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R.D. Sharma Solutions Class 9th: Ch 20 Surface Area and Volume of a Right Circular Cone Exercise 20.2

September 1, 2018, 12:36 pm
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Chapter 20 Surface Area and Volume of a Right Circular Cone R.D. Sharma Solutions for Class 9th Exercise 20.2

Exercise 20.2

1. Find the volume of a right circular cone with:
(i) Radius 6 cm, height 7 cm.
(ii) Radius 3.5 cm, height 12 cm
(iii) Height 21 cm and slant height 28 cm.

Solution

2.  Find the capacity in litres of a conical vessel with:
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm.
  
Solution


3. Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Find the ratio of their volumes.

Solution

4. The radius and the height of a right circular cone are in the ratio 5 : 12. If its volume is 314 cubic metre, find the slant height and the radius .  (Use π = 3.14) .

Solution


5. The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone.
(Use π = 3.14).

Solution


6. The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2 : 3, Find the ratio of their vertical heights.

Solution

7. A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3 : 1.

Solution


8. If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone ?

 Solution


9.  A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14).

 Solution


10. Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.

Solution


11.  A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.

Solution


12. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm. 

Solution

13. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find:
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.

 Solution


14. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Solution


15. Monica has a piece of Canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m2. Find the volume of the tent that can be made with it.

 Solution

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Extra Questions for Class 9th: Ch 1 French Revolution History

September 1, 2018, 11:13 pm
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Extra Questions for Class 10th: Ch 1 French Revolution Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. When did the French Revolution begin?
Answer

The French Revolution began on 14th July 1789 with the storming of Bastille fortress prison.

(Para - 1, Page No. - 3)
2. Which ruler came to power in France in 1774?
Answer

Louis XVI of Bourbon family came to power in France in 1774.

(Para – 1, Page No. 4)
3. Whom did Louis XVI get married to?
Answer

Louis XVI get married the Austrian princess Marie Antoinette.

(Para – 1, Page No. 4)
4. Which classes formed the privileged estates?
Answer

The clergy and the nobility classes formed the privileged estates.

(Para – 3, Page No. 4)
5. Which estate of French society paid all taxes?
Answer
Third estate of French society paid all taxes.

(Para – 3, Page No. 4)
6. Which was treasury empty when Louis XVI ascended the throne?
Answer
Long years of war had drained the financial resources of France therefore, Louis XVI found an empty treasury when he ascended the throne.

(Para – 1, Page No. 4)
7. What was the name of direct tax collected by the state from the peasants in the 18th Century of France?
Answer

Tax named Taille was collected by the state from the peasants in the 18th Century of France.

(Para – 4, Page No. 4)
8. What was ‘tithe’?
Answer

Tithe was a tax levied by the church, comprising one-tenth of the agricultural produce.

(Para – 4, Page No. 4)
9. What do you mean by ‘Subsistence Crisis’?
Answer

Subsistence crisis is an extreme situation where the basic means of livelihood are endangered.

(New Words, Page No. 5)
10. Which social group emerged in the 18th Century in France?
Answer

In the 18th Century, social group termed as middle class emerged in France.

(Para – 2, Page No. 6)

Short Answer Questions (SAQs):

1. Describe the division of French society before French revolution?

Answer

Before French revolution, French society was divided into three Estates:
(i) First Estate (Clergy): It comprised of persons who were involved in the functions of church. They were exempted from paying taxes and enjoyed certain privileges by birth.
(ii) Second Estate (Nobility): It comprised those persons who had high social and political rank. They enjoyed certain privileges based on birth and also exempted from paying taxes. They also enjoyed feudal privileges. They extract feudal dues from the peasants.
(iii) Third Estate: It comprises of big businessmen, merchants, court officials, lawyers, peasants, artisans, small peasants, landless labourers and servants. They pay direct tax to state called taille and a number of indirect taxes, levied on articles of everyday consumption. Peasants also pay called tithes to church.

(Source of Answer: Para – 3 and 4, Fig. 2, Page No. 4)

2. How did the political system work in France under the constitution of 1791?

Answer

(i) The Constitution of 1791 gave the power to make laws in the National Assembly, which was indirectly elected. 
(ii) Citizens were given the right to vote for an electoral group which in turn, elected the Assembly. 
(iii) But all the citizens were not given this right. Only those men older than 25 years who paid taxes equal to minimum 3 days of a labourer's wage were given this right. 
(iv) Rest of the men and women were classed as passive citizens and not given right to vote.

(Para – 2, Page No. 10)

3. How was slavery abolished in France?

Answer

(i) The National Assembly held long debates about whether the rights of man should be extended to all French subjects including those in the colonies but it did not pass any laws, fearing opposition from businessmen whose incomes depended on the slave trade. 
(ii) It was finally the Convention which in 1794 legislated to free all slaves in the French overseas possessions. 
(iii) However, ten years later, Napoleon reintroduced slavery.
(iv) Slavery was finally abolished in French colonies in 1848.

(Para – 2, Page No. 21)

4. Describe the concept of active and passive citizens of France.

Answer

(i) Active citizens were those who had the right to vote. 
(ii) Only men above 25 years of age who paid taxes equal to at least 3 days of a labourer’s wage were given the status of active citizens.
(iii) Passive citizens were the remaining men and all women. They had no right to vote.

(Para – 2, Page No. 10)

5. What was the importance of the Declaration of the Rights of Man?

Answer

(i) The Declaration of the Rights of Man did away all the privileges based on the birth which was prevailing in the old regime. 
(ii) It considered rights such as the right to life, freedom of speech, freedom of opinion, equality before law as basic and natural rights that belonged to each human being by birth and could not be taken away. 
(iii) It was the duty of the state to protect each citizen’s natural rights.

(Para – 1, Page No. 11)

6. What was the Estates General? Which demand of the Third Estate did Louis XVI reject?

Answer

The Estates General was a political body to which the three estates i.e., the clergy, the nobility and the third estate sent their representatives.
(i) In the past, voting in the Estates General had been conducted according to the principle that each estate had one vote. 
(ii) But the third estate demanded that voting now be conducted by the assembly as a whole, where each member would have one vote.

(Para – 1 and 3, Page No. 11)

7. Describe briefly the contribution of Mirabeau in the formation of National Assembly.

Answer

(i) Mirabeau was born in a noble family but was convinced of the need to do away with a society of feudal privilege. 
(ii) On 20 June, representatives of the third estate led by Mirabeau and Abbé Sieyès, assembled in the hall of an indoor tennis court in the grounds of Versailles.
(iii) Mirabeau brought out a journal and delivered powerful speeches to the crowds assembled at Versailles.

(Para – 4, Page No. 8| Para – 1, Page No. 9)

8. How did peasants protest against the feudal lords or nobles in the countryside of France?

Answer

(i) In the countryside rumours spread from village to village that the lords of the manor had hired people who were on their way to destroy the ripe crops. 
(ii) Due to fear, peasants in several districts seized hoes and pitchforks and attacked chateaux.
(iii) They looted hoarded grain and burnt down documents containing records of manorial dues. 
(iv) A large number of nobles fled from their homes, many of them migrating to neighbouring countries.

(Para – 3, Page No. 9)

Long Answer Questions (LAQs):

1. Explain the impact of French Revolution on France in everyday life of people.

Answer

(i) In the Old Regime all written material and cultural activities could be published or performed only after they had been approved by the censors of the king but after the Declaration of the Rights of Man and Citizen proclaimed freedom of speech and expression to be a natural right.
(ii) Newspapers, pamphlets, books and printed pictures flooded the towns of France from where they travelled rapidly into the countryside. 
(iii) They all described and discussed the events and changes taking place in France.
(iv) Freedom of the press also meant that opposing views of events could be expressed. 
(v) Plays, songs and festive processions attracted large numbers of people.

(Para – 2, Page No. 22)

2. Why was the reign of Robespierre termed as ‘reign of terror’ despite various reforms introduced by him?

Answer

(i) Robespierre’s government adopted various reforms such as maximum ceiling on wages and prices, rationed meat and bread, fixed prices of grains, made whole-wheat bread compulsory for all and converted buildings of churches into barracks or offices.
(ii) However, his period from 1793 to 1794 is referred to as the Reign of Terror because Robespierre followed a policy of severe control and punishment.
(iii) All those whom he saw as being ‘enemies’ of the republic – ex-nobles and clergy, members of other political parties, even members of his own party who did not agree with his methods – were arrested, imprisoned and then tried by a revolutionary tribunal.
(iv) If the court found them ‘guilty’ they were guillotined.
(v) Robespierre pursued his policies so relentlessly that even his supporters began to demand moderation.

(Para – 1, 2 and 3, Page No. 16)

3. What measures were taken by the Robespierre to bring about equality in the French society?

Answer

(i) Robespierre’s government issued laws placing a maximum ceiling on wages and prices. 
(ii) Meat and bread were rationed. 
(iii) Peasants were forced to transport their grain to the cities and sell it at prices fixed by the government.
(iv) The use of more expensive white flour was forbidden and all citizens were required to eat a loaf made of wholewheat. 
(v) Instead of the traditional Monsieur (Sir) and Madame (Madam) all French men and women were henceforth Citoyen and Citoyenne (Citizen). 
(vi) Churches were shut down and their buildings converted into barracks or offices.
(Para – 2, Page No. 16)

4. Explain the role of philosophers in the French Revolution.

Answer

(i) The philosophers presented idea of a society based on freedom and equal laws and opportunities for all.
(ii) John Locke in his book ‘Two Treatises of Government’, sought to refute the doctrine of the divine and absolute right of the monarch. 
(iii) Rousseau carried the idea forward, proposing a form of government based on a social contract between people and their representatives. 
(iv) In ‘The Spirit of the Laws’, Montesquieu proposed a division of power within the government between the legislative, the executive and the judiciary.
(v) The ideas of these philosophers were discussed intensively in salons and coffee-houses and spread among people through books and newspapers.

(Para – 2, Page No. 6| Para – 1 and 2, Page No. 7)

HOTS:

1. ‘While the National Assembly was busy at Versailles drafting a constitution, the rest of France seethed with turmoil.’ Elucidate.

Answer

(i) While drafting constitution, a severe winter resulted in bad harvest. The price of bread rose, often bakers exploited the situation and hoarded supplies.
(ii) Crowds of angry women stormed into the shops after spending hours in long queues at the bakery
(iii) At the same time, the king ordered troops to move into Paris. On 14 July, the agitated crowd stormed and destroyed the Bastille.
(iv) In the countryside rumours spread that the lords of the manor had hired people who were on their way to destroy the ripe crops. 
(v) Due to fear, peasants attacked castles of noblemen and looted hoarded grain and burnt down documents containing records of manorial dues.

(Para – 2 and 3, Page No. 9)

VBQs:

1. Emergence of middle class and their belief gave last blow to monarchy rule in France. Explain.

Answer

(i) In the past, peasants and workers had participated in revolts against increasing taxes and food scarcity. But they lacked the means and programmes to carry out full-scale measures.
(ii) The middle class earned their wealth through an expanding overseas trade and from the manufacture of goods.
(iii) All of these were educated and believed that no group in society should be privileged by birth. Rather, a person’s social position must depend on his merit. 
(iv) These wanted a society based on freedom and equal laws and opportunities for all which were put forward by philosophers. Thus, they revolted against the cruel regime.

(Para – 1 and 2, Page No. 6)



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R.D. Sharma Solutions Class 9th: Ch 20 Surface Area and Volume of a Right Circular Cone MCQ's

September 2, 2018, 8:16 am
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Chapter 20 Surface Area and Volume of a Right Circular Cone R.D. Sharma Solutions for Class 9th MCQ's

Mark the correct alternative in each of the following:

1. The number of surfaces of a cone has, is
(a) 1
(b) 2
(c) 3
(d) 4

Solution

The surfaces or faces that a cone has are :
(1) Base
(2) Slanted Surface
So, the number of surfaces that a cone has is 2.
Hence the correct choice is (b).

2. The area of the curved surface of a cone of radius 2r and slant height l/2, is
(a) πrl
(b) 2πrl
(c) 1/2 πrl
(d) π(r+l)r

Solution

The formula of the curved surface area of a cone with base radius ‘r’ and slant height ‘l’ is given as
Curved Surface Area = πrl
Here the base radius is given as ‘l/2’ and the slant height is given as ‘’
Substituting these values in the above equation we have
Curved Surface Area = (π)(2)(r)(l)/2
= πrl
Hence the correct choice is (a).

3.The total surface area of a cone of radius r/2 and length 2l, is
(a) 2πr (1+r)
(b) πr(1+r/4)
(c) πr (1+r)
(d) 2πrl

 Solution


4. A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio
(a) 9 : 1
(b) 1 : 9
(c) 3 : 1
(d) 1 : 3

 Solution



5. The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increased by
(a) 10 %
(b) 12.1 %
(c) 20 %
(d) 21 %

Solution

The formula of the curved surface area of a cone with base radius ‘r’ and slant height ‘l’ is given as
Curved Surface Area = πrl
Now, it is said that the slant height has increased by 10%. So the new slant height is ‘1.1l’
So, now
New Curved Surface Area = 1.1πrl
We see that the percentage increase of the Curved Surface Area is 10%
Hence the correct option is (a).

6. The height of a solid cone is 12 cm and the area of the circular base is 64πcm2. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the areas of the base of the new cone so formed is
(a) 9π cm2
(b) 16π cm2
(c) 25π cm2
(d) 36π cm2

Solution

If a cone is cut into two parts by a plane parallel to the base, the portion that contains the base is called the frustum of a cone


7. If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is
(a) 1/3πr3
(b) 2/3πr3
(c) 3πr3
(d) 9πr3 

Solution


8. If the volume of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is
(a) 1 : 5
(b) 5 : 4
(c) 5 : 16
(d) 25 : 64

Solution


9. The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
(a) 2 : 1
(b) 4 : 1
(c) 8 : 1
(d) 1 : 1

Solution


10. If the height and radius of a cone of volume V are doubled, then the volume of the cone, is
(a) 3 V
(b) 4 V
(c) 6 V
(d) 8 V

 Solution


11. The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is
(a) 1 : 3
(b) 3 : 1
(c) 4 : 3
(d) 3 : 4

Solution


12. A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is
(a) 3 : 5
(b) 2 : 5
(c) 3 : 1
(d) 1 : 3

 Solution


13. If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately
(a) 60
(b) 68
(c) 73
(d) 78

Solution


14. The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is
(a) 4 : 5
(b) 25 : 16
(c) 16 : 25
(d) 5 : 4

Solution


15. If h, S and V denote respectively  the height, curved surface area and volume of a right circular cone, then 3πVh3  − S2h2 + 9V2 is equal to
(a) 8
(b) 0
 (c) 4π
(d) 32π2

Solution

Here we are asked to find the value for a given specific equation which is in terms of V, h and S representing the volume, vertical height and the Curved Surface Area of a cone.

16. If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the axis, the ratio of the volumes of upper and lower part is
(a) 1 : 2
(b) 2 : 1
(c) 1: 7
(d) 1 : 8

Solution

If a cone is cut into two parts by a plane parallel to the base, the portion that contains the base is called the frustum of a cone .


17. If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 1

Solution

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R.D. Sharma Solutions Class 9th: Ch 21 Volume of a Sphere Exercise 21.1

September 3, 2018, 4:42 am
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Chapter 21 Volume of a  Sphere R.D. Sharma Solutions for Class 9th Exercise 21.1

Exercise 21.1

1. Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm

Solution

(i) Given Radius= 10.5 cm
Surface area= 4πr2
= 4×22/7×(10.5)2
= 1386cm2

(ii) Given radius= 5.6cm
Surface area= 4πr2
= 4×22/7×(5.6)2
= 394.24cm2

(iii) Given radius= 14cm
Surface area= 4πr2
= 4×22/7×(14)2
= 2464cm2

2. Find the surface area of a sphere of diameter.
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm

Solution

(i) Given Diameter = 14 cm
Radius = Diameter/2
= 14/2 = 7cm
Surface area = 4πr2
= 4×22/7×(7)2
= 616cm2

(ii) Given Diameter = 21cm
Radius = Diameter/2
= 2/12 = 10.5cm
Surface area = 4πr2
= 4×22/7×(10.5)2
= 1386cm2

(iii) Given diameter = 3.5cm
Radius = Diameter/2
= 3.5/2 = 1.75cm
Surface area = 4πr2
= 4×22/7×(1.75)2
= 38.5cm

3. Find the total surface area of a hemisphere and a solid hemisphere each of radius 10cm (π =3.14)

Solution

The surface area of the hemisphere= 2πr2
= 2×3.14×(10)2
= 628cm2
The surface area of solid hemisphere=  2πr2
= 3×3.14×(10)2
= 942cm2


4. The surface area of a sphere is 5544 cm2, find its diameter .

Solution


5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 4 per 100 cm2.

Solution

Inner diameter of hemispherical bowl= 10.5cm
Radius= 10.5/2 = 5.25cm2
Surface area of hemispherical bowl= 2πr2
=2×3.14×(5.25)2
=173.25cm2
Cost of tin plating 100cm2 area= Rs.4
Cost of tin plating 173.25cm2 area= Rs. 4×173.25/100 = Rs. 6.93
Thus, the cost of tin plating the inner side of hemispherical bowl is Rs.6.93

6. The dome of a building is in the form of a hemisphere. Its radius is 63 dm. Find the cost of painting it at the rate of Rs 2 per sq. m.

Solution


Dome radius= 63dm= 6.3m
Inner surface area of dome= 2πr2
= 2×3.14×(6.3)2
= 249.48m2
Now, cost of 1m2= Rs.2
Therefore cost of 249.48m2 = Rs. (249.48×2) = Rs.498.96 .

7. Assuming the earth to be a sphere of radius 6370 km, how many square kilometres is area of the land, if three-fourth of the earth's surface is covered by water?

Solution

34th of earth surface is covered by water
Therefore 1/4th of earth surface is covered by land
Therefore Surface area covered by land= 1/4×4πr2
= 1/4×4×22/7×(6370)2
=127527400km2

8. A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the shape if the length of the shape be 7 cm.

 Solution


Given length of the shape= 7cm
But length = r+r
2r = 7cm
r = 3.5cm
Also; h = r
Total surface area of shape = 2πrh+2πr2
= 2πrr+2πr2
= 2πr2+2πr2
= 4×2/27×(3.5)2 = 154cm2

9. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at Rs 7 per 100 cm2.

Solution



10. A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder by 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

Solution

Diameter of a cylinder= 1.4m
Therefore radius of cylinder= 1.4/2 = 0.7m
Height of cylinder = 8m
Therefore surface area of tank= 2πrh+2πr2
= (2 × 2/27 × 0.7 × 8) + [2 × 22/7 × (0.7)2]
=176/5+77/25 = 38.28m2
Now cost of 1m2 = Rs.10
Therefore, cost of 38.28m2 = Rs. 382.80

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R.D. Sharma Solutions Class 9th: Ch 21 Volume of a Sphere Exercise 21.2

September 3, 2018, 5:21 am
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Chapter 21 Volume of a  Sphere R.D. Sharma Solutions for Class 9th Exercise 21.2

Exercise 21.2

1. Find the volume of a sphere whose radius is:
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm

Solution

(i) Radius(r)= 2cm
Therefore volume=4/3πr3
= 4/3×22/7×(2)3
= 33.52cm3

(ii) Radius(r)= 3.5cm
Therefore volume=4/3πr3
= 4/3×22/7×(3.5)3 = 179.666cm3

(iii) Radius(r)= 10.5cm
Therefore volume=4/3πr3
= 4/3×22/7×(10.5)3 = 4851cm3

2. Find the volume of a sphere whose diameter is:
 (i) 14 cm
 (ii) 3.5 dm
 (iii) 2.1 m

Solution

(i) Diameter=14cm, Radius(r)= 14/2= 7cm
Therefore volume=4/3πr3
= 4/3×22/7×(7)3 = 1437.33cm3

(ii) Diameter=3.5dm, Radius(r)= 3.5/2= 1.75dm
Therefore volume=4/3πr3
=4/3×22/7×(1.75)3
=22.46dm3

(iii) Diameter = 2.1m, Radius(r)= 2.1/2 = 1.05m
Therefore volume=4/3πr3
= 4/3×22/7×(1.05)3 = 4.851m3

3. A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.

Solution

Radius of the tank= 2.8m
Therefore Capacity= 2/3πr3
=2/3×22/7×(2.8)3 = 45.994m3
1m3=1000l
Therefore capacity in litres = 45994 litres .

4. A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. find the volume of steel used in making the bowl.

Solution

Inner radius = 5cm
Outer radius = 5 + 0.25 = 5.25
Volume of steel used= Outer volume-Inner volume
= 2/3×π×((5.25)3− (5)3)
= 2/3×22/7×((5.25)3− (5)3)
= 41.282cm3

5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter ?

Solution

6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made.

Solution


7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2 cm and 2 cm, find the diameter of the third ball.

Solution


8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.

Solution


9. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere ?

Solution


10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution


11. A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.

Solution


12. A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Solution


13. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder. 

Solution


14. A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball ?

 Solution


15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use π =22/7).

Solution



16. The diameter of a coper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

Solution

Given that diameter of a copper sphere =18cm
Radius of the sphere = 9cm
Length of the wire =108m =10800cm
Volume of cylinder = Volume of sphere


17. A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters? 


Solution
Given that,
Radius of the cylinder jar = 6cm = r1
Level to be rised = 2cm Radius of each iron sphere =1.5cm = r2 


18. A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar ?

Solution

Given that,
Diameter of jar = 10cm
Radius of jar = 5cm
Let the level of water be raised by h
Diameter of the spherical bowl = 2cm
Radius of the ball = 1cm
Volume of jar = 4(Volume of spherical ball)
 

19. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.

Solution 


20. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 22/3cm. Find the diameter of the cylinder.

Solution 

Given that,
Internal radius of the sphere =3cm = r1
External radius of the sphere =5cm = r2
Height of the cylinder = 8/3cm = h
Volume of the spherical shell = Volume of cylinder .

21. A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

Solution

Given
Radius of the hemisphere = Volume of cone


22. A hollow sphere of internal and external radii 2cm and 4 cm respectively is melted into a cone of base radius 4cm. Find the height and slant height of the cone.

Solution

Given that
Hollow sphere external radii = r2 = 4cm
Internal radii = r1= 2cm
Cone base radius(R) = 4cm
Height = h
Volume of cone = Volume of sphere

23. A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

Solution

Given that
Metallic sphere of radius = 10.5cm
Cone radius =3.5cm
Height of radius = 3cm
Let the number of cones obtained be x

24. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution

Given that
A cone and a hemisphere have equal bases and volumes

25. A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.

Solution

Given that
A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight
We know that

26. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Solution

Radius of cylindrical tub =12cm
Depth =20cm
Let r be the radius of the ball
Then
Volume of the ball= Volume of water raised

27. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

Solution

Side of cube =10.5cm
Volume of sphere = v
Diameter of the largest sphere =10.5cm
2r = 10.5
 r = 5.25cm

28. A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.


Solution 

Let r be the common radius
Height of the cone = height of the cylinder = 2r
Let

29. A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between. 

Solution

It is given that
Cube side = 4cm
Volume of cube = (4cm)3 = 64cm3
Diameter of the sphere= Length of the side of the cube=4cm
Therefore radius of the sphere = 2cm .

30. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.


Solution 

Inner radius of the hemispherical tank = 1m = r1
Thickness of the hemispherical tank = 1cm = 0.01m
Outer radius of hemispherical tank = (1+0.01) = 1.01m =r2
Volume of iron used to make the tank =

31. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (mm3) is needed to fill this capsule?


Solution 

Given that
Diameter of capsule = 3.5mm
Radius = 3.5/2 = 1.75 mm
Volume of spherical sphere =


32. The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon ?


Solution 

↧
↧

R.D. Sharma Solutions Class 9th: Ch 21 Surface Area and Volume of a Sphere MCQ's

September 5, 2018, 12:24 am
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Chapter 21 Surface Area and Volume of a Sphere R.D. Sharma Solutions for Class 9th MCQ's

Multiple Choice Questions

1. In a sphere the number of faces is
(a) 1
(b) 2
(c) 3
(d) 4

Solution

The number of faces is 1.

2. The total surface area of hemisphere of radius r.
(a) πr2
(b) 2πr2
(c) 3πr2
(d) 4πr2

Solution
So, here we have to find total surface area, which is sum of curved surface area and top area. So total surface area is 2πr2 + πr2 = 3πr2

3. The ratio of the total surface area of a sphere and a hemisphere of same radius is
(a) 2:1
(b) 3:2
(c) 4:1
(d) 4:3

Solution

Total Surface Area of sphere = 4πr2
Total Surface Area of hemisphere = 3πr2
The Ratio is (4πr2)/(3πr2) = 4/3
So, option (d) is correct

4. A Sphere and a cube are of the same height. The ratio of their volumes is
(a) 3:4
(b) 21:11
(c) 4:3
(d) 11:21

Solution


5. The largest sphere is cut off from a cube of side 6 cm. The volume of the sphere will be
(a) 27π cm3
(b) 36π cm3
(c) 108 cm3
(d) 12π cm3

Solution


6. A cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. The number of balls will be
(a) 4
(b) 3
(c) 6
(d) 8

Solution


7. If the ratio of volumes of two sphere is 1:8, then ratio of their surface area is
(a) 1:2
(b) 1:4
(c) 1:8
(d) 1:16

Solution


8. If the surface area of a sphere is 144π m2, then its volume in (m3) is
(a) 288π
(b) 316 π
(c) 300π
(d) 188π

Solution


9. If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then the surface area of each ball (in sq.cm) is
(a) 100
(b) 75π
(c) 60π
(d) 50π

Solution


10. The ratio between volume of a sphere and volume of circumscribing right circular cylinder is
(a) 2:1
(b) 1:1
(c) 2:3
(d) 1:2

Solution


11. If a sphere is inscribed in a cube, then the ratio of the volume of the sphere to the volume of cube is
(a) π :2
(b) π:3
(c) π:4
(d) π:6

Solution


12. If a solid sphere of radius r is melted and cast into the shape of a solid cone of height r, then the radius of the base of the cone
(a) 2r
(b) 3r
(c) r
(d) 4r

Solution


13. A sphere is placed inside a right circular cylinder so as to touch the top, base and lateral surface of the cylinder. If the radius of sphere is r, the volume of cylinder is
(a) 4πr3
(b) 8/3 πr3
(c) 2πr3
(d) 8πr3

Solution


14. A cone and a hemisphere have equal bases and equal volumes, the ratio of their heights is
(a) 1:2
(b) 2:1
(c) 4:1
(d) √2: 1

Solution


15. A cone and a hemisphere and a cylinder stand on the equal bases and have the same height . The ratio of their volumes is
(a)1:2:3
(b) 2:1:3
(c) 2:3:1
(d) 3:2:1

Solution

↧

R.D. Sharma Solutions Class 9th: Ch 22 Tabular Representation of Statistical Data Exercise 22.1

September 5, 2018, 4:20 am
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Chapter 22 Tabular Representation of Statistical Data R.D. Sharma Solutions for Class 9th Exercise 22.1

Exercise 22.1

1. What do you understand by the word ''statistics'' in
(i) Singular form ?    
(ii) Plural form ?

Solution

(i) In singular form statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data.
(ii) In plural form statistics means numerical facts or observations collected with definite purpose.
For examples, the income and expenditure of persons in a particular locality, number of males and females in a particular town are statistics.

2. Describe some fundamental characteristic of statistics.

Solution

The plural form statistics has the simplest structure and the singular form statistics has many components. There is only structural difference between singular and plural form statistics. Some of the characteristics of a statistics are .
1. Statistics is a collection of observations. So, clearly a single observation cannot form a statistics.
2. Statistics are collected with definite purpose.
3. Statistics are comparable and classified into various types depending on their properties.
4. Statistics are expressed quantitatively and not qualitatively.

3. What are (i) primary data ? (ii) secondary data ? Which of the two - the primary or the secondary data − is more reliable and why ?

Solution

(i) When an investigator collects data himself with a definite plan or designs in his (her) mind, it is called primary data.
(ii) Data which are not originally collected rather obtained from published or unpublished sources are known as secondary data.

4. Why do we group data ?

Solution

To study the features of a collected data, the data must be arranged in a condensed form. There are a number of ways to arrange the data in condensed form, namely,
1. Serial order or alphabetical order
2. Ascending order
3. Descending order
But, if the number of observations is large, then arranging data in ascending or descending or serial order is a tedious job and it does not tell us much except perhaps the minimum(s) and maximum(s) of data. So to make it easily understandable and clear we condense the data into groups or table form.

5. Explain the meaning of the following terms:
(i) Variate
(ii) Class-integral
(iii) Class-size
(iv) Class-mark
(v) Frequency
(vi) Class limits
(vii) True class limits.

Solution

(i) Variate - Any character that can vary from one individual to another is called variate.
(ii) Class interval - In the data each group into which raw data is considered is called a class-interval.
(iii) Class-size - The difference between the true upper limit and lower limit is called the class size of that class size.
(iv) Class-mark -The middle value of the class is called as the class mark.Class mark = upper limit + lower limit / 2
(v) Frequency - The number of observations corresponding to class is called its frequency.
(vi) Class limits - Each class is bounded by two figures ,called the class limits .The figures on the left side of the classes are called lower limits while figures on the right side are called upper limits.
(vii) True class limits  - If classes are inclusive.Eg:15-19,20-24,25-29….
Then, true lower limit of class = upper limit of class-0.5
Eg:- True limits of the class is 15-19 are 14.5 and 19.5
But if classes are exclusive like 10-20,20-30,30-40…
Here class limits and true class limits are the same.

6. The ages of ten students of a group are given below. The ages have been recorded in years and months:
8 - 6, 9 - 0, 8 - 0,4, 9 - 3, 7 - 8, 8 - 11, 8 - 7, 9 - 2, 7 - 10, 8 - 8
(i) What is the lowest age?
(ii) What is the highest age?
(iii) Determine the range?

Solution

The ages of ten students of a group are given below
8-6,9-0,8-4,9-3,7-8,8-11,8-7,9-2,7-10,8-8.
(i) Lowest age is 7 years 8months
(ii) Highest age is  9 years,3 months
(iii) Range = Highest age-lowest age = 9 year,3 months,7 years,8 months = 1 year 7 months

7. The monthly pocket money of six friends is given below:
Rs 45, Rs 30, Rs 50, Rs 25, Rs 45
(i) What is the highest pocket money?
(ii) What is the lowest pocket money?
(iii) What is the range?
(iv) Arrange the amounts of pocket money in ascending order.

Solution

The monthly pocket money of six friends is given below:
Rs 45,Rs 30, Rs 40, Rs 25, Rs 45.
(i) Highest pocket money = Rs 50
(ii) Lowest pocket money = Rs 25
(iii) Range = 50-25 = 25
(iv) The amounts of pocket money in an ascending order is: Rs 25,Rs 30, Rs 40, Rs 45, Rs 45,Rs 50.

8. Write the class-size in each of the following:
(i) 0-4, 5-9, 10-14
(ii) 10-19, 20-29, 30-39
(iii) 100-120, 120-140, 160-180
(iv) 0-0.25, 0.25-0, 0.50-0.75
(v) 5-5.01, 5.01-5.02, 5.02-5.03

Solution

(i) 0-4 ,5-9, 10-14
True class limits are 0.5-4.5,4.5-9.5,9.5-14.5
Therefore class size = 14.5-9.5 = 5

(ii) 10-19 , 20-29 , 30-39
True class limitsà 19.5-19.5,19.5-29.5,29.5-29.5
Class size =39.5-29.5 = 10

(iii) 100-120 , 120-140, 160-180
Here the class limits and true class limits are the same
Therefore class size = 120-100 = 20

(iv) 0-0.25,0.25-00.50,0.50-0.75
Here the class limits and true class limits are the same
Therefore class size = 0.25-0 = 0.25 .

(v) 5-5.01,5.01-5.02,5.02-5.03.
Here the class limits and true class limits are the same
Therefore class size = 5.01-5.0 = 0.01.

9. The final marks in mathematics of 30 students are as follows:
53, 61, 48, 60, 78, 68, 55, 100, 67, 90
75, 88, 77, 37, 84, 58, 60, 48, 62, 56
44, 58, 52, 64, 98, 59, 70, 39, 50, 60

(i) Arrange these marks in the ascending order, 30 to 39 one group, 40 to 49 second group, etc.
Now answer the following:
(ii) What is the highest score ?
(iii) What is the lowest score ?
(iv) What is the range ?
(v) If 40 is the pass mark how many have failed ?
(vi) How many have scored 75 or more ?
(vii) Which observations between 50 and 60 have not actually appeared ?
(viii) How many have scored less than 50 ?

Solution

(ii) The highest score is 100.
(iii) The lowest score is 37.
(iv) The range of the scores is
 HighestScore - LowestScore = 100 - 37 = 63
(v) If 40 is the pass mark, then the number of students failed is 2.
(vi) The number of students scored more than 75 is 8.
(vii) The observations 51, 54 and 57 in between 50-60 have not actually appeared.
(viii) Number of students scored less than 50 is 5.

10. The weights of new born babies are as follows: 2.3,2.2,2.1,2.7,2.6,2.5,3.0,2.8,2.8,2.9,3.1,2.5,2.8,2.7,2.9,2.4.
(i) Rearrange the weights in descending order.
(ii) What is the highest weight?
(iii) What is the lowest weight?
(iv) Determine the range?
(v) How many babies were born on that day?
(vi) How many babies weigh below 2.5 Kg?
(vii) How many babies weigh more than 2.8 Kg?
(viii)How many babies weigh 2.8 Kg?

Solution

The weights of new born babies(in kg) are as follows
2.3,2.2,2.1,2.7,2.6,2.5,3.0,2.8,2.8,2.9,3.1,2.5,2.8,2.7,2.9,2.4.
(i) The weights in descending order
3.1,3.0,2.9,2.9,2.8,2.8,2.7,2.7,2.6,2.5,2.5,2.4,2.3,2.2,2.1.
(ii) The highest weight = 3.1 Kg
(iii) The lowest weight = 2.1 Kg
(iv) Range = 3.1-2.1 = 1.0 Kg
(v) 15 babies were born on that particular day.
(vi) 4 babies weight below 2.5 Kg.
(vii) Weight more than 2.8 Kg are 4 babies.
(viii) Weightà2 babies

11. The number of runs scored by a cricket player in 25 innings are as follows:
26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47
(i) Rearrange these runs in ascending order.
(ii) Determine the player, is highest score.
(iii) How many times did the player not score a run ?
(iv) How many centuries did he score ?
(v) How many times did he score more than 50 runs ?

Solution

The numbers of runs scored by a player in 25 innings are
26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,15,34,6,71,0,64,15,34,15,34,67,0,42,124,84,54,48,139,64,47.
(i) Runs in an ascending order are
0,0,0,0,6,15,15,15,15,26,34,34,34,34,35,39,42,42,47,48,48,53,54,64,64,64,67,71,71,82,90,124,139.
(ii) The highest number =139
(iii) The player did not score any runs 3 times
(iv) He scored 3 centuries.
(v) He scored more than 50 runs 12 times.

12. The class size of a distribution is 25 and the first class-interval is 200-224. There are seven class-intervals.
(i) Write the class-intervals.
(ii) Write the class-marks of each intervals.

Solution

Given
Class size = 25
First class interval = 200-224
(i) Seven class interval are:
200-240,225-249,250-274,275-299,300-324,325-349,350-374.

(ii) Class mark 200-224 = 200+224/2
= 424/2
= 212
Class mark 225-249 = 225+249/2
= 479/2
= 237
Class mark 250-274/2
= 524/2
= 287
Class mark 300-324 = 300+324/2
= 642/2
= 312
Class mark 325-349 = 325+349/2
= 674/2
= 337
Class mark 350-374 = 350+374/2
= 724/2
= 362

13. Write the class size and class limits in each of the following:
(i) 104, 114, 124, 134, 144, 154 and 164
(ii) 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 97 and 102
(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5

 Solution



14. Following data gives the number of children in 40 families:
1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.
Represent it in the form of a frequency distribution.

Solution

Here, the maximum and minimum values of the variate are 6 and 0 respectively.
So the range = 6 - 0 = 6
Here, we will take class size 1. So we must have 6 classes each of size 1.
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

15.  The marks scored by 40 students of class IX in mathematics are given below:
81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54
Prepare a frequency distribution with class size of 10 marks.

Solution

Here, the maximum and minimum values of the variate are 95 and 29 respectively.
So the range = 95 – 29 = 66
Here, we will take class size 10. So we must have 7 classes each of size 10
Lower limit of first class interval is;

16. The heights (in cm) of 30 students of class IX are given below:
155, 158, 154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157, 162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153
Prepare a frequency distribution table with 160-164 as one of the class intervals.

Solution

One of the class intervals is 160–164. This means that class size is 4
Here, the maximum and minimum values of the variate are 163 and 147 respectively.
So the range = 163 – 147 = 16
Here, we will take class size 4. So we must have 5 classes each of size 4
Lower limit of first class interval is;


17 . The monthly wages of 30 workers in a factory are given below:
83.0, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 836, 878, 840, 868, 890, 806, 840, 890.
Represent the data in the form of a frequency distribution with class size 10.

Solution

Here, the maximum and minimum values of the variate are 898 and 804 respectively.
So the range = 898 – 804 = 94
Here, we will take class size 10. So we must have 94/10 i.e. 10 classes each of size 10.
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:
Lower limit of first class interval is; 

18. The daily maximum temperatures (in degree celsius) recorded in a certain city during the month of November are as follows:
25.8, 24.5, 25.6, 20.7, 21.8, 20.5, 20.6, 20.9, 22.3, 22.7, 23.1, 22.8, 22.9, 21.7, 21.3, 20.5, 20.9, 23.1, 22.4, 21.5, 22.7, 22.8, 22.0, 23.9, 24.7, 22.8, 23.8, 24.6, 23.9, 21.1
Represent them as a frequency distribution table with class size 1°C.

Solution

Here, the maximum and minimum values of the variate are 25.8 and 20.5 respectively.
So the range = 25.8 – 20.5 = 5.3
Here, we will take class size 1.
Lower limit of first class interval is;

19. Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included):
220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.

Solution

Here, the maximum and minimum values of the variate are 320 and 210 respectively.
So the range = 320 – 210 = 110
Here, we will take class size 20 (As one class interval is 210–230).
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

20. The daily minimum temperatures in degrees Celsius recorded in a certain Arctic region are as follows:
− 12.5, −10.8, −18.6, −8.4, −10.8, −4.2, −4.8, −6.7, −13.2, −11.8, −2.3, 1.2, 2.6, 0, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −2.4, 0, 3.2, 2.7, 3.4, 0, −2.4, −5.8, −8.9, −14.6, −12.3, −11.5, −7.8, −2.9.
Represent them as frequency distribution table taking −19.9 to − 15 as the first class interval. 

Solution

Since the first class is –19.9 to –15 .
Therefore, the frequency distribution in which the lower limit is included and upper limit excluded is:

21.  The blood groups of 30 students of class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Find out which is the most common and which is the rarest blood group among these students

Solution

It can be observed that 9 students have their blood group as A, 6 as B, 3 as AB, and 12 as O.
Therefore, the blood group of 30 students of the class can be represented as follows. 
It can be observed clearly that the most common blood group and the rarest blood group among these students is O and AB respectively as 12 (maximum number of students) have their blood group as O, and 3 (minimum number of students) have their blood group as AB.

22. Three coins were tossed 30 times. Each time the number of heads occurring was noted down as follow:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.

Solution



23. Thirty children were asked about the number of hours they watched T.V. programmes in the previous week. The results were found as follows:
1, 6, 2, 3, 5, 12, 5, 8, 4, 8
10, 3, 4, 12, 2, 8, 15, 1, 17, 6
3, 2, 8, 5, 9, 6, 8, 7, 14, 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5-10.
(ii) How many children watched television for 15 or more hours  a week?

Solution

(i) Our class intervals will be 0 − 5, 5 − 10, 10 −15…..
The grouped frequency distribution table can be constructed as follows.


(ii) The number of children who watched TV for 15 or more hours a week is 2 (i.e., the number of children in class interval 15 − 20). 
↧

R.D. Sharma Solutions Class 9th: Ch 22 Tabular Representation of Statistical Data Exercise 22.2

September 7, 2018, 3:18 am
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Chapter 22 Tabular Representation of Statistical Data R.D. Sharma Solutions for Class 9th Exercise 22.2

Exercise 22.2

1. Define cumulative frequency distribution.

Solution

Cumulative frequency distribution is a table which displays the manner in which cumulative frequencies are distributed over various classes. In a grouped frequency distribution the cumulative frequency of a class is the total of all frequencies upto and including that particular class. Make sure that, for calculating cumulative frequencies, the classes should be written in ascending order.
For example, the following table gives the cumulative frequency distribution of marks scored by 55 students in a test :

Cumulative frequency distributions are of two types, namely, less than and greater than or more than. For less than cumulative frequency distributions we add up the frequencies from the above and for greater than cumulative frequencies we add up the frequencies from below.

2. Explain the difference between a frequency distribution and a cumulative frequency distribution.

Solution

Frequency distribution displays the frequencies of the corresponding class-intervals. But, the cumulative frequency distribution displays the cumulative frequencies of the corresponding classes. For example, the following table gives the frequency and cumulative frequency distribution of marks scored by 55 students in a test:
In a frequency distribution, the sum of all the frequencies is equal to the total number of observations. But, in the cumulative frequency distribution, the last cumulative frequency is same as the total number of observations. For example, in the above table the sum of all the frequencies is 55, which is same as the total number of students and the last cumulative frequency is 55, which is same as the total number of students.

3. The marks scored by 55 students in a test are given below:
Prepare a cumulative frequency table.

Solution


4. Following are the ages of 360 patients getting medical treatment in a hospital on a day:
Age (in years):10-2020-3030-4040-5050-6060-70
No.of Patients:905060805030
Construct a cumulative frequency distribution.

Solution

The ages in years of 360 patients are given as

5. The water bills (in rupees) of 32 houses in a certain street for the period 1.1.98 to 31.3.98 are given below:
56, 43, 32, 38, 56, 24, 68, 85, 52, 47, 35, 58, 63, 74, 27, 84, 69, 35, 44, 75, 55, 30, 54, 65, 45, 67, 95, 72, 43, 65, 35, 59.
Tabulate the data and present the data as a cumulative frequency table using 70-79 as one of the class intervals.

Solution

The minimum and maximum bills are 24 Rs. and 95 Rs.
The range 95-24 = 71 is.
Given that 70-79 is a class-interval. So, the class size 79 - 70 = 9 is.
Now calculate
Range/Class - size
= 71/9 = 7.89
Thus,the number of classes is 8
The cumulative frequency distribution is the following :


6. The number of books in different shelves of a library are as follows:
30, 32, 28, 24, 20, 25, 38, 37, 40, 45, 16, 20
19, 24, 27, 30, 32, 34, 35, 42, 27, 28, 19, 34,
38, 39, 42, 29, 24, 27, 22, 29, 31, 19, 27, 25
28, 23, 24, 32, 34, 18, 27, 25, 37, 31, 24, 23,
43, 32, 28, 31, 24, 23, 26, 36, 32, 29, 28, 21
Prepare a cumulative frequencies distribution table using 45-49 as the last class-interval.

Solution

The minimum and maximum numbers of books in shelves are 16 and 45.
The range is 45 - 16 = 29.
Given that 45-49 is the last class-interval. So, the class size is 49 - 45 = 4.
Now calculate
Range/Class-size
= 29/4
= 7.25
Thus,the number of classes is 8.
The cumulative frequency distribution is the following :


7. Given below are the cumulative frequencies showing the weights of 685 students of a school. Prepare a frequency distribution table. 

Solution

We make class intervals 0-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55, 55-60, 60-65 and 65-70.
We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.
The frequency of the class 0-25 is 0.
The frequency of the class 25-30 is 24-0 = 24.
The frequency of the class 30-35 is 78-24 = 54
The frequency of the class 35-40 is 183-78 = 105.
The frequency of the class 40-45 is 294-183 = 111.
The frequency of the class 45-50 is 408-294 = 114.
The frequency of the class 50-55 is 543-408 = 135.
The frequency of the class 55-60 is 621-543 = 78.
The frequency of the class 60-65 is 674-621 = 53.
The frequency of the class 65-70 is 685-674 = 11.
Here is the cumulative frequency distribution table.

8. The following cumulative frequency distribution table shows the daily electricity consumption (in KW) of 40 factories in an industrial state:
(i) Represent this as a frequency distribution table.
(ii) Prepare a cumulative frequency table.

Solution

We make class intervals 0-240, 240-270, 270-300, 300-330, 330-360, 360-390 and 390-420.
(i) We will now find the frequencies of the different class-intervals.
Just proceed reverse to you proceed to generate cumulative frequency
Here is the frequency distribution table.

(ii) We are given the cumulative frequency table and we are asked to prepare new one.
So we will generate in some other fashion. Concentrate on frequency table and prepare the table. Here is the cumulative frequency distribution table.

9. Given below is a cumulative frequency distribution table showing the ages of people living in a locality:
 
Prepare a frequency distribution table.

Solution

We make class intervals 0-12, 12-24, 24-36, 36-48, 48-60, 60-72, 72-84, 84-96 and 96-108.
We will now find the frequencies of the different class-intervals from the cumulative frequency distribution table.

The frequency of the class 0-12 is 1026−112 = 98
The frequency of the class 12-24 is1026−809 = 217
The frequency of the class 24-36 is 809−427 = 382
The frequency of the class 36-48 is 427−158 = 269
The frequency of the class 48-60 is 158−20 = 138
The frequency of the class 60-72 is 20−5 = 15
The frequency of the class 72-84 is 5−3 = 2
The frequency of the class 84-96 is 3−1 = 2
The frequency of the class 96-108 is 1−0 = 1
Here is the cumulative frequency distribution table.

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R.D. Sharma Solutions Class 9th: Ch 22 Tabular Representation of Statistical Data MCQ's

September 7, 2018, 6:44 am
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Chapter 22 Tabular Representation of Statistical Data R.D. Sharma Solutions for Class 9th MCQ's

Mark the correct alternative in each of the following:

1. Tally marks are used to find
(a) Class intervals
(b) Range
(c) Frequency
(d) Upper limits

Solution

Tally marks are used to find the frequencies.
Hence, the correct choice is (c).

2. The difference between the highest and lowest values of the observations is called
(a) Frequency
(b) Mean
(c) Range
(d) Class-intervals

Solution

The difference between the highest and lowest values of the observations is called the range. Hence, the correct choice is (c).

3. The difference between the upper and the lower class limits is called
(a) Mid-points
(b) Class size
(c) Frequency
(d) Mean

Solution

The difference between the upper and the lower class limits is called the class size. Hence, the correct choice is (b).

4. In the class intervals 10-20, 20-30, 20 is taken in
(a) The interval 10-20
(b) The interval 20-30
(c) both intervals 10-20, 20-30
(d) none of the intervals

Solution

The given class intervals are 10-20, 20-30. In these class intervals the value 20 is lies in the class interval 20-30. Hence, the correct choice is (b).

5. In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is
(a) 10
(b) 12
(c) 13
(d) 14

Solution

6. The mid-value of a class interval is 42. If the class size is 10, then the upper and lower limits of the class are:
(a) 47 and 37
(b) 37 and 47
(c) 37.5 and 47.5
(d) 47.5 and 37.5

Solution

Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is l+m/2 and the class-size is (l-m).
Given that the mid-value of the class is 42 and the class-siz is 10. Therefore, we have two equations
l+m/2 = 42
⇒ l+m = 84,
m-l = 10
Adding the above two equations, we have
(l+m)+(m-l)= 84 + 10
⇒ l+m+m-l = 94
⇒ 2m = 94
⇒ m = 47
Substituting the value of m in the first equation, we have
l+47 = 84
⇒ l = 84-47
⇒ l = 37
Hence, the upper and lower limits of the class are 47 and 37 respectively . Thus, the correct choice is a.

7. The number of times a particular item occurs in a given data is called its
(a) variation
(b) frequency
(c) cumulative frequency
(d) class-size

Solution

The number of times a particular item occurs in a given data is called the frequency of the item. Hence, the correct choice is (b).

8. The width of each of nine classes in a frequency distribution is 2.5 and the lower class boundary of the lowest class 10.6. Then the upper class boundary of the highest class is
(a) 35.6
(b) 33.1
(c) 30.6
(d) 28.1

Solution

The number of classes is 9 and the uniform class size is 2.5. The lower limit of the lower class (first class) is 10.6. Therefore, the upper limit of the last class is
10.6 + (9×2.5)
= 10.6 + 22.5
= 33.1
Hence, the correct choice is (b).

9. Let l be the lower class limit of a class-interval in a frequency distribution and m be the mid point of the class. Then, the upper class limit of the class is
(a) m + (l+m)/2
(b) l + (m+l)/2
(c) 2m - 1
(d) m - 2l

Solution
Given that, the lower class limit of a class-interval is l and the mid-point of the class is m. Let u be the upper class limit of the class-interval. Therefore, we have
m = l+u/2
⇒ l+u = 2m
⇒ u = 2m - l
Thus the upper class limit of the class is (2m-l). Hence, the correct choice is (c).

10. The following marks were obtained by the students in a test:
81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62
The range of the marks is
(a) 9
(b) 17
(c) 27
(d) 33

Solution

The marks obtained by the students are 81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79 and 62.
The highest and lowest marks are 95 and 62 respectively. Therefore, the range of marks is
95-62 = 33
Hence, the correct option is (d).

11. Tallys are usually marked in a bunch of
(a) 3
(b) 4
(c) 5
(d) 6

Solution

Tallies are usually marked in a bunch of 4. Hence, the correct option is (b). 
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Extra Questions for Class 9th: Ch 2 Socialism in Europe and the Russian Revolution History

September 8, 2018, 10:27 am
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Extra Questions for Class 10th: Ch 2 Socialism in Europe and the Russian Revolution Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. Who were Liberals?

Answer

Liberals was a group of those people who wanted a nation which tolerated all religions. They opposed the uncontrolled power of dynastic rulers.

(Para – 4, Page No. 25| Para – 1, Page No. 26)

2. Which group supported women’s suffragette movements?

Answer

Radicals supported women’s suffragette movements.

(Para – 2, Page No. 26)

3. Who were jadidists?

Answer

The Muslim reformers within the Russian empire who wanted modernised Islam to lead their societies.

(Para – 1, Page No. 33)

4. Who sought build a cooperative community called ‘New Harmony’?

Answer

Robert Owen south to build a cooperative community called ‘New Harmony’.

(Para – 5, Page No. 28)

5. How were the liberals different from the democrats?

Answer

Liberals argued for a representative, elected parliamentary government but unlike democrats, they did not believe in universal adult franchise, that is, the right of every citizen to vote.

(Para – 1, Page No. 26)

6. Who wanted revolutions to put an end to monarchical system?

Answer

Some nationalists, liberals and radicals wanted revolutions to overthrow monarchical system.

(Para – 2, Page No. 27)

7. By the nineteenth century what changes occurred in the idea of conservatives?

Answer

By the nineteenth century, conservative accepted that some change was necessary but believed that the past had to be respected and change had to be brought about through a slow process.

(Para – 3, Page No. 26)

8. Who was Louis Blanc?

Answer

Louis Blanc was a socialist of France who wanted the government to encourage cooperatives and replace capitalist enterprises.

(Para – 5, Page No. 28)

9. Why were socialist against private property?

Answer

Because the propertied were concerned only with personal gain and not with the welfare of those who made the property productive.

(Para -3, Page No. 28)

10. According to Karl Marx, how workers could free themselves from capitalist exploitation?

Answer

By constructing a radically socialist society where all property was socially controlled.

(Para – 6, Page No. 28)

Short Answer Questions (SAQs):

1. Differentiate between the ideas of the liberals and radicals in Europe.

Answer

(i) Liberals wanted a nation that tolerated all religions while radicals wanted a government based on majority.
(ii) Liberals opposed to uncontrolled power of dynastic rulers while radicals were opposed to privileges of land owners and wealthy factory owners.
(iii) Liberals wanted representative elected parliamentary government and did not believe in Adult Franchise while radicals supported women's suffragette movements.

(Para – 4, Page No. 25| Para – 1, Page No. 26)

2. Mention any three features of socialism.

Answer

(i) Socialists were against private property.
(ii) Society as a whole should control property collectively.
(iii) Socialists regarded the private property as the root cause of all social evils.

(Para – 3, Page No. 28| Para – 6, Page No. 28)

3. Mention any three factors responsible for the Russian Revolution of 1905.

Answer

(i) Political: The rule of Tsar was autocratic as he was not subject to parliament.
(ii) Economical: In the year 1904, prices of essential goods rose so quickly that real wages declined by 20 percent.
(iii) Immediate: The Bloody Sunday incident took place in which more than 100 workers were killed.
(iv) Poor condition of the working class.

(Para – 6, Page No. 32| Para – 1, 2 and 3, Page No. 33)

4. Who was Karl Marx? What were his views about capitalism?

Answer

(i) Karl Marx was socialist.
(ii) He was against capitalism.
(iii) According to Marx, the profit of capitalists was produced by workers.
(iv) He wanted to construct a radically socialist society.

(Para – 6, Page No. 28)

5. Describe the economic condition of the workers at the beginning of the twentieth century.

Answer

(i) Most industry was the private property of industrialists.
(ii) Most of the workers were working for about 10 to 12 hours, a day.
(iii) The working conditions were also very poor.
(iv) Women workers made up about 31% of the factory labour, but they were pair less than men.

(Para – 3 and 4, Page No. 31)

6. Why were socialists against private property and how it could be improved?

Answer

(i) According to socialists, individuals who owned the property that gave employment but the propertied were concerned only with personal gain.
(ii) They were not concerned with the welfare of those who made the property productive.
(iii) Socialists wanted that society as a whole should control property so that more attention would be paid to collective social interests.

(Para – 3, Page No. 28)

7. Describe the viewpoint of radicals.

Answer

(i) Radicals wanted a nation in which government was based on the majority of a country’s population.
(ii) Many supported women’s suffragette movements.
(iii) They opposed the privileges of great landowners and wealthy factory owners.
(iv) They were not against the existence of private property but disliked concentration of property in the hands of a few.

(Para – 2, Page No. 26)

8. Describe the events that happened in 1905 revolution in Russia?

Answer

(i) The attack of police on the procession of workers at the Winter Palace started a series of events that became known as the 1905 Revolution.
(ii) Strikes took place all over the country and universities closed down when student bodies staged walkouts, complaining about the lack of civil liberties.
(iii) Lawyers, doctors, engineers and other middle-class workers established the Union of Unions and demanded a constituent assembly.

(Para – 3, Page No. 33)

Long Answer Questions (LAQs):

1. Explain the views of socialists on private property. Which international body was set up by socialists to spread and coordinate their efforts?

Answer

(i) Socialists were against private property and saw it as a root of all social evils.
(ii) They argued that if society as a whole rather than single individuals controlled property, more attention would be paid to collective social interests.
(iii) Some believed in the idea of cooperatives were to be associations of people who produced goods together and divided the profits according to the work done by members.
(iv) Socialists like Karl Marx believed that workers had to overthrow capitalism and had to construct a radically socialist society where all property was socially controlled to improve their conditions.
A body named 'Second International' was set up by socialists to coordinate their efforts.

(Para – 3 and 6, Page No. 28)

2. Explain the major events that were responsible for the Russian Revolution of 1905.

Answer

(i) Autocratic Rule: The rule of Tsar was autocratic as he was not subject to parliament.
(ii) Role of Liberals and Socialists: Liberals in Russia campaigned to end dynastic rule. Together with the Social Democrats and Socialist Revolutionaries, they worked with peasants and workers during the revolution of 1905 to demand a constitution.
(iii) Liberals were supported in the empire by nationalists and in Muslim-dominated areas by jadidists who wanted modernised Islam to lead their societies.
(iv) Uprising of workers: In 1904, the prices of essential goods rose so quickly that real wages declined by 20%. Also, the working conditions was very poor. Thus, workers demanded reduction in working hours, an increase in wages and improvement in the working conditions.
(v) Bloody Sunday: The incident of Bloody Sunday in which more than 100 workers were killed fuelled the revolution as it lead to an all Russia strike. Lawyers, doctors, engineers and others demanded constituent assembly.

(Para -6, Page No. 32| Para – 1, 2 and 3, Page No.33)

3. Explain the collectivisation programme of Stalin.

Answer

(i) From 1929, Stalin forced all peasants to cultivate in collective farms, kolkhoz.
(ii) The bulk of land and implements were transferred to the ownership of collective farms.
(iii) Peasants worked on the land, and the kolkhoz profit was shared.
(iv) Those who resisted collectivisation were severely punished. Many were deported and exiled.
(v) As they resisted collectivisation, peasants argued that they were not rich and they were not against socialism. They merely did not want to work in collective farms for a variety of reasons.
(vi) Stalin’s government allowed some independent cultivation, but treated such cultivators unsympathetically.

(Para - 4, Page No. 44)

4. Mention the main events of the February Revolution of 1917.

Answer

Main events of the February Revolution were:
(i) In February 1917, there was a severe food shortage in the workers’ quarters.
(ii) All the factories and workers’ quarters were located on the right bank of the River Neva. On the left bank, there were Winter Palace, official buildings and the palace where Duma met.
(iii) On 22nd February, a lockout took place at a factory leading to a strike by the workers. The next day, workers in fifty factories called a strike in sympathy.
(iv) The demonstrating workers ultimately crossed the river and surrounded the official buildings in Petrograd.
(v) The Government imposed a curfew and called out the cavalry and police to keep check on them.

(Topic- The February Revolution in Petrograd, Page No. 35 and 36)

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R.D. Sharma Solutions Class 9th: Ch 23 Graphical Representation of Statistical Data Exercise 23.1

September 9, 2018, 2:42 am
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Chapter 23 23 Graphical Representation of Statistical Data R.D. Sharma Solutions for Class 9th Exercise 23.1

Exercise 23.1

1. If the height of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Solution

Given,
The heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm
∴ Mean Weight = sum of heights / total no. of persons
= 140+150+152+158+161/5
= 761/5 = 152.2

2. Find the mean of 994, 996, 998, 1002, 1000.

Solution

Given,
Numbers are 994,996,998,1000,1002
∴ Mean Weight = sum of heights/total no. of persons
= 994+996+998+1000+1002/5
= 4990/5 = 998
Mean = 998

3. Find the mean of first five natural numbers.

Solution

The first five odd numbers are 1 , 2 , 3 , 4 , 5.
∴ Mean = sum of numbers/total numbers
= 1+2+3+4+5/5
= 15/5 = 3
Mean = 3

4. Find the mean of all factors of 10.

Solution

All factors of 6 are 1 , 2 , 5 , 10.
∴ Mean = sum of factors/ total factors
= 1+2+5+10/4 = 4.5
Mean = 4.5

5. Find the mean of first 10 even natural numbers.

Solution

The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20
∴ Mean = sum of numbers / total numbers
= 2+4+6+8+10+12+14+16+18+20/10 = 11
Mean = 11

6.  Find the mean of x, x+2, x+4, x+6, x+8.

Solution

Numbers are x , x + 2 , x + 4 , x + 6 , x + 8.
∴ Mean = sum of numbers total numbers
= x + x + 2 + x + 4 + x + 6 + x + 85
= 5x + 20/5
= 5(x+4/5)
= x + 4

7. Find the mean of first five multiples of 3.

Solution

First five multiples of 3 are 3 , 6 , 9 , 12 , 15.
∴ Mean = sum of numbers/total numbers
= 3+6+9+12+15/5
= 9
Mean = 9

8. Following are the weights of 10 new born babies in a hospital on a particular day : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean.

Solution

The weights (in kg) of 10 new born  babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6
∴ Mean Weight = sum of weights/total no.of babies
= 3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.610
= 4 kg

9. The percentage marks obtained by students of a class in mathematics are as follows: 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1 . Find their mean.

Solution

The percentage marks obtained by students are 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1
∴ Mean marks = sum of marks/total numbers of marks
= 64+36+47+23+0+19+81+93+72+35+3+1/5 =39.5
Mean Marks = 39.5

10. The numbers of children in 10 families of a locality are 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5 . Find the mean number of children per family.

Solution

The numbers of children  in 10 families  are : 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5
∴ Mean = total no.children/total families
= 2+4+3+4+2+3+5+1+1+5/10 = 3

11. If M is the mean of x1, x2, x3, x, x5 and x6, prove that
(x1 − M) + (x2 − M) (x3 − M) (x4 − M) + (x5 − M) + (x6 − M) = 0.

Solution 


12. Durations of sunshine (in hours) in Amritsar to first 10 days of August 1997 as reported by the Meteorological Department are given below:
9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
(i) Find the mean x̅ .

Solution 

Given that the duration of sunshine (in hours) in Amritsar for 10 days are 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, and 10.9.
(i) We have to find their mean.
Remember the definition of mean of n values x1, x2… xn is

13. Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) subtracting a constant k from each them, (iii) multiplying each term by a constant k and (iv) dividing each term by a non-zero constant k.

Solution

Let say numbers are 3 , 4 , 5
∴ Mean = sum of numbers total numbers
= 3+4+5/3 = 4 .

(i) Adding constant term k = 2 in each term.
New numbers are = 5 , 6 , 7
∴ Mean = sum of numbers total numbers
= 5+6+7/3
= 6 = 4 + 2
∴ new  mean will be 2 more than the original mean.

(ii) Subtracting constant term k = 2 in each term.
New numbers are = 1 , 2 , 3
∴ Mean    = sum of numbers total numbers
= 1+2+3/3
= 2 = 4 – 2
∴ new  mean will be 2 less than the original mean 


(iii) Multiplying by constant term k = 2 in each term.
New numbers are = 6 , 8 , 10
∴ Mean    = sum of numbers total numbers
= 6+8+10/3
= 8 = 4×2
∴ new  mean will be 2 times of  the original mean.

(iv) Divide the constant term k =2 in each term.
New numbers are = 1.5 , 2 , 2.5.
∴ Mean    = sum of numbers/total numbers
= 1.5+2+2.53
= 2 = 42
∴ new  mean will be half  of  the original mean.


14. The mean of marks scored by 100 students was found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.

Solution

Mean marks of 100 students = 40
Sum of marks of 100 students = 100×40
= 4000
Correct value = 53
Incorrect value = 83
Correct sum = 4000 – 83 + 53 = 3970
∴ correct mean = 3970/100 = 39.7

15. The traffic police recorded the speed (in km/hr) of 10 motorists as 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded 5 km/hr less in each case.

Solution

The speed of 10 motorists are 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 .
Later on it was discovered that the instrument recorded 5 km/hr less than in each case
∴ correct values are = 52 , 58 , 54 , 65 , 44 , 47 , 60 , 62 , 57 , 53.
∴ correct mean = 52+58+54+65+44+47+60+62+57+53/10
= 552/10 =55.2 km/hr .

16. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Solution

The mean of five numbers is 27
The sum of five numbers = 5×27 = 135
If one number is excluded , the new mean is 25
∴ Sum of 4 numbers = 4×25 = 100
∴ Excluded number = 135 – 100 = 35

17. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Solution

The mean weight per student in a group of 7 students = 55 kg
Weight of 6 students (in kg) = 52 , 54 , 55 , 53 , 56 and 54
Let the weight of seventh student = x kg
∴ Mean Weight = sum of weights/total no.of students
⇒ 55 = (52+54+55+53+56+54+x)/ 7
⇒ 385 = 324 + x ⇒ x = 385−324
⇒ x = 61 kg
∴ weight of seventh student = 61 kg.

18. The mean weight of 8 numbers is 15. If each number is multiplied by 2 , what will be the new mean ?

Solution

We have ,
The mean weight of 8 numbers is 15
Then , the sum of 8 numbers = 8×15 = 120
If each number is multiplied by 2
Then , new mean = 120×2 = 240
∴ New  mean = 240/8 = 30.

19. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Solution

The mean of 5 numbers is 18
Then , the sum of 5 numbers = 5×18 = 90
If one number is excluded
Then , the mean of 4 numbers = 16
∴ sum of 4 numbers = 4×16 = 64
Excluded number = 90 – 64 = 26.

20. The mean of 200 items was 50. Later on, it was on discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution

The mean of 200 items = 50
Then the sum of 200 items = 200×50 = 10,000
Correct values = 192 and 88.
Incorrect values = 92 and 8.
∴ correct sum = 10000 - 92 - 8 + 192 + 88 = 10180
∴ correct mean = 10180/200 = 101.8/2 = 50.9 .

21. Find the values of n and X in each of the following cases:

Solution


22. The sum of the deviations of a set of n values x1, x2,...., xn, measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

Solution


23. Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Solution


24. If x̅ is the mean of the ten natural numbers x1, x1, x1 ....,x1, Show that (x1 - x̅) + (x2 - x̅) + ... + (x10 - x̅) = 0.

 Solution


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R.D. Sharma Solutions Class 9th: Ch 23 Graphical Representation of Statistical Data Exercise 23.2

September 9, 2018, 8:05 am
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Chapter 23 Graphical Representation of Statistical Data R.D. Sharma Solutions for Class 9th Exercise 23.2

Exercise 23.2


1. Explain the reading and interpretation of bar graphs .

Solution

A bar graph is a diagram consisting of a sequence of vertical or horizontal bars or rectangles, each of which represents an equal interval of the values of a variable, and has the height proportional to the quantities of the phenomenon under consideration in that interval. A bar graph may also be used to illustrate discrete data, in which case each bar represents a distinct circumstance.

While drawing a bar graph, we keep in mind that:

1. The width of the bars should be uniform throughout.
2. The gap between any two bars should be uniform throughout.
3. Bars may be either horizontal or vertical.
Each bar must be of the same width and the gap between them must be uniform. Make sure that the width of the bars and the gap between them should not be necessarily same.

2. Read the following bar graph and answer the following questions:
(i)What information is given by the bar graph ?
(ii) In which year the export is minimum ?
(iii)In which year the import is maximum ?
(iv) In which year the difference of the values of export and import is maximum ?

(i) The bar graph represents the import and export (in 100 Crores of rupees) from 1982-83 to 1986-87.
(ii) The export is minimum in the year 1982-83 at the height of the bar corresponding to export is minimum in the year 1982-83.
(iii) The import is maximum in the year 1986-87 as the height of the bar corresponding to import is maximum in the year 1986-87.
(iv) The bars of export and import are side by side. Clearly, it is seen from the bar graph that the difference between the values of export and import is maximum in the year 1986-87.
It is seen from the bar graph that the height of the 3s bar from the left is least, which is corresponding to DCE. Hence, the requirement is least in DCE.

3. The following bar graph shows the results of an annual examination in a secondary school. Read the bar graph (Fig. 23.28) and choose the correct alternative in each of the following:
(i) The pair of classes in which the result of boys and girls are inversely proportional are:
(a) VI, VIII
(b) VI, IX
(c) VIII, IX
(d) VIII, X

(ii) The class having the lowest failure rate of girls is
(a) VII
(b) X
(c) IX
(d) VIII

(iii) The class having the lowest pass rate of students is
(a) VI
(b) VII
(c) VIII
(d) IX

Solution

(i) The pair of classes in which the results of boys and girls are inversely proportional are VI and IX.

(ii) The lowest failure rate of girls is same to the highest pass rate. Hence, the class having the lowest failure rate of girls is VII (the height of the bar corresponding to girls for this class is maximum).

(iii) The sum of the heights of the bars for boys and girls in class VII is minimum, which is 95 + 40 = 135. Hence, the class having the lowest pass rate is VII. Hence, the correct choice is (b).

4. The following data gives the number (in thousands) of applicants registered with an
Year199519961997199819992000
Number of applicants registered (in thousands)182024283034
Construct a bar graph to represent the above data.
Solution 

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the number of applicants registered in thousands respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the number of applicants registered.
The vertical bar graph of the given data is following:

5. The production of salable steel in some of the steel plants our country during 1999 is given below:
PlantBhilaiDurgapurRourkelaBokaro
Production (in thousand tonnes)16080200150
Construct a bar graph to represent the above data on a graph paper by using the scale 1 big divisions = 20 thousand tonnes.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the plants and the production in thousand tonnes respectively. We have to draw 4 bars of different lengths given in the table.
The scale 1 big divisions must be 20 thousand tonnes. So, first find the heights of the bars corresponding to different plants. After that, we follow the well known procedure.

The heights of the different bars are:-
(i) The height of the bar corresponding to Bhilai is 160/20 = 8 big division .
(ii) The height of the bar corresponding to Durgapur is 80/20 = 4 big divisions.
(iii) The height of the bar corresponding to Rourkela is 200/20 = 10 big divisions.
(iv) The height  of the bar corresponding to Bokaro is 150/20 = 7.5 big division.

At first we mark 4 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the productions.
The vertical bar graph of the given data is following :


Note that the one big division in the vertical axis is equivalent to 20 thousand tonnes. 

6. The following table gives the route length (in thousand kilometres) of the Indian Railways in some of the years:
Year1960-611970-711980-811990-912000-2001
Rough length (in thousand km)5060617498

Represent the above data with the help of a bar graph.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the route lengths in thousand km respectively. We have to draw 5 bars of different lengths given in the table.
At first we mark 5 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the route lengths.
The vertical bar graph of the given data is following:


7. The following data gives the amount of loans (in crores of rupees) disbursed by a bank during some years:
Year19921993199419951996
Loan (in crores rupees)2833555580

(i) Represent the above data with the help of a bar graph.
(ii) With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the amount of loan in Crores of rupees respectively. We have to draw 5 bars of different lengths given in the table.
At first we mark 5 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the amount of loan disbursed by the bank.
(i) The vertical bar graph of the given data is following:
(ii) It is seen from the bar graph that the heights of the bars in the years 1994 and 1995 are same. Hence, the amount of loan is not increased in the year 1995 over the preceding year 1994.

8. The following table shows the interest paid by a company (in lakhs):
Year1995-961996-971997-981998-991999-2000
Interest (in lakhs rupees)2025151830
Draw the bar graph to represent the above information.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the interests in lakhs of rupees respectively. We have to draw 5 bars of different lengths given in the table.
At first we mark 5 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the interests paid by the company.
The vertical bar graph of the given data is following:
9. The following data shows the average age of men in various countries in a certain year:
Represent the above information by a bar graph.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the countries and the average age of men’s respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the average age of men’s in different countries.
The vertical bar graph of the given data is following:

10. The following data gives the production of foodgrains (in thousand tonnes) for some years:
Represent the above with the help of a bar graph.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the production of food grains in thousand tonnes respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the production of food grains. The vertical bar graph of the given data is following:
11. The following data gives the amount of manure (in thousand tonnes) manufactured by a company during some years:
(i) Represent the above data with the help of a bar graph.
(ii) Indicate with the help of the bar graph the year in which the amount of manufactured by the company was maximum.
(iii) Choose the correct alternative:

The consecutive years during which there was maximum decrease in manure production are:
(a) 1994 and 1995
(b) 1992 and 1993
(c) 1996 and 1997
(d) 1995 and 1996

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the amount of manure in thousand 9 ones respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the amount of manures manufactured by the company.

(i) The vertical bar graph of the given data is following:
(2) It is seen from the bar graph that the height of the 3rd bar from the left is maximum, which is corresponding to the year 1994. So in 1994 the quantity manufactured by the company was maximum.
(3) It is seen from the bar graph that the manure production is decreased in the years 1995 (1.5 scale divisions) and 1997 (2 full scale divisions). So, the maximum decrease is in the year 1997.
Hence, the correct choice is (c).

12. The following data gives the demand estimates of the Government of India, Department of Electronics for the personnel in the Computer sector during the Eighth Plan period (1990-95):
Represent the data with the help of a bar graph. Indicate with the help of the bar graph the course where estimated requirement is least.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the qualifications and the personnel required in hundreds respectively. We have to draw 5 bars of different lengths given in the table.
At first we mark 5 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the number of personnel required.
The vertical bar graph of the given data is following:


It is seen from the bar graph that the height of the 3rd bar from the left is least, which is corresponding to DCE. Hence, the requirement is least in DCE.

13. The income and expenditure for 5 years of a family is given in the following data:
Represent the above data by a gar graph.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the income or expenditure in thousand rupees respectively. We have to draw 5 bars for each income and expenditure side by side of different lengths given in the table. At first we mark 5 points for each income and expenditure in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the corresponding income or expenditures. The vertical bar graph of the given data is following:-



14. The investment (in ten crores of rupees) of Life Insurance Corporation of India in different sectors are given below:
Represent the above data with the help of bar graph.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the sectors and the investment in ten Crores of rupees respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the investments of Life Insurance Corporation of India. The vertical bar graph of the given data is following:



The short forms used in the graph are 
(a) C.G.S. : Central Government Securities
(b) S.G.S. : State Government Securities
(c) S.G.G. : Securities Guaranteed by Government
(d) P.S. : Private Sectors
(e) S.O.S.(P) : Socially Oriented Sectors (Plan)
(f) S.O.S.(NP) : Socially Oriented Sectors (Non-Plan)  

15. The following data gives the value (in crores of rupees) of the Indian export of cotton textiles for different years:
Represent the above data with the help of a bar graph. Indicate with the help of a bar graph the year in which the rate of increase in exports is maximum over the preceding year.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the value of Indian export of cotton textiles in Crores of rupees respectively. We have to draw 5 bars of different lengths given in the table.  At first we mark 5 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the values of Indian export of cotton textiles in different years. The vertical bar graph of the given data is as follows:

16. The following tables gives the quantity of goods (in crore tonnes)
Represent this information with the help of a bar graph.
Explain through the bar graph if the quantity of goods carried by the Indian Railways in 1965-66 is more than double the quantity of goods carried in the year 1950-51.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the years and the quantity of goods in crores tonnes respectively. We have to draw 6 bars of different lengths given in the table.
The heights of the rectangles are proportional to the quantity of goods carried by Indian railways in different years.
The vertical bar graph of the given data is as follows:
It is seen from the bar graph that the quantity of goods carried in the years 1950-51 and 1965-66 are 20 Crores tonnes and 9 Crores tonnes. Clearly 20 is more than 2 multiplied by 9.
Hence, the statement is true.

17. The production of oil (in lakh tonnes) in some of the refineries in India during 1982 was given below:
Construct a bar graph to represent the above data so that the bars are drawn horizontally.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the vertical and horizontal axes representing the refineries and the production of oil in lakh tonnes respectively. We have to draw 5 bars of different lengths given in the table.
At first we mark 5 points in the vertical axis at equal distances and erect rectangles of the same width at these points. The lengths of the rectangles are proportional to the productions of oil.
The horizontal bar graph of the given data is following:



18. The expenditure (in 10 crores of rupees) on health by the Government of India during the various five year plans is shown below:
Construct a bar graph to represent the above data.

Solution

To represent the given data by a vertical bar graph, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes representing the years and the expenditures on health in 10 crores of rupees respectively. We have to draw 6 bars of different lengths given in the table.
At first we mark 6 points in the horizontal axis at equal distances and erect rectangles of the same width at these points. The heights of the rectangles are proportional to the expenditures on health by the government of India in different years.
The vertical bar graph of the given data is following:



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R.D. Sharma Solutions Class 9th: Ch 23 Graphical Representation of Statistical Data Exercise 23.3

September 11, 2018, 5:28 am
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Chapter 23 Graphical Representation of Statistical Data R.D. Sharma Solutions for Class 9th Exercise 23.3

Exercise 23.3

1. Construct a histogram for the following data:

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. It should be noted that the scale for horizontal axis may not be same as the scale for vertical axis. Let us take one vertical division is equal to 3 rupees.

The heights of the different rectangles are as following .

1. The height of the rectangle corresponding to the class-interval 30-60 is 5/3 = 1.66 big divisions.
2. The height of the rectangle corresponding to the class-interval 60-90 is 12/3 = 4 big divisions.
3. The height of the rectangle corresponding to the class-interval 90-120 is 14/3 = 4.66 big divisions.
4. The height of the rectangle corresponding to the class-interval 120-150 is 18/3 = 6 big divisions.
5. The height of the rectangle corresponding to the class-interval 150-180 is 10/3 = 3.33 big divisions.
6. The height of the rectangle corresponding to the class-interval 180-210 is 9/3 = 3 big divisions.
7. The height of the rectangle corresponding to the class-interval 210-240 is 4/3 = 1.33 big divisions.

The histogram of the given data is the following:-


2. The distribution of heights (in cm) of 96 children is given below. Construct a histogram and a frequency polygon on the same axes.

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.

The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights.
To draw the frequency polygon of the given data using histogram, obtain the mid-points of the upper horizontal side of each rectangle and then join these mid-points of the adjacent rectangles of the histogram by line segments. Obtain the mid-points of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Complete the polygon by joining the mid-points of first and last class-intervals to the mid-points of imagined class-intervals adjacent to them. Let us take one vertical division is equal to 4.

The heights of the different rectangles are as following:-

1. The height of the rectangle corresponding to the class-interval 124-128 is 5/4 = 1.25 big divisions.
2. The height of the rectangle corresponding to the class-interval 128-132 is 8/4 = 2 big divisions .
3. The height of the rectangle corresponding to the class-interval 132-136 is 17/4 = 4.25 big divisions.
4. The height of the rectangle corresponding to the class-interval 136-140 is 24/4 = 6 big divisions .
5. The height of the rectangle corresponding to the class-interval 140-144 is 16/4 = 4 big divisions.
6. The height of the rectangle corresponding to the class-interval 144-148 is 12/4 = 3 big divisions .
7. The height of the rectangle corresponding to the class-interval 148-152 is 6/4 = 1.5 big divisions .
8. The height of the rectangle corresponding to the class-interval 152-156 is 4/4 = 1 big divisions .
9. The height of the rectangle corresponding to the class-interval 156-160 is 3/4 = 0.75 big divisions .
10.The height of the rectangle corresponding to the class-interval 160-164 is 1/4 = 0.25 big divisions 

The histogram and frequency polygon of the given data is the following:

3. The time taken, in seconds, to solve a problem by each of 25 pupils is as follows:
16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20
(a) Construct a frequency distribution for these data, using a class interval of 10 seconds.
(b) Draw a histogram to represent the frequency distribution. 

Solution

Given that the times (in seconds) taken to solve a problem by each of 25 pupils are 16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 43, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52 and 20. The minimum and maximum time values are 16 and 64 respectively.

(a) At first construct the following frequency distribution for the given data. Since, the lowest value is 16; we start with the class-interval 15-25, as the class size must be 10.

(b) To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The above data is a continuous grouped frequency distribution with equal class-intervals, which is 10. Construct rectangles with class-intervals as bases and respective frequencies as heights. 

The histogram of the data in part (a) is as follows: 


4. Draw, in the same diagram, a histogram and a frequency polygon to represent the following data which shows the monthly cost of living index of a city in a period of 2 years:

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.

The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. It should be noted that the scale for horizontal axis may not be same as the scale for vertical axis. To draw the frequency polygon of the given data using histogram, obtain the mid-points of the upper horizontal side of each rectangle and then join these mid-points of the adjacent rectangles of the histogram by line segments. Obtain the mid-points of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Complete the polygon by joining the mid-points of first and last class-intervals to the mid-points of imagined class-intervals adjacent to them. Let us take one vertical division is equal to 1 month. 

The heights of the different rectangles are as follows: 


1. The height of the rectangle corresponding to the class-interval 440-460 is 2 big divisions.
2. The height of the rectangle corresponding to the class-interval 460-480 is 4 big divisions.
3. The height of the rectangle corresponding to the class-interval 480-500 is 3 big divisions.
4. The height of the rectangle corresponding to the class-interval 500-520 is 5 big divisions.
5. The height of the rectangle corresponding to the class-interval 520-540 is 3 big divisions.
6. The height of the rectangle corresponding to the class-interval 540-560 is 2 big divisions.
7. The height of the rectangle corresponding to the class-interval 560-580is 1 big division.
8. The height of the rectangle corresponding to the class-interval 580-600 is 4 big divisions.

The histogram and frequency polygon of the given data is as follows:-

5. The following is the distribution of total household expenditure (in Rs.) of manual worker in a city:
Draw a histogram and a frequency polygon representing the above data.

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. It should be noted that the scale for horizontal axis may not be same as the scale for vertical axis. To draw the frequency polygon of the given data using histogram, obtain the mid-points of the upper horizontal side of each rectangle and then join these mid-points of the adjacent rectangles of the histogram by line segments. Obtain the mid-points of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Complete the polygon by joining the mid-points of first and last class-intervals to the mid-points of imagined class-intervals adjacent to them. Let us take one vertical division is equal to 5 workers.

The heights of the different rectangles are as follows:-

 1. The height of the rectangle corresponding to the class-interval 100-150 is 25/5 = 5 big divisions.
2. The height of the rectangle corresponding to the class-interval 150-200 is 40/5 = 8 big divisions.
3. The height of the rectangle corresponding to the class-interval 200-250 is 33/5 = 6.6 big divisions.
4. The height of the rectangle corresponding to the class-interval 250-300 is 28/5 = 28/5 = 5.6 big divisions.
5. The height of the rectangle corresponding to the class-interval 300-350 is 30/5 = 30/5 = 6 big divisions.
6 The height of the rectangle corresponding to the class-interval 350-400  is 22/5 = 4.4 big divisions.
7. The height of the rectangle corresponding to the class-interval 400-450 is 16/5 = 3.2 big divisions.
8. The height of the rectangle corresponding to the class-interval 450-500 is 8/5 = 1.6 big divisions.

The histogram of the given data is as follows:


6. The following table gives the distribution of IQ's (intelligence quotients) of 60 pupils of class V in a school:
Draw a frequency polygon for the above data.

Solution

We first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. To draw the frequency polygon of the given data without using histogram, obtain the class-limits of the class intervals. Obtain the class-limits of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Then plot the frequencies against class-limits.

The following table is useful to draw the frequency polygon of the given data.
We represent class marks on X-axis on a suitable scale and the frequencies on Y-axis on a suitable scale.

To obtain the frequency polygon we plot the points (66, 1), (73, 3), (80, 5), (87, 15), (94, 12), (101, 10), (108, 6), (115, 4), (122, 3), (129, 1).
Now we join the plotted points by line segments. The end points (66, 1) and (129, 1) are joined to the mid points (59, 0) and ( 136, 0) respectively of imagined class intervals to obtain the frequency polygon.


7. Draw a histogram for the daily earnings of 30 drug stores in the following table:

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-intervals and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. The scale for horizontal axis may not be same as the scale for vertical axis. Let us take one vertical division is equal to 1 unit.

The heights of the different rectangles are as follows:

1. The height of the rectangle corresponding to the class-interval 450-500 is 16 big divisions.
2. The height of the rectangle corresponding to the class-interval 500-550 is 10 big divisions.
3. The height of the rectangle corresponding to the class-interval 550-600 is 7 big divisions.
4. The height of the rectangle corresponding to the class-interval 600-650 is 3 big divisions.
5. The height of the rectangle corresponding to the class-interval 650-700is 1 big divisions.

The histogram of the given data is as follows:


8. The monthly profits (in Rs.) of 100 shops are distributed as follows:
Draw a histogram for the data and show the frequency polygon for it.

Solution

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-intervals and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. The scale for horizontal axis may not be same as the scale for vertical axis. Let us take one vertical division is equal to 3 shops.

The heights of the different rectangles are as following

1.The height of the rectangle corresponding to the class-interval 0-50 is 12/3 = 4 big divisions.
2.The height of the rectangle corresponding to the class-interval 50-100 is 18/3 = 6 big divisions.
3.The height of the rectangle corresponding to the class-interval 100-150 is 27/3 = 9 big divisions.
4.The height of the rectangle corresponding to the class-interval 150-200 is 20/3 = 6.67 big divisions.
5.The height of the rectangle corresponding to the class-interval 200-250 is 17/3 = 5.67 big divisions.
6.The height of the rectangle corresponding to the class-interval 250-300 is 6/3 = 2 big divisions .

The histogram of the given data is as follows:
For frequency polygon, first we will obtain the class marks as given in the following table. 



We plot the points (25, 12), (75, 18), (125, 27), (175, 20), (225, 17) and (275, 6).
Now, we join the plotted points by line segments . The end points (25, 12) and (275, 6) are joined to the mid-points (−25, 0) and (325, 0) respectively of imagined class-intervals to obtain the frequency polygon.

The frequency polygon of the given data is as follows:

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R.D. Sharma Solutions Class 9th: Ch 23 Graphical Representation of Statistical Data MCQ's

September 11, 2018, 5:49 am
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Chapter 23 Graphical Representation of Statistical Data R.D. Sharma Solutions for Class 9th MCQ's

Multiple Choice Questions

1. Which one of the following is not the graphical representation of statistical data:
(a) Bar graph
(b) Histogram
(c) Frequency polygon
(d) Cumulative frequency distribution

Solution

(d) Cumulative frequency distribution
We know that bar graph, histogram and frequency polygons are all graphical representation of statistical data.
Hence, the correct answer is option (d).

2. In a frequency distribution, ogives are graphical representation of
(a) Frequency
(b) Relative frequency
(c) Cumulative frequency
(d) Raw data

Solution

(c) Cumulative frequency
In a frequency distribution, ogives are graphical representation of cumulative frequency.
Hence, the correct answer is option (c).

3. A frequency polygon is constructed by plotting frequency of the class interval and the
(a) Upper limit of the class
(b) Lower limit of the class
(c) Mid value of the class
(d) Any values of the class

Solution

(c) mid value of the class
Frequency polygon is the plot of frequencies vs. the mid values of the classes.
Hence, the correct answer is option (c)

4. In a histogram the area of each rectangle is proportional to
(a)  the class mark of the corresponding class interval
(b) the class size of the corresponding class interval
(c) frequency of the corresponding class interval
(d) cumulative frequency of the corresponding class interval

Solution

(c) frequency of the corresponding class interval
In a histogram the area of each rectangle is proportional to the frequency of the corresponding class interval.
Hence, the correct answer is option (c).

5.  In the 'less than' type of ogive the cumulative frequency is plotted against

(a) the lower limit of the concerned class interval
(b) the upper limit of the concerned class interval
(c) the mid-value of the concerned class interval
(d) any value of the concerned class interval

Solution

(b) the upper limit of the concerned class interval
In the less than type of ogive the cumulative frequency is plotted against the upper limit of the concerned class interval.
Hence, the correct answer is option (b).

6. In a histogram the class intervals or the group are taken along
(a) Y-axis
(b) X-axis
(c) both of X-axis and Y-axis
(d) in between X and Y axis

Solution

(b) X-axis
In a histogram the class intervals or the groups are taken along the horizontal axis or X-axis.
Hence, the correct answer is option (b).


7. A histogram is a pictorial representation of the grouped data in which class intervals and frequency are respectively taken along
(a) vertical axis and horizontal axis
(b) vertical axis only
(c) horizontal axis only
(d) Horizontal axis and vertical axis

Solution

(d) Horizontal axis and vertical axis
In a histogram the class intervals and frequencies are taken along horizontal and vertical axes respectively.
Hence, the correct option is (d).

8. In a histogram, each class rectangle is constructed with base as
(a) Frequency
(b) Class-intervals
(c) Range
(d) Size of the class

Solution

(b) Class-intervals
In a histogram, the class rectangles are constructed with base as the class-intervals.
Hence, the correct answer is option (b).

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Extra Questions for Class 9th: Ch 3 Nazism and the Rise of Hitler History

September 12, 2018, 5:11 am
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Extra Questions for Class 10th: Ch 3 Nazism and the Rise of Hitler Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. Who was propaganda minister of Hitler?
Answer

Goebbels

(Para – 3, Page No. 49)
2. Which court was set up at the end of the Second World War to prosecute Nazi war criminals?
Answer

An International Military Tribunal at Nuremberg

(Para – 3, Page No. 49)
3. What do you mean by Genocidal war?
Answer

Genocidal war means killing on large scale leading to destruction of large sections of people.

(New Words, Page No. 50)
4. What do you mean by Reichstag?
Answer

Reichstag is the name of the German Parliament.

(Para – 2, Page No. 51)
5. Which treaty was signed by Germany after its defeat In the First World war?
Answer

The Treaty of Versailles was signed by Germany after its defeat in the First World War.

(Para – 1, Page No. 52)
6. Who were mockingly called 'November Criminals'?
Answer

Those who supported the Weimar Republic, mainly Socialists, Catholics and Democrats, were mockingly called the ‘November Criminals'.

(Para – 2, Page No. 52)
7. How were the deputies of the Reichstag appointed?
Answer
The deputies of the Reichstag were elected on the basis of universal votes cast by all adults including women.

(Para – 2, Page No. 51)
8. Which article of the Weimar Constitution gave the President the powers to impose emergency, suspend civil rights and rule by decree in Germany?
Answer

Article 48 of the Weimar Constitution gave the President the powers to impose emergency, suspend civil rights and rule by decree in Germany.

(Para – 2, Page No. 55)
9. What do you mean by Free Crops?
Answer
The war veterans organisation through which the Weimar Republic crushed the uprising of the Spartacist League are called Free Corps.

(Para – 1, Page No. 53)
10. Why did Nazis hold massive rallies and public meeting in Germany?
Answer
Nazis held massive rallies and public meetings in Germany to demonstrate the support for Hitler and instil a sense of unity among the people.

(Para – 2, Page No. 57)

Short Answer Questions (SAQs):

1. Describe the events that happened during Great Economic Depression in the USA?

Answer

• The Wall Street Exchange crashed in 1929 and due to fear a fall in prices, people made frantic efforts to sell their shares. 
• Over the next three years, between 1929 and 1932, the national income of the USA fell by half. 
• Factories shut down, exports fell, farmers were badly hit and speculators withdrew their money from the market.

(Para – 3, Page No. 54)

2. Describe any three inherent defects in the Weimar Constitution that made it vulnerable to dictatorship.
OR
Mention three reasons responsible for the failure of the Weimar Republic.
OR
State any three factors which made the Weimer Republic politically fragile.

Answer

• The constitution provided that government must be based proportional representation which made achieving a majority by any one party a near impossible.
• The coalitions governments were not stable and keeps changing.
• The constitution had Article 48, which gave the President the powers to impose emergency, suspend civil rights and rule by decree.

(Para – 2, Page No. 55)

3. What promises did Hitler make to the Germans when he came to power? 

Answer

• He promised to build a strong nation and undo the justice of Treaty of Versailles and restore the dignity of the German people. 
• He promised employment for those looking for work. 
• He promised to remove all foreign influences and resist all foreign conspiracies against Germany.

(Para – 1, Page No. 57)

4. Explain any three effects of the Treaty of Versailles over Germany. 

Answer

• Germany lost its overseas colonies. 
• It was demilitarized. 
• The allied armies occupied resource-rich Rhineland. 
• It lost 75 percent of its iron and 26 percent of its coal to France, Poland, Denmark & Lithuania.

(Para – 1, Page No. 52)

5. Describe the political impact of defeat of the Imperial Germany.

Answer

• The defeat of Imperial Germany and the abdication of the emperor gave an opportunity to recast German polity. 
• A National Assembly met at Weimar and established a democratic constitution with a federal structure. 
• Deputies were now elected to the German Parliament or Reichstag, on the basis of equal and universal votes cast by all adults including women.

(Para – 2, Page No. 51)

6. Why did Weimar Republic set up in Germany after the First World War become unpopular? Give three reasons. 

Answer

• The Weimar Republic was politically too fragile which created instability in Germany.
• The Weimar Republic was held responsible for signing the Treaty of Versailles and accept its harsh terms.
• The Weimar Republic did not take any steps to improve the economic conditions of the people.

(Para – 3, Page No. 51| Para – 1, Page No. 52| Para – 1 and 2, Page No. 55)

7. How did Nazis demonstrate support for Hitler?

Answer

• Nazis held massive rallies and public meetings to demonstrate the support for Hitler and instil a sense of unity among the people.
• The Red banners with the Swastika, the Nazi salute, and the ritualised rounds of applause after the speeches were all part of this spectacle of power.
• Nazi propaganda skilfully projected Hitler as a messiah, a saviour, as someone who had arrived to deliver people from their distress.

(Para – 2, Page No. 57| Para – 1, Page No. 58)

8. What was the verdict of Nuremberg Tribunal? Why did the Allies avoid harsh punishment on Germany?

Answer

• The Nuremberg Tribunal sentenced only eleven leading Nazis to death for the mass murder of selected groups of innocent civilians of Europe.
• Many others were imprisoned for life.
• The Allies did not want to be as harsh on defeated Germany as they had been after the First World War which led to the rise of Nazi Germany.

(Para – 2, Page No. 50)

Long Answer Questions (LAQs):

1. Mention any five effects of the ‘Great Economic Depression’ on the economy of Germany?

Answer

• The industrial production was reduced to 40 percent of the 1929 level.
• Workers lost their jobs or were paid reduced wages.
• Unemployed youth took to criminal activities and total despair became commonplace.
• The middle classes, especially salaried employees and pensioners, saw their savings diminish when the currency lost its value.
• Small businessmen, the self-employed and retailers suffered as their businesses got ruined.

(Para – 4, Page No. 54)

2. Why was the 'Treaty of Versailles' treated as harsh and humiliating to people of Germany? Explain.

Answer

• Germany lost its overseas colonies, a tenth of its population, 13 percent of its territories, 75 percent of its iron and 26 percent of its coal to France, Poland, Denmark and Lithuania. 
• The Allied Powers demilitarised Germany to weaken its power. 
• The War Guilt Clause held Germany responsible for the war and damages the Allied countries suffered. 
• Germany was forced to pay compensation amounting to £6 billion. 
• The Allied armies also occupied the resource-rich Rhineland for much of the 1920s.

(Para – 1, Page No. 52)

3. The First World War left a deep imprint on the European society and polity'. Elaborate the given statement. 

Answer

• Soldiers came to be placed above civilians. 
• Politicians and publicists laid great stress on the need for the men to be aggressive, strong and masculine. 
• Media glorified trench life. 
• Aggressive war propaganda and national honour occupied centre stage. 
• Popular support grew for conservative dictatorships.

(Para – 3, Page No. 52)


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R.D. Sharma Solutions Class 9th: Ch 24 Measure of Central Tendency Exercise 24.1

September 12, 2018, 5:51 am
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Chapter 24 Measure of Central Tendency R.D. Sharma Solutions for Class 9th Exercise 24.1

Exercise 24.1

1. If the height of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Solution

Given,
The heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm
∴ Mean Weight = sum of heights / total no. of persons
= 140+150+152+158+161/5
= 761/5 = 152.2

2. Find the mean of 994, 996, 998, 1002, 1000.

Solution

Given,
Numbers are 994,996,998,1000,1002
∴ Mean Weight = sum of heights/total no. of persons
= 994+996+998+1000+1002/5
= 4990/5 = 998
Mean = 998

3. Find the mean of first five natural numbers.

Solution

The first five odd numbers are 1 , 2 , 3 , 4 , 5.
∴ Mean = sum of numbers/total numbers
= 1+2+3+4+5/5
= 15/5 = 3
Mean = 3

4. Find the mean of all factors of 10.

Solution

All factors of 6 are 1 , 2 , 5 , 10.
∴ Mean = sum of factors/ total factors
= 1+2+5+10/4 = 4.5
Mean = 4.5

5. Find the mean of first 10 even natural numbers.

Solution

The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20
∴ Mean = sum of numbers / total numbers
= 2+4+6+8+10+12+14+16+18+20/10 = 11
Mean = 11

6.  Find the mean of x, x+2, x+4, x+6, x+8.

Solution

Numbers are x , x + 2 , x + 4 , x + 6 , x + 8.
∴ Mean = sum of numbers total numbers
= x + x + 2 + x + 4 + x + 6 + x + 85
= 5x + 20/5
= 5(x+4/5)
= x + 4

7. Find the mean of first five multiples of 3.

Solution

First five multiples of 3 are 3 , 6 , 9 , 12 , 15.
∴ Mean = sum of numbers/total numbers
= 3+6+9+12+15/5
= 9
Mean = 9

8. Following are the weights of 10 new born babies in a hospital on a particular day : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean.

Solution

The weights (in kg) of 10 new born  babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6
∴ Mean Weight = sum of weights/total no.of babies
= 3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.610
= 4 kg

9. The percentage marks obtained by students of a class in mathematics are as follows: 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1 . Find their mean.

Solution

The percentage marks obtained by students are 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1
∴ Mean marks = sum of marks/total numbers of marks
= 64+36+47+23+0+19+81+93+72+35+3+1/5 =39.5
Mean Marks = 39.5

10. The numbers of children in 10 families of a locality are 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5 . Find the mean number of children per family.

Solution

The numbers of children  in 10 families  are : 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5
∴ Mean = total no.children/total families
= 2+4+3+4+2+3+5+1+1+5/10 = 3

11. If M is the mean of x1, x2, x3, x, x5 and x6, prove that
(x1 − M) + (x2 − M) (x3 − M) (x4 − M) + (x5 − M) + (x6 − M) = 0.

Solution 


12. Durations of sunshine (in hours) in Amritsar to first 10 days of August 1997 as reported by the Meteorological Department are given below:
9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9
(i) Find the mean x̅ .

Solution 

Given that the duration of sunshine (in hours) in Amritsar for 10 days are 9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, and 10.9.

(i) We have to find their mean.
Remember the definition of mean of n values x1, x2… xn is

13. Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) subtracting a constant k from each them, (iii) multiplying each term by a constant k and (iv) dividing each term by a non-zero constant k.

Solution

Let say numbers are 3 , 4 , 5
∴ Mean = sum of numbers total numbers
= 3+4+5/3 = 4 .

(i) Adding constant term k = 2 in each term.
New numbers are = 5 , 6 , 7
∴ Mean = sum of numbers total numbers
= 5+6+7/3
= 6 = 4 + 2
∴ new  mean will be 2 more than the original mean.

(ii) Subtracting constant term k = 2 in each term.
New numbers are = 1 , 2 , 3
∴ Mean    = sum of numbers total numbers
= 1+2+3/3
= 2 = 4 – 2
∴ new  mean will be 2 less than the original mean 


(iii) Multiplying by constant term k = 2 in each term.
New numbers are = 6 , 8 , 10
∴ Mean    = sum of numbers total numbers
= 6+8+10/3
= 8 = 4×2
∴ new  mean will be 2 times of  the original mean.

(iv) Divide the constant term k =2 in each term.
New numbers are = 1.5 , 2 , 2.5.
∴ Mean    = sum of numbers/total numbers
= 1.5+2+2.53
= 2 = 42
∴ new  mean will be half  of  the original mean.


14. The mean of marks scored by 100 students was found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean.

Solution

Mean marks of 100 students = 40
Sum of marks of 100 students = 100×40
= 4000
Correct value = 53
Incorrect value = 83
Correct sum = 4000 – 83 + 53 = 3970
∴ correct mean = 3970/100 = 39.7

15. The traffic police recorded the speed (in km/hr) of 10 motorists as 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 . Later on, an error in recording instrument was found. Find the correct average speed of the motorists if the instrument is recorded 5 km/hr less in each case.

Solution

The speed of 10 motorists are 47 , 53 , 49 , 60 , 39 , 42 , 55 , 57 , 52 , 48 .
Later on it was discovered that the instrument recorded 5 km/hr less than in each case
∴ correct values are = 52 , 58 , 54 , 65 , 44 , 47 , 60 , 62 , 57 , 53.
∴ correct mean = 52+58+54+65+44+47+60+62+57+53/10
= 552/10 =55.2 km/hr .

16. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Solution

The mean of five numbers is 27
The sum of five numbers = 5×27 = 135
If one number is excluded , the new mean is 25
∴ Sum of 4 numbers = 4×25 = 100
∴ Excluded number = 135 – 100 = 35

17. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Solution

The mean weight per student in a group of 7 students = 55 kg
Weight of 6 students (in kg) = 52 , 54 , 55 , 53 , 56 and 54
Let the weight of seventh student = x kg
∴ Mean Weight = sum of weights/total no.of students
⇒ 55 = (52+54+55+53+56+54+x)/ 7
⇒ 385 = 324 + x ⇒ x = 385−324
⇒ x = 61 kg
∴ weight of seventh student = 61 kg.

18. The mean weight of 8 numbers is 15. If each number is multiplied by 2 , what will be the new mean ?

Solution

We have ,
The mean weight of 8 numbers is 15
Then , the sum of 8 numbers = 8×15 = 120
If each number is multiplied by 2
Then , new mean = 120×2 = 240
∴ New  mean = 240/8 = 30.

19. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Solution

The mean of 5 numbers is 18
Then , the sum of 5 numbers = 5×18 = 90
If one number is excluded
Then , the mean of 4 numbers = 16
∴ sum of 4 numbers = 4×16 = 64
Excluded number = 90 – 64 = 26.

20. The mean of 200 items was 50. Later on, it was on discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution

The mean of 200 items = 50
Then the sum of 200 items = 200×50 = 10,000
Correct values = 192 and 88.
Incorrect values = 92 and 8.
∴ correct sum = 10000 - 92 - 8 + 192 + 88 = 10180
∴ correct mean = 10180/200 = 101.8/2 = 50.9 .

21. Find the values of n and X in each of the following cases:


Solution



22. The sum of the deviations of a set of n values x1, x2,...., xn, measured from 15 and -3 are -90 and 54 respectively. Find the value of n and mean.

Solution


23. Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean.

Solution


24. If x̅ is the mean of the ten natural numbers x1, x1, x1 ....,x1, Show that (x1 - x̅) + (x2 - x̅) + ... + (x10 - x̅) = 0.

 Solution


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Extra Questions for Class 9th: Ch 4 Forest Society and Colonialism History

September 13, 2018, 4:35 am
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Extra Questions for Class 10th: Ch 4 Forest Society and Colonialism Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. What is meant by 'sleepers' ?

Answer

Sleepers are wooden planks laid across railway tracks to hold the tracks in position. 

(New Words, Page No. 79)

2. When did the railway network expand rapidly in India?

Answer 

From the 1860s. 

(Para – 3, Page No. 80)

3. What is meant by scientific forestry?

Answer 

Scientific forestry seeks to ensure that the different types of trees in a forest are cut down and replaced by only type of tree planted in straight rows. 

(New Words, Page No. 84)

4. Who was Dietrich Brandis?

Answer 

Dietrich Brandis was the first Inspector General of Forests in India. 

(Para- 1, Page No. 83)

5. Who started the Bastar Rebellion?

Answer 

The Bastar rebellion first started in the Kanger forest area and soon spread to other parts of the state. 

(Para – 3, Page No. 91)

6. When was the Indian Forest Service set up?

Answer 

The Indian Forest Service was set up in 1864. 

(Para – 1, Page No. 84)

7. Where was the Imperial Forest Research Institute set up?

Answer 

The Imperial Forest Research Institute was established at Dehradun. 

(Para – 1, Page No. 84)

8. What is shifting cultivation called in Sri Lanka?

Answer 

It is known as chena in Sri Lanka. 

(Para – 2, Page No. 86)

9. Where is Bastar located?

Answer 

Bastar is located in Chhattisgarh. 

(Para – 2, Page No. 90)

10. When did the first rebellion take place in Bastar?

Answer 

The first rebellion took place in Bastar in 1910. 

(Para – 1, Page No. 90)

Short Answer Questions (SAQs):

1. Why did the British appoint the Inspector General of Forests in India? Explain any three reasons. 

Answer

• The British needed forests in order to build ships and railways. 
• They were worried that the use of forests by local people and the reckless felling of trees by traders would destroy forests.
• They wanted forests for the development of plantations.

(Para – 1, Page No. 83| Para – 2, Page No. 84)

2. Explain the system of scientific forestry. 

Answer

• Scientific forestry seeks to ensure that the different types of trees in a forest are cut down and replaced by only one type of tree planted in straight rows. 
• Forest officials surveyed the forests estimated the area under different types of trees and planned how much of the plantation area to be cut every year. 
• The area cut was then to be replanted so that it was ready to be cut again in some years. 

(Para – 2, Page No. 84)

3. Who was Dietrich Brandis? Why was he invited to India? Mention his two major contributions.

Answer 

Dietrich Brandis was a German expert in forestry. He was invited to India by the British for advice and help who were worried that the use of forests by local people and the reckless felling of trees by traders would destroy forests. 
His two major contributions are as follows: 
• Scientific Forestry was introduced. 
• He introduced a proper system to manage the forests.
• Rules about the use of forest resources were also laid down.

(Para – 1, Page No. 83| Para – 2, Page No. 84) 

4. State the reasons why Shifting cultivation was banned under European colonialism in India. 

Answer

• European foresters felt that land used for cultivation every few years could not support trees like Sal and Oak for railway timber. 
• Also, when a forest was burnt, there was a danger of the flames spreading and burning valuable timber. 
• Shifting cultivation also made it harder for the government to calculate taxes.

(Para – 4, Page No. 86| Para – 1, Page No .87)

5. How did the forest acts affect the lives of foresters and villagers? 

Answer

• The daily practices of villagers such as cutting wood for their houses, hunting, fishing and collecting fruits now become illegal. 
• People were forced to steal wood from the forests and if caught were at the mercy of the forest guards who would take bribes from them.
• It was common for police constables and forest guards to harass people.

(Para – 3, Page No. 85| Para – 1, Page No. 86)

6. When was the Forest Act enacted during the British period? How many times and when was it amended? Name the three categories of forests according to the Forest Act?

Answer

• Forest Act was enacted in 1865. 
• It was amended twice, once in 1878 and then in 1927. 
• The 1878 Act divided the forests into three categories: reserved, protected and village forests.

(Para – 3, Page No. 84)

7. Mention any three provisions of forest laws passed by the Dutch. 

Answer

• The access to the forests was restricted for the villagers. 
• Timber was now allowed to cut only for specific purposes like making boats. 
• Forests were kept under strict supervision. Those villagers who grazed their cattle in the forests were severely punished.

(Para – 3, Page No. 93)

8. What contribution did Dietrich Brandis make towards the development and preservation of forest?   
Answer

• Brandis believed that a proper system had to be introduced to manage the forests and people had to be trained in the science of conservation. 
• This system would need legal sanction and rules about the use of forest resources had to be framed.
• Felling of trees and grazing had to be restricted so that forests could be preserved for timber production. 
• Anyone who cut trees without following the regulations had to be punished.

(Para – 2, Page No. 83)

Long Answer Questions (LAQs):

1. "Deforestation became more systematic and extensive under the colonial rule." Explain this statement with suitable examples. 

Answer

• With the increase in population, the demand for food went up thus, peasants extended boundaries for cultivation by clearing forest. 
• British encouraged production of commercial crops like jute, sugar, wheat and cotton as the demand for these crops increased in nineteenth-century Europe.
• British thought forests are unproductive thus they had to be brought under cultivation so that the land could yield agricultural products and revenue.
• Oak forests disappeared and problem of timber supply for the Royal Navy started. Hence, trees were felled at a massive scale.
• The spread of railways demanded more sleepers which was fulfilled by felling trees.
• A large number of natural forests were also cleared to make way for plantation agriculture such as tea, coffee, etc.

(Topic: Why Deforestation, Page No. 78 to 82)

2. Why were forests important to the villagers? 

Answer

• Fruits and tubers were eaten because they were nutritious and herbs were used for medicinal purposes. 
• Bamboo was used to make fences, baskets and umbrellas. 
• The wood was used to make agricultural implements like yokes and ploughs. 
• A dried out gourd was used as a portable water bottle.
• The creeper can be used to make ropes, and the thorny bark of the semur tree is used to grate vegetables.
• Oil for cooking and lighting lamps was acquired from the fruit of the Mahua tree. 

(Para – 2, Page No. 85)

3. Describe some of the common customs and beliefs of the Baster people. 

Answer

• The people believed that each village was given its land by the Earth and thus they look after the earth by making some offerings at each agricultural festival. 
• Respect is also shown to the spirits of the river, the forest and the mountain. 
• As each village was aware of their boundaries, all the natural resources within that boundary were looked after by the local people.
• If people from a village want to take some wood from the forests of another village, they pay a small fee called dand or man. 
• Some villages also protect by engaging watchmen and every household contributes some grain to pay them.

(Para – 1, Page No. 91)

4. What is shifting cultivation? Why did the British Government ban it ? Give any three reasons. 

Answer

Shifting agriculture is a traditional agricultural practice in many parts of Asia, Africa and South America. 
In shifting cultivation, parts of the forest are cut down and burnt in rotation. Seeds are sown in the ashes after the first monsoon rains and the crop is harvested by October - November. 
Colonial impacts on shifting agriculture: 
• Europeans regarded this practice harmful for the forests. They felt that the burning down forest would destroy timber and the dangerous flames would spread and burn valuable timber. 
• Shifting cultivation made it harder for the government to calculate taxes, so British government decided to ban shifting cultivation. 
• As a result, many communities were forcibly displaced from their homes in forest.
• Some had to change occupations, while some resisted through large and small rebellions.
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Extra Questions for Class 9th: Ch 5 Pastoralists in the Modern World History

September 14, 2018, 1:03 am
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Extra Questions for Class 10th: Ch 5 Pastoralists in the Modern World Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. What are Bugyals?

Answer

Pastures lands on above 12000 feet high mountains.

(Fig.1, Page No. 97)

2. Gaddi were an important pastoral community of which state?

Answer 

Himachal Pradesh

(Para – 2, Page No. 98)

3. Define the term ‘Pastoral Nomads’.

Answer 

Pastoral nomads are those who move from place to place with their cattle.

(Para – 1, Page No.97)

4. Why did the pastoral nomads raise cattle?

Answer 

For the sale of milk and other products of their cattle like leather, wool etc.

5. Who are Bhotiyas, Sherpas and Kinnauris?

Answer 

Pastoral communities of the Himalayas

(Para – 3, Page No. 99)

6. Why nomadic tribes need to move from one place to another place?

Answer 

In search of pastures

7. Raika pastoral community belongs to which state?

Answer 

Rajasthan

(Para – 3, Page No. 101)

8. Where is Serengiti Park located?

Answer 

Tanzania

(Para – 3, Page No. 110)

9. What are Dhars?

Answer 

They are high meadows.

(Fig. 4, Page No. 99)

10. Where is the Samburu National Park located?

Answer 

Kenya

(Para – 3, Page No. 110)

Short Answer Questions (SAQs):

1. Describe the life of Dhangars of Maharashtra.

Answer

• The Dhangar shepherds stay in the central plateau of Maharashtra during the monsoon. By October, they harvest their bajra and move west to Konkan.
• The Dhangar flocks manure the fields and feed on stubble.
• The Konkani peasants give them rice which they take to the plateau as grain is scarce there. With the onset of monsoon they leave Konkan and return to the dry plateau.

(Para – 2, Page No. 100)

2. List any three factors that the pastoral groups have to consider to sustain their life.

Answer

• How long the herds could stay in one area.
• Where they could find water and pasture.
• Calculation of timing and assessment of their movements,
• Relationship with farmers on the way, so that the herds could graze in harvested fields and manure the soil.

(Para – 1, Page No. 102)

3. Describe how the movement of the Kurumas and Kurubas is defined by the requirement of their cattle.

Answer 

• Alternation of monsoons and dry season in dry central plateau of Karnataka and Andhra Pradesh defines the movement of the Kurumas and Kurubas.
• They move from the cultivated patches near the woods to the coastal tract during the dry season.
• They leave the coastal area for the dry plateau when the rain arrives and the herd has to be shifted. Their buffaloes like the wet conditions of the monsoon month.

(Para - 1, Page No. 101)

4. Who were Gujjar Bakarwals? How did they earn their livelihood?

Answer 

Gujjar Bakarwals were the herders of goats and sheep living in the region of Jammu and Kashmir.
Their herds moved out of this area between summer and winter and travelled in groups called Kafilas.
• They earned their livelihood by selling milk, ghee, and other products of their herds.
• Women went to the markets and sold home-made products like pots filled with buttermilk, honey, etc. while the men took the cattle to graze.

(Para – 1, Source – A, Page No. 98)

5. Describe the life of Raikas community of Rajasthan.

Answer 

• Raikas lived in the desert of Rajasthan.
• The rainfall in the region was meagre and uncertain. On cultivated land, harvest fluctuated every year. Over vast stretches, no crop could be grown. So, the Raikas combined cultivation with pastoralism.
• During the monsoon, the Raikas of Barmer, Jaisalmer, Jodhpur and Bikaner stayed in their home villages, where pasture was available.
• By October, when these grazing grounds were dry and exhausted, they moved out in search of other pastures and water, and returned again during the next monsoon.

(Para – 3, Page No. 101| Para – 1, Page No. 102)

6. How did the life of the pastoralists change under the colonial rule?

Answer 

Under the colonial rule, the life of pastoralists changed in the following ways:

• Grazing grounds shrank as it was transformed into cultivated farms which meant the decline of pastures.
• Forest Acts enacted, pastoralists were prevented from entering many forests that had earlier provided valuable fodder for their cattle.
• In 1871, the colonial government, in India passed the Criminal Tribes Act by which many pastoral communities were classified as criminal tribes.
• Taxes were imposed on land, canal water, on salt, on trade of goods and even on animals. Pastoralists had to pay tax on every animal they possessed and also the land they used to graze on the pastures.

(Topic- Colonial Rule and Pastoral Life, Page No. 104 and 105)

Long Answer Questions (LAQs):

1. Who are the pastoral nomads? Describe any four features. 

Answer 

Pastoral nomads are those groups who earn their livelihood by subsistence farming and cattle rearing.
• They move from place to place with their cattle in search of pastures, farming and cattle rearing.
• Their movement is seasonal and is guided by the need of their flock.
• They raise cattle, camels, goats, sheep, etc.
• They sell milk, meat, animal skin and wool.
• Some also earn through trade and transport.
• Others combine pastoral activity with agriculture.

(Topic – Pastoral Nomads and their movements, Page No. 98 to 103)

2. Explain the cycle of seasonal movement of Gaddi Shepherds of Himachal Pradesh. 

Answer

Seasonal movement of Gaddi Shepherds of the Himachal Pradesh:
• They spend their winter in the lower hills of Shiwalik Range, grazing their flocks in scrub forests.
• By April, they moved to north and spent their summer in Lahaul and Spiti.
• When the snow melted and the high passes were cleared, many of them moved onto higher mountain meadows.
• By September, they began their return movement.
• On the way, they stopped once again in the village of Lahaul and Spiti reaping the summer crop and sowing their winter crop.

(Para – 2, Page No. 98| Para – 1, Page No. 99)

3. Why were the forest lands considered as waste lands? Why did the British want to transform these lands into cultivated farms? 

Answer 

The forest lands were considered as waste lands because:
• These lands did not yield agricultural product nor any other revenue.
• They considered these lands as unproductive and referred to them as waste lands.
Reasons to transform these lands into cultivated farms:
• By expanding cultivated land, the British wanted to increase their revenue.
• For increasing production of commercial crops like jute, cotton, wheat, etc.
• Since uncultivated land appeared to be unproductive and Britishers thought it to be unproductive and as a waste land.

(Para – 2, Page No. 104) 

4. Describe five main features of the Criminal Tribes Act introduced by the colonial government in India. 

Answer

• By this Act, many communities of craftsmen, traders and pastoralists were classified as Criminal Tribes. They were stated to be criminal by nature and birth.
• These communities were expected to live only in notified village settlements.
• They were not allowed to move out without permit.
• The village police kept a continuous watch on them.
• Settled groups were considered peaceful and law abiding.
• Nomads were considered criminals.
• Because of their movement, the nomads could not be identified or controlled.
• Because of their movement, the nomads could not be taxed.

(Para – 2, Page No. 105)
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Extra Questions for Class 9th: Ch 6 Peasants and Farmers History

September 14, 2018, 8:03 pm
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Extra Questions for Class 10th: Ch 6 Peasants and Farmers Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. How did turnip and clover help in increasing the fertility of soil?

Answer 

These crops had the capacity to increase the nitrogen content of the soil and nitrogen was important for crop growth and soil fertility.

(Para – 1, Page No. 122)

2. What was the use of the common land in England?

Answer 

The farmers pastured their cows and grazed their sheep, collected fuel wood for fire and berries and fruit for food.

(Para – 3, Page No. 118)

3. What did the enclosure imply?

Answer 

Piece of land enclosed from all sides.

(Para – 3, Page No. 119)

4. The Great Agrarian Depression of the 1930s was caused by?

Answer 

Overproduction of wheat

(Para – 4, Page No. 129)

5. Which natural event proved to be a nightmare for American peasants?

Answer

Blizzards

(Para – 5, Page No. 129)

6. What is the advantage of Enclosure Movement to landowners?

Answer 

Enclosures allowed the rich farmers to expand the land under their control.

(Para – 3, Page No. 119)

7. What kind of new ploughs were designed for the prairies in USA?

Answer 

New ploughs were devised locally, some of them 12 feet long. Their front rested on small wheels and they were hitched on to six yokes of oxen or horses.

(Para – 4, Page No. 127)

8. What was the occupation of the Native Americans in the 18th century?

Answer 

Several of them were nomadic, some were settled.

(Para – 1, Page No. 124)

9. What made America’s dream of land of plenty turn into a nightmare?

Answer 

Terrifying dust storms.

(Para – 2, Page No. 130)

10. What do you mean by Manchus?

Answer 

Chinese rulers

(Para – 6, Page No. 131)

Short Answer Questions (SAQs):

1. Why were the poor farmers of England against the threshing machines? What was the Captain Swing Movement? 

Answer 

The poor farmers felt the threshing machines would replace people, would deprive them of their livelihood and render them jobless. 
Captain Swing was a mythical name used in threatening letters, written by workmen against the use of threshing machines by rich farmers.

(Topic - The Introduction of Threshing Machines, Page No. 123)

2. Explain three factors which led to the Enclosure Movement in England after the mid-eighteenth century. 

Answer 

• Rapid expansion of population from 7 million in 1750 to 21 million in 1850 and 30 million in 1900. 
• Increased demand for foodgrains to feed the growing population. 
• War with France disrupted trade and import of foodgrains from Europe. Prices in England skyrocketed, encouraging landowners to enclose lands and enlarge the area under grain cultivation.

(Para – 3, Page No. 119)

3. Why did the farmers of East Kent destroyed their own threshing machines? 

Answer

• On the night of 28th August 1830, a threshing machine of a farmer was destroyed by labourers in East Kent in England. 
• In the subsequent two years, riots spread over southern England and about 387 threshing machines were broken. During this period, farmers received threatening letters urging them to stop using machines that deprived workmen of their livelihood. 
• Most of these letters were signed in the name of Captain Swing.

(Para – 1, Page No. 118)

4. Why did the landlords pressurise the British Parliament to pass the Enclosure Act? Explain. 

Answer 

• France was at war with England which disrupted trade and import of food grains from Europe. 
• Prices of food grains in England shot up, thereby encouraging landlords to enclose lands. 
• They had enlarged the area under grain cultivation. Profits flowed in and landowners pressurised the Parliament to pass the Enclosure Acts.

(Para – 2, Page No. 120)

5. Why did the farmers in England start growing turnip and clover? 

Answer 

• In early eighteenth century, farmers began cultivating turnip and clover regularly. These crops became part of the cropping system. 
• Later findings showed that these crops had the capacity to increase the nitrogen content of the soil. Nitrogen was important for crop growth. 
• Cultivation of the same soil over a few years depleted the nitrogen in the soil and reduced its fertility. By restoring nitrogen, turnip and clover made the soil fertile once again.

(Para – 1, Page No. 122)

6. Who was Captain Swing? What did the name symbolize or represent? 

Answer 

Captain Swing was a mythic name used in the threatening letters issued after the attack on threshing machines in England and the symbol of danger for landlords. Captain Swing represented the deprived workmen who were struggling for their livelihood because of the introduction of machines in agriculture.

(Topic - The Introduction of Threshing Machines, Page No. 123)

Long Answer Questions (LAQs):

1. Mention any five uses of commons for the villagers of England. 

Answer

• All villagers had access to the commons. Here they pastured their cows and grazed their sheeps.
• They collected fuel, wood for fire and berries and fruits for food.
• They fished in the rivers and ponds and hunted rabbit in common forests.
• For the poor, the common land was essential for survival. They gathered fruits and other forest products.
• It supplemented their meagre income, sustained their cattle and help them to tide over bad times when crops failed.

(Topic - The Time of Open fields and Commons, Page No. 118 and 119)

2. How did peasants of England get land for cultivation in the early 18th century? Explain.

Answer 

• In the beginning of the 18th century, in large parts of England, the countryside was open. It was not partitioned into enclosed lands privately owned by landlords.
• Peasants cultivated on strips of land around the village they lived in. At the beginning of each year, at a public meeting, each villager was allocated a number of strips to cultivate.
• Usually, these strips were of varying quality and often located in different places, not next to each other.
• The effort was to ensure that everyone had a mix of good and bad land.

(Para – 3, Page No. 118)

3. Who was Captain Swing? Who were the Swing rioters? Describe the result of these riots.

Answer 

Captain Swing was a mythic name used in the threatening letters issued after attack on the threshing machines in England. Labourers of England were the Swing rioters. The result of these riots were:
• Due to these riots, many threshing machines were destroyed by the troop of labourers and many machines were destroyed by even farmers themselves.
• Regarding these riots, the government action was severe.
• Those suspected of rioting were rounded up. 1,976 prisoners were tried, nine men were hanged, 505 men were transported. 

(Para – 1, Page No. 118)

5. How did the ‘Enclosure Movement’ start in England? Write any three advantages of enclosures. 

Answer 

The early enclosures were usually created by individual landlords. They were not supported by the State or the Church. After the mid-eighteenth century, the Enclosure Movement swept through the countryside and changed the English landscape forever. Advantages of enclosures were as follows:
• Enclosures had become necessary to move longterm investments on land and plan crop rotation to improve the soil.
• Enclosures also allowed the rich farmers to expand the land under their control.
• They would produce more for the market to earn more profit.

(Para – 3, Page No. 119)

6. Why did the demand of food grains increase after 1750 in England? How was production of food grains increased rapidly after 1780 in England? Explain. 

Answer 

The demand of food grains increased due to the following reasons: English population increased rapidly between 1750 to 1900. It mounted over four times from 7 million to 30 million. This led to an increased demand for food grains to feed population. The production of food grains increased rapidly in the following ways:
• At this time, Britain was industrialising. More people began to live and work in urban areas.
• By the end of the eighteenth century, France was at war with England. This disrupted trade and the import of food grains from Europe.
• Food grain production increased by agricultural technology.
• By bringing new lands under cultivation, landlords sliced up pasture lands curved up open fields, cut up forest, commons took over marshes, and turned larger areas into agricultural fields.
• By simple innovations in agriculture.

(Para – 1 and 2, Page No. 120)
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Extra Questions for Class 10th: Ch 1 Power Sharing Civics

September 16, 2018, 7:26 am
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Extra Questions for Class 10th: Ch 1 Power Sharing Social Studies (S.St) Important Questions Answer Included

Very Short Answer Questions (VSAQs): 

1. What is the state religion of Sri Lanka?
Answer

Buddhism is the state religion of Sri Lanka.

2. Which measure adopted by the democratically elected government to establish Sinhala supremacy?
Answer

Majoritarian measure was adopted by the democratically elected government to establish Sinhala supremacy.

3. What is meant by ‘vertical division of power’?
Answer

The division of powers between higher and lower levels of government is called vertical division of power.

4. Which community was rich in Belgium?
Answer

The minority French-speaking community was rich in Belgium.

5. Which system of power sharing called, ‘Checks and Balances’?
Answer

Horizontal distribution of power sharing is called system of ‘Checks and Balances’.

6. Who elects the community government in Belgium?
Answer

The ‘community government’ is elected by people belonging to one language community – Dutch, French and German-speaking – no matter where they live.

Short Answer Questions (SAQs):

1. Power sharing is the true spirit of democracy. Justify this statement.

Answer

Power sharing is the very spirit of democracy: 

• A democratic rule involves sharing power with those affected by its exercise, and who have to live with its effects. 

• People have a right to be consulted on how they are to be governed.  A legitimate government is one where citizens, through participation, acquire a stake in the system.

• It helps to reduce the possibility of conflict between social groups.

2. Describe any three forms of power sharing in modern democracies.

Answer

Three forms of power sharing with examples:

• Power is shared among different organs of government, such as the legislature, executive and judiciary.

• Power can be shared among governments at different levels – a general government for the entire country and governments at the provincial or regional level. 

• Power may also be shared among different social groups such as the religious and linguistic groups. ‘Community government’ in Belgium is a good example of this arrangement.

3. ‘Both Belgium and Sri Lanka are democracies but they follow different systems of power sharing’. Support the statement by giving three points of difference. 

Answer

• In Belgium, communities have equal share in the government while in Sri Lanka leaders of the Sinhala community sought to secure dominance over government by virtue of their majority.

• In Belgium, there is provision of special government called ‘Community government’ to look after cultural, educational and language-related issues while In Sri Lanka, none of the major political parties led by the Buddhist Sinhala leaders was sensitive to language and culture of Tamils.

• In Belgium, there is no discrimination between various religions while In Sri Lanka, Buddhism is the official religion of the country.

4. Write one prudential and one moral reason for power sharing.

Answer

• Prudential reason for power sharing: Power sharing reduces social conflicts among social groups with different interests and aspirations. Social conflicts lead to violence and instability. Power sharing ensures the stability of political order.

• Moral reason for power sharing: Power sharing is the very spirit and essence of democracy. Democracy involves sharing power with those affected by its exercise, and who have to live with its effects. People have the right to be consulted on how they are to be governed.

Long Answer Questions (LAQs):

1. What is majoritarianism? How has it increased the feelings of alienation among Sri Lankan Tamils? Explain with examples.

Answer 

‘Majoritarianism’ refers to the policy of domination over minority, just by the virtue of being in majority.

In Sri Lanka, it has led to the feelings of alienation among the Tamils. They feel that none of the major political parties, led by Buddhist Sinhala leaders are sensitive to their language and culture. According to them, the Constitution and the government policies have denied them equal political rights and discriminated against them in getting jobs and other opportunities and ignored their interests.

This alienation among the Tamils is due to the following reasons:

• In 1956, an Act was passed to recognise Sinhala as the only official language, thus disregarding Tamil.

• The government framed policies favouring Sinhala people for government jobs.

• The new Constitution specified that the state shall protect and foster Buddhism.



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