NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 8 Introduction To Trigonometry

FAQ on Chapter 6 Triangles

How many exercises in Chapter 6 Triangles

There are total 6 exercises in the Chapter 5 Class 10. Here, we have provided solutions of evrry questions in detailed way. These NCERT Solutions for Chapter 6 Triangles Class 10 Maths will help you in scoring more marks in the examinations.

What do you mean by AA or AAA Similarity Criterion?

If two angles of one triangle are respectively equal to the two angles of another triangle, then the third angles of the two triangles are necessarily equal, because the sum of three angles of a triangle is always 180°.

What is RHS Similarity Criterion?

If in two right triangles, hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle, then the two triangles are similar by RHS similarity criterion.

What is Converse of Pythagoras Theorem?

In a triangle if the square of one side is equal to the sum of the square of the other two sides, then the angle opposite to the first side is a right angle.

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NCERT Solutions for Class 10 Maths Ch 5 Arithmetic Progressions

October 14, 2020, 2:19 am

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NCERT Solutions for Class 10 Maths Ch 5 Arithmetic Progressions

NCERT Solutions Chapter 5 Arithmetic Progressions Class 10 Maths is given here which will be useful in completing homework on time and improving your problem solving skills. Our experts have prepared detailed solutions of every question so a student can understand them easily. As finding free and accurate NCERT Solutions for Class 10 Maths can be sometimes very challenging that is why we have provided them. It will be really helpful for those students who want to improve their marks.

NCERT Solutions for Class 10th: Ch 5 Arithmetic Progressions Math

Page No: 99

Exercise 5.1

1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.

Answer

It can be observed that
Taxi fare for 1st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39

Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Answer

Let the initial volume of air in a cylinder be V litres. In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time. In other words, after every stroke, only 1 - 1/4 = 3/4th part of air will remain.
Therefore, volumes will be V, 3V/4 , (3V/4)² , (3V/4)³...
Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.

Answer

Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.

(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer

We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

2. Write first four terms of the A.P. when the first term a and the common differenced are given as follows
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = - 3
(iv) a = -1 d = 1/2
(v) a = - 1.25, d = - 0.25

Answer

(i) a = 10, d = 10
Let the series be a₁, a₂, a₃, a₄, a₅…
a₁ = a = 10
a₂ = a₁ + d = 10 + 10 = 20
a₃ = a₂ + d = 20 + 10 = 30
a₄ = a₃ + d = 30 + 10 = 40
a₅ = a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) a = - 2, d = 0
Let the series be a₁, a₂, a₃, a₄…
a₁ = a = -2
a₂ = a₁ + d = - 2 + 0 = - 2
a₃ = a₂ + d = - 2 + 0 = - 2
a₄ = a₃ + d = - 2 + 0 = - 2
Therefore, the series will be - 2, - 2, - 2, - 2 …
First four terms of this A.P. will be - 2, - 2, - 2 and - 2.

(iii) a = 4, d = - 3
Let the series be a₁, a₂, a₃, a₄…
a₁ = a = 4
a₂ = a₁ + d = 4 - 3 = 1
a₃ = a₂ + d = 1 - 3 = - 2
a₄ = a₃ + d = - 2 - 3 = - 5
Therefore, the series will be 4, 1, - 2 - 5 …
First four terms of this A.P. will be 4, 1, - 2 and - 5.

(iv) a = - 1, d = 1/2
Let the series be a₁, a₂, a₃, a₄…a₁ = a = -1
a₂ = a₁ + d = -1 + 1/2 = -1/2
a₃ = a₂ + d = -1/2 + 1/2 = 0
a₄ = a₃ + d = 0 + 1/2 = 1/2
Clearly, the series will be-1, -1/2, 0, 1/2
First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) a = - 1.25, d = - 0.25
Let the series be a₁, a₂, a₃, a₄…
a₁ = a = - 1.25
a₂ = a₁ + d = - 1.25 - 0.25 = - 1.50
a₃ = a₂ + d = - 1.50 - 0.25 = - 1.75
a₄ = a₃ + d = - 1.75 - 0.25 = - 2.00
Clearly, the series will be 1.25, - 1.50, - 1.75, - 2.00 ……..
First four terms of this A.P. will be - 1.25, - 1.50, - 1.75 and - 2.00.

3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, - 1, - 3 …
(ii) -5, - 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9 …

Answer

(i) 3, 1, - 1, - 3 …
Here, first term, a = 3
Common difference, d = Second term - First term
= 1 - 3 = - 2

(ii) - 5, - 1, 3, 7 …
Here, first term, a = - 5
Common difference, d = Second term - First term
= ( - 1) - ( - 5) = - 1 + 5 = 4
(iii) 1/3, 5/3, 9/3, 13/3 ....
Here, first term, a = 1/3

Common difference, d = Second term - First term

= 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9 …
Here, first term, a = 0.6
Common difference, d = Second term - First term
= 1.7 - 0.6
= 1.1

4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ....
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, - 6, - 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, - 4, - 8, - 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ....
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a…
(xi) a, a², a³, a⁴…
(xii) √2, √8, √18, √32 ...
(xiii) √3, √6, √9, √12 ...
(xiv) 1², 3², 5², 7²…
(xv) 1², 5², 7², 7³…

Answer

(i) 2, 4, 8, 16 …
Here,
a₂ - a₁ = 4 - 2 = 2
a₃ - a₂ = 8 - 4 = 4
a₄ - a₃ = 16 - 8 = 8
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(ii) 2, 5/2, 3, 7/2 ....
Here,

a₂ - a₁ = 5/2 - 2 = 1/2
a₃ - a₂ = 3 - 5/2 = 1/2
a₄ - a₃ = 7/2 - 3 = 1/2
⇒ a_n+1 - a_n is same every time.
Therefore, d = 1/2 and the given numbers are in A.P.
Three more terms are
a₅ = 7/2 + 1/2 = 4
a₆ = 4 + 1/2 = 9/2
a₇ = 9/2 + 1/2 = 5

(iii) -1.2, - 3.2, -5.2, -7.2 …
Here,
a₂ - a₁ = ( -3.2) - ( -1.2) = -2
a₃ - a₂ = ( -5.2) - ( -3.2) = -2
a₄ - a₃ = ( -7.2) - ( -5.2) = -2
⇒ a_n+1 - a_n is same every time.
Therefore, d = -2 and the given numbers are in A.P.
Three more terms are
a₅ = - 7.2 - 2 = - 9.2
a₆ = - 9.2 - 2 = - 11.2
a₇ = - 11.2 - 2 = - 13.2

(iv) -10, - 6, - 2, 2 …
Here,
a₂ - a₁ = (-6) - (-10) = 4
a₃ - a₂ = (-2) - (-6) = 4
a₄ - a₃ = (2) - (-2) = 4
⇒ a_n+1 - a_n is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Three more terms are
a₅ = 2 + 4 = 6
a₆ = 6 + 4 = 10
a₇ = 10 + 4 = 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
Here,
a₂ - a₁ = 3 + √2 - 3 = √2
a₃ - a₂ = (3 + 2√2) - (3 + √2) = √2
a₄ - a₃ = (3 + 3√2) - (3 + 2√2) = √2
⇒ a_n+1 - a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = (3 + √2) + √2 = 3 + 4√2
a₆ = (3 + 4√2) + √2 = 3 + 5√2
a₇ = (3 + 5√2) + √2 = 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a₂ - a₁ = 0.22 - 0.2 = 0.02
a₃ - a₂ = 0.222 - 0.22 = 0.002
a₄ - a₃ = 0.2222 - 0.222 = 0.0002
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(vii) 0, -4, -8, -12 …
Here,
a₂ - a₁ = (-4) - 0 = -4
a₃ - a₂ = (-8) - (-4) = -4
a₄ - a₃ = (-12) - (-8) = -4
⇒ a_n+1 - a_n is same every time.
Therefore, d = -4 and the given numbers are in A.P.
Three more terms are
a₅ = -12 - 4 = -16
a₆ = -16 - 4 = -20
a₇ = -20 - 4 = -24

(viii) -1/2, -1/2, -1/2, -1/2 ....
Here,
a₂ - a₁ = (-1/2) - (-1/2) = 0
a₃ - a₂ = (-1/2) - (-1/2) = 0
a₄ - a₃ = (-1/2) - (-1/2) = 0
⇒ a_n+1 - a_n is same every time.
Therefore, d = 0 and the given numbers are in A.P.
Three more terms are
a₅ = (-1/2) - 0 = -1/2
a₆ = (-1/2) - 0 = -1/2
a₇ = (-1/2) - 0 = -1/2

(ix) 1, 3, 9, 27 …
Here,
a₂ - a₁ = 3 - 1 = 2
a₃ - a₂ = 9 - 3 = 6
a₄ - a₃ = 27 - 9 = 18
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(x) a, 2a, 3a, 4a …
Here,
a₂ - a₁ = 2a - a = a
a₃ - a₂ = 3a - 2a = a
a₄ - a₃ = 4a - 3a = a
⇒ a_n+1 - a_n is same every time.
Therefore, d = a and the given numbers are in A.P.
Three more terms are
a₅ = 4a + a = 5a
a₆ = 5a + a = 6a
a₇ = 6a + a = 7a

(xi) a, a², a³, a⁴ …
Here,
a₂ - a₁ = a²- a = (a - 1)
a₃ - a₂ = a³-a²= a²(a - 1)
a₄ - a₃ = a⁴ - a³= a³(a - 1)
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xii) √2, √8, √18, √32 ...
Here,
a₂ - a₁ = √8 - √2 = 2√2 - √2 = √2
a₃ - a₂ = √18 - √8 = 3√2 - 2√2 = √2
a₄ - a₃ = 4√2 - 3√2 = √2
⇒ a_n+1 - a_n is same every time.
Therefore, d = √2 and the given numbers are in A.P.
Three more terms are
a₅ = √32 + √2 = 4√2 + √2 = 5√2 = √50
a₆ = 5√2 +√2 = 6√2 = √72
a₇ = 6√2 + √2 = 7√2 = √98

(xiii) √3, √6, √9, √12 ...
Here,
a₂ - a₁ = √6 - √3 = √3 × 2 -√3 = √3(√2 - 1)
a₃ - a₂ = √9 - √6 = 3 - √6 = √3(√3 - √2)
a₄ - a₃ = √12 - √9 = 2√3 - √3× 3 = √3(2 - √3)
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xiv) 1², 3², 5², 7²…

Or, 1, 9, 25, 49 …..
Here,
a₂− a₁ = 9 − 1 = 8
a₃− a₂= 25 − 9 = 16
a₄− a₃ = 49 − 25 = 24
⇒ a_n+1 - a_n is not the same every time.

Therefore, the given numbers are forming an A.P.

(xv) 1², 5², 7², 73 …
Or 1, 25, 49, 73 …
Here,
a₂− a₁ = 25 − 1 = 24
a₃− a₂= 49 − 25 = 24
a₄− a₃ = 73 − 49 = 24
i.e., a_k₊₁− a_k is same every time.
⇒ a_n+1 - a_n is same every time.
Therefore, d = 24 and the given numbers are in A.P.
Three more terms are
a₅ = 73+ 24 = 97
a₆ = 97 + 24 = 121
a₇= 121 + 24 = 145

Page No: 105

Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the n^th term of the A.P.

	a	d	n	a_n
(i)	7	3	8	…...
(ii)	− 18	…..	10	0
(iii)	…..	− 3	18	− 5
(iv)	− 18.9	2.5	…..	3.6
(v)	3.5	0	105	…..

Answer

(i) a = 7, d = 3, n = 8, a_n = ?
We know that,
For an A.P. a_n = a + (n− 1) d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence, a_n = 28

(ii) Given that
a = −18, n = 10, a_n = 0, d = ?
We know that,
a_n = a + (n− 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = 18/9 = 2
Hence, common difference, d = 2

(iii) Given that
d = −3, n = 18, a_n = −5
We know that,
a_n = a + (n− 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a = 51 − 5 = 46
Hence, a = 46

(iv) a = −18.9, d = 2.5, a_n = 3.6, n = ?
We know that,
a_n = a + (n− 1) d
3.6 = − 18.9 + (n− 1) 2.5
3.6 + 18.9 = (n− 1) 2.5
22.5 = (n− 1) 2.5
(n - 1) = 22.5/2.5
n - 1 = 9
n = 10
Hence, n = 10

(v) a = 3.5, d = 0, n = 105, a_n = ?
We know that,
a_n = a + (n− 1) d
a_n = 3.5 + (105 − 1) 0
a_n = 3.5 + 104 × 0
a_n = 3.5
Hence, a_n = 3.5

Page No: 106

Choose the correct choice in the following and justify
(i) 30^th term of the A.P: 10, 7, 4, …, is
(A) 97 (B) 77 (C) −77 (D.)−87

(ii) 11^thterm of the A.P. -3, -1/2, ,2 .... is
(A) 28 (B) 22 (C) - 38 (D)

Answer

(i) Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a₂− a₁= 7 − 10 = −3
We know that, a_n = a + (n− 1) d
a₃₀ = 10 + (30 − 1) (−3)
a₃₀ = 10 + (29) (−3)
a₃₀ = 10 − 87 = −77
Hence, the correct answer is option C.

(ii) Given that A.P. is -3, -1/2, ,2 ...
First term a = - 3
Common difference, d = a₂ − a₁ = (-1/2) - (-3)
= (-1/2) + 3 = 5/2
We know that, a_n = a + (n − 1) d
a₁₁ = 3 + (11 -1)(5/2)
a₁₁ = 3 + (10)(5/2)
a₁₁ = -3 + 25
a₁₁ = 22
Hence, the answer is option B.

3. In the following APs find the missing term in the boxes.

Answer

(i) For this A.P.,

a = 2
a₃ = 26
We know that, a_n = a + (n − 1) d
a₃ = 2 + (3 - 1) d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 - 1) 12
= 14
Therefore, 14 is the missing term.

(ii) For this A.P.,
a₂ = 13 and
a₄ = 3
We know that, a_n = a + (n − 1) d
a₂ = a + (2 - 1) d
13 = a + d ... (i)
a₄ = a + (4 - 1) d
3 = a + 3d ... (ii)
On subtracting (i) from (ii), we get
- 10 = 2d
d = - 5
From equation (i), we get
13 = a + (-5)
a = 18
a₃ = 18 + (3 - 1) (-5)
= 18 + 2 (-5) = 18 - 10 = 8
Therefore, the missing terms are 18 and 8 respectively.

(iii) For this A.P.,
a = 5 and
a₄ = 19/2
We know that, a_n = a + (n − 1) d
a₄ = a + (4 - 1) d
19/2 = 5 + 3d
19/2 - 5 = 3d3d = 9/2
d = 3/2

a₂ = a + (2 - 1) d
a₂ = 5 + 3/2
a₂ = 13/2

a₃ = a + (3 - 1) d

a₃ = 5 + 2×3/2

a₃ = 8

Therefore, the missing terms are 13/2 and 8 respectively.

(iv) For this A.P.,
a = −4 and
a₆ = 6
We know that,
a_n = a + (n− 1) d
a₆ = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2 (2) = 0
a₄ = a + 3d = − 4 + 3 (2) = 2
a₅ = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.

(v)
For this A.P.,
a₂ = 38
a₆ = −22
We know that
a_n = a + (n− 1) d
a₂ = a + (2 − 1) d
38 = a + d ... (i)
a₆ = a + (6 − 1) d
−22 = a + 5d ... (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a₂− d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

4. Which term of the A.P. 3, 8, 13, 18, … is 78?

Answer

3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂− a₁ = 8 − 3 = 5
Let n^th term of this A.P. be 78.
a_n = a + (n− 1) d
78 = 3 + (n− 1) 5
75 = (n− 1) 5
(n− 1) = 15
n = 16
Hence, 16^th term of this A.P. is 78.

5. Find the number of terms in each of the following A.P.
(i) 7, 13, 19, …, 205
(ii) 18,

, 13,...., -47

Answer

(i) For this A.P.,
a = 7
d = a₂− a₁ = 13 − 7 = 6
Let there are n terms in this A.P.
a_n = 205
We know that
a_n = a + (n− 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n− 1) 6
33 = (n− 1)
n = 34
Therefore, this given series has 34 terms in it.

(ii) For this A.P.,
a = 18

Let there are n terms in this A.P.
a_n = 205

a_n = a + (n − 1) d

-47 = 18 + (n - 1) (-5/2)

-47 - 18 = (n - 1) (-5/2)
-65 = (n - 1)(-5/2)
(n - 1) = -130/-5
(n - 1) = 26
n = 27
Therefore, this given A.P. has 27 terms in it.

6. Check whether -150 is a term of the A.P. 11, 8, 5, 2, …

Answer

For this A.P.,
a = 11
d = a₂− a₁ = 8 − 11 = −3
Let −150 be the n^th term of this A.P.
We know that,
a_n = a + (n − 1) d
-150 = 11 + (n - 1)(-3)
-150 = 11 - 3n + 3
-164 = -3n
n = 164/3
Clearly, n is not an integer.
Therefore, - 150 is not a term of this A.P.

7. Find the 31^st term of an A.P. whose 11^th term is 38 and the 16^th term is 73.

Answer

Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
a_n = a + (n− 1) d
a₁₁ = a + (11 − 1) d
38 = a + 10d ... (i)
Similarly,
a₁₆ = a + (16 − 1) d
73 = a + 15d ... (ii)
On subtracting (i) from (ii), we get
35 = 5d
d = 7
From equation (i),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a₃₁ = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210
= 178
Hence, 31^st term is 178.

8. An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term.

Answer

Given that,
a₃ = 12
a₅₀ = 106
We know that,
a_n = a + (n− 1) d
a₃ = a + (3 − 1) d
12 = a + 2d ... (i)
Similarly, a₅₀= a + (50 − 1) d
106 = a + 49d ... (ii)
On subtracting (i) from (ii), we get
94 = 47d
d = 2
From equation (i), we get
12 = a + 2 (2)
a = 12 − 4 = 8
a₂₉ = a + (29 − 1) d
a₂₉ = 8 + (28)2
a₂₉ = 8 + 56 = 64
Therefore, 29^th term is 64.

9. If the 3^rd and the 9^th terms of an A.P. are 4 and − 8 respectively. Which term of this A.P. is zero.

Answer

Given that,

a₃ = 4
a₉ = −8
We know that,
a_n = a + (n− 1) d
a₃ = a + (3 − 1) d
4 = a + 2d ... (i)
a₉ = a + (9 − 1) d
−8 = a + 8d ... (ii)
On subtracting equation (i) from (ii), we get,
−12 = 6d
d = −2
From equation (i), we get,
4 = a + 2 (−2)
4 = a− 4
a = 8
Let n^th term of this A.P. be zero.
a_n= a + (n − 1) d
0 = 8 + (n− 1) (−2)
0 = 8 − 2n + 2
2n = 10
n = 5
Hence, 5^th term of this A.P. is 0.

10. If 17^th term of an A.P. exceeds its 10^th term by 7. Find the common difference.

Answer

We know that,
For an A.P., a_n = a + (n− 1) d
a₁₇ = a + (17 − 1) d
a₁₇ = a + 16d
Similarly, a₁₀ = a + 9d
It is given that
a₁₇− a₁₀ = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

11. Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54^th term?

Answer

Given A.P. is 3, 15, 27, 39, …
a = 3
d = a₂− a₁ = 15 − 3 = 12
a₅₄ = a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^th term be 771.
a_n = a + (n− 1) d
771 = 3 + (n− 1) 12
768 = (n− 1) 12
(n− 1) = 64
n = 65
Therefore, 65^th term was 132 more than 54^th term.

Or

Let n^th term be 132 more than 54^th term.
n = 54 + 132/2
= 54 + 11 = 65^th term

12. Two APs have the same common difference. The difference between their 100^th term is 100, what is the difference between their 1000^th terms?

Answer

Let the first term of these A.P.s be a₁ and a₂ respectively and the common difference of these A.P.s be d.
For first A.P.,
a₁₀₀ = a₁ + (100 − 1) d
= a₁ + 99d
a₁₀₀₀ = a₁ + (1000 − 1) d
a₁₀₀₀ = a₁ + 999d
For second A.P.,
a₁₀₀ = a₂ + (100 − 1) d
= a₂ + 99d
a₁₀₀₀ = a₂ + (1000 − 1) d
= a₂ + 999d
Given that, difference between
100^th term of these A.P.s = 100
Therefore, (a₁ + 99d) − (a₂ + 99d) = 100
a₁− a₂ = 100 ... (i)
Difference between 1000^th terms of these A.P.s
(a₁ + 999d) − (a₂ + 999d) = a₁− a₂
From equation (i),
This difference, a₁− a₂= 100
Hence, the difference between 1000^th terms of these A.P. will be 100.

13. How many three digit numbers are divisible by 7?

Answer

First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …, 994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_n = 994
n = ?
a_n = a + (n− 1) d
994 = 105 + (n− 1) 7
889 = (n− 1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

Or

Three digit numbers which are divisible by 7 are 105, 112, 119, .... 994 .
These numbers form an AP with a = 105 and d = 7.
Let number of three-digit numbers divisible by 7 be n, a_n = 994
⇒ a + (n - 1) d = 994
⇒ 105 + (n - 1) × 7 = 994
⇒7(n - 1) = 889
⇒ n - 1 = 127
⇒ n = 128

14. How many multiples of 4 lie between 10 and 250?

Answer

First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …, 248
Let 248 be the n^th term of this A.P.
a = 12
d = 4
a_n = 248
a_n = a + (n - 1) d
248 = 12 + (n - 1) × 4
236/4 = n - 1
59 = n - 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250.

Or

Multiples of 4 lies between 10 and 250 are 12, 16, 20, ...., 248.
These numbers form an AP with a = 12 and d = 4.
Let number of three-digit numbers divisible by 4 be n, a_n = 248
⇒ a + (n - 1) d = 248
⇒ 12 + (n - 1) × 4 = 248
⇒4(n - 1) = 248
⇒ n - 1 = 59
⇒ n = 60

15. For what value of n, are the n^th terms of two APs 63, 65, 67, and 3, 10, 17, … equal?

Answer

63, 65, 67, …
a = 63
d = a₂− a₁ = 65 − 63 = 2
n^th term of this A.P. = a_n = a + (n− 1) d
a_n= 63 + (n− 1) 2 = 63 + 2n− 2
a_n = 61 + 2n ... (i)
3, 10, 17, …
a = 3
d = a₂− a₁ = 10 − 3 = 7
n^th term of this A.P. = 3 + (n− 1) 7
a_n= 3 + 7n− 7
a_n = 7n− 4 ... (ii)
It is given that, n^th term of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n− 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13^th terms of both these A.P.s are equal to each other.
16. Determine the A.P. whose third term is 16 and the 7^th term exceeds the 5^th term by 12.

Answer

a₃ = 16

a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a₇ − a₅ = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

Page No: 107

17. Find the 20^th term from the last term of the A.P. 3, 8, 13, …, 253.

Answer

Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 5
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a₂₀ = a + (20 − 1) d
a₂₀ = 253 + (19) (−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20^th term from the last term is 158.

18. The sum of 4^th and 8^th terms of an A.P. is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the A.P.

Answer

We know that,
a_n = a + (n − 1) d
a₄ = a + (4 − 1) d
a₄ = a + 3d
Similarly,
a₈ = a + 7d
a₆ = a + 5d
a₁₀ = a + 9d
Given that, a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ... (i)
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ... (ii)
On subtracting equation (i) from (ii), we get,
2d = 22 − 12
2d = 10
d = 5
From equation (i), we get
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a₂ = a + d = − 13 + 5 = −8
a₃ = a₂ + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Answer

It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after n^th year, his salary be Rs 7000.
Therefore, a_n = a + (n− 1) d
7000 = 5000 + (n− 1) 200
200(n− 1) = 2000
(n− 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the n^th week, her week, her weekly savings become Rs 20.75, find n.

Answer

Given that,
a = 5
d = 1.75
a_n= 20.75
n = ?
a_n = a + (n− 1) d
20.75 = 5 + (n - 1) × 1.75
15.75 = (n - 1) × 1.75
(n - 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n - 1 = 9
n = 10
Hence, n is 10.

Page No: 112

Exercise 5.3

1. Find the sum of the following APs.
(i) 2, 7, 12 ,…., to 10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, ...... , to 11 terms

Answer

(i) 2, 7, 12 ,…, to 10 terms
For this A.P.,
a = 2
d = a₂− a₁ = 7 − 2 = 5
n = 10
We know that,
S_n = n/2 [2a + (n - 1) d]
S₁₀ = 10/2 [2(2) + (10 - 1) × 5]
= 5[4 + (9) × (5)]
= 5 × 49 = 245

(ii) −37, −33, −29 ,…, to 12 terms
For this A.P.,
a = −37
d = a₂− a₁ = (−33) − (−37)
= − 33 + 37 = 4
n = 12
We know that,
S_n = n/2 [2a + (n - 1) d]
S₁₂ = 12/2 [2(-37) + (12 - 1) × 4]
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6(-30) = -180

(iii) 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
a = 0.6
d = a₂− a₁ = 1.7 − 0.6 = 1.1
n = 100
We know that,
S_n = n/2 [2a + (n - 1) d]
S₁₂ = 50/2 [1.2 + (99) × 1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505

(iv) 1/15, 1/12, 1/10, ...... , to 11 terms
For this A.P.,

2. Find the sums given below
(i) 7 +

+ 14 + .................. +84
(ii)+ 14 + ………… + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)

Answer

(i) For this A.P.,
a = 7
l = 84
d = a₂ − a₁ =

- 7 = 21/2 - 7 = 7/2
Let 84 be the n^thterm of this A.P.
l = a (n - 1)d
84 = 7 + (n - 1) × 7/2
77 = (n - 1) × 7/2
22 = n− 1
n = 23
We know that,
S_n = n/2 (a + l)
S_n = 23/2 (7 + 84)
= (23×91/2) = 2093/2
=

(ii) 34 + 32 + 30 + ……….. + 10
For this A.P.,
a = 34
d = a₂− a₁ = 32 − 34 = −2
l = 10
Let 10 be the n^th term of this A.P.
l = a + (n − 1) d
10 = 34 + (n− 1) (−2)
−24 = (n − 1) (−2)
12 = n− 1
n = 13
S_n = n/2 (a + l)
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286

(iii) (−5) + (−8) + (−11) + ………… + (−230)For this A.P.,
a = −5
l = −230
d = a₂− a₁ = (−8) − (−5)
= − 8 + 5 = −3
Let −230 be the n^th term of this A.P.
l = a + (n − 1)d
−230 = − 5 + (n− 1) (−3)
−225 = (n− 1) (−3)
(n− 1) = 75
n = 76
And,
S_n = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930

3. In an AP
(i) Given a = 5, d = 3, a_n = 50, find n and S_n.
(ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(v) Given d = 5, S₉ = 75, find a and a₉.
(vi) Given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) Given a = 8, a_n = 62, S_n = 210, find n and d.
(viii) Given a_n = 4, d = 2, S_n = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.

Answer

(i) Given that, a = 5, d = 3, a_n = 50
As a_n = a + (n − 1)d,
⇒ 50 = 5 + (n - 1) × 3
⇒ 3(n - 1) = 45
⇒ n - 1 = 15
⇒ n = 16
Now, S_n = n/2 (a + a_n)
S_n = 16/2 (5 + 50) = 440

(ii) Given that, a = 7, a₁₃ = 35
As a_n = a + (n − 1)d,⇒ 35 = 7 + (13 - 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, S_n = n/2 (a + a_n)
S₁₃ = 13/2 (7 + 35) = 273

(iii)Given that, a₁₂ = 37, d = 3As a_n = a + (n− 1)d,
⇒ a₁₂ = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
S_n = n/2 (a + a_n)
S_n = 12/2 (4 + 37)
= 246

(iv) Given that, a₃ = 15, S₁₀ = 125
As a_n = a + (n− 1)d,
a₃ = a + (3 − 1)d
15 = a + 2d ... (i)
S_n = n/2 [2a + (n - 1)d]
S₁₀ = 10/2 [2a + (10 - 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ... (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ... (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a− 2
a = 17
a₁₀ = a + (10 − 1)d
a₁₀ = 17 + (9) (−1)
a₁₀ = 17 − 9 = 8

(v) Given that, d = 5, S₉ = 75
As S_n = n/2 [2a + (n - 1)d]
S₉ = 9/2 [2a + (9 - 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = -35/3
a_n = a + (n− 1)d
a₉ = a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3

(vi) Given that, a = 2, d = 8, S_n = 90
As S_n = n/2 [2a + (n - 1)d]
90 = n/2 [2a + (n - 1)d]
⇒ 180 = n(4 + 8n - 8) = n(8n - 4) = 8n² - 4n
⇒ 8n² - 4n - 180 = 0
⇒ 2n² - n - 45 = 0
⇒ 2n² - 10n + 9n - 45 = 0
⇒ 2n(n -5) + 9(n - 5) = 0
⇒ (2n - 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a₅= 8 + 5 × 4 = 34

(vii) Given that, a = 8, a_n = 62, S_n = 210
As S_n = n/2 (a + a_n)
210 = n/2 (8 + 62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8 + 5d
⇒ 5d = 62 - 8 = 54
⇒ d = 54/5 = 10.8

(viii) Given that, a_n = 4, d = 2, S_n = −14
a_n = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n ... (i)
S_n = n/2 (a + a_n)
-14 = n/2 (a + 4)
−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n² + 10n
2n²− 10n− 28 = 0
n²− 5n −14 = 0
n²− 7n + 2n− 14 = 0
n (n− 7) + 2(n− 7) = 0
(n− 7) (n + 2) = 0
Either n− 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 − 2n
a = 6 − 2(7)
= 6 − 14
= −8

(ix) Given that, a = 3, n = 8, S = 192
As S_n = n/2 [2a + (n - 1)d]
192 = 8/2 [2 × 3 + (8 - 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6

(x) Given that, l = 28, S = 144 and there are total of 9 terms.
S_n = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4

Page No: 113

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer

Let there be n terms of this A.P.
For this A.P., a = 9
d = a₂− a₁ = 17 − 9 = 8
As S_n = n/2 [2a + (n - 1)d]
636 = n/2 [2 × a + (8 - 1) × 8]
636 = n/2 [18 + (n- 1) × 8]
636 = n [9 + 4n− 4]
636 = n (4n + 5)
4n² + 5n− 636 = 0
4n² + 53n− 48n− 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n− 12) = 0
Either 4n + 53 = 0 or n− 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

Given that,
a = 5
l = 45
S_n = 400
S_n = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n− 1) d
45 = 5 + (16 − 1) d
40 = 15d
d = 40/15 = 8/3

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Given that,

a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
S_n = n/2 (a + l)
S₃₈ = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22^nd term is 149.

Answer

d = 7
a₂₂ = 149
S₂₂ = ?
a_n = a + (n− 1)d
a₂₂ = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
S_n = n/2 (a + a_n)
= 22/2 (2 + 149)
= 11 × 151
= 1661

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer

Given that,
a₂ = 14
a₃ = 18
d = a₃− a₂ = 18 − 14 = 4
a₂ = a + d
14 = a + 4
a = 10
S_n = n/2 [2a + (n - 1)d]
S₅₁ = 51/2 [2 × 10 + (51 - 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer

Given that,

S₇ = 49
S₁₇ = 289
S₇ = 7/2 [2a + (n - 1)d]
S₇ = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S₁₇ = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
S_n = n/2 [2a + (n - 1)d]
= n/2 [2(1) + (n - 1) × 2]
= n/2 (2 + 2n - 2)
= n/2 (2n)
= n²
10. Show that a₁, a₂… , a_n , … form an AP where a_n is defined as below
(i) a_n = 3 + 4n
(ii) a_n = 9 − 5n
Also find the sum of the first 15 terms in each case.

Answer

(i) a_n = 3 + 4n
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 3 + 8 = 11
a₃ = 3 + 4(3) = 3 + 12 = 15
a₄ = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a₂− a₁ = 11 − 7 = 4
a₃− a₂ = 15 − 11 = 4
a₄− a₃ = 19 − 15 = 4
i.e., a_k_{+ 1}− a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
S_n = n/2 [2a + (n - 1)d]
S₁₅= 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) a_n = 9 − 5n
a₁ = 9 − 5 × 1 = 9 − 5 = 4
a₂ = 9 − 5 × 2 = 9 − 10 = −1
a₃ = 9 − 5 × 3 = 9 − 15 = −6
a₄ = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a₂− a₁ = − 1 − 4 = −5
a₃− a₂ = − 6 − (−1) = −5
a₄− a₃ = − 11 − (−6) = −5
i.e., a_k_{+ 1}− a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
S_n = n/2 [2a + (n - 1)d]
S₁₅= 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)
= -465

11. If the sum of the first n terms of an AP is 4n− n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly find the 3^rd, the10^th and the n^th terms.

Answer

Given that,
S_n = 4n− n²
First term, a = S₁ = 4(1) − (1)² = 4 − 1 = 3
Sum of first two terms = S₂
= 4(2) − (2)² = 8 − 4 = 4
Second term, a₂ = S₂− S₁ = 4 − 3 = 1
d = a₂− a = 1 − 3 = −2
a_n = a + (n− 1)d
= 3 + (n− 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1
a₁₀ = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

12. Find the sum of first 40 positive integers divisible by 6.

Answer

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S₄₀ = ?
S_n = n/2 [2a + (n - 1)d]
S₄₀= 40/2 [2(6) + (40 - 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

13. Find the sum of first 15 multiples of 8.

Answer

The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S₁₅ = ?
S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960

14. Find the sum of the odd numbers between 0 and 50.

Answer

The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 … 49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n− 1) d
49 = 1 + (n− 1)2
48 = 2(n− 1)
n− 1 = 24
n = 25
S_n = n/2 (a + l)
S₂₅ = 25/2 (1 + 49)
= 25(50)/2
=(25)(25)
= 625

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

Answer

It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀
= 30/2 [2(200) + (30 - 1) 50]

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Answer

Let the cost of 1^st prize be P.
Cost of 2^nd prize = P− 20
And cost of 3^rd prize = P− 40
It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
a = P
d = −20
Given that, S₇ = 700
7/2 [2a + (7 - 1)d] = 700

a + 3(−20) = 100
a− 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

Answer

It can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 − 1 = 1
S_n = n/2 [2a + (n - 1)d]
S₁₂ = 12/2 [2(1) + (12 - 1)(1)]
= 6 (2 + 11)
= 6 (13)
= 78
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3 × 78 = 234
Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Answer

perimeter of semi-circle = πr
_P1 = π(0.5) = π/2 cm
_P2 = π(1) = π cm
_P3 = π(1.5) = 3π/2 cm
_P1, P₂, P₃ are the lengths of the semi-circles

π/2, π, 3π/2, 2π, ....
P1 = π/2 cm
_P2 = π cm
d = P2- P1 = π - π/2 = π/2
First term = P1 = a = π/2 cm
S_n = n/2 [2a + (n - 1)d]
Therefor, Sum of the length of 13 consecutive circles
S₁₃ = 13/2 [2(π/2) + (13 - 1)π/2]
= 13/2 [π + 6π]
=13/2 (7π) = 13/2 × 7 × 22/7
= 143 cm

Page No: 114

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a₂− a₁ = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_n = 200
S_n = n/2 [2a + (n - 1)d]
S₁₂ = 12/2 [2(20) + (n - 1)(-1)]
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n− n²
n²− 41n + 400 = 0
n²− 16n− 25n + 400 = 0
n (n− 16) −25 (n− 16) = 0
(n − 16) (n− 25) = 0
Either (n− 16) = 0 or n− 25 = 0
n = 16 or n = 25
a_n = a + (n− 1)d
a₁₆ = 20 + (16 − 1) (−1)
a₁₆ = 20 − 15
a₁₆ = 5
Similarly,
a₂₅ = 20 + (25 − 1) (−1)
a₂₅ = 20 − 24
= −4
Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

Answer

The distances of potatoes from the bucket are 5, 8, 11, 14…
Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept because first she has to first pick the potato and again return back to the same place in order to start picking the second potato.. Therefore, distances to be run are
10, 16, 22, 28, 34,……….
a = 10
d = 16 − 10 = 6
S₁₀ =?
S₁₀ = 10/2 [2(20) + (n - 1)(-1)]
= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4 (Optional)

1. Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find n for a_n < 0]

Answer

We have the A.P. having a = 121 and d = 117 - 121 = - 4

∴ an = a + (n - 1) d

= 121 + (n - 1) × (- 4)

= 121 - 4n + 4

= 125 - 4n

For the first negative term, we have

an < 0

⇒ (125 - 4n) < 0

⇒ 125 < 4n

⇒ 125/4 <n

⇒ n > 31 1⁄4

Thus, the first negative term is 32nd term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Answer

3. A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 21⁄2 m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = 250/25 + 1]

Answer

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

[Hint : S_x-1 = S₄₉ – S_x]

Answer

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1⁄4 m and a tread of 1⁄2 m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace.

[Hint : Volume of concrete required to build the first step = 1/4 × 1/2 × 50m³]

Answer

Go Back To NCERT Solutions for Class 10th Maths

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions

Chapter 5 Arithmetic Progressions NCERT Solutions is provided here which will be essential in knowing the topics that could come in the examinations. A sequence of numbers in which the successive terms increase or decrease by a constant number is called an Arithmetic Progression (AP).

There are total 4 sections in this chapter that will widen your perspective. It will develop you thinking skills and application of formulas to yield results.

• Introduction: This section will introduce you to some real life examples where arithmetic progressions can be practically used.
• Arithmetic Progressions: You will find definition of Arithmetic progressions and terms like general form of an AP, finite and infinite Arithmetic Progressions.
• nth Term of an AP: We will learn to calculate nth term of an AP. The nth term a n of the AP with first term a and common difference d is given by a_n = a + (n – 1) d.
• Sum of First n Terms of an AP: We will know how to find the sum of the first n terms of an AP is which is given by S = n/2 [2a + (n-1)d]. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP which is given by S = n/2(a+l).

There are total 4 exercises in the chapter in which last one is optional which can be useful in revising the chapter. We have also provided exercisewise NCERT Solutions of chapter 5 Arithmetic Progressions.

NCERT Solutions for Class 10 Maths Chapters:

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 8 Introduction To Trigonometry

CBSE Notes for Class 10 Maths Chapter 7 Coordinate Geometry

FAQ on Chapter 5 Arithmetic Progressions

How many exercises in Chapter 5 Arithmetic Progressions

There are only 4 exercises in the Chapter 5 Class 10. You will find the answers of every questions in detailed so you can complete your work on time and crosscheck your answers. These NCERT Solutions for Chapter 5 Arithmetic Progresssions Class 10 Maths will help you a lot in board exams.

What do you mean by Sequence?

When some numbers are arranged in a definite order, according to a definite rule, they are said to form a sequence.
The number occurring at the 1st place is called the 1st term, denoted by T₁.
The number occurring at the nth place is called the nth term, denoted by T_n.

If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.

Let ‘a’ be the first term and ‘d’ be the common difference.
Now, Using T_n = a + (n - 1) d, we have
T₉ = a + 8d ⇒ a + 8d = 0 ...(1) [∵ T9 = 0 Given]
T₁₉ = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d ...(2)
[∵ a + 8d = 0]
T₂₉ = a + 28d
= (a + 8d) + 20d
= 0 + 20d = 20d [∵ a + 8d = 0]
= 2 × (10d) = 2 (T₁₉) [∵ T₁₉ = 10d]
⇒ T₂₉ = 2 (T₁₉)
Thus, the 29th term of the A.P. is double of its 19th term.

What is Common Difference?

The constant difference between two consecutive terms of an AP is called “Common Difference”.

↧

MCQ Questions for Class 12 Business Studies: Ch 5 Organising

October 14, 2020, 3:52 am

≫ Next: MCQ Questions for Class 12 Business Studies: Ch 6 Staffing

≪ Previous: NCERT Solutions for Class 10 Maths Ch 5 Arithmetic Progressions

MCQ Questions for Class 12 Business Studies: Ch 5 Organising

1. Centralisation refers to

(a) Opening new centers or branches

(b) Retention of decision making authority

(d) Creating divisions as profit centers

► (b) Retention of decision making authority

2. The possibilities of inadequate profits or even losses due to uncertainties are known as ________

(a) Business risks

(b) Business contingencies

(d) None of the above

► (a) Business risks

3. Grouping of activities on the basis of functions is a part of

(a) Decentralised organisation

(b) Centralised organisation

(d) Divisional organisation

► (c) Functional organisation

4. Properly perform the assigned duty, is ___________

(a) Accountability

(b) Responsibility

(d) None of the above

► (b) Responsibility

5. _________ is helpful in reducing the burden of the manager

(a) Decentralisation

(b) Centralisation

(d) None of the above

► (c) Delegation

6. Meaning of authority is__________________

(a) Subordinate to Superior relationship

(b) Right to avoid taxes

(d) None of the above

► (c) Right to Command

7. Span of management refers to

(a) Number of managers

(b) Number of subordinates under a superior

(d) Number of members in top management

► (b) Number of subordinates under a superior

8. What is the first step in organising process?

(a) Establishing reporting relationship

(b) Assignment of duties

(d) Identification and division of work

► (d) Identification and division of work

9. Which of the following is not an element of delegation?

(a) Authority

(b) Responsibility

(d) Informal organization

► (d) Informal organization

10. A tall structure has a

(a) Narrow span of management

(b) Wide span of management

(d) Less levels of management

► (a) Narrow span of management

11. For delegation to be effective it is essential that responsibility be accompanied with necessary

(a) Manpower

(b) Incentives

(d) Authority

► (d) Authority

12. Right of an individual to command his subordinates is ______________

(a) Accountability

(b) Authority

(d) None of the above

► (b) Authority

13. Authority delegated to a subordinate can be taken back under __________

(a) Centralisation

(b) Delegation

(d) None of the above

► (b) Delegation

14. Grouping of activities on the basis of product lines is a part of

(a) Delegated organisation

(b) Functional organisation

(d) Divisional organisation

► (d) Divisional organisation

15. What is the last step in organising process?

(a) Identification and division of work

(b) Departmentalisation

(d) Establishing reporting relationship

► (d) Establishing reporting relationship

16. The form of organisation known for giving rise to rumors is called

(a) Centralised organisation

(b) Formal organization

(d) Informal organisation

► (d) Informal organisation

17. Organisation establishes relationship between

(a) People, work and resources

(b) Customer, work and resources

(d) Customer, work and management

► (a) People, work and resources

18. ________ cannot be entirely delegated

(a) Responsibility

(b) Accountability

(d) None of the above

► (a) Responsibility

19. Which of the following is arises from delegation of authority?

(a) Authority

(b) Responsibility

(d) None of the above

► (b) Responsibility

20. Nature of informal organisation is _______.

(a) Mandatory

(b) Rigid

(d) None of the above

► (c) Flexible

21. Under Accountability a person is answerable for the ___________

(a) Final outcome of the assigned task

(b) Delegation of Authority

(d) None of the above

► (a) Final outcome of the assigned task

22. _______ is helpful in increasing the role of the subordinates in the organisation

(a) Delegation

(b) Decentralisation

(d) None of the above

► (b) Decentralisation

23. Authority, Responsibility and Accountability are the elements of ______________

(a) Decentralisation

(b) Informal Organisation

(d) None of the above

► (c) Delegation

24. Product Specialisation is concerned with_____________

(a) Functional Structure

(b) Divisional Structure

(d) None of the above

► (b) Divisional Structure

25. A network of social relationship that arises spontaneously due to interaction at work is called:

(a) Informal organisation

(b) Formal organisation

(d) Decentralisation

► (a) Informal organisation

↧

MCQ Questions for Class 12 Business Studies: Ch 6 Staffing

October 14, 2020, 5:18 am

≫ Next: MCQ Questions for Class 12 Business Studies: Ch 7 Directing

≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 5 Organising

MCQ Questions for Class 12 Business Studies: Ch 6 Staffing

1. Staffing is __________ function

(a) Top Level Function

(b) Planning

(d) Managerial Function

► (d) Managerial Function

2. Which of the following is not concerned with staffing?

(a) Recruitment

(b) Training

(d) Selection

► (c) Publicity

3. Face to face conversation between employer and applicant is known as:

(a) Selection

(b) Orientation

(d) Campus recruitment

► (c) Interview

4. Training is a process by which __________ of employees is increased.

(a) Knowledge

(b) Aptitudes

(d) All of the above

► (d) All of the above

5. Which method of training is concerned with familiarize the new employee to the organization?

(a) Induction training

(b) Apprenticeship training

(d) Job rotation

► (a) Induction training

6. The main purpose of Selection is _______________

(a) To Select board members

(b) To Select Recruiters

(d) To Select most suitable candidate

► (d) To Select most suitable candidate

7. Estimating Manpower Requirement is a ________

(a) Job of Supervisor

(b) Job of Directors

(d) Budgetary Technique

► (c) Workload analysis

8. Lectures and Conferences are method of __________

(a) Training

(b) Development

(d) Selection

► (a) Training

9. Staffing is very important function these days because of ______________

(a) Advancement of technology

(b) Increasing size of organisation

(d) All of the above

► (d) All of the above

10. Expand HRM __________

(a) Human Responsibility Management

(b) Human Resource Maintenance

(d) None of the above

► (c) Human Resource Management

11. Recruitment is the process of _____________

(a) Promotion of employees

(b) Training to employees

(d) Searching for prospective employees

► (d) Searching for prospective employees

12. Staffing function begins with ____________

(a) Selection

(b) Estimating manpower requirements

(d) Promotion

► (b) Estimating manpower requirements

13. Which test is helpful in assessing the individuals? capacity for learning new skills?

(a) Trade Test

(b) Aptitude Test

(d) Intelligence Test

► (b) Aptitude Test

14. Under which method of training, employees are shifted from one job position to another:

(a) Promotion

(b) Transfer

(d) Job Rotation

► (d) Job Rotation

15. Training of employees is a ________

(a) Short Term Process

(b) Day to Day Process

(d) Ongoing Process

► (a) Short Term Process

16. Estimating Manpower Requirements include ________

(a) Selection Expenses

(b) Financial Expenses

(d) Number and types of persons required

► (d) Number and types of persons required

17. HRM is much broader concept than ______________

(a) Staffing

(b) Organising

(d) Controlling

► (a) Staffing

18. Development of employee is concerned with

(a) Personality

(b) Maturity

(d) None of the above

► (c) Bother personality and Maturity

19. Human Resource Management includes many specialised activities except____

(a) Collecting information about jobs

(b) Staring a company

(d) Searching for qualified people

► (b) Staring a company

20. Selection is done after the __________

(a) Placement

(b) Recruitment

(d) Performance Appraisal

► (b) Recruitment

21. Selection is a _______

(a) Day-to-day Process

(b) Negative Process

(d) None of the above

► (b) Negative Process

22. Staffing function of management is concerned with ____________

(a) Organising

(b) Right Man on Right Job

(d) None of the above

► (b) Right Man on Right Job

23. Recruitment is a ___________

(a) Day-to-day Process

(b) Negative Process

(d) None of the above

► (c) Positive process

24. Main aim of Recruitment is _____________

(a) To attract more and more candidates

(b) To attract less candidates

(d) None of the above

► (a) To attract more and more candidates

25. Selection is the process of __________

(a) Inviting Applications

(b) Identifying different sources of recruitment

(d) None of the above

► (c) Identifying and choosing best candidate

↧

MCQ Questions for Class 12 Business Studies: Ch 7 Directing

October 14, 2020, 9:07 am

≫ Next: MCQ Questions for Class 12 Business Studies: Ch 8 Controlling

≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 6 Staffing

MCQ Questions for Class 12 Business Studies: Ch 7 Directing

1. Bonus, Promotion and Recognition are the example of _____________

(a) Controlling

(b) Positive Motivation

(d) Planning

► (b) Positive Motivation

2. The process of converting the message into communication symbols is known as-

(a) Decoding

(b) Encoding

(d) Media

► (b) Encoding

3. Esteem needs are concerned with ____________

(a) Affection

(b) Prestige and Self-respect

(d) Acceptance

► (b) Prestige and Self-respect

4. The motivation theory which classifies needs in hierarchical order is developed by

(a) Fred Luthans

(b) Abraham Maslow

(d) Peter F. Drucker

► (b) Abraham Maslow

5. Punishment and stopping increments etc. are the examples of __________

(a) Communication

(b) Positive Motivation

(d) Negative Motivation

► (d) Negative Motivation

6. Directing is concerned with the following elements _______________

(a) Communication

(b) Supervision and Motivation

(d) All of the above

► (d) All of the above

7. Two way communication technique is used under _________________

(a) Laissez Faire

(b) Democratic style

(d) None of the above

► (b) Democratic style

8. Hierarchy Theory of Motivation was given by ________

(a) Abraham Maslow

(b) Peter Maslow

(d) David Maslow

► (a) Abraham Maslow

9. Which of the following is not a leadership style?

(a) Free-rein

(b) Autocratic

(d) Dictatorship

► (d) Dictatorship

10. Which of the following is a financial incentive?

(a) Job Security

(b) Promotion

(d) Stock Incentive

► (d) Stock Incentive

11. Basic Physiological Needs are concerned with ____________

(a) Shelter

(b) Hunger

(d) All of the above

► (d) All of the above

12. Supervision means _________________

(a) Planning for the future

(b) Starting a business

(d) Overseeing what is being done by subordinates

► (d) Overseeing what is being done by subordinates

13. One way communication system is used under ____________

(a) Authoritative Style of Leadership

(b) Free-rein

(d) None of the above

► (a) Authoritative Style of Leadership

14. Workers chit-chating in a canteen about the behaviour of supervisor is ______ type of communication.

(a) Pure Official

(b) Formal

(d) None of the above

► (c) Informal

15. Which of the following is not an element of communication process?

(a) Communication

(b) Channel

(d) Decoding

► (a) Communication

16. Social/Affiliation/Belongingness needs are concerned with __________

(a) Salary

(b) Pension

(d) Affection, Acceptance and friendship

► (d) Affection, Acceptance and friendship

17. Which of the following is a function of supervisor?

(a) Accounting

(b) Staffing

(d) Planning

► (c) Ensures performance of work

18. Which one of the following is not an element of direction?

(a) Delegation

(b) Communication

(d) Motivation

► (a) Delegation

19. Which of the following is not a financial incentive ______________

(a) Profit Sharing

(b) Recognition

(d) Bonus

► (b) Recognition

20. Grapevine is

(a) Formal communication

(b) Informal communication

(d) Barrier to communication

► (b) Informal communication

21. Under which style of leadership there is no delegation of authority?

(a) Democratic

(b) Autocratic

(d) None of the above

► (b) Autocratic

22. Leadership is the art or process of ______________

(a) Making people responsible without providing them any authority

(b) Influencing the behaviour of people

(d) Delegating Authority

► (b) Influencing the behaviour of people

23. Which of the following is an example of Non financial incentives?

(a) Recognition

(b) Job Security

(d) All of the above

► (d) All of the above

24. The software company promoted by Narayana Murthy is

(a) Infosys

(b) HCL

(d) Wipro

► (a) Infosys

25. Status comes under the following type of barriers

(a) Psychological barrier

(b) Semantic barrier

(d) Organisational barrier

► (d) Organisational barrier

↧

MCQ Questions for Class 12 Business Studies: Ch 8 Controlling

October 14, 2020, 8:52 pm

≫ Next: NCERT Solutions for Class 10 Maths Ch 7 Coordinate Geometry

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MCQ Questions for Class 12 Business Studies: Ch 8 Controlling

1. Controlling is __________ aspect of management.

(a) Physical

(b) Practical

(d) Theoretical

► (b) Practical

2. What is the last step in controlling process?

(a) Setting Performance Standards

(b) Taking corrective action

(d) Measurement of actual performance

► (b) Taking corrective action

3. What will be the corrective action for defective machinery?

(a) Sell at the book value

(b) Sell at a loss

(d) Repair or Replace the machine

► (d) Repair or Replace the machine

4. Which of the following is not a technique of controlling?

(a) Break even analysis

(b) Budget

(d) Cash Flow Statement

► (d) Cash Flow Statement

5. What is the first step in controlling process?

(a) Comparing actual performance with standards

(b) Analysing deviations

(d) Measurement of actual performance

► (c) Setting performance standards

6. What are the techniques of measurement of performance?

(a) Performance reports

(b) Sample Checking

(d) All of the above

► (d) All of the above

7. Technological upgradation is the best corrective action in case of _________

(a) Defective machinery

(b) Defective material

(d) None of the above

► (c) Obsolete Machinery

8. What will be the corrective action for defective material?

(a) Change in Quantity

(b) Change in Price

(d) None of the above

► (c) Change in Quality Specifications for the material used

9. An efficient control system helps to

(a) Judges accuracy of standards

(b) Accomplishes organisational objectives

(d) All of the above

► (d) All of the above

10. Planning without controlling is ________

(a) Cheap

(b) Effective

(d) Costly

► (c) Meaningless

11. If there is lack of motivation among employees, suggest one better corrective action

(a) New Recruitment

(b) Providing Suitable incentives

(d) Warning to employees

► (b) Providing Suitable incentives

12. Why actual performance is compared with standards?

(a) To take corrective action

(b) To improve the quality of material

(d) None of the above

► (a) To take corrective action

13. Controlling is _______________

(a) Economical

(b) Costly

(d) Very Cheap

► (b) Costly

14. Which of the following is a limitation of controlling?

(a) Controlling improves employee motivation

(b) Controlling is goal oriented

(d) Controlling ensures order and discipline

► (c) Controlling is expensive

15. Which of the following is a technique of controlling?

(a) Managerial Audit

(b) Break even analysis

(d) All of the above

► (d) All of the above

16. Controlling function of an organisation is

(a) Forward as well as backward looking

(b) Backward looking

(d) None of the above

► (a) Forward as well as backward looking

17. When actual performance is better than the standard performance it is called _____

(a) Negative Deviation

(b) Poor Deviation

(d) None of the above

► (c) Positive Deviation

18. Controlling is necessary for ____________

(a) Medium Level Enterprise

(b) Small Scale Enterprise

(d) All of the above

► (d) All of the above

19. There is one popular saying that Planning is looking _________ while controlling is looking ________

(a) Back, Ahead

(b) Ahead, Back

(d) None of the above

► (b) Ahead, Back

20. Budgetary control requires the preparation of

(a) Budgets

(b) Network diagram

(d) Training schedule

► (a) Budgets

21. Which of the following is not correct?

(a) Controlling is goal oriented

(b) Controlling is pervasive function

(d) Controlling ensures order and discipline

► (c) Controlling can prevent deviation

22. Which of the following is not applicable to responsibility accounting

(a) Accounting centre

(b) Cost centre

(d) Investment centre

► (a) Accounting centre

23. Management control is done by the __________________

(a) Managers at Top Level

(b) Managers at Middle Level

(d) All of the above

► (d) All of the above

24. The main focus of controlling is on ________________

(a) Competition

(b) External Forces

(d) Business Environment

► (c) Results

25. Management audit is a technique to keep a check on the performance of

(a) Company

(b) Management of the company

(d) Customers

► (b) Management of the company

↧

NCERT Solutions for Class 10 Maths Ch 7 Coordinate Geometry

October 14, 2020, 10:46 pm

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≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 8 Controlling

NCERT Solutions for Class 10 Maths Ch 7 Coordinate Geometry

Chapter 7 Coordinate Geometry NCERT Solutions for Class 10 Maths is available here which will be vey helpful during the preparation of examinations. It will be also useful in completing your homework and understanding concepts clearly. Class 10 Maths NCERT Solutions for Chapter 7 Coordinate Geometry is very essential for scoring good marks in the class test or board examinations.

Page No: 161

Exercise 7.1

1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)

Answer

(i) Distance between the points is given by

(ii) Distance between (−5, 7) and (−1, 3) is given by

(iii) Distance between (a, b) and (− a, − b) is given by

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.

Answer

Distance between points (0, 0) and (36, 15)

Yes, Assume town A at origin point (0, 0).
Therefore, town B will be at point (36, 15) with respect to town A.
And hence, as calculated above, the distance between town A and B will be 39 km.

3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

Answer

Let the points (1, 5), (2, 3), and (- 2,-11) be representing the vertices A, B, and C of the given triangle respectively.Let A = (1, 5), B = (2, 3) and C = (- 2,-11)

Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and ( - 2, - 11) are not collinear.

4. Check whether (5, - 2), (6, 4) and (7, - 2) are the vertices of an isosceles triangle.

Answer

Let the points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively.

Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.

5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don't you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.

Answer

Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).

By using distance formula, we get

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa was correct

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, - 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Answer

Let the points ( - 1, - 2), (1, 0), ( - 1, 2), and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square.

(ii) Let the points ( - 3, 5), (3, 1), (0, 3), and ( - 1, - 4) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc.

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

7. Find the point on the x-axis which is equidistant from (2, - 5) and (- 2, 9).

Answer

We have to find a point on x-axis. Therefore, its y-coordinate will be 0.
Let the point on x-axis be (x,0)

(x - 2)² + 25 = (x - 2)² + 81
x² + 4 - 4x + 25 = x² + 4 + 4x + 81
8x = 25 -81
8x = -56
x = -7
Therefore, the point is ( - 7, 0).

8. Find the values of y for which the distance between the points P (2, - 3) and Q (10, y) is 10 units.

Answer

It is given that the distance between (2, - 3) and (10, y) is 10.

64 + (y + 3)² = 100
(y +3)² = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = -6
Therefore, y = 3 or -9

9. If Q (0, 1) is equidistant from P (5, - 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Answer

PQ = QR

41 = x² + 25
16 = x²
x = ±4
Therefore, point R is (4, 6) or ( - 4, 6).
When point R is (4, 6),

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Answer

Point (x, y) is equidistant from (3, 6) and ( - 3, 4).

x² + 9 - 6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y
36 - 16 = 6x + 6x + 12y - 8y
20 = 12x + 4y
3x + y = 5
3x + y - 5 = 0

Exercise 7.2

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2:3.

Answer

Let P(x, y) be the required point. Using the section formula, we get

Therefore, the point is (1, 3).

2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Answer

Let P (x₁, y₁) and Q (x₂, y₂) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flagexactly halfway between the line segment joining the two flags, where should she post her flag?

Answer

It can be observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1×100/4)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1×100/5) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR

Therefore, Rashmi should post her blue flag at 22.5m on 5th line.

4. Find the ratio in which the line segment joining the points (-3, 10) and (6, - 8) is divided by (-1, 6).

Answer

Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k:1.
Therefore, -1 = 6k-3/k+1
-k - 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.

5. Find the ratio in which the line segment joining A (1, - 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Answer

Let the ratio in which the line segment joining A (1, - 5) and B ( - 4, 5) is divided by x-axis be k:1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).

We know that y-coordinate of any point on x-axis is 0.

∴ 5k-5/k+1 = 0

Therefore, x-axis divides it in the ratio 1:1.

6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer

Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.

Mid point of diagonal AC is

and Mid point of Diagonal BD is

Since the diagonals of a parallelogram bisect each other, the mid point of AC and BD are same.

∴ x+1/2 = 7/2 and 4 = 5+y/2

⇒ x + 1 = 7 and 5 + y = 8

⇒ x = 6 and y = 3

7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, - 3) and B is (1, 4).

Answer

Let the coordinates of point A be (x, y).
Mid-point of AB is (2, - 3), which is the center of the circle.

⇒ x + 1 = 4 and y + 4 = -6

⇒ x = 3 and y = -10
Therefore, the coordinates of A are (3,-10).

8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

Answer

The coordinates of point A and B are (-2,-2) and (2,-4) respectively.

Since AP = 3/7 AB

Therefore, AP:PB = 3:4

Point P divides the line segment AB in the ratio 3:4.

9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.

Answer

From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]

Answer

Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD.

Page No: 170

Exercises 7.3

1. Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

Answer

(i) Area of a triangle is given by

Area of triangle = 1/2 {x₁ (y₂ - y₃)+ x₂ (y₃ - y₁)+ x₃ (y₁ - y₂)}
Area of the given triangle = 1/2 [2 { 0- (-4)} + (-1) {(-4) - (3)} + 2 (3 - 0)]
= 1/2 {8 + 7 + 6}
= 21/2 square units.

(ii) Area of the given triangle = 1/2 [-5 { (-5)- (4)} + 3(2-(-1)) + 5{-1 - (-5)}]
= 1/2{35 + 9 + 20}
= 32 square units

2. In each of the following find the value of 'k', for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

Answer

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, -2) (5, 1), and (3, k), area = 0
1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) + 1}] = 0
7 - 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, - 4), and (2, - 5), area = 0
1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 - 6k + 10 = 0
6k = 18
k = 3

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (0+2/2 , -1+1/2) = (1,0)

E = (0+0/2 , -3-1/2) = (0,1)

F = (2+0/2 , 1+3/2) = (1,2)

Area of a triangle = 1/2 {x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)}
Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}
= 1/2 (1+1) = 1 square units
Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]
= 1/2 {8} = 4 square units
Therefore, the required ratio is 1:4.

4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Answer

Let the vertices of the quadrilateral be A ( - 4, - 2), B ( - 3, - 5), C (3, - 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 {x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)}
Area of ΔABC = 1/2 [(-4) {(-5) - (-2)} + (-3) {(-2) - (-2)} + 3 {(-2) - (-5)}]
= 1/2 (12+0+9)
= 21/2 square units
Area of ΔACD = 1/2 [(-4) {(-2) - (3)} + 3{(3) - (-2)} + 2 {(-2) - (-2)}]
= 1/2 (20+15+0)
= 35/2 square units
Area of ☐ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units = 28 square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, - 6), B (3, - 2) and C (5, 2).

Answer

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = (3+5/2, -2+2/2) = (4,0)

Area of a triangle = 1/2 {x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)}

Area of ΔABD = 1/2 [(4) {(-2) - (0)} + 3{(0) - (-6)} + (4) {(-6) - (-2)}]

= 1/2 (-8+18-16)
= -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
Area of ΔABD = 1/2 [(4) {0 - (2)} + 4{(2) - (-6)} + (5) {(-6) - (0)}]
= 1/2 (-8+32-30)
= -3 square units
However, area cannot be negative. Therefore, area of ΔABD is 3 square units.
The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.

Go Back To NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Chapter 7 Class 10 Maths NCERT Solutions is given here which will help you in revising the chapter properly and knowing application formulae. The chapter is divided into 4 topics that will improve your knowledge regarding Coordinate Geometry.

• Introduction: Coordinate geometry is an algebraic tool for studying geometry of figures. In earlier class, we studied how to locate the position of a point on a plane and various terms related to it.

• Distance Formula: You will learn how to find the distance between the two points whose coordinates are given.

• Area of the Triangle: You will learn to find the area of the triangle formed by three given points.

• Section Formula: You will study how to find the coordinates of the point which divides a line segment joining two given points in a given ratio.

You can also access other study resources prepared by Studyrankers experts:

MCQ Questions for Classs 10 Maths Chapter 7 Coordinate Geometry

There are total 4 exercises in the NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry which are very helpful in increasing concentration among students. If you want exercisewise NCERT Solutions then you can get them below.

NCERT Solutions for Class 10 Maths Chapters:

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 8 Introduction To Trigonometry

FAQ on Chapter 7 Coordinate Geometry

How many exercises in Chapter 7 Coordinate Geometry

There are only 4 exercises in the Chapter 7 Class 10 Maths Coordinate Geometry in which last one is optional. These answers are prepared by experts of Studyrankers who have vast experience in teaching. These will help the students in learning the concepts efficiently.

What is Coordinate Geometry?

Coordinate Geometry is the branch of Mathematics in which algebraic methods are used to solve geometrical problems is known as coordinate geometry.

Find the point on y-axis which is equidistant from the points (5, − 2) and (− 3, 2).

Let the required point be P (0, y) The given points are A (5, − 2) and B (− 3, 2)

∴ PA = PB

⇒ PA² = PB²

∴(5 − 0)2 + (− 2 − y)² =(− 3 − 0)² + (2 − y)²

⇒ 5² + (− 2 − y)² = (− 3)² + (2 − y)²

⇒ 25 + 4 + y² + 4y = 9 + 4 + y²− 4y

⇒ 25 + 4y = 9 − 4y

⇒ 8 y = − 16 ⇒ y = − 2

Thus, the required point is (0, − 2)

If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x.

Let O (2, 3) be the centre of the circle.

∴ OA = OB ⇒ OA² = OB²

⇒ (4 − 2)² + (3 − 3)² =(x − 2)² + (5 − 3)²

⇒ 22 =(x − 2)² + 2²

⇒ (x − 2)² = 0

⇒ x −2 = 0

⇒ x = 2

Thus, the required value of x is 2.

↧

MCQ Questions for Class 12 Business Studies: Ch 9 Financial Management

October 14, 2020, 11:26 pm

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MCQ Questions for Class 12 Business Studies: Ch 9 Financial Management

1. Financial leverage is called favourable if:

(a) Return on Investment is lower than the cost of debt

(b) If the degree of existing financial leverage is low

(d) ROI is higher than the cost of debt

► (d) ROI is higher than the cost of debt

2. Short-term Investment Decision is also known as ____

(a) Working capital

(b) Dividend Decision

(d) None of the above

► (a) Working capital

3. The primary goal of the financial management is ____________.

(a) To maximize the return

(b) To minimize the risk

(d) To maximize profit

► (c) To maximize the wealth of owners

4. Which of the following affects the Dividend Decision of a company?

(a) Taxation Policy

(b) Cash Flow Position

(d) All of the above

► (d) All of the above

5. A fixed asset should be financed through:

(a) A long-term liability

(b) A short-term liability

(d) None of the above

► (a) A long-term liability

6. Long term investment decision is also known as _____________

(a) Capital Budgeting

(b) Working Capital

(d) None of the above

► (a) Capital Budgeting

7. Which of the following is not concerned with the Long term investment decision

(a) Management of fixed capital

(b) Inventory management

(d) Opening a new branch

► (b) Inventory management

8. Favourable financial leverage is a situation where _____

(a) ROI is higher than the rate of interest on debt

(b) ROI is Equal to the Rate of interest on debt

(d) None of the above

► (a) ROI is higher than the rate of interest on debt

9. 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

(d) None of the above

► (a) Make the debt relatively cheaper

10. The main objective of financial planning is to ensure that_________

(a) Enough funds are available at the right time

(b) Dividend is paid to shareholders on the right time

(d) Purchase of fixed assets

► (a) Enough funds are available at the right time

11. Financial planning arrives at:

(a) Doing only what is possible with the funds that the firms has at its disposal

(b) Entering that the firm always have significantly more funds than required so that there is no paucity of funds

(d) Ensuring that the firm faces neither a shortage nor a glut of unusable funds

► (d) Ensuring that the firm faces neither a shortage nor a glut of unusable funds

12. Current assets of a business firm should be financed through:

(a) Current liability only

(b) Long-term liability only

(d) None of the above

► (c) Both of the above

13. Companies with a higher growth pattern are likely to:

(a) Dividends are not affected by growth considerations

(b) Pay higher dividends

(d) None of the above

► (c) Pay lower dividends

14. Which of the following is not a financial Decision?

(a) Financing Decision

(b) Investment Decision

(d) Dividend Decision

► (c) Staffing Decision

15. The cheapest source of finance is:

(a) Preference share

(b) Retained earning

(d) Debenture

► (b) Retained earning

16. A decision to acquire a new and modern plant to upgrade an old one is a:

(a) Investment decision

(b) Working capital decision

(d) None of the above

► (a) Investment decision

17. Shareholders funds refer to ________________

(a) Share capital

(b) Surpluses and Retained Earnings

(d) All of the above

► (d) All of the above

18. Which of the following affects the Dividend Decision of a company?

(a) Earnings

(b) Cash Flow Position

(d) All of the above

► (d) All of the above

19. Short term investment decisions affect the ___________

(a) Purchase of fixed assets

(b) Long term profitability

(d) Large amount of funds for future

► (c) Day to Day working of business

20. Financial Management is mainly concerned with ______________.

(a) All aspects of acquiring and utilizing financial resources for firms activities.

(b) Arrangement of funds.

(d) Profit maximization

► (a) All aspects of acquiring and utilizing financial resources for firms activities.

21. Higher debt-equity ratio results in:

(a) Higher degree of financial risk

(b) Higher degree of operating risk

(d) Lower financial risk

► (a) Higher degree of financial risk

22. Cost of advertising and printing prospectus is called__________

(a) Floatation cost

(b) Debt cost

(d) Dividend cost

► (a) Floatation cost

23. Which of the following affects capital budgeting decision?

(a) Investment Criteria and interest rate

(b) Rate of Return

(d) All of the above

► (d) All of the above

24. Higher working capital usually results in:

(a) Higher equity, lower risk and lower profits

(b) Lower current ratio, higher risk and profits

(d) Higher current ratio, higher risk and higher profits

► (d) Higher current ratio, higher risk and higher profits

25. Higher dividend per share is associated with:

(a) High earnings, high cash flows, stable earnings and high growth opportunities

(b) High earnings, high cash flows, stable earnings and lower growth opportunities

(d) High earning, high cash flows, unstable earnings and higher growth opportunities

► (b) High earnings, high cash flows, stable earnings and lower growth opportunities

↧

MCQ Questions for Class 12 Business Studies: Ch 10 Financial Market

October 15, 2020, 2:34 am

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MCQ Questions for Class 12 Business Studies: Ch 10 Financial Market

1. Instruments traded in capital market are__________

(a) Bonds

(b) Equity Shares and Preference Shares

(d) All of the above

► (d) All of the above

2. OTCEI was started on the lines of:

(a) NSE

(b) NYSE

(d) NASAQ

► (c) NASDAQ

3. Primary and secondary markets:

(a) Control each other

(b) Complement each other

(d) Function independently

► (b) Complement each other

4. The settlement cycle in NSE is:

(a) T + 2

(b) T + 5

(d) T + 1

► (a) T + 2

5. Only institutional investors can participate in __________

(a) Loan Market

(b) Money Market

(d) Capital Market

► (b) Money Market

6. Money market deals in ________

(a) Long term Securities

(b) Medium term securities

(d) None of the above

► (c) Short term Securities

7. A Treasury Bill is basically:

(a) An instrument to borrow short-term funds

(b) An instrument to borrow long-term funds

(d) None of the above

► (a) An instrument to borrow short-term funds

8. What is meant by Demat Account?

(a) Development Market Account

(b) Depository Participant Account

(d) Demand Depository Account

► (c) Dematerialisation of of Securities

9. To be listed on OTCEI, the minimum capital requirement for a company is:

(a) Rs. 10 crores

(b) Rs. 3 crores

(d) Rs. 5 crores

► (b) Rs. 3 crores

10. Clearing and settlement operations of NSE are carried out by:

(a) NSDL

(b) SBI

(d) CDSL

► (c) NSCCL

11. NSE commenced futures trading in the year:

(a) 2000

(b) 2001

(d) 1999

► (a) 2000

12. The National Stock Exchange of India was recognized as stock exchange in the year:

(a) 1995

(b) 1992

(d) 1993

► (d) 1993

13. _________ is not a participant in money market

(a) SEBI

(b) NBFCs

(d) Mutual Funds

► (a) SEBI

14. The total number of Stock Exchanges in India is:

(a) 22

(b) 21

(d) 20

► (b) 21

15. When a trade bill is accepted by a commercial bank, it is known as a _____

(a) Certificate of deposit

(b) Commercial Bill

(d) None of the above

► (b) Commercial Bill

16. Securities Exchange Board of India (SEBI) was established in __

(a) 1992

(b) 1956

(d) 1984

► (a) 1992

17. A Treasury bill is an instrument of _____________

(a) Long term debt

(b) Short term debt

(d) Dividend

► (b) Short term debt

18. Capital market deals in ___________

(a) Medium and long term securities

(b) Very short term securities

(d) None of the above

► (a) Medium and long term securities

19. What is the meaning of right issue?

(a) Company sells the securities to some selected institutions

(b) Company offers new shares to its existing shareholders

(d) None of the above

► (b) Company offers new shares to its existing shareholders

20. _______ Market instruments enjoy higher degree of liquidity

(a) Both money and capital market

(b) Money market

(d) None of the above

► (b) Money market

21. At present only two depositories are registered with SEBI _____

(a) NSDL and HDFC

(b) NSDL and CDSL

(d) NEFT and TDS

► (b) NSDL and CDSL

22. Which of the following is not a part of capital market?

(a) Banks

(b) Financial Institutions

(d) RBI

► (d) RBI

23. Treasury bills are also known as _____________

(a) Fixed interest Bonds

(b) Low Interest Bonds

(d) Zero Coupon Bonds

► (d) Zero Coupon Bonds

24. Which of the following is a method of floatation?

(a) Offer for sale

(b) Private Placement

(d) All of the above

► (d) All of the above

25. What type of instruments are traded in a Money Market?

(a) Treasury bills

(b) Commercial bills

(d) All of the above

► (d) All of the above

↧

MCQ Questions for Class 12 Business Studies: Ch 11 Marketing Management

October 15, 2020, 3:44 am

≫ Next: MCQ Questions for Class 12 Business Studies: Ch 12 Consumer Protection

≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 10 Financial Market

MCQ Questions for Class 12 Business Studies: Ch 11 Marketing Management

1. Which of the following is not a function of marketing?

(a) Bending the customers according to product

(b) Marketing planning

(d) Gathering and analysing market information

► (a) Bending the customers according to product

2. Providing information to the customers about product, its features and quality etc., are part of :

(a) Production

(b) Pricing

(d) None of the above

► (c) Promotion

3. Which of the following is the part of marketing mix?

(a) Price

(b) Product

(d) All of the above

► (d) All of the above

4. Which of the following is not a part of 'Place and Promotion'?

(a) Warranties

(b) Advertising

(d) Sales Promotion

► (a) Warranties

5. Which of the following is not concerned with the 'Price'?

(a) Advertising

(b) Discounts

(d) Margins

► (a) Advertising

6. _________ refers to further packaging components necessary for storage and transportation.

(a) Transportation package

(b) Primary package

(d) None of the above

► (a) Transportation package

7. Secondary package refers to _________

(a) First layer of protection

(b) Transportation package

(d) Immediate container

► (c) Additional layers of protection

8. ______ refers to the act of designing and producing the container or wraper of a product.

(a) Trade Mark

(b) Brand Name

(d) Packaging

► (d) Packaging

9. Under marketing mix which of the following is not a part of ?Product? :

(a) List price

(b) Quality

(d) Brand Name

► (a) List price

10. Discounts, coupons, contests, free samples and offering extra quantity etc., are the examples of __________

(a) Channels of distribution

(b) Personal Selling

(d) None of the above

► (c) Sales Promotion

11. The word 'Market' has come from the Latin word _____________

(a) Marketo

(b) Marketio

(d) Matico

► (c) Marcatus

12. Which of the following is a brand name?

(a) Nike

(b) Woodland

(d) All of the above

► (d) All of the above

13. Which of the following is a direct channel of distribution?

(a) Manufacturer, wholesaler, retailer, customer

(b) Manufacturer to customer

(d) None of the above

► (b) Manufacturer to customer

14. The Basic role of marketer is to _________

(a) Sell the product

(b) Advertise only

(d) Identify a need and fill it

► (d) Identify a need and fill it

15. Which of the following will not affect the choice of channel of distribution?

(a) Market factors

(b) Environmental factors

(d) Competitive factors

► (c) Trade mark

16. ________ refers to the product?s immediate container

(a) Transportation package

(b) Secondary package

(d) None of the above

► (c) Primary Package

17. ________ refers to designing the label to be put on the package

(a) Pricing

(b) Packaging

(d) Labelling

► (d) Labelling

18. Which of the following is a function of packaging?

(a) Product promotion

(b) Product identification

(d) All of the above

► (d) All of the above

19. SWOT Analysis is a strong part of ________

(a) Marketing

(b) Selling

(d) None of the above

► (a) Marketing

20. Which of the following is a good quality of a good salesperson?

(a) Mental quality

(b) Honesty

(d) All of the above

► (d) All of the above

21. Advertising lacks ______________

(a) Reaching to a large number of people

(b) Covering market in short time

(d) Impersonal communication

► (c) Direct Feedback

22. Which of the following starts from (starting point) factory_____________

(a) Product concept

(b) Production concept

(d) All of the above

► (d) All of the above

23. Scope of _______ is limited but scope of _________ is wider

(a) Marketing, Selling

(b) Selling, Marketing

(d) Selling, Manufacturing

► (b) Selling, Marketing

24. Airtight containers and packets used for chips, biscuits and jams etc., are the example of

(a) Product differentiation

(b) Product identification

(d) Product Protection

► (d) Product Protection

25. __________ Activities start after the product has been developed

(a) Selling

(b) Publicity

(d) None of the above

► (a) Selling

↧

MCQ Questions for Class 12 Business Studies: Ch 12 Consumer Protection

October 15, 2020, 5:24 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 1 The Cold War Era

≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 11 Marketing Management

MCQ Questions for Class 12 Business Studies: Ch 12 Consumer Protection

1. Which act provides safeguards and reliefs to the buyers of the goods in case the goods purchased do not match with express or implied conditions or warranties?

(a) Sale of goods Act, 1930

(b) Consumer protection Act, 1986

(d) None of the above

► (a) Sale of goods Act, 1930

2. ________A standardised mark is printed on jewellery.

(a) Agmark

(b) ISO 2009

(d) Hallmark

► (d) Hallmark

3. Which of the following statements is not true with regard to the National Commission?

(a) It consists of a President and at least five other members, one of whom should be a woman.

(b) The members are appointed by the Central Government.

(c) A complaint can be made to the National Commission when the value of the goods or services in question, along with the compensation claimed, exceeds ₹1 crore.

(d) Where the aggrieved party was not satisfied with the order of the National Commission, the case can be taken to the Supreme Court of India.

► (a) It consists of a President and at least five other members, one of whom should be a woman.

4. ___ means any person who buys any goods for personal use or hires or avails of any services for a consideration.

(a) Consumer

(b) Manufacturer

(d) Service Provider

► (a) Consumer

5. Rights of a consumer are protected under __________________

(a) Consumer Protection Act, 1984

(b) Consumer Protection Act, 1986

(d) Consumer Protection Act, 1982

► (b) Consumer Protection Act, 1986

6. A complaint can be filed against ___ who adopt(s) unfair trade practices.

(a) Trader only

(b) Manufacturer only

(d) Consumer

► (c) Trader and Manufacturer

7. A District Forum has jurisdiction to entertain a consumer complaint involving product value:

(a) Up to ₹ 20 lakh

(b) Up to ₹ 10 lakh

(d) Above ₹ 1 crore

► (a) Up to ₹ 20 lakh

8. Which one of the following is not a consumer right?

(a) Right to safety

(b) Right to seek redressal

► (c) Right to get discount

9. In case an aggrieved consumer is not satisfied with the decision of the State Commission, he can make a further appeal in

(a) National Commission

(b) District Forum

(d) All of the above

► (a) National Commission

10. In case a consumer is not satisfied with the order passed in the State Commission, he can further make an appeal in the National Commission within a time period of

(a) 10 days

(b) 20 days

(d) 45 days

► (c) 30 days

11. A consumer has a right to file a complaint and to be heard in case of dissatisfaction with goods or services according to the ___________

(a) Right to Seek redressal

(b) Right to choose

(d) None of the above

► (c) Right to be heard

12. _______ is set up by the Central Government

(a) State commissions

(b) District forum

(d) None of the above

► (c) National commission

13. ______ mark is used for electrical products.

(a) ISI

(b) FPO

(d) Agmark

► (a) ISI

14. _____ are set up in each district by the State Government concerned.

(a) District Forums

(b) National Commission

(d) None of the above

► (a) District Forums

15. The state commissions are set up in each state by ________

(a) Supreme court

(b) High court

(d) State Government

► (d) State Government

16. NGOs are __________

(a) Government organisations

(b) Non profit organisations

(d) Service providing organisations

► (b) Non profit organisations

17. Rights of a consumer are protected under _________

(a) Consumer Protection Act, 1982

(b) Consumer Protection Act, 1986

(d) Consumer Protection Act, 1984

► (b) Consumer Protection Act, 1986

18. Consumers can ask anything regarding date of manufacture, price, quantity etc., is an example of _____________

(a) Right to be informed

(b) Right to choose

(d) Right to be heard

► (a) Right to be informed

19. Who can file a complaint under the Consumer Protection Act, 1986?

(a) A legal her of a deceased consumer

(b) Any registered consumers association

(d) All of the above

► (d) All of the above

20. The consumer has a right to get relief in case of defective goods or deficient services as per the________________

(a) Right to be informed

(b) Right to choose

(d) Right to seek redressal

► (d) Right to seek redressal

21. Consumers should use ISI marked electrical appliance is an example of

(a) Right to be informed

(b) Right to safety

(d) Right to be heard

► (b) Right to safety

22. As per the _____________ the consumer has right to acquire knowledge and to be a well informed consumer through life.

(a) Right to consumer education

(b) Right to be choose

(d) Right to be informed

► (a) Right to consumer education

↧

MCQ Questions for Class 12 Political Science: Ch 1 The Cold War Era

October 15, 2020, 8:50 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 2 The End of Bipolarity

≪ Previous: MCQ Questions for Class 12 Business Studies: Ch 12 Consumer Protection

MCQ Questions for Class 12 Political Science: Ch 1 The Cold War Era

1. The event that took place in 1961 was

(a) The construction of the Berlin wall.

(b) Soviet intervention in Afghanistan.

(d) The unification of Germany.

► (a) The construction of the Berlin wall.

2. After second World War, the world was divided into the blocks of

(a) U.S.A and U.K.

(b) USSR and China.

(d) U.S.A and USSR.

► (d) U.S.A and USSR.

3. The person that opposed neo-colonialism was

(a) Churchill.

(b) Brezhnev.

(d) Kwame Nkrumah.

► (d) Kwame Nkrumah.

4. The Warsaw pact was led by _________.

(a) Japan

(b) Britain

(d) Soviet Union

► (d) Soviet Union

5. Which of the following countries was not a NATO member?

(a) Germany

(b) Czech Republic

(d) India

► (d) India

6. The Western Alliance was headed by

(a) United States.

(b) Brazil.

(d) Balkan countries.

► (a) United States.

7. The island country that received both diplomatic and financial aid from the Soviet Union was

(a) Japan.

(b) Cuba.

(d) Norway.

► (b) Cuba.

8. The rivalry between USA & USSR during the post-second world war period is known as _____________.

(a) Apartheid

(b) Cold War

(d) Hot war

► (b) Cold War

9. Korea was a colony of

(a) USA.

(b) Japan.

(d) Poland.

► (b) Japan.

10. In which country are the cities of Hiroshima and Nagasaki located?

(a) China

(b) Japan

(d) Italy

► (b) Japan

11. When USSR placed nuclear missiles in Cuba?

(a) 1960

(b) 1961

(d) 1963

12. What is the full form of ‘SEATO’?

(a) South East Asian Task Organisation

(b) South East Asian Treaty Organisation

(d) South East Asian Team Organisation

► (b) South East Asian Treaty Organisation

13. What is the full form of ‘CENTO’?

(a) The Central Task Organisation

(b) The Central Treaty Organisation

(d) The Central Team Organisation

► (b) The Central Treaty Organisation

14. Which two ideologies were involved in a conflict during the Cold War era?

(a) Communism and Capitalism

(b) Monarchism and Capitalism

(d) None of the above

► (a) Communism and Capitalism

15. Which organisation came into existence in April 1949?

(a) SEATO

(b) CENTO

(d) Warsaw Pact

► (c) NATO

16. Which one of the following is an agreement signed between the two superpowers starting in the 1960s?

(a) Berlin Pact

(b) Limited Test Ban Treaty.

(d) Warsaw Pact

► (b) Limited Test Ban Treaty.

↧

MCQ Questions for Class 12 Political Science: Ch 2 The End of Bipolarity

October 15, 2020, 10:25 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 3 US Hegemony in World Politics

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 1 The Cold War Era

MCQ Questions for Class 12 Political Science: Ch 2 The End of Bipolarity

1. Which was the first Soviet Republic to declare its independence from Soviet Russia?

(a) Lithuania.

(b) Moldova.

(d) Georgia.

► (a) Lithuania.

2. Bolshevik Communist party was founded by

(a) Vladimir Lenin.

(b) Stalin.

(d) Leonid Brezhnev.

► (a) Vladimir Lenin.

3. The most severe conflict took place in the Balkan republics of:

(a) Bulgaria

(b) Greece

(d) Macedonia

► (c) Yugoslavia

4. The Berlin wall fall in _________.

(a) November 1989

(b) December 1989

(d) February 1990

► (a) November 1989

5. When did the Soviet Union collapse?

(a) 1989

(b) 1990

(d) 1992

6. The country that gave the aid and technical assistance for steel plants in India is

(a) The United States.

(b) USSR.

(d) Japan.

► (b) USSR.

7. The USSR came into being after the Socialist Revolution in Russia in the year

(a) 1915.

(b) 1916.

(d) 1918.

8. The Russian Revolution in 1917 was led by

(a) Joseph Stalin.

(b) Nikita Khrushchev.

(d) Vladimir Ilyich Lenin.

► (d) Vladimir Ilyich Lenin.

9. The Chechens are

(a) Bhuddhist group.

(b) Christain ethnic group.

(d) Muslim ethinic group.

► (d) Muslim ethinic group.

10. The Soviet Union invaded Afghanistan in the year

(a) 1975.

(b) 1977.

(d) 1981.

11. Who became General Secretary of the Communist Party of the Soviet Union in 1985?

(a) Joseph Stalin.

(b) Nikita Khrushchev.

(d) Vladimir Ilyich Lenin.

► (b) Nikita Khrushchev.

12. Which country in Central Asia witnessed a civil war that went on for ten years?

(a) Azerbaijan

(b) Tajikistan

(d) Turkmenistan

► (b) Tajikistan

↧

MCQ Questions for Class 12 Political Science: Ch 3 US Hegemony in World Politics

October 15, 2020, 10:31 pm

≫ Next: MCQ Questions for Class 12 Political Science: Ch 4 Alternative Centres of Power

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 2 The End of Bipolarity

MCQ Questions for Class 12 Political Science: Ch 3 US Hegemony in World Politics

1. The US Hegemony began in

(a) 1990.

(b) 1991.

(d) 1993.

► (b) 1991.

2. The first business school in the world is

(a) Harvard School.

(b) Oxford School.

(d) Stanford School.

► (c) Wharton School.

3. The root of the word 'hegemony' lies in the

(a) Latin.

(b) Classical Greeks.

(d) Spanish.

► (b) Classical Greeks.

4. The duration of Cold War was_______.

(a) 1914 - 1919

(b) 1939 - 1945

(d) 1965 – 1991

5. United States was founded in which year?

(a) 1774

(b) 1776

(d) 1780

► (b) 1776

6. The hegemony that shapes minds through ideological forces like values, beliefs, etc., is referred as

(a) Hegemony as unlimited power.

(b) Hegemony as structural power.

(d) Hegemony as soft power.

► (d) Hegemony as soft power.

7. The 'Operation Iraqi Freedom' was launched on __________.

(a) 19th March, 2003

(b) 19th March, 2004

(d) 21st March, 2004

► (a) 19th March, 2003

8. The headquarters of the US defence department is at

(a) New York.

(b) Washington.

(d) Virginia.

► (d) Virginia.

9. The World Trade Center and Pentagon buildings of USA were attacked by:

(a) Zamait–e-Islami.

(b) Al-Qaeda.

(d) Hamas.

► (b) Al-Qaeda.

10. US embassies were bombed by Al-Qaeda in

(a) Dublin.

(b) New Delhi.

(d) Zimbabwe.

► (c) Dar-es-Salaam.

11. The US operation in Iraq was called:

(a) Operation Desert Suicide

(b) Operation Desert Storm

(d) Operation Desert Hamas

► (b) Operation Desert Storm

12. Why was the 'First Gulf War' called the ‘Video Game War’?

(a) Use of military's in large number.

(b) Use of Mass Destruction Weapons

(d) None of the above

► (c) Widespread television coverage

13. When Iraq invaded Kuwait?

(a) June1990

(b) July1990

(d) September1990

► (c) August 1990

14. ‘What was ‘Operation Iraqi Freedom’?

(a) A Military operation

(b) An Economic operation

(d) An Economic Pact

► (a) A Military operation

↧

MCQ Questions for Class 12 Political Science: Ch 4 Alternative Centres of Power

October 15, 2020, 11:37 pm

≫ Next: NCERT Solutions for Class 10 Maths Ch 8 Introduction to Trigonometry

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 3 US Hegemony in World Politics

MCQ Questions for Class 12 Political Science: Ch 4 Alternative Centres of Power

1. EURATOM is known as

(a) Europe Atom Efficiency Community.

(b) Europe Aviation Energy Center.

(d) European Ariel Energetic Center.

► (c) European Atomic Energy Community.

2. ASEAN stands for ___________.

(a) Alliance of Southeast Asian Nations

(b) Alliance of Southern Asian Nations

(d) Association of South East Asian Nations

► (d) Association of South East Asian Nations

3. The major landmark in the Indo-China relations was

(a) Indira Gandhi's visit to China.

(b) A.B. Vajpayee's visit to China.

(d) Rajiv Gandhi's visit to China.

► (d) Rajiv Gandhi's visit to China.

4. WTO stands for

(a) World Trade Organisation.

(b) World Territorial Organisation.

(d) World Trade Organ.

► (a) World Trade Organisation.

5. In January 2007, the two countries that joined EU were

(a) Britain and Denmark.

(b) Italy and France.

(d) Bulgaria and Romania.

► (d) Bulgaria and Romania.

6. What is the name given to the 13 members of the European Union who have adopted Euro?

(a) Euro zone

(b) Euro alliance

(d) Euro league

► (a) Euro zone

7. Which of the following is the only country that suffered the destruction caused by nuclear bombs?

(a) China

(b) India

(d) Japan

► (d) Japan

8. In which year ASEAN was established?

(a) 1967

(b) 1965

(d) 1960

► (a) 1967

9. The second largest contributor to the regular budget of the UN is

(a) Japan.

(b) China.

(d) Britain.

► (a) Japan.

10. The European Economic Community was formed in 1958 by

(a) 10 countries.

(b) 12 countries.

(d) 6 countries.

► (d) 6 countries.

11. In which year OEEC was established?

(a) 1946

(b) 1947

(d) 1949

12. When was European Union eastablished?

(a) 1990

(b) 1991

(d) 1993

13. When was The ASEAN Regional Forum eastablished?

(a) 1991

(b) 1992

(d) 1994

► (d) 1994

14. When was People’s Republic of China eastablished?

(a) 1947

(b) 1948

(d) 1950

15. Who proposed the ‘four modernisations’ in China?

(a) Mao Zedong

(b) Zhou Enlai

(d) None of the above

► (b) Zhou Enlai

16. When China became member of WTO?

(a) 1991

(b) 1998

(d) 2005

↧

NCERT Solutions for Class 10 Maths Ch 8 Introduction to Trigonometry

October 16, 2020, 12:32 am

≫ Next: MCQ Questions for Class 12 Political Science: Ch 5 Contemporary South Asia

≪ Previous: MCQ Questions for Class 12 Political Science: Ch 4 Alternative Centres of Power

NCERT Solutions for Class 10 Maths Ch 8 Introduction to Trigonometry

Chapter 8 Introduction to Trigonometry Class 10 Maths NCERT Solutions will be helpful in completing homework in no time. These NCERT Solutions for Class 10 Maths will be useful in increasing concentration among students. These answers are detailed and prepared according to the latest marking scheme release by CBSE 2020-21.These NCERT Solutions will enhance knowledge of students and they can be able to solve difficult problems in less time.

Exercise 8.1

Page No: 181

1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Answer

In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get

AC² = AB² + BC²= (24)² + 7² = (576+49) cm² = 625 cm²
⇒ AC = 25

(i) sin A = BC/AC = 7/25 cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25
cos C = BC/AC = 7/25

2. In Fig. 8.13, find tan P – cot R.

Answer

By Applying Pythagoras theorem in ΔPQR , we get
PR² = PQ² + QR² = (13)² = (12)² + QR²=169 = 144 + QR²
⇒ QR² = 25 ⇒ QR = 5 cm

Now,
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
A/q
tan P – cot R = 5/12 - 5/12 = 0

3. If sin A =3/4, calculate cos A and tan A.

Answer

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sin A = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,
AC² = AB² + BC²
(4k)² = AB² + (3k)²
16k² - 9k² = AB²
AB²=7k²
AB = √7 k

cos A = AB/AC = √7 k/4k = √7/4

tan A = BC/AB = 3k/√7 k = 3/√7

4. Given 15 cot A = 8, find sin A and sec A.

Answer

Let ΔABC be a right-angled triangle, right-angled at B.

We know that cot A = AB/BC = 8/15 (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.

By Pythagoras theorem we get,
AC² = AB² + BC²
AC² = (8k)² + (15k)²
AC² = 64k² + 225k²
AC²=289k²
AC = 17 k

sin A = BC/AC = 15k/17k = 15/17

sec A = AC/AB = 17k/8 k = 17/8

5. Given sec θ = 13/12, calculate all other trigonometric ratios.

Answer

Let ΔABC be a right-angled triangle, right-angled at B.

We know that sec θ = OP/OM = 13/12 (Given)
Let OP be 13k and OM will be 12k where k is a positive real number.

By Pythagoras theorem we get,
OP² = OM² + MP²
(13k)² = (12k)²+ MP²
169k² - 144k² = MP²
MP²=25k²

MP = 5

Now,

sin θ = MP/OP = 5k/13k = 5/13

cos θ = OM/OP = 12k/13k = 12/13

tan θ = MP/OM = 5k/12k = 5/12

cot θ = OM/MP = 12k/5k = 12/5

cosec θ = OP/MP = 13k/5k = 13/5

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

Answer

Let ΔABC in which CD ⊥ AB.

A/q,

cos A = cos B

⇒ AD/AC = BD/BC

⇒ AD/BD = AC/BC

Let AD/BD = AC/BC = k

⇒ AD = kBD .... (i)

⇒ AC = kBC .... (ii)

By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD² = AC² - AD² …. (iii)
and also CD² = BC² - BD² …. (iv)
From equations (iii) and (iv) we get,
AC² - AD² = BC² - BD²
⇒ (kBC)² - (k BD)² = BC² - BD²
⇒ k² (BC² - BD²) = BC² - BD²
⇒ k² = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B (Angles opposite to equal sides of a triangle are equal-isosceles triangle)

7. If cot θ =7/8, evaluate :

(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)

(ii) cot²θ

Answer

Let ΔABC in which ∠B = 90º and ∠C = θ

A/q,

cot θ = BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC²
AC² = (8k)² + (7k)²
AC² = 64k² + 49k²
AC²=113k²
AC = √113 k

sin θ = AB/AC = 8k/√113 k = 8/√113

and cos θ = BC/AC = 7k/√113 k = 7/√113

(i) (1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ) = (1-sin²θ)/(1-cos²θ) = {1 - (8/√113)²}/{1 - (7/√113)²}

= {1 - (64/113)}/{1 - (49/113)} = {(113 - 64)/113}/{(113 - 49)/113} = 49/64

(ii) cot²θ = (7/8)² = 49/64

8. If 3cot A = 4/3 , check whether (1-tan²A)/(1+tan²A) = cos²A – sin²A or not.

Answer

Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC²
AC² = (4k)² + (3k)²
AC² = 16k² + 9k²
AC²=25k²
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan²A)/(1+tan²A) = 1- (3/4)²/1+ (3/4)²= (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos²A – sin²A = (4/5)²- (3/4)² =(16/25) - (9/25) = 7/25
R.H.S. = L.H.S.
Hence, (1-tan²A)/(1+tan²A) = cos²A – sin²A

9. In triangle ABC, right-angled at B, if tan A =1/√3 find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Answer

Let ΔABC in which ∠B = 90º,
A/q,

tan A = BC/AB = 1/√3
Let AB = √3 k and BC = k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC² = AB² + BC²
AC² = (√3 k)² + (k)²
AC² = 3k² + k²
AC²=4k²
AC = 2k
sin A = BC/AC = 1/2 cos A = AB/AC = √3/2 ,
sin C = AB/AC = √3/2 cos A = BC/AC = 1/2
(i) sin A cos C + cos A sin C = (1/2×1/2) + (√3/2×√3/2) = 1/4+3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2×1/2) - (1/2×√3/2) = √3/4 - √3/4 = 0

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer

Given that, PR + QR = 25 , PQ = 5
Let PR be x. ∴ QR = 25 - x

By Pythagoras theorem ,
PR² = PQ² + QR²
x² = (5)² + (25 - x)²
x² = 25 + 625 + x² - 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 - 13) cm = 12 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

11. 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) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.

Answer

(i) False.

In ΔABC in which ∠B = 90º,

AB = 3, BC = 4 and AC = 5

Value of tan A = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as

it will follow the Pythagoras theorem.

AC² = AB² + BC²
5² = 3² + 4²
25 = 9 + 16
25=25

(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC² = AB² + BC²
(12k)² = (5k)² + BC²
BC²+ 25k²= 144k²
BC²= 119k²

Such a triangle is possible as it will follow the Pythagoras theorem.

(iii) False.

Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.

(iv) False.

cot A is not the product of cot and A. It is the cotangent of ∠A.

(v) False.

sin θ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.

Excercise 8.2

Page No: 187

1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan²45° + cos²30° – sin²60°
(iii) cos 45°/(sec 30° + cosec 30°) (iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos²60° + 4sec²30° - tan²45°)/(sin²30° + cos²30°)

Answer

(i) sin 60° cos 30° + sin 30° cos 60°
= (√3/2×√3/2) + (1/2×1/2) = 3/4 + 1/4 = 4/4 = 1

(ii) 2 tan²45° + cos²30° – sin²60°
= 2×(1)²+ (√3/2)² - (√3/2)²= 2

(iii) cos 45°/(sec 30° + cosec 30°)
= 1/√2/(2/√3 + 2) = 1/√2/{(2+2√3)/√3)
= √3/√2×(2+2√3) = √3/(2√2+2√6)
= √3(2√6-2√2)/(2√6+2√2)(2√6-2√2)
= 2√3(√6-√2)/(2√6)²-(2√2)²
= 2√3(√6-√2)/(24-8) = 2√3(√6-√2)/16
= √3(√6-√2)/8 = (√18-√6)/8 = (3√2-√6)/8

(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
= (1/2+1-2/√3)/(2/√3+1/2+1)
= (3/2-2/√3)/(3/2+2/√3)
= (3√3-4/2√3)/(3√3+4/2√3)
= (3√3-4)/(3√3+4)
= (3√3-4)(3√3-4)/(3√3+4)(3√3-4)
= (3√3-4)²/(3√3)²-(4)²
= (27+16-24√3)/(27-16)
= (43-24√3)/11]

(v) (5cos²60° + 4sec²30° - tan²45°)/(sin²30° + cos²30°)
= 5(1/2)²+4(2/√3)²-1²/(1/2)²+(√3/2)²
= (5/4+16/3-1)/(1/4+3/4)
= (15+64-12)/12/(4/4)
= 67/12

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan²30° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan²45°/1+tan²45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan²30° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Answer

(i) (A) is correct.
2tan 30°/1+tan²30° = 2(1/√3)/1+(1/√3)²
= (2/√3)/(1+1/3) = (2/√3)/(4/3)

= 6/4√3 = √3/2 = sin 60°

(ii) (D) is correct.

1-tan²45°/1+tan²45° = (1-1²)/(1+1²)

= 0/2 = 0

(iii) (A) is correct.

sin 2A = 2 sin A is true when A =

= As sin 2A = sin 0° = 0
2 sin A = 2sin 0° = 2×0 = 0

or,

sin 2A = 2sin A cos A

⇒2sin A cos A = 2 sin A

⇒ 2cos A = 2 ⇒ cos A = 1

⇒ A = 0°

(iv) (C) is correct.

2tan30°/1-tan²30° = 2(1/√3)/1-(1/√3)²

= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°

3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.

Answer

tan (A + B) = √3

⇒ tan (A + B) = tan 60°

⇒ (A + B) = 60° ... (i)

tan (A – B) = 1/√3

⇒ tan (A - B) = tan 30°
⇒ (A - B) = 30° ... (ii)

Adding (i) and (ii), we get

A + B + A - B = 60° + 30°

2A = 90°

A= 45°

Putting the value of A in equation (i)

45° + B = 60°

⇒ B = 60° - 45°

⇒ B = 15°

Thus, A = 45° and B = 15°

4. State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cot A is not defined for A = 0°.

Answer

(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°

= 1/2 + √3/2 = 1+√3/2

(ii) True.

sin 0° = 0

sin 30° = 1/2

sin 45° = 1/√2

sin 60° = √3/2

sin 90° = 1

Thus the value of sin θ increases as θ increases.

(iii) False.

cos 0° = 1

cos 30° = √3/2

cos 45° = 1/√2

cos 60° = 1/2

cos 90° = 0

Thus the value of cos θ decreases as θ increases.

(iv) True.

cot A = cos A/sin A

cot 0° = cos 0°/sin 0° = 1/0 = undefined.

Excercise 8.3

Page No : 189

1. Evaluate :

(i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° – sin 42° (iv) cosec 31° – sec 59°

Answer

(i) sin 18°/cos 72°

= sin (90° - 18°) /cos 72°

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

= tan (90° - 36°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° - sin 42°

= cos (90° - 42°) - sin 42°

= sin 42° - sin 42° = 0

(iv) cosec 31° - sec 59°

= cosec (90° - 59°) - sec 59°
= sec 59° - sec 59° = 0

2. Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Answer

(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° - 42°) tan (90° - 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° - sin 38° sin 52°
= cos (90° - 52°) cos (90°-38°) - sin 38° sin 52°
= sin 52° sin 38° - sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Answer

A/q,
tan 2A = cot (A- 18°)
⇒ cot (90° - 2A) = cot (A -18°)
Equating angles,
⇒ 90° - 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°

4. If tan A = cot B, prove that A + B = 90°.

Answer

A/q,

tan A = cot B
⇒ tan A = tan (90° - B)
⇒ A = 90° - B
⇒ A + B = 90°

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

Answer

A/q,
sec 4A = cosec (A - 20°)
⇒ cosec (90° - 4A) = cosec (A - 20°)

Equating angles,
90° - 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°

Page No : 190

6. If A, B and C are interior angles of a triangle ABC, then show that

sin (B+C/2) = cos A/2

Answer

In a triangle, sum of all the interior angles

A + B + C = 180°

⇒ B + C = 180° - A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

⇒ sin (B+C)/2 = sin (90°-A/2)

⇒ sin (B+C)/2 = cos A/2

7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Answer

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

Excercise 8.4

Page No : 193

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer

cosec²A- cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ 1/sin²A = 1 + cot²A
⇒sin²A = 1/(1+cot²A)

Now,
sin²A = 1/(1+cot²A)
⇒ 1 - cos²A = 1/(1+cot²A)
⇒cos²A = 1 - 1/(1+cot²A)
⇒cos²A = (1-1+cot²A)/(1+cot²A)
⇒ 1/sec²A = cot²A/(1+cot²A)
⇒ secA = (1+cot²A)/cot²A

also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A

2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Answer

We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos²A + sin²A = 1
⇒ sin²A = 1 - cos²A
⇒ sin²A = 1 - (1/sec²A)
⇒ sin²A = (sec²A-1)/sec²A

also,
sin A = 1/cosec A
⇒cosec A = 1/sin A

Now,
sec²A - tan²A = 1
⇒ tan²A = sec²A + 1

also,
tan A = 1/cot A
⇒ cot A = 1/tan A

3. Evaluate :
(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
(ii)  sin 25° cos 65° + cos 25° sin 65°

Answer

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
= [sin²(90°-27°) + sin²27°]/[cos²(90°-73°) + cos²73°)]
= (cos²27°+ sin²27°)/(sin²27° + cos²73°)
= 1/1 =1 (∵ sin²A + cos²A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos²65°+ sin²65° = 1

4. Choose the correct option. Justify your choice.
(i) 9 sec²A - 9 tan²A =
(A) 1   (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ - cosec θ)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (secA + tanA) (1 - sinA) =
(A) secA   (B) sinA (C) cosecA (D) cosA

(iv) 1+tan²A/1+cot²A =

(A) sec²A (B) -1 (C) cot²A (D) tan²A

Answer

(i) (B) is correct.

9 sec²A - 9 tan²A

= 9 (sec²A - tan²A)
= 9×1 = 9 (∵ sec2 A - tan2 A = 1)

(ii) (C) is correct

(1 + tan θ + sec θ) (1 + cot θ - cosec θ)

= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ - 1/sin θ)

= (cos θ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ)²-1²/(cos θ sin θ)

= (cos²θ + sin²θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ)

= (2cos θ sin θ)/(cos θ sin θ) = 2

(iii) (D) is correct.

(secA + tanA) (1 - sinA)

= (1/cos A + sin A/cos A) (1 - sinA)

= (1+sin A/cos A) (1 - sinA)

= (1 - sin²A)/cos A

= cos²A/cos A = cos A

(iv) (D) is correct.

1+tan²A/1+cot²A

= (1+1/cot²A)/1+cot²A

= (cot²A+1/cot²A)×(1/1+cot²A)

= 1/cot²A = tan²A

5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.

(i) (cosec θ - cot θ)²= (1-cos θ)/(1+cos θ)

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

[Hint : Write the expression in terms of sin θ and cos θ]

(iv) (1 + sec A)/sec A = sin²A/(1-cos A)

[Hint : Simplify LHS and RHS separately]

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A.

(vii) (sin θ - 2sin³θ)/(2cos³θ-cos θ) = tan θ
(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² =tan²A

Answer

(i) (cosec θ - cot θ)²= (1-cos θ)/(1+cos θ)
L.H.S. = (cosec θ - cot θ)²
= (cosec²θ + cot²θ - 2cosec θ cot θ)
= (1/sin²θ + cos²θ/sin²θ - 2cos θ/sin²θ)
= (1 + cos²θ - 2cos θ)/(1 - cos²θ)
= (1-cos θ)²/(1 - cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.

(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos²A + (1+sin A)²]/(1+sin A)cos A
= (cos²A + sin²A + 1 + 2sin A)/(1+sin A)cos A
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.

(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]
= sin²θ/[cos θ(sin θ-cos θ)] - cos²θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin²θ/cos θ) - (cos²θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin³θ - cos³θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.

(iv) (1 + sec A)/sec A = sin²A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin²A/(1-cos A)
= (1 - cos²A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.

(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec²A = 1+cot²A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
= (cot A - 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A - cosec²A + cot²A + cosec A)/(cot A+ 1 – cosec A) (using cosec²A - cot²A = 1)
= [(cot A + cosec A) - (cosec²A - cot²A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) - (cosec A + cot A)(cosec A - cot A)]/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A = R.H.S.

Dividing Numerator and Denominator of L.H.S. by cos A,

= sec A + tan A = R.H.S.

(vii) (sin θ - 2sin³θ)/(2cos³θ-cos θ) = tan θ
L.H.S. = (sin θ - 2sin³θ)/(2cos³θ - cos θ)
= [sin θ(1 - 2sin²θ)]/[cos θ(2cos²θ- 1)]
= sin θ[1 - 2(1-cos²θ)]/[cos θ(2cos²θ -1)]
= [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]
= tan θ = R.H.S.

(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7+tan²A+cot²A
L.H.S. = (sin A + cosec A)²+ (cos A + sec A)²
= (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
= (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A
= 1 + 2 + 2 + 2 + tan²A + cot²A
= 7+tan²A+cot²A = R.H.S.

(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A - sin A)(1/cos A - cos A)
= [(1-sin²A)/sin A][(1-cos²A)/cos A]
= (cos²A/sin A)×(sin²A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin²A+cos²A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.

(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² =tan²A
L.H.S. = (1+tan²A/1+cot²A)
= (1+tan²A/1+1/tan²A)
= 1+tan²A/[(1+tan²A)/tan²A]
= tan²A

Go Back To NCERT Solutions for Class 10 Matha

NCERT Solutions for Class 10 Maths Chapter 8 Coordinate Geometry

Chapter 8 Class 10 Maths NCERT Solutions will help students in preparing well before the examinations and guide students in a better way. There are total 5 topics in the chapter which will enhance your knowledge regarding trigonometric identities.

• Introduction: Trigonometry is the study of relationships between the sides and angles of a triangle. In today's world, most of the technologically advanced methods used in Engineering and Physical Sciences are based on trigonometrical concepts.

• Trigonometric Ratios: We will study some ratios of the sides of a right triangle with respect to its acute angles, called trigonometric ratios of the angle. We will limit our study to acute angles only.

• Trigonometric Ratios of Some Specific Angles: We will be finding the values of the trigonometric ratios for 30°, 45°, 60° and also 0°.

• Trigonometric Ratios of Complementary Angles: Two angles are said to be complementary

if their sum equals 90°. You need to find the vale of specific angle through the situation given.

• Trigonometric Identities: An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved.

You can also access other study resources prepared by Studyrankers experts:

CBSE Notes for Class 10 Maths Chapter 8 Introduction to Trigonometry

MCQ Questions for Classs 10 Maths Chapter 8 Introduction to Trigonometry

There are total 4 exercises in the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry which are great source of inculcating correct learning habits among students. You can also find exercisewise NCERT Solutions for Chapter 8 Class 10 Maths by clicking on the links given below.

Exercise 8.1
Exercise 8.2
Exercise 8.3
Exercise 8.4

NCERT Solutions for Class 10 Maths Chapters:

Chapter 3 Pair of Linear Equations in Two Variables

CBSE Notes for Class 10 Maths Chapter 9 Some Applications of Trigonometry

FAQ on Chapter 8 Introduction to Trigonometry

How many exercises in Chapter 8 Introduction to Trigonometry

There are only 4 exercises in the Chapter 8 Class 10 Maths. The answers provided here are accurate and will provide opportunities to learn new things.

If tan A = cot B, prove that A + B = 90°.

∵ tan A = cot B

∴ tan A = tan (90° − B)

⇒ A = 90° − B

⇒ A + B = 90°.

What is the value of sinθ. cos(90° − θ) + cosθ . sin(90° − θ)?

sinθ · cos(90° − θ) + cosθ · sin(90° − θ) = sinθ · sinθ + cosθ · cosθ

[∵ cos(90° − θ) = sinθ , sin(90° − θ) = cos θ]

= sin2 θ + cos2 θ

= 1

Express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°.

∵ 67° = 90° − 23° and 75° = 90° − 15°

∴ sin 67° + cos 75° = sin (90° − 23°) + cos (90° − 15°) = cos 23° + sin 15°.

↧

MCQ Questions for Class 12 Political Science: Ch 5 Contemporary South Asia

October 16, 2020, 12:51 am

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MCQ Questions for Class 12 Political Science: Ch 5 Contemporary South Asia

1. In Nepal there was violent conflict between the armed forces of the King and

(a) Maoist guerrillas.

(b) Political activists.

(d) Armed military forces.

► (a) Maoist guerrillas.

2. Maoist means

(a) Terrorist groups.

(b) Freedom fighters.

(d) Communists who believe in the ideology of Mao.

► (d) Communists who believe in the ideology of Mao.

3. The country that is Bhutan’s biggest source of development aid is

(a) The USA.

(b) Russia.

(d) Maldives.

► (c) India.

4. The South Asian leaders signed the SAARC Charter on:

(a) 7th November 1985.

(b) 8th December 1985.

(d) 5th December, 1988.

► (b) 8th December 1985.

5. In the political field Pakistan lacks

(a) A courageous and relatively free press.

(b) A strong human rights movement.

(d) Genuine international support for democracy.

► (d) Genuine international support for democracy.

6. A land locked country of South Asia with Maoist guirellas

(a) Maldives.

(b) Bangladesh.

(d) Pakistan.

► (c) Nepal.

7. Nepal was a constitutional monarchy till

(a) 2006.

(b) 2005.

(d) 2003.

► (a) 2006.

8. The Seven Party Alliance is a coalition of

(a) Maoist parties only.

(b) Monarchial forces.

(d) Non-governmental organisations in Nepal.

► (c) Nepali political parties.

9. IPKF stands for

(a) Indian Peaceful Knight Foundation.

(b) Indian Peace Keeping Force.

(d) Indian Peace Keeping Forum.

► (b) Indian Peace Keeping Force.

10. In East Pakistan the popular struggle against West Pakistan domination was led by

(a) Sheikh Mujib Ur Rahman.

(b) Sheikh Mujib Mohammed.

(d) Sheikh Mujib Ali Rahman.

► (a) Sheikh Mujib Ur Rahman.

11. Expand LTTE.

(a) Liberation Tactics of Tamil Eelam

(b) Liberation Tigers of Tamil Eelam

(d) Liberation Team of Tamil Eelam

► (b) Liberation Tigers of Tamil Eelam

12. Which party dominates the political affairs of Maldives?

(a) Maldivian Democratic Party

(b) Maldivian Socialist Party

(d) Maldivian Republic Party

► (a) Maldivian Democratic Party

13. Under which military ruler Pakistan faced Military Crisis?

(a) Yahya Khan

(b) Ayub Khan

(d) Nawaz Sharif

► (a) Yahya Khan

14. When was LTTE vanquished?

(a) 2006

(b) 2008

(d) 2010

15. When India conducted nuclear explosion in Pokaran?

(a) 1996

(b) 1997

(d) 1999

16. Expand SAFTA.

(a) South Asian Free Trade

(b) South Asian Free Tax

(d) South Asian Free Toll

► (a) South Asian Free Trade

↧

MCQ Questions for Class 12 Political Science: Ch 6 International Organisations

October 16, 2020, 2:48 am

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MCQ Questions for Class 12 Political Science: Ch 6 International Organisations

1. The International Court of Justice popularly known as the

(a) World Court.

(b) Supreme Court.

(d) Subordinate Courts.

► (a) World Court.

2. Trygre Lie was the first Secretary General from

(a) Germany.

(b) Norway.

(d) Italy.

► (b) Norway.

3. The two aspects of Human Rights are

(a) Social and legal aspects.

(b) Natural and legal aspects.

(d) Economic and natural aspects.

► (b) Natural and legal aspects.

4. The international organisation that works for the protection of human rights all over the world is

(a) Amnesty International.

(b) International Monetary Fund.

(d) International Court.

► (a) Amnesty International.

5. The original member nations that signed the charter of the United Nations back in 1945 were

(a) 58.

(b) 49.

(d) 45.

► (c) 51.

6. The UNs most visible public figure and representative head is the

(a) Chief Executive Officer.

(b) Secretary General.

(d) Deputy General.

► (b) Secretary General.

7. The United Nations was founded on

(a) 24th October, 1945.

(b) 30th October, 1945.

(d) 26th January, 1942.

► (a) 24th October, 1945.

8. Kofi Annan created the Global Fund to fight AIDS, tuberculosis and

(a) Diarrhea.

(b) Datigue.

(d) Smallpox.

► (c) Malaria.

9. The main objectives of International Atomic Energy Agency are to promote the peaceful use of

(a) Nuclear energy.

(b) Global trade.

(d) Veto power.

► (a) Nuclear energy.

10. The more weightage to India's proposal for permanent membership in the Security Council is

(a) Nuclear capability.

(b) Located in Asia.

(d) India's growing economic power and stable political system.

► (d) India's growing economic power and stable political system.

11. How many judges are there in the International Court of Justice?

(a) Fifteen judges

(b) Sixteen judges

(d) Eighteen judges

► (a) Fifteen judges

12. How many non-permanent members does the UN Security Council have?

(a) Eight

(b) Nine

(d) Eleven

► (c) Ten

13. Expand IAEA?

(a) International Atomic Energy Act

(b) International Atomic Energy Accord

(d) International Atomic Eastern Agency

► (c) International Atomic Energy Agency

14. When was World Trade Organisation was set up?

(a) 1993

(b) 1994

(d) 1996

↧

NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

October 16, 2020, 3:17 am

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NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

Here you will find Chapter 9 Some Application of Trigonometry Class 10 Maths NCERT Solutions that will be useful in knowing the important points about the chapter. These NCERT Solutions for Class 10 Maths are updated according to the latest pattern of Syllabus. By regular practise of NCERT Solutions for Class 10, you will be able to solve the difficult problems in a given exercise.

Exercise: 9.1

Page No: 203

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

Answer

Let AB be the vertical pole Ac be 20 m long rope tied to point C.
In right ΔABC,
sin 30° = AB/AC
⇒ 1/2 = AB/20

⇒ AB = 20/2

⇒ AB = 10

The height of the pole is 10 m.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer

Let AC be the broken part of the tree.

∴ Total height of the tree = AB+AC

In right ΔABC,

cos 30° = BC/AC

⇒ √3/2 = 8/AC

⇒ AC = 16/√3

Also,

tan 30° = AB/BC

⇒ 1/√3 = AB/8

⇒ AB = 8/√3

Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer

There are two slides of height 1.5 m and 3 m. (Given)

Let AB is 1.5 m and PQ be 3 m slides.

ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at

60° with length PR.

A/q,

In right ΔABC,

sin 30° = AB/AC

⇒ 1/2 = 1.5/AC

⇒ AC = 3m

also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3 m

Hence, length of the slides are 3 m and 2√3 m respectively.

Page No: 204

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

A/q,

In right ΔABC,

tan 30° = AB/BC
⇒ 1/√3 = AB/30

⇒ AB = 10√3
Thus, the height of the tower is 10√3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer

Let BC be the height of the kite from the ground,
AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.

A/q,

In right ΔABC,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m
Thus, the length of the string from the ground is 40√3 m.

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to
the point X with inclination 60°.
∴ XY is the distance he walked towards the building.
also, XY = CD.
Height of the building = AZ = 30 m

AB = AZ - BZ = (30 - 1.5) = 28.5 m
A/q,

In right ΔABD,

tan 30° = AB/BD
⇒ 1/√3 = 28.5/BD

⇒ BD = 28.5√3 m
also,
In right ΔABC,

tan 60° = AB/BC
⇒ √3 = 28.5/BC

⇒ BC = 28.5/√3 = 28.5√3/3 m
∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.
Thus, the distance boy walked towards the building is 57/√3 m.

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer

Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC - BC
A/q,

In right ΔBCD,

tan 45° = BC/CD
⇒ 1 = 20/CD

⇒ CD = 20 m
also,

In right ΔACD,

tan 60° = AC/CD
⇒ √3 = AC/20

⇒ AC = 20√3 m
Height of transmission tower = AB = AC - BC = (20√3 - 20) m = 20(√3 - 1) m.

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer

Let AB be the height of statue.
D is the point on the ground from where the elevation is taken.

Height of pedestal = BC = AC - AB

A/q,

In right ΔBCD,
tan 45° = BC/CD
⇒ 1 = BC/CD

⇒ BC = CD.
also,

In right ΔACD,

tan 60° = AC/CD
⇒ √3 = AB+BC/CD

⇒ √3CD = 1.6 m + BC
⇒ √3BC = 1.6 m + BC
⇒ √3BC - BC = 1.6 m
⇒ BC(√3-1) = 1.6 m
⇒ BC = 1.6/(√3-1) m
⇒ BC = 0.8(√3+1) m
Thus, the height of the pedestal is 0.8(√3+1) m.

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer

Let CD be the height of the tower equal to 50 m (Given)
Let AB be the height of the building.
BC be the distance between the foots of the building and the tower.
Elevation is 30° and 60° from the tower and the building respectively.
A/q,

In right ΔBCD,

tan 60° = CD/BC
⇒ √3 = 50/BC

⇒ BC = 50/√3
also,

In right ΔABC,

tan 30° = AB/BC
⇒ 1/√3 = AB/BC

⇒ AB = 50/3
Thus, the height of the building is 50/3.

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer

Let AB and CD be the poles of equal height.
O is the point between them from where the height of elevation taken.
BD is the distance between the poles.
A/q,
AB = CD,
OB + OD = 80 m
Now,

In right ΔCDO,

tan 30° = CD/OD
⇒ 1/√3 = CD/OD

⇒ CD = OD/√3 ... (i)
also,

In right ΔABO,

tan 60° = AB/OB
⇒ √3 = AB/(80-OD)

⇒ AB = √3(80-OD)
AB = CD (Given)
⇒ √3(80-OD) = OD/√3
⇒ 3(80-OD) = OD
⇒ 240 - 3 OD = OD
⇒ 4 OD = 240
⇒ OD = 60
Putting the value of OD in equation (i)
CD = OD/√3 ⇒ CD = 60/√3 ⇒ CD = 20√3 m
also,
OB + OD = 80 m ⇒ OB = (80-60) m = 20 m
Thus, the height of the poles are 20√3 m and distance from the point of elevation are 20 m and 60 m respectively.

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer

Here, AB is the height of the tower.
CD = 20 m (given)
A/q,

In right ΔABD,

tan 30° = AB/BD
⇒ 1/√3 = AB/(20+BC)

⇒ AB = (20+BC)/√3 ... (i)
also,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC

⇒ AB = √3 BC ... (ii)
From eqn (i) and (ii)
AB = √3 BC = (20+BC)/√3
⇒ 3 BC = 20 + BC
⇒ 2 BC = 20 ⇒ BC = 10 m
Putting the value of BC in eqn (ii)
AB = 10√3 m
Thus, the height of the tower 10√3 m and the width of the canal is 10 m.

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer

Let AB be the building of height 7 m and EC be the height of tower.
A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°
EC = DE + CD
also, CD = AB = 7 m.
and BC = AD
A/q,
In right ΔABC,
tan 45° = AB/BC
⇒ 1= 7/BC
⇒ BC = 7 m = AD
also,
In right ΔADE,
tan 60° = DE/AD
⇒ √3 = DE/7
⇒ DE = 7√3 m
Height of the tower = EC = DE + CD
= (7√3 + 7) m = 7(√3+1) m.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer

Let AB be the lighthouse of height 75 m.
Let C and D be the positions of the ships.
30° and 45° are the angles of depression from the lighthouse.
A/q,
In  right ΔABC,
tan 45° = AB/BC
⇒ 1= 75/BC
⇒ BC = 75 m
also,
In  right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = 75/BD
⇒ BD = 75√3  m

The distance between the two ships = CD = BD - BC = (75√3 - 75) m = 75(√3-1) m.

Page No: 205

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m
from the ground. The height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Answer

Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height = 88.2 m - 1.2 m = 87 m
Distance travelled by the balloon =
DE = CE - CD
A/q,
In right ΔBEC,
tan 30° = BE/CE
⇒ 1/√3= 87/CE
⇒ CE = 87√3 m
also,
In right ΔADC,
tan 60° = AD/CD
⇒ √3= 87/CD
⇒ CD = 87/√3 m = 29√3 m
Distance travelled by the balloon = DE = CE - CD = (87√3 - 29√3) m = 29√3(3 - 1) m = 58√3 m.

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer

Let AB be the tower.
D is the initial and C is the final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In right ΔABC,
tan 60° = AB/BC
⇒ √3 = AB/BC
⇒ BC = AB/√3 m
also,
In right ΔABD,
tan 30° = AB/BD
⇒ 1/√3 = AB/(BC + CD)
⇒ AB√3 = BC + CD
⇒ AB√3 = AB/√3 + CD
⇒ CD = AB√3 - AB/√3
⇒ CD = AB(√3 - 1/√3)
⇒ CD = 2AB/√3
Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC = 6/2 = 3 sec.

16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer

Let AB be the tower.
C and D be the two points with distance 4 m and 9 m from the base respectively.
A/q,
In  right ΔABC,
tan x = AB/BC
⇒ tan x = AB/4
⇒ AB = 4 tan x ... (i)
also,
In  right ΔABD,
tan (90°-x) = AB/BD
⇒ cot x = AB/9
⇒ AB = 9 cot x  ... (ii)
Multiplying eqn (i) and (ii)
AB² = 9 cot x × 4 tan x
⇒ AB² = 36
⇒ AB = ± 6
Height cannot be negative. Therefore, the height of the tower is 6 m. Hence, Proved.

Go Back To NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Ch 9 Some Applications of Trigonometry

These Chapter 9 Some Applications of Trigonometry Class 10 Maths NCERT Solutions that will be helpful in Grasping the key concepts inside the chapter properly. There are only 2 topics and only one exercise which will be enhance your knowledge about the applications of trigonometry in real life.

• Introduction: Trigonometry is used in our daily lives. It is used in astronomy, geography and in navigation. The knowledge of trigonometry is used to construct maps, determine the position of an island in relation to the longitudes and latitudes.

• Heights and Distances: In this topic, we will understand about various terms such as line of sight, angle of elevation and angle of depression and discuss various problems.

You can also access other study resources prepared by Studyrankers experts:

MCQ Questions for Classs 10 Maths Chapter 9 Some Applications of Trigonometry

You can also find exercisewise NCERT Solutions for Chapter 9 Some Applications of Trigonometry by clicking on the given below.

Exercise 9.1

NCERT Solutions for Class 10 Maths Chapters:

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 8 Introduction to Trigonometry