, from the table we see that N = 57. 
So, N/2 = 57/2 = 28.5
28.5 lies in the class 11.5 – 17.5
The upper limit of the interval 11.5-17.5 is 17.5
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