Ameerun

Chapter 4 Mixture

Mixture Introduction

Different types of problems in arithmetic can be solved by the application of the ratio and proportion methods. The mixture is one of them. The concept of ratio and proportion is directly applied to the problems with the mixture. So we may consider the present chapter as an exercise of our previous knowledge. In fact, in this chapter, our aim is to investigate several types of problems on a mixture and to obtain their solutions by the application of ratio and proportion. 

4.2 Mixture

First of all, let us try to find what a mixture means.

In our practical life sometimes we need to mix two or more articles of different values and in different quantities to produce an article of a different value and quality. This is known as a mixture.

Thus a mixture is defined as a combination of two or more articles or ingredients. Diagrammatically, a mixture can be represented as:

Ingredient #1 +  Ingredient # 2

+………..= Mixture

While analyzing mixture problems, the following cases may arise : 

1. In some cases we are to determine the value of a mixture when the quantity and value of each ingredient are known. 

2. Sometimes we are to find the ratio in which two or more ingredients are to be mixed to obtain a mixture of a given value. 

3. Also there are some cases where we are to increase or decrease the quantities of the ingredients of a mixture and to find the new ratio of the ingredients of the mixture.

4.3 Different types of mixture 

In your physical science book, you will study, in detail, the different types of mixtures. Here we shall mainly confine our attention to the following two types of mixture.

1. Mixture of solid in solid.

2. Mixture of liquid in liquid.

Read And Learn More WBBSE Solutions For Class 8 Maths

Chapter 4 Mixture Some problems with a mixture

Example 1

In a mixed fertilizer the ratio of urea to potash is 3: 7. If 8 kg of urea is further added to 40 kg of this type of fertilizer, then find the new ratio of urea to potash. 

Solution:

Given:

In a mixed fertilizer, the ratio of urea to potash is 3: 7

If 8 kg of urea is further added to 40 kg of this type of fertilizer

Let, in 40 kg of the fertilizer the quantity of urea be 3x kg and that of potash be 7x kg. 

∴ 3x+7x=40

or, 10x = 40

or, x = 40/10 = 4

∴ quantity of urea in the mixed fertilizer

3 x 4 kg = 12 kg and quantity of potash 

=7×4 kg 28 kg.

If 8 kg of urea is added to this mixture; then the new mixture will contain, (12 +8) kg or 20 kg of urea and 28 kg of potash. Hence, in the new mixture, the ratio of urea to potash = 20: 28 = 5:7. 

The new ratio of urea to potash is 5:7.

Example 2

The ratio of water and Dettol in 36 liters of Dettol-water is 5: 1. What volume of Dettol should be added to the mixture so that the ratio of water and Dettol becomes 3:1?

Solution:

Given:

The ratio of water and Dettol in 36 liters of Dettol water is 5: 1.

In 36 liters of Dettol water,

volume of water = 36 x 5/6 litres = 30 litres

volume of Dettol = 36 × 1/6 litres = 6 litres

Let, if x liters of Dettol be mixed with the solution; 

then the ratio of water and Dettol will be 3: 1.

∴ 30 / 6+x

= 3/1

or, 3x+18= 30 

or, 3x= 30-18

or, 3x= 12 

or, x = 12/3 

= 4

4 volume of  Dettol should be added to the mixture so that the ratio of water and Dettol becomes 3:1

Example 3

49 kg of blended tea contains Assam and Darjeeling tea in a ratio of 5: 2. What quantity of Darjeeling tea is to be added to the mixture to make the ratio of Assam to Darjeeling tea 2:1? 

Solution:

Given:

49 kg of blended tea contains Assam and Darjeeling tea in a ratio of 5: 2.

Let, in the 49 kg of blended tea, the quantity of Assam tea be 5x kg and that of Darjeeling tea be 2x kg.

∴ 5x+2x=49

or, 7x=49

or, x = 49/7 = 7

∴ Quantity of Assam tea = 5 x 7 kg = 35 kg 

Quantity of Darjeeling tea = 2 × 7 kg = 14 kg 

Let y kg of Darjeeling tea is to be added to the mixture so that the ratio of Assam tea. to Darjeeling tea becomes 2: 1.

∴ 35/14+y

= 2/1

Or. 2y + 28 = 35

2y= 35-28 = 7

or, y = 7/2

= 3.5

The required quality of Darjeeling tea is 3.5 kg.

Example 4

In a certain type of brass the ratio of copper and zinc is 5:2. What will be the ratio of copper and zinc in 28 kg of such brass if 4 kg of copper is added to it? 

Solution:

Given:

In a certain type of brass the ratio of copper and zinc is 5:2.

In 28 kg of brass

weight of copper = 28/7 × 5 kg 

= 20 kg

weight of zinc = 28/7 x 2 kg 

= 8 kg

If 4 kg of copper is mixed with this brass the ratio of copper and zinc becomes 

= 20+4/8

= 24/8

= 3:1

The new ratio of copper and zinc is 3:1.

Example 5

Two containers contain 16 liters and 20 liters of mixtures of milk and water. The ratios of water and milk in the two containers are 3: 1 and 4: 1 respectively. If the entire mixture of both containers is poured into a third container, what will be the new ratio of milk and water in that container?

Solution:

Given:

Two containers contain 16 liters and 20 liters of mixtures of milk and water.

The ratios of water and milk in the two containers are 3: 1 and 4: 1 respectively.

In the first container, the ratio of water and milk is 3:1 and the total quantity of mixture= is 16 liters.

∴ Quantity of water = 3/4 x 16 liters = 12 liters[ 3 parts out of 4 are water]

Quantity of milk = 1/4 x 16 litres = 4 liters

In the second container, the ratio of water and milk is 4: 1 and the total quantity of the mixture = 20 liters.

∴ Quality of water = 4/5 x 20 liters = 16 liters

Quantity of milk = 1/5 x 20 liters = 4 liters

Hence, the ratio of milk and water in the third container

= 4+4 / 12+16

= 8/28

= 2/7

= 2:7

The new ratio of milk and water is 2:7

Example 6

In a phenyl water solution of 60 liters the ratio of phenyl and water is 2:23. How much phenyl should be added to this solution so that the ratio of phenyl and water becomes 9:46?

Solution:

Given:

In a phenyl water solution of 60 liters the ratio of phenyl and water is 2:23.

In 60 liters of phenyl water solution,

volume of phenyle= 60/25 x 2 litres

= 24/5 liters

volume of water = 60/25 x 23 liters

= 276/5 liters

Let, if x liters of phenyle be mixed with the solution then the ratio of phenyle and

water will be 9:46.

∴ 24/5 + x / 276/5 = 9/46

or, 24+5x / 276 = 9/46

or, 230x + 1104

= 2484

or, 230x = 2484 – 1104

= 1380

or, x = 1380/230

= 6

The required quality of phenyle is 6 liters

Example 7

In an alloy of brass the ratio of copper to zinc is 7: 4. If 6 kg of zinc be added to 33 kg of such a quality of brass, then find the new ratio of copper to zinc.

Solution:

Given:

In an alloy of brass the ratio of copper to zinc is 7: 4.

In that alloy

weight of copper = (33 x 7/11)kg

= 21 kg

weight of zinc = (33 x 44/11)kg

= 12 kg

If 6 kg of zinc is added then the weight of zinc becomes (12+6)kg

= 18kg

:. New ratio of copper to zinc = 21/18

= 7/6

= 7:6.

The new ratio of copper to zinc is 7: 6.

Example 8

A mason has prepared a masonry mixture with sand and cement in a ratio of 7:1. But after the brick work done, it is seen that 72 kg of the mixture remains. He added more cement to this mixture and the ratio of sand and cement became 6: 1. What quantity of cement was mixed?

Solution :

Given:

A mason has prepared a masonry mixture with sand and cement in a ratio of 7:1.

But after the brick work done, it is seen that 72 kg of the mixture remains.

He added more cement to this mixture and the ratio of sand and cement became 6: 1.

In 72 kg of the mixture,

weight of sand = 72/8 x 7kg

= 63kg

weight of cement= 72/8 x 1 kg

= 9 kg

Let, if x kg of cement is mixed then the ratio of sand and cement will be 6:1.

∴ 63 / 9+x

= 6/1

or, 6x + 54

= 63

or, 6x = 63-54

= 9

or, x = 9/6

= 3/2

= 1.5

1.5 kg of comment was mixed.

Example 9

The ratio of the volumes of water and glycerine in 240 cc. of glycerine mixed with water is 1: 3. What additional volume of water and glycerine become 2: 3?

Solution:

Given:

The ratio of the volumes of water and glycerine in 240 cc. of glycerine mixed with water is 1: 3.

In that mixture

volume of water = (240 x 1/4)cc.

= 60cc.

volume of glycerine = (240 x 3/4)cc.

= 180cc.

Let, x cc. of water is to be added so that the ratio of water and glycerine becomes 2 : 3.

∴ 60+x / 180

= 2/3

or, 3x + 180 = 360

or, 3x = 360 – 180

or, 3x = 180

or, x = 180/3

= 60

The required quantity of water is 60cc.

Example 10

The ratio of copper, zinc, and nickel in a type of German silver is 4:3:2. How many 54 kg of zinc should be added to 54 kg of this type of German silver so that their ratio becomes 6:5:3?

Solution :

Given:

The ratio of copper, zinc, and nickel in a type of German silver is 4:3:2.

In 54 kg of German silver,

weight of copper = 54/9 x 4kg

= 24 kg

weight of zinc = 24/9 x 3 kg

= 18 kg

Let, x kg of zinc is to be added.

∴ 24/18+x = 6/5

or, 6x +108 = 120

or, 6x=120 – 108

= 12

or, x= 12/6

= 2

The required quantity of zinc is 2 kg.

Example 11

In a glass of beverage, syrup, and water are in the ratio 3: 1. What fraction of the quantity of beverage should be replaced by water to make the ratio 1: 1?

Solution :

Given

In a glass of beverage, syrup, and water are in the ratio 3: 1.

In that beverage

the syrup is 3/4 part and the water is 1/4 part.

In order to make the ratio 1: 1 syrup will be 1/2 part and water will be 1/2 part

.. syrup to be withdrawn = (3/4 – 1/2) part

= 1/4 part

Now,

3 parts of syrup are in 4 parts of the beverage

1 part of syrup is in 4/3 parts of the beverage

1/4 part of syrup is in 4/3 x 1/4 parts of the beverage

= 1/3 parts of the beverage

1/3 part of the beverage should be replaced.

Example 12

In two different kinds of washing powder the ratios of soda and soap powder are 2 : 3 and 4: 5. Find the part of soap powder in the new washing powder which is prepared by mixing 10 kg of first washing powder with 18 kg of second washing powder.

Solution: 

Given:

In two different kinds of washing powder the ratios of soda and soap powder are 2 : 3 and 4: 5.

In 10 kg of first washing powder,

weight of soda = 10 /5 x 2 kg = 4 kg

weight of soap powder = 10/5 x 3 kg

= 6kg

In 18 kg of second washing powder,

weight of soda = 18/9 x 4 kg

= 8 kg

weight of soap powder = 18/9 x 5 kg

= 10 kg

∴ Total weight of washing powder

= (10 + 18) kg = 28 kg

Total weight of soap powder

= (6 + 10) kg = 16 kg

∴ In the new washing powder, the part of soap powder =16/28

= 4/7

There is 4/7 part of soap powder in the new washing powder.

Example 13

Syrup and water are in the ratio 3: 2 and 45 respectively in the two vessels. What the volume of the second mixture is to be mixed with 3 liters of first mixture so that the volume of syrup and water in the new mixture becomes equal?

Solution:

Given:

Syrup and water are in the ratio 3: 2 and 45 respectively in the two vessels.

Let, x liters of the second mixture is to be

mixed with 3 liters of the first mixture.

In the 3 liters of the first mixture

the volume of syrup = 3 x 3/5 liters

= 9/5 liters

the volume of water = 3 x 2/5 liters

= 6/5 liters

In the x liters of the second mixture

volume of syrup = x x 4/9 liters

= 4x/9 liters

volume of water = x x 5/9 liters

=5x/9 liters

∴ According to the problem,

4x/9 + 9/5 =  5x/9 + 6/5

or, 5x/9 – 4x/9 = 9/5 – 6/5

or, x/9 = 3/5

or, x = 27/5

= 5 2/5

The required volume of 2nd mixture is 5 2/5

Example 14

3 similar glasses of equal size are filled with beverages. The ratios of water and syrup in these three glasses are 3:1, 5:3, and 9:7 respectively. The beverage of three glasses is poured into a big vessel. What is the ratio of water and syrup in the new vessel?

Solution:

Given:

3 similar glasses of equal size are filled with beverages.

The ratios of water and syrup in these three glasses are 3:1, 5:3, and 9:7 respectively.

The beverage of three glasses is poured into a big vessel.

In the first glass, water is 3/4 part and syrup is 1/4 part

In the second glass, water is 5/8 part and syrup is 3/8 part

In the third glass, water is 9/16 part and syrup is 7/16 part

∴ Total water

= (3/4 +5/8 +9/16) part

= 12+10+9 / 16 part

= 31/16 part.

Total syrup

= (1/4 + 3/8 + 7/16) part

= 4+6+7 / 16 part

= 17/16 part

∴ water: syrup = 31/16:17/16

= 31:17

The ratio of water and syrup in the new vessel is 31:17.

Example 15

In a mixture, the ratio of the first and second liquid is 2: 3 and that in another mixture is 5: 4. Find the ratio in which these two mixtures should be mixed so that in the new mixture the two liquids be in equal volume.

Solution:

Given:

In a mixture, the ratio of the first and second liquid is 2: 3 and that in another mixture is 5: 4.

Let, x liters of the first mixture be mixed

with y liters of the second mixture so that

in the new mixture, the two liquids be in equal volume.

In x liters of the first mixture

the volume of the first liquid = x x 2/5 liter

= 2x/5 liter

In y liters of the second mixture

the volume of the first liquid = y x 4/9 liter

= 4y/9 liter

According to the question,

2x/5 + 5y/9 = 3x/5 + 4y/9

or, 5y/9 – 4y/9

= 3x/5 – 2x/5

or, y/9=x/5

or, x/y = 5/9

or, x:y = 5:9

The required ratio is 5:9.

Example 16

In a vessel of beverage, the ratio of syrup and water is 5:2. What part of the beverage should be removed and replaced by water so that the volume of syrup and water becomes equal?

Solution :

Given:

In a vessel of beverage, the ratio of syrup and water is 5:2.

In that beverage, the syrup is 5/7 part and the water is 2/7 part.

In order to make the quality of syrup and water equal, the syrup will be 1/2 part and the water will be 1/2 part.

.. Syrup to be withdrawn = (5/7 – 1/2) part

= 10-7 / 14 part

= 3/14 part

Now, 5 parts of syrup are in 7 parts of the beverage

1 part of syrup is in 7/5 parts of the beverage

3/14 parts of syrup is in 7/5 x 3/14 parts of beverage = 3/10 parts of beverage

3/10 part of the beverage should be removed.

Example 17

The ratio of water and syrup in three glasses of equal volume is 5: 1, 5:3, and 5: 7 respectively. If the three mixtures be poured in a big vessel find the ratio of syrup and water in that vessel.

Solution :

Given:

The ratio of water and syrup in three glasses of equal volume is 5: 1, 5:3, and 5: 7 respectively.

Let, the volume of each glass of equal

volume be x c.c.

In the first glass, the volume of water is 5x/6c.c.

and that of syrup is x/6c.c.

In the second glass, the volume of water is 5x/8 c.c and that of syrup is 7x/12 c.c.

Hence, in the big vessel, the ratio of syrup and water

= \(\frac{\frac{x}{6}+\frac{3 x}{8}+\frac{7 x}{12}}{\frac{5 x}{6}+\frac{5 x}{8}+\frac{5 x}{12}}=\frac{\frac{4 x+9 x+14 x}{24}}{\frac{20 x+15 x+10 x}{24}}\)

= \(\frac{27 x}{24} \times \frac{24}{45 x}=\frac{27}{45}=\frac{3}{5}=3: 5\)

The ratio of syrup and water is 3:5.

Example 18

There are three kinds of liquid in a beverage of 700 liters. The ratio of the first and second liquids is 2: 3 and the ratio of the second and the third liquid is 4: 5. What quantity of the first and the second liquid be mixed with the third beverage so that the ratio of the three liquids will be 6:5:3?

Solution:

Given:

There are three kinds of liquid in a beverage of 700 liters.

The ratio of the first and second liquids is 2: 3 and the ratio of the second and the third liquid is 4: 5.

In 700 liters of beverage,

the first liquid second liquid

= 2:3= (2x 4): (3 x 4) = 8:12

the second liquid third liquid

= 4:5= (4 x 3): (5 x 3) = 12:15

∴ First liquid: Second liquid: Third liquid

= 8:12:15

∴ The volume of the first liquid = 700/35 x 8 liters

= 160 liters

Volume of second liquid = 700/35 x 12 liters

= 240 liters

Volume of third liquid = 700/35 x 115 liters

300 liters

Let, x liters of the first liquid is to be mixed.

∴ 160 + x / 300

= 6/3

= 2

or, x + 160 = 600.

or, x = 600 – 160

= 440

Again. let y liters of the second liquid is to be mixed.

∴ 240 + y / 300 = 5/3

or, 3y + 720 = 1500

or, 3y + 720 = 1500

or, 3y = 1500 – 720

= 780

or, y = 780/3

= 260

The first liquid is 440 liters and the second liquid is 260 liters.

Example 19

In two types of mixed tea, the ratios of Darjeeling and Assam tea are 27 and 1:5 respectively. Find the ratio in which these two types of tea should be mixed so that the new ratio of Darjeeling and Assam tea will be 1: 4.

Solution:

Given:

In two types of mixed tea, the ratios of Darjeeling and Assam tea are 27 and 1:5 respectively.

Let, x kg of the first type of tea should be mixed

with y kg of the second type of tea, so that the In the first bottle, phenyle is the ratio of Darjeeling and Assam tea in the

mixed tea will be 1: 4.

Now, in x kg of the first type of tea

weight of Darjeeling tea = x x 2/9 kg

= 2x/9kg

and weight of Assam tea = x x 7/9 kg

= 7x/9 kg

In y kg of second type tea

weight of  Darjeeling tea = y x 1/6 kg

= y/6 kg

and weight of Assam tea = y x 5/6 kg

= 5y/6kg

According to the question,

\(\frac{\frac{2 x}{9}+\frac{y}{6}}{\frac{7 x}{9}+\frac{5 y}{6}}=\frac{1}{4} \quad \text { or, } \frac{\frac{4 x+3 y}{18}}{\frac{14 x+15 y}{18}}=\frac{1}{4}\)

 

or, 4x + 3y / 14x + 15y = 1/4

or, 16x + 12y = 14x + 15y

or, 16x – 14x = 15y – 12y

or, 2x = 3y

or, x:y = 3:2

The required ratio = 3:2.

Example 20

The ratio of the volumes of the three bottles is 5:3:2. These three bottles are filled with a solution of phenyle and water. The ratio of phenyle and water in the three bottles is 2:3, 1:2, and 1:3 respectively.2/3 part of the first bottle, 1/2 part of the second bottle, and 2/3 part of the third bottle are mixed together. Find the ratio of phenyle and water in the new mixture.

Solution:

Given:

The ratio of the volumes of the three bottles is 5:3:2.

These three bottles are filled with a solution of phenyle and water.

The ratio of phenyle and water in the three bottles is 2:3, 1:2, and 1:3 respectively.

2/3 part of the first bottle, 1/2 part of the second bottle, and 2/3 part of the third bottle are mixed together.

In the first bottle, phenyle is 2/5 part and water is 3/5 part.

In the second bottle, phenyle is 1/3 part and water is 2/3 part.

In the third bottle, phenyle is 1/4 part and water is 3/4 part.

Since, the ratio of the volumes of the three bottles is 5:3:2 and 1/3 part of the first bottle, 1/2 part of the second bottle and 2/3 part of the third bottle are mixed together, therefore,

total phenyle

\(=\left(\frac{5}{10} \times \frac{1}{3} \times \frac{2}{5}+\frac{3}{10} \times \frac{1}{2} \times \frac{1}{3}+\frac{2}{10} \times \frac{2}{3} \times \frac{1}{4}\right) \text { part }\) \(=\left(\frac{1}{15}+\frac{1}{20}+\frac{1}{30}\right) \text { part }\) \(=\frac{4+3+2}{60} \text { part }=\frac{9}{60} \text { part }\) \(=\left(\frac{5}{10} \times \frac{1}{3} \times \frac{3}{5}+\frac{3}{10} \times \frac{1}{2} \times \frac{2}{3}+\frac{2}{10} \times \frac{2}{3} \times \frac{3}{4}\right) \text { part }\) \(=\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\right) \text { part }=\frac{3}{10} \text { part }\) \(=\frac{9}{60}: \frac{3}{10}=\frac{9}{60} \times \frac{10}{3}=\frac{1}{2}=1: 2\)

The ratio of phenyle and water is 1:2.

Example 21

In two types of brass, the ratios of copper to zinc are 8: 3 and 15: 7 respectively. If the two types of brass be mixed in the ratio 5: 2 a new type of brass is obtained. Find the ratio of copper to zinc in this new type of brass.

Solution:

Given

In two types of brass, the ratios of copper to zinc are 8: 3 and 15: 7 respectively.

If the two types of brass be mixed in the ratio 5: 2 a new type of brass is obtained.

Let, 5x kg of the first type of brass be mixed

with 2x kg of the second type of brass. In 5x kg of the first type of brass

weight of copper = (5x x 8/11)kg

= 40x/11 kg

and weight of zinc = (5x x 3/11)kg

= 15x/11 kg

In 2x kg of the second type of brass

weight of copper = (2x x 15/22)kg

= 15x/11 kg

and weight of zinc = (2x x 7/22) kg

7x/11 kg

In the new type of brass, the ratio of copper to zinc

\(=\frac{\frac{40 x}{11}+\frac{15 x}{11}}{\frac{15 x}{11}+\frac{7 x}{11}}=\frac{\frac{55 x}{11}}{\frac{22 x}{11}}=\frac{55 x}{11} \times \frac{11}{22 x}=5: 2\)

The ratio of copper to zinc is 5:2

Example 22

In two types of stainless steel, the ratios of chromium and steel are 2 11 and 5: 21 respectively. In what ratio should the two types of stainless steel be mixed so that the ratio of chromium to steel in the mixed type is 7:32? 

Solution:

Given:

In two types of stainless steel, the ratios of chromium and steel are 2: 11 and 5: 21 respectively.

Let, x kg of the first type of stainless steel be mixed with y kg of the second type. In the x kg of the first type of stainless steel-

weight of chromium = x x2/12 kg

= 2x/13kg

weight of steel = x x11/13 kg

= 11x / 13 kg

In the y kg of the second type of stainless steel 

weight of chromium = y x 5/26 kg

= 5y/26 kg

weight of steel = y x 21/26 kg

= 21y/26 kg

According to the question

\(\frac{\frac{2 x}{13}+\frac{5 y}{26}}{\frac{11 x}{13}+\frac{21 y}{26}}=\frac{7}{32} \quad \text { or, } \frac{\frac{4 x+5 y}{26}}{\frac{22 x+21 y}{26}}=\frac{7}{32}\)

or, \(\frac{4 x+5 y}{22 x+21 y}=\frac{7}{32}\)

or, 128x + 160y = 154x + 147y

or, 154x – 128x = 160y – 147y

or, 26x = 13y

or, \(\frac{x}{y}=\frac{13}{26} \quad \text { or, } x: y=1: 2\)

The ratio is 1:2

Example 23

1/3rd and 1/4th parts of two vessels of equal volume are filled with fruit juice. Water is now added to the vessels to full capacity and they are mixed in a third vessel. What is the ratio of fruit juice to water in the third vessel?

Solution:

Given

1/3rd and 1/4th parts of two vessels of equal volume are filled with fruit juice. Water is now added to the vessels to full capacity and they are mixed in a third vessel.

In the first vessel

1/3rd part is filled with fruit juice and 

therefore (1-1/3)

or, 2/3rd part is filled with water.

In the second vessel –

1/4th past is filled with fruit juice and 

Therefore (1-1/4)

or, the 3/4th part is filled with water.

In the third vessel the ratio of fruit juice to water

\(=\frac{\frac{1}{3}+\frac{1}{4}}{\frac{2}{3}+\frac{3}{4}}=\frac{\frac{4+3}{12}}{\frac{8+9}{12}}=\frac{7}{12} \times \frac{12}{17}=7: 17\)

The ratio of fruit juice to water is 7: 17.

Example 24

Pure milk contains 89% water. If in a sample of milk, there is 90% water; then find the quantity of water mixed in 22 liters of such milk. 

Solution:

Given:

Let, x liters of water have been mixed. Now, in this 22 liters of milk mixed with 90

water, the quantity of water =22 x 90/100

litres = 99/5 litres.

Now, there were (22-x) liters of pure milk before water was mixed.

The quantity of water in this pure milk of (22-x) liters is, (22-x) x89/100 liters = (979/50 – 89x/100) liters.

So, 979/50 – 89x/100 + x = 99/5

or, x – 89x/100 = 99/5 – 979/50

or, 11x/100 = 11/50

or, x = 11/50 x 100/11

= 2

The required quantity of water is 2 liters.

Example 25

The ratio of land-region and water-region of the earth is 1: 2. If this ratio be 2: 3 in the Northern Hemisphere find it in the Southern Hemisphere.

Solution:

Given:

The ratio of land-region and water-region of the earth is 1: 2.

If this ratio be 2: 3 in the Northern Hemisphere.

1/3rd of the earth is land-region

and 2/3rd of the earth is water-region 

Land-region in the Northern Hemisphere

= 2/5th of the Northern Hemisphere

= (1/2 x 2/5) th of the earth = 1/5th of the earth water-region in the Northern Hemisphere

= 3/5th of the Northern Hemisphere

= (1/2 x 3/5)th of the earth = 3/10th of the earth Land-region in the Southern Hemisphere

=(1/3-1/5)th of the earth = 2/15th of the earth

= 4/15th of the Southern Hemisphere water region in the Southern Hemisphere

=(2/3 – 3/10)th of the earth = 11/30th of the earth = 11/15th of the Southern Hemisphere.

∴ In the Southern Hemisphere the ratio of land-region and water-region

= 4/15:11/15 = 4:11

The ratio is 4:11.

Chapter 3 Rule Of Three

Introduction

There are some problems in arithmetic that may be solved by more than one method. For example, the problems which can be solved by the unitary method can also be solved by applying the idea of proportion. So in this chapter, our aim is to solve those problems which we have solved earlier by using the unitary method. This new technique of solving arithmetical problems on the basis of the idea of proportion is the ‘Rule of Three’.

The Rule of Three

You know that when three terms of a proportion are known we can find the fourth term. Since the fourth term of a proportion can be identified when any three terms of the said proportion are given, this method is known as the ‘Rule of Three’. In fact, this method is nothing but finding a fourth proportional to the three given quantities.

When 5, 7, or more odd numbers of quantities in any proportion are given and we are to find from them the 6th, 8th, etc. quantity then this method is known as the multiple rules of three.

Read And Learn More WBBSE Solutions For Class 8 Maths

How to solve a problem by the Rule of Three

1. The numerical value of the required quantity is taken as x, and it is placed in the fourth position of the proportion. Let us take a problem. If 5 bottles of medicines cost 1000, then how much would 8 bottles cost? In this problem, the cost of 8 bottles is the required

quantity to be found out. Hence the numerical value of the required quantity (i.e. the cost of 8 bottles) is to be taken as x and it is to be placed in the fourth position of the proportion. 

2. The given similar quantity is placed in the third position of the proportion. In the above problem, a similar quantity as x is the given cost of 5 bottles. Hence, 1000 is to be placed in the third position of the proportion.

3. We have to determine from the subject- a matter of the problem whether the required quantity will be less than or greater than the given similar quantity. These two cases are better understood by the concept of variation.

1. Direct variation: In a direct variation, the increase in one quantity (the number) causes an increase in the other quantity (the cost), and a decrease in one quantity causes a decrease in the other quantity.

Our problem with the cost of bottles of medicines is of this type since the more the number of bottles, the more the cost, and the less the number of bottles, the less the cost. The required quantity of cost of 8 bottles is more than the cost of 5 bottles. 

2. Inverse variation In an inverse variation the increase in one quantity causes a decrease in the other quantity and a decrease in one quantity causes an increase in the other quantity. Let us take another example. If 8 men can do a piece of work in 12 days then how long will 16 men take to do the same work?

Let the required time be x days. In this problem, since the number of men increases, therefore, the number of days would decrease. Hence, the value of x will be less than 12 days. 

4. When x is smaller than the quantity placed in the third position then between the other two quantities the greater one is placed in the first position and the smaller one is placed in the second position of the proportion.

5. When x is greater than the quantity placed in the third position then between the other two quantities the smaller one is placed in the first position and the greater one is placed in the second position of the proportion.

Some problems with the Rule of  Three

Example 1

12 men can do a piece of work in 15 days. How long will 10 men take to do the same work?

Solution :

Given:

12 men can do a piece of work in 15 days.

Let, the required time be x days. Then the data can be arranged as,

Number of men                   Number of pens

12                                               8

10                                             12

Here, since the number of men decreases, therefore the number of days will increase. Hence, the value of x will be greater than 15

∴ 10/12 = 15/x

or, 10x = 15 x 13

or, x = 15 x 12 / 10

= 18

The required time is 18 days

18 days will 10 men take to do the same work

Example 2

A man bought 15 kg of rice for 390. How much he would require if he would have bought 17 kg of rice?

Solution:

Given:

A man bought 15 kg of rice for 390.

Let, the required price be ₹ x. Then the data can be arranged as

Quantity of rice (in kg)        Expenditure (in ₹)

    15                                             360                                                                                   

     17                                              x

In this case, the quantity of rice has increased, therefore expenditure will also increase. Hence, the value of x will be greater than 390.

∴ 15/17 = 390/x

or, x = 390 x 17 / 17

= 442

The required expenditure is ₹ 442

Example 3

The price of 8 pens is 120. Find the price of 12 such pens. 

Solution :

Given:

The price of 8 pens is 120.

 Let, the required expenditure be ₹ x. Then the data can be arranged as,

Number of pens                       Price(in ₹)

8                                                     120

12                                                       x

Here, since the number of pens increases, therefore the price will also increase.

Hence, the value of x will be greater than 1120

∴ 8/12 = 120/x

or, 8x = 12 x 120

or, x = 12 x 120 / 8

= 180

The required price is ₹ 180.

Example 4

A man will make 4 shirts of the same size with 20 meters of cloth. How many meters of cloth he will have to purchase for making 12 such shirts?

Solution:

Given:

A Man Will Make 4 Shirts Of The Same Size With 20 Meters Of Cloth.

Let, he will have to purchase x meters of cloth. Then the data can be arranged as, 

No. of shirts                   Cloth (in meters)

4                                             20

12                                            X

Since the number of shirts has increased, therefore the quantity of cloth will also increase. Hence, the value of x will be greater than 20.
∴ 4/12 = 20/x

or, x =  20 x 12 / 4

= 60

He will have to purchase 60 meters of clothes.

Example 5

If the price of 7 meters of cloth is 126, then find how long cloth can be obtained by ₹ 81.

Solution:

Given:

If the price of 7 meters of cloth is 126.

Let, the required length of the cloth be x meters. Then the data can be arranged as,

Price (in ₹)                             Length of cloth (in meters)

126                                                 7

81                                                   x

Here, since the price decreases, therefore the length of the cloth will also decrease.

Hence, the value of x will be less than 7.

∴ 126/81 = 7/x

or, x x 126 = 81 x 7

or, x = 81 x 7 / 126

= 4.5

4.5 meters long cloth can be obtained.

Example 6

It took 15 days for 30 laborers to dig a pond. In how many days 25 labourers could have completed the said work?

Solution:

Given:

It took 15 days for 30 laborers to dig a pond

Let, the required time be x days. Then the data can be arranged as,

No. of laborers No. of days

30                                           15

25                                            x

Here, since the no. of laborers has decreased therefore no. of days will increase. Hence, the value of x will be greater than 15.

∴ 25/30 = 15/x

or, x = 30 x 15 / 25

= 18

The required time is 18 days.

Example 7

12 men can do a piece of work in 30 days. How many more men will be required to do it in 20 days?

Solution: 

Given:

12 men can do a piece of work in 30 days.

Let, x men will do the work in 20 days.

Number of days                        Number of men

30                                                      12

20                                                        x

Here, since the number of days decreases, therefore the number of men will increase.

Hence, the value of x will be greater than 12.

∴ 20/30 = 12/x

or, 20x = 30 x 12

or, x = 30 x 12 / 20

= 18

Since 12 men were already there, si (18-12) or, 6 more men will be required.

6 more men will be required.

Example 8

A man reached a certain place in 5 hours by driving the car at a speed of 40 km/hr. How long it would have taken to reach the same place if he would drive the car at 50 km/hr?

Solution:

Let, the required time be x hours. Then the data can be arranged as,

Speed of the car (km/hr)                                 Time (hour)

40                                                                           5

50                                                                          x

Since the speed of the car has increased, therefore, less time will be required.

Hence, the value of x will be less than 5.

∴ 50/40 = 5/x

or, x = 5 x 40 / 50

= 4

The Required time is 4 hours.

Example 9

2 men or 3 boys can finish a work in 48 days. In how many days will 4 men and 6 boys finish the work?

Solution:

 Work of 2 men = Work of 3 boys

∴ Work of 1 men = work of 3/2 boys

Work of 4 men = work of 3/2 x 4 boys

= work of 6 boys

∴ Work of 4 men and 6 boys (6+6) boys = Work of 12 boys. 

Now, the problem is, if 3 boys can finish a work in 48 days, then in how many days will 12 boys finish the work?

Let, the required number of days be x.

Number of boys                           Number of days 

3                                                            48

12                                                            x

Here, since the number of boys increases, therefore the number of days will decrease.

Hence, the value of x will be less than 48.

∴ 12/3 = 48/x

or, 12 x x = 48 x 3

or, x = 48 x 3 /12

= 12

The required time is 12 days.

Example 10

There was a stock of food for 9 days to cater for the needs of 4000 people at a shelter camp. After 3 days 1000 people left the camp for another place. In how many days the rest of the people will consume the remaining food?

Solution:

After 3 days there was food for 6 days and the number of people reduced to 3000. Let, the required no. of days be x. Then the data can be arranged as,

No. of people                      No. of days 6

4000                                              6

3000                                              x

Since the number of people has reduced therefore the number of days will increase.

Hence, the value of x will be greater than 6.

∴ 3000/4000 = 6/x

or, x = 6 x 4000 / 3000

= 8

The required no .of days = is 8 days.

Example 11

In a camp of 4000 men, there was a stock of food for 190 days. After 30 days 800 men left the camp. How many days will the rest of the food last?

Solution :

After 30 days, the number of days left = (190-30)

= 160 

and present number of men = (4000-800)

= 3200.

Now the problem is if there is a stock of food for 4000 men for 160 days, then how many days will it last for 3200 men.

Let the number of days be x. Then,

Number of men                          Number of days

4000                                                 160

3200                                                   x

Here, since the number of men decreases, therefore the number of days will increase.

Hence, the value of x will be greater than 160.

∴ 3200/4000 = 160/x

or, x x 3200 = 160 x 4000

or, x = 160 x 4000/3200

=200

The rest of the food will last for 200

Example 12

42 laborers of a farm can cultivate the entire land of the farm in 24 days. But suddenly 6 laborers became ill. How many days it will take to cultivate the entire land of the farm by the rest of the laborers?

Solution:

Due to the illness of 6 laborers the number of laborers became 42 – 6 = 36. Let, the required time be x days. Then the data can be arranged as,

No. of laborers No. of days

42                                                              24

36                                                               x

Since the number of laborers has decreased, therefore the number of days will increase. Hence, the value of x will be greater than 24.

∴ 36/42 = 24/x

or, x = 42 x 24 / 36

= 28

The required time is 28 days

Example 13

15 men can earn ₹ 1200 in 30 days. How much will 75 men earn in 5 days?

Solution:

Given:

15 men can earn ₹ 1200 in 30 days

Let the required money be ₹ x. Then the problem is

Number of men                Number of days          Earned money

15                                            30                                ₹ 1200

75                                             5                                ₹ x

Since the earned money increases with the increase in the number of men and decreases with the decrease in the number of days.

∴ \(\left.\begin{array}{l}
15: 75 \\
30: 5
\end{array}\right\}:: 1200: x\)

or, x = 75 x 5 x 1200 / 15 x 30

= 1000

They will earn ₹ 1000.

Example 14

It takes 27 days to make 1000 spare parts by 16 numbers of machines. How many days will it take to make the same number of spare parts if additional two machines are installed? 

Solution:

Given:

27 days to make 1000 spare parts by 16 numbers of machines.

Since two machines are installed the number of machines becomes 18.

Let, the required time be x days. Then the data can be arranged as,

No. of machines                            Time (in days)

16                                                          27

18                                                            x

Since the number of machines has increased therefore less time will be required.

Hence the value of x will be less than 27.

∴ 18/16 = 27/x

or, x = 27 x 16 / 18

= 24

The required time is 24 days.

Example 15

A contractor undertook a 12 km long road construction job scheduled to be completed in 350 days. After employing 45 men for 200 days, he found that only 4½ km of road was completed. How many additional men must be engaged to finish the work in time?

Solution:

Given:

A contractor undertook a 12 km long road construction job scheduled to be completed in 350 days

After employing 45 men for 200 days, he found that only 4½ km of road was completed.

After 200 days, the number of days left = (350-200) = 150.

The length of the road remained unfinished = (12 – 9/2)km

= 15/2 km.

Now let us calculate, if 45 men can finish 9/2 km of road in 200 days then how many men will finish 15/2 km of road in 150

Number of days              Length of road                   Number of men

200                                       9/2km                                    45

150                                       15/2km                                  x

Since, the number of men increases with the increase in the length of the road, the decrease in the number of days and it increases in the length of the road,

\(\left.\begin{array}{c}
150: 200 \\
\frac{9}{2}: \frac{15}{2}
\end{array}\right\}:: 45: x\)

or, x = 200 x 445 x 15/2 / 150 x 9/2

= 200 x 45 x 15 / 150 x 9

= 100

∴ Additional men required = (100-45)

= 55

55 more men are required.

 Example 16

A pump of 5 H.P. can lift 36000 liters of water in 8 hours. How long it will take to lift 63000 liters of water by a 7 H.P. pump? 

Solution:

Given:

A pump of 5 H.P. can lift 36000 liters of water in 8 hours.

Let, the required time be x hours. Then the data can be arranged as,

H.P. of pump           Volume of water                 Time

5                                   36000 liters                    8 hours

7                                   63000 litres                   x hours

Since, if the H.P. of the pump increases, then the time decreases, and if the volume of water increases the time increases, therefore,

\(\left.\begin{array}{c}
7: 5 \\
36000: 63000
\end{array}\right\}:: 8: x\)

or, x = 5 x 63000 x 8 / 7 x 36000

= 10

The required time is 10 hours.

Example 17

There are two motors of 5 H.P. and 3 H.P. in a factory. The 5 H.P. motor requires 20 units of electricity in 8 hours. How many units will be required if 3 H.P. motor works for 10 hours?

Solution :

Given:

There are two motors of 5 H.P. and 3 H.P. in a factory.

The 5 H.P. motor requires 20 units of electricity in 8 hours.

Let, x units of electricity will be required. Then the data can be arranged as, 

H.P. of motor                Time                      Electricity required

5                                   8 hours                       20 units

3                                   10 hours                     x units

Since if the H.P. of the motor decreases less electricity is required and if time is increased more electricity is required, therefore,

\(\left.\begin{array}{l}
5: 3 \\
8: 10
\end{array}\right\}:: 20: x\)

or, x = 20 x 3 x 10 / 5 x 8

= 15

15 units will be required.

Example 18

In a loom, 14 weavers can weave 210 sarees in 12 days. How many extra weavers have to be employed to weave 300 sarees in 10 days?

Solution:

Given:

In a loom, 14 weavers can weave 210 sarees in 12 days

Let x number of weavers can weave 300 sarees in 10 days. Then the data can be

No. of days                        No. of sarees                        No. of weavers

12                                             210                                      14

10                                             300                                        x

Since the quantity of work done increases if the number of people increases and the number of sarees increases if the number of weavers increases, therefore,

\(\left.\begin{array}{l}
10: 12 \\
210: 300
\end{array}\right\}:: 14: x\)

or, x = 14 x 12 x 300 / 10 x 210

= 24

∴ Extra weavers have to be employed

= (24-14)

= 10

10 extra weavers have to be employed.

Example 19

A company has got the work of unloading goods from a ship in 10 days. 280 people have been employed for this purpose. After 3 days it is seen that 1/4th of the work has been completed. How many extra people are to be engaged to complete the work in time? 

Solution:

Given:

A company has got the work of unloading goods from a ship in 10 days.

280 people have been employed for this purpose. After 3 days it is seen that 1/4th of the work has been completed.

Work left = (1-1/4) = 3/4th part and number of days left = (10-3) days = 7 days. 

Let, x people are to be engaged to complete the work in time. Then the data can be arranged as,

Quantity of work                  No. of days                  No. of people

1/4                                               3                                    280

3/4                                               7                                      x

Since the number of days decreases if the number of weavers increases and the number of days increases if the number of people decreases,

\(\left.\begin{array}{c}
\frac{1}{4}: \frac{3}{4} \\
7: 3
\end{array}\right\}:: 280: x\)

or, x = 3/4 x 3 x 280 / 1/4x 7

= 360

∴ Extra people will be required

= (360 – 280)

= 80

80 Extra people will be required.

Example 20

A power-loom is 2 1/4  times more powerful than a hand-loom. 12 hand-looms weave 1080 meters in length of cloth in 18 days. How many power looms will be required to weave 2700 meters length of cloth in 15 days? 

Solution:

Given:

A power-loom is 2 1/4  times more powerful than a hand-loom. 12 hand-looms weave 1080 meters in length of cloth in 18 days.

Let, the number of handlooms is x. 

Then the data can be arranged as,

Length of cloth                           No.of days                       No.of hand-looms

1080 meters                                  18                                           12

2700 meters                                   15                                           x

Since, the length of cloth increases if the number of looms increases and the number of days decreases if the number of looms increases, therefore,

\(\begin{aligned}
& 1080: 2700 \\
& 15: 18 \quad\}:: 12: x
\end{aligned}\)

or, x = 2700 x 18 x 12 / 1080 x 15

= 36

9/4 hand-looms = 1 power-loom

1 hand-loom = 4/9 power-loom

36 hand-looms = 4/9 x 36 power-looms

= 16 power-looms

Required no of power looms is 16.

Example 21

2400 bighas of land of a cooperative society can be cultivated by 25 farmers in 36 days. It was seen that half of the land of the society could be cultivated in 30 days after purchasing a tractor. Find the power of the tractor equivalent to the number of farmers.

Solution:

Given:

2400 bighas of land of a cooperative society can be cultivated by 25 farmers in 36 days

It was seen that half of the land of the society could be cultivated in 30 days after purchasing a tractor

Let, the power of the tractor is equivalent to x number of farmers. The data can be arranged as,

Quality of land                      No.of days                         No.of farmers

2400 bighas                              36                                        25

1200 bighas                               30                                         x

Since, if the quantity of land increases the number of farmers increases and if the number of days decreases, then the number of farmers increases, therefore,

\(\left.\begin{array}{l}
2400: 1200 \\
30: 36
\end{array}\right\}:: 25: x,\)

or, x = 1200 x 36 x 25 / 2400 x 30

= 15

1 tractor is equivalent to 15 farmers.

Example 22

A ship takes 25 days to sail from Kolkata to Cochin. The ship started with 36 sailors for each of which 850 gm of food was allotted per day. But the ship rescued 15 sailors from another sinking ship after 13 days of journey and the ship reached Cochin in 10 days increasing the speed. What will be the quantity of food each sailor required to reach Cochin safely and the entire storage of food would be consumed by this time? 

Solution:

Given:

If the 15 sailors are rescued then the number of sailors becomes (36 +15)= 51. 

If 13 days go out of 25 days the number of days remaining = (25-13) = 12.

Now, the food which lasts 12 days for 36

sailors will have to last 10 days for 51 sailors. 

The data can be arranged as,

No. of sailors                       No. of days                 Food allotted per head per day

36                                           12                                        850 gm

51                                            10                                        x gm(say)

If the number of sailors increases the food allotted per head per day decreases and if the number of days decreases the food allotted per head per day increases, therefore,

\(\left.\begin{array}{l}
51: 36 \\
10: 12
\end{array}\right\}:: 850: x\)

 

or, x = 850 × 36 × 12 / 51 x 10

= 720

The required quantity of food is 720 gm.

Example 23

36 people of a certain village can construct 120 meters of road in 8 days by working 6 hours daily. Another 6 people were involved in this work and the duration of work each day was increased by 2 hours. Find the length of the road to be constructed now in 9 days. 

Solution:

Given:

36 people of a certain village can construct 120 meters of road in 8 days by working 6 hours daily.

Another 6 people were involved in this work and the duration of work each day was increased by 2 hours.

Let x meters of the road may be constructed. 

The data may be arranged as,

No.of people                    Daily time                 No.of days                    Length of road

36                                    6 hr                             8                                 120 meters

42                                    8hr                              9                                   x meters

If the number of people increases, daily time increases, and if the number of days increases, then the length of the road increases, therefore,

\(\left.\begin{array}{rl}
36 & : 42 \\
6 & : 8 \\
8 & : 9
\end{array}\right\}:: 120: x\)

or, x = 42 x 8 x 9 x 120 / 36 x 6 x 8

= 120

The length of the road is 210 meters.

Example 24

250 people can excavate a pond of size 50 meters long, 35 meters wide, and 5.2 meters deep in 18 days by working 10 hours daily. How many days it will take for 300 people to excavate a pond of size 65 meters long, 40 meters wide, and 5.6 meters deep by working 8 hours daily?

Solution:

Given:

250 people can excavate a pond of size 50 meters long, 35 meters wide, and 5.2 meters deep in 18 days by working 10 hours daily.

Volume of the first pond = 50 x 35 x 5.2 cu m

= 9100 cu m

Volume of the second pond = 65 × 40 × 5.6 cu m

= 14560 cu m

Let, the number of days is = x. Then the data can be arranged as,

No.of people                       Volume of pond                  Daily working time                  No.of days

250                                         9100 cu m                             20 hours                                    18

300                                         14560 cu m                            8 hours                                      x

250 Since, the number of people increases if the number of days decreases and the volume of the pond increases if the number of days increases and if daily working time increases the number of days decreases, therefore,

\(\left.\begin{array}{rl}
300 & : 250 \\
9100 & : 14560 \\
8 & : 10
\end{array}\right\}:: 18: x\)

 

or, x = 18 x 250 x 14560 x 10 / 300 x 9100 x 8

= 30

The required no. of days is 30 days.

Example 25

24 men take 6 days more to finish a piece of work than 33 men take to finish it. How long will 44 men take to finish the work? 

Solution:

Given:

24 men take 6 days more to finish a piece of work than 33 men take to finish it.

Let, 33 men take x days to finish the work. 

Then 24 men take (x+6) days to finish it. 

No. of men                      No. of days

33                                         x

24                                      x+6

Since, the number of days increases with the decrease in the number of men, therefore,

24/33 = x/x+6

or, 33x = 24x + 144

or, 33x – 24x = 144

or, 9x = 144

or, x = 144/9

= 16

∴ 33 men do the work in 16 days

Let, 44 men do the work in y days

No. of men                                    No. of days

33                                                    16

44                                                     y

∴  44/33 = 16/y

or, 4/3 = 16/y

or, 4y= 48

or, y = 12

They will finish the work in 12 days.

Example 26

In a camp, there is a stock of food for 20 days. After one week 100 more people took shelter in the camp and the food lasted for 11 days. In the beginning how many people were there? 

Solution:

Given:

In a camp, there is a stock of food for 20 days.

After one week 100 more people took shelter in the camp and the food lasted for 11 days.

Let, in the beginning, x people were there. 

After 1 week (207) 

or, 13 days remain.

No.of people                                 No.of days

x                                                        13

(100 + x)                                            11

Since the number of days decreases as the number of men increases, therefore,

100 + x / x = 13/11

or, 13x = 11x + 1100

or, 13x – 11x = 1100

or, 2x = 1100 

or, x = 550

In the beginning, there were 550 people.

Example 27

24 men take 12 days to prepare a pond. How many more men will be required to prepare a pond in 8 days?

Solution:

Given:

24 men take 12 days to prepare a pond.

Let, x men prepare the pond in 8 days.

No.of days                               No.of men

12                                             24

8                                               x

Here, since, the number of days has decreased, therefore, more men will be required. Therefore, the value of x will be greater than 24.

∴ 8/12 = 24/x

or, 8x = 24 x 12

or, x = 24 x 12 / 8

= 36

Since there were already 24 men, therefore (3624) or 12 more men will be required.

12 more men will be required.

Example 28

In a book written by hand, there are 105 pages and 25 lines on each page, and 8 words in each line. If that book is so printed that there will be 30 lines on each page and 10 words on each line, then what will be the number of pages of the book? 

Solution:

Given:

In a book written by hand, there are 105 pages 25 lines on each page, and 8 words in each line.

If that book is so printed that there will be 30 lines on each page and 10 words on each line.

Let the number of pages of the book be x.

No.of lines on each page                         No.of words in each line                     No.of pages

25                                                                            8                                                105

30                                                                           10                                                 x

Since, as the number of lines on each page will increase, the number of pages will decrease and as the number of words in each line will increase, the number of pages will decrease.

Therefore,

\(\left.\begin{array}{l}
30: 25 \\
10: 8
\end{array}\right\}:: 105: x\)

or, \(x=\frac{25 \times 8 \times 105}{30 \times 10}=70\)

 

The number of pages of the book will be 70.

Chapter 2 Pictograph And Pie Chart

Pictograph and Pie Chart Introduction

When we receive some data in a haphazard way we feel difficulty in studying and comparing it. But when the data are arranged in systematic order and represented through figures/pictures we can understand it easily. For this purpose, drawing pictographs plays an important role in communicating various information as well as comparing different data with others. Pictograph is used in magazine and newspapers to make the representation interesting to the readers. Let us draw some pictographs.

Drawing of pictographs

To draw pictographs, certain symbols, representing particular data are used. Depending upon the extent of the data received, we use the symbols to have a particular meaning or designation. In other words, depending on the size of the data we assign a pre-determined symbol to indicate a particular number so that the data supplied in a problem may be represented by that symbol. Sometimes, half of such a symbol indicates half of a number valid for a total/full symbol.

Read And Learn More WBBSE Solutions For Class 8 Maths

Chapter 2 Pictograph And Pie Chart Some Examples

Example 1

You have worked out the following number of sums during the seven days of a week.

Monday- 8,

Tuesday- 5,

Wednesday- 0,

Thursday- 6,

Friday- 5,

Saturday- 4,

Sunday- 7.

Describe the above data with the help of a pictograph.

From the above pictograph we can conclude the following:

1. You worked out the highest number of sums on Monday.

2. You did not work out any sum on Wednesday.

3. You have worked out an equal number of sums on Tuesday and Friday.

Example 2

A toy manufacturing company has prepared toys as shown by the following pictograph.

Class 8 Maths Solutions Wbbse

Watch the following questions and answers

Question 1. How many toys have been manufactured on Friday?

Answer.

On Friday

5 x 10 toys

= 50 toys.

50 toys have been manufactured on Friday

Question 2. On which day the maximum number of toys have been manufactured and how many? 

Answer: On Saturday. The number of toys manufactured is 15 x 10 toys = 150 toys.

Question 3. On which days an equal number of toys have been manufactured?

Answer: On Friday and Sunday.

Question 4. How many more toys have been manufactured on Saturday than on Thursday?

Answer: On Saturday 15 x 10, toys = 150 toys have been manufactured. 

On Thursday (12×10+1×5) toys = 125 toys have been manufactured.

∴On Saturday (150 – 125) or 25 more toys have been manufactured.

Example 3

The following pictograph shows the number of students who passed the different classes in the Annual Examination 2012.

From the above pictograph we may conclude the following:

Solution:

1. Number of students passed in class two = 4 x 10 + 1 x 5 = 40+ 5 = 45

2. The highest number of students passed in class four and their number = 5 x 10 = 50

3. Number of students who passed in class three and class five is equal. This number = 3 x 10 = 30

Example 4

A Pictograph of the number of trees planted by the students of your class is given below :

Watch the following questions and answers

Question 1. How many trees have been Q. planted on Thursday?

Answer: (3 × 10+ 1x 5) trees = 35 trees.

Question 2. On which day the maximum number of trees were planted and how many?

Answer: On Wednesday. The number of trees planted = 7 x 10 = 70 trees.

Question 3: On which days an equal number of trees were planted?

Answer: On Monday and Friday. 

Question 4. How many more trees were planted on Wednesday than on Tuesday?

Answer: On Wednesday 7 x 10 = 70, trees were planted. 

On Tuesday 3 x 10, 30 trees were planted.

∴ On Wednesday (70-30) or 40 more trees were planted.

Horizontal Bar graph

Sometimes data may be represented by a horizontal bar graph. For example, if we express example 4 given above by horizontal bar graph it looks like this

Example :

A horizontal bar graph is drawn showing how many students out of a total of 40 students help with their household work (in hours) during holidays. From this bar graph answer the following questions 

1. Write how many students in your class do the household work on every holiday from the horizontal bar graph.

2. How many students help in the household work for the maximum time? 

3. How many students help with their household work for two hours on every holiday?

Solution:

1. 8 students do household work for 2 hours on every holiday.

12 Students do household work for 3 hours on every holiday.

14 Students do household work for 4 hours on every holiday.

6 Students do household work for 5 hours on every holiday.

2. 6 students do household work in holidays for maximum time.

3. On each holiday 8 students do household work for 2 hours.

Representation by tally mark

If the number of observations is large, then data can be tabulated in the form of a table as given below:

Example 1

15 students prepare dolls as follows: 5, 3, 2, 1, 2, 4, 4, 4, 2, 2, 3, 4, 4, 5, 5. Prepare a frequency distribution chart with the tally marks for this raw data.

Solution:

Given:

15 students prepare dolls as follows: 5, 3, 2, 1, 2, 4, 4, 4, 2, 2, 3, 4, 4, 5, 5.

In the first column of the table, data pertaining to the number of dolls prepared are written from the lowest to the highest. Corresponding to the first value in the given raw data, a bar (or vertical line) is put in the second column opposite to it. Next, we see the second value in the given raw data and put a bar opposite to it in the second column. This process is repeated till all the observations in the given raw data are exhausted.

The bars drawn in the second column are known as tally marks. In general, we record tally marks in bunches of five, the fifth tally mark is drawn diagonally across the first four. For example, W = 8. Finally, the number of tally marks are counted corresponds to each observation and written in the third column. Representation by tally mark thus makes data easily understandable and clear.

Example 2

In a certain month, 22 days were denoted to academics in your school. You have written the number of boys presents in those 22 days in your class. 

They are :

Prepare a frequency distribution chart with tally marks from this raw data:

Pie Chart

A pie chart is a pictorial representation of the numerical data by non-intersecting adjacent sectors of a circle such that the area of each sector is proportional to the magnitude of the data represented by the sector. The pie chart is very useful to express the relationships among different parts of data and also the relation between the whole data and a part of it.

In the pie chart to express the whole data, a complete circle is used. This circle is subdivided into sectors by radii in such a manner that the areas of different sectors are proportional to the values of the corresponding component parts of data. Thus pie charts are very useful to show percentage breakdowns. For example, pie charts can depict various percentage shares of students of a school who like various activities like singing, dancing, recitation, drama, etc.

Example 1

On a particular day, 70% of the students of your school are present and 30% of students are absent. Then it can be expressed by pie chart as follows:

Example 2

50% students of in your class play football, 30% play cricket and 20% play hockey. It may be represented by a pie chart.

Let us briefly discuss the construction of a pie chart based on the above examples. The circle is divided into as many sectors as there are components in the data. Examples 1 and 2 have 2 and 3 component parts of data respectively. The area of each sector is proportional to the component value of the component (class) represented by the sector. But, we know that the area of a sector is proportional to the angle subtended at the center by its arc. Therefore, sector angles or central angles are proportional to the component values of the components (classes) represented by the sectors.

Thus, the central angle of a component

= (Value of the component (or %)/sum of the component values (or 100%) x 360°)

Finally, the descriptive labels shall be put either inside or beside each sector to make them easily identified. When completed, the entire graph or chart looks like a pie and the components resemble the slices cut from a pie. 

2.7 Reading of Pie chart 

The process of obtaining various information related to given data from the given pie chart is known as a reading of the pie chart.

The following formulae will be useful to find

the values of various computer nts of data from its pie chart:

Central angle for a component

 = (Value of the component/sum of the component values x 360°)

or, Value of the component 

= (Central angle for the component x Sum of the component values / 360°)

The percentage value of a component

(Central angle of the component  / 360° x100)

Example 3

From the pie chart answer the questions:

The pie chart of which types of songs the audience likes to listen to:

1. What part of the total circular region is the sector of an audience of the folk song?

2. Write from the pie chart which type of song has the most number of listeners.

3. Write which type of song has the least number of listeners.

Solution: 

1. The sector of audience of folk songs is = 20% 

= 20/100 part 

= 1/5 part of the folk song of the total circular region.

2. Audience of the modern songs is maximum.

3. Audience of the classical songs is minimum. 

Example 4

From the pie chart answer the questions: The pie chart of what kind of programs the audience likes:

1. In the pie chart what part of the total circular region is the section of the audience who watch the news?

2. What kind of program gets the most audience?

3. What kind of program gets the least audience?

4. What part of the total audience watches the programs on sports?

Solution: 

1. The sector of the audience who watch the news is the 20/360 part

= 1/18 total circular region.

2. Entertainment-based program gets the most audience.

3. Information Information News-based program gets the audience. 

4. 90/360 part = 1/4 part of the total audience watching programs on sports.

Illustrative Examples on PIE chart

Example 5

The percentage of marks which Ram has secured on the total marks in the final examination of class VII is given below:

Prepare a pie chart of this information and write the central angle of each sector. 

Solution:

At first, a circle of any radius is drawn. Then constructing the radius OA, 

∠AOB = 54°, 

∠BOC= 72°, 

∠COD = 108°, 

∠DOE = 54° 

and ∠EOA = 72° are drawn.

The produced is a pie chart of the given data.

Example 6

The list of various types of things which are sold in a shop of man for a particular day is as follows:

Draw a pie chart of this data.

Solution: Total sales on that day (320+ 100+ 160+ 140)=720.

At first, a circle of any radius is drawn. 

Then constructing the radius OA, 

∠AOB = 160°, 

∠BOC = 50°, 

∠COD = 80° 

and ∠DOA = 70° are drawn.

The figure produced is the pie chart of the given data.

Example 7

A list is made of things that the students of class VIII of both sections like to do during their leisure. (One student can like only one subject).

Work out from this data what parts of the total students like which subjects. Find the central angle of each sector and make a pie chart accordingly.

Solution: Here total number of students = (20+25+27 +28+ 20)

= 120

At first, a circle of any radius is drawn. 

Then constructing the radius OA, 

∠AOB = 60°, 

∠BOC= 75°, 

<COD = 81°, 

∠DOE= 84° 

and ∠EOA = 60° are drawn.

The produced is a pie chart of the given data.

Example 8

A pie chart is made by asking for the name of the favorite season for a group of 180 students. Answer the questions from the pie chart. 

1. Which season is liked by most of the students and how many like it? 

2. Which season is liked by the least number of students and how many like it?

3. How many students like summer? 

4. How many students like spring? 

5. Which season has been expressed by the least sector?

Solution:

1. Most of the students like winter = 180 x 40/100

= 72 students.

2. The least number of students during the rainy season = 180 x 10/100 

= 18 students.

3. The number of students in summer is 180 x =  x 20/100 

= 36 students.

4. The number of students in spring = 180 x 30/100

= 54 students.

5. Rainy season has been expressed by the least sector.

Example 9

A pie chart is made representing the expenditures made by a family on different items in a certain month. Find the percentage expenditures on different items from the given pie chart.

Solution:

For this problem, the percentage expenditure made on an item

(Central angle of the corresponding sector / 360° x 100)

Using the above formula, percentage expenditures on various items are computed as shown in the table

Chapter 1 Arithmetic Ratio And Proportions

Arithmetic Ratio and Proportions Introduction

In Class 7 you have studied the problems on ratio and proportion, time and distance, percentage, simple interest, profit and loss, square root of fractions and decimals, and also square measure. Here we shall discuss those topics in brief for recapitulations.

Ratio and proportion

Ratio: If x and y are two quantities of the same kind and the numerical value of y ≠ 0 then the fraction x/y is called the ratio of x to y and is written as x : y.

This ratio is the quotient of two quantities of the same kind.

Hence ratio is an abstract number and has no unit.

The quantities forming the ratio are called the terms of the ratio, in which the first quantity is called the antecedent while the second is called the consequent.

Proportion: A proportion is the equality of two ratios.

We say that one ratio is proportional to the other and the four numbers forming the two ratios are in proportion.

Read And Learn More WBBSE Solutions For Class 8 Maths

Some problems with ratio and proportion

Example 1

Find the ratio of 1.50 to 3.50. 

Solution:

Given 1.50 And 3.50

Required ratio = ₹1.50/₹3.50

= 150P/350P

= 3/7

= 3:7

The ratio of 1.50 to 3.50 = 3:7

Example 2

Find the value of x from the relation 4:7 = 12:x.

Solution:

Given

The Relation 4:7 = 12:x

From the given relation, we get,

4/7 = 12/x

or, 4 x x = 12 × 7

or, x = 12 x 7/4

= 21

The value of x is 21 

Example 3

Distinguish between x ÷ y and x: y. 

Solution. x ÷ y is defined even when x and y are not of the same kind.

But x:y is defined only when x and y are of the same kind. 

Example 4 If A: B = 3:4 and B: C = 6: 5, then find the value of A: B: C. 

Solution:

Given A: B = 3:4 And B: C = 6: 5

A: B=3:4=(3×3): (4 x 3)=9:12 

B:C = 6:5

= (6x²): (5 x 2)

= 12: 10 

A: B: C= 9:12:10

The value of A: B: C= 9: 12:10 

Example 5 

The ratio of two quantities is 5: 7. If the antecedent is 40 kg, find the consequent. 

Solution:

Given

The ratio of two quantities is 5: 7

Let the consequent be x kg.

∴ 40 kg/ x kg  = 5/7 

or, 40/x = 5/ 7

or, x= 40 x 7 / 5

= 8 x 7

= 56

The consequent is 56 kg.

Example 6

The ratio of the ages of Ram and Shyam is 8: 7. If the age of Shyam is 21 years, find the age of Ram.

Solution:

Given

The ratio of the ages of Ram and Shyam is 8: 7

The age of Shyam is 21 years

Let the age of Ram be x years.

∴ x/21 = 8/7

or, 7x=21 x 8

or, x = 21 x 8 / 7

=24

The age of Ram is 24 years. 

Example 7

The ratio of prices of two books the price of the first book is ₹ 75. If the price of the first book was ₹ 25 more, what would be the ratio of their prices? 

Solution:

Given

The ratio of prices of two books the price of the first book is ₹ 75

The price of the first book was ₹ 25

Let the price of the second book be by ₹ x.

∴ 75/x = 3/1

or, x = 75/3

= 25

If the price of the first book was ₹ 25 more, then the required ratio

= 75+25/25

= 10025

= 4/1

= 4:1

∴ The required ratio of their prices would be 4:1

Example 8

The ratio of daily work of A and B is 4: 3. If A can do a work in 12 days, then find the number of days required by B to do the work. 

Solution:

Given:

The ratio of daily work of A and B is 4: 3

A can do a work in 12 days

Since the ratio of the daily work of A and B is 4: 3, therefore the work done by A in 3 days is equal to the work done by B in 4 days.

∴ work done by A in 1 day

= work done by B in 4/3 days

∴ work done by A in 12 days

work done by B in 4/3 x 12 days 

= 16 days

B can do work in 16 days

Example 9

Ram earns ₹ 125 in 8 days and Shyam earns ₹ 140 in 10 days. Find the ratio of their earnings.

Solution:

Given:

Ram earns ₹ 125 in 8 days and Shyam earns ₹ 140 in 10 days

The earning of Ram for 8 days = ₹ 125

∴ The earning of Ram for 1 day = ₹125/8

The earning of Shyam for 10 days = ₹140

The earning of Shyam for  1 day = 140/10

∴ The ratio of their earnings = 125/8 : 140/10

= 125/8 x 10/140 

= 125:112

The ratio of their earnings is 125:112.

Example 10

A policeman starts to chase a thief. When the thief goes 5 steps, the policeman moves 6 steps. Again, 3 steps of the policeman is equal to 6 steps of the thief. Find the ratio of their speeds.

Solution:

Given:

A policeman starts to chase a thief . When the thief goes 5 steps, the policeman moves 6 steps. Again, 3 steps of the policeman is equal to 6 steps of the thief.

3 steps of the police = 6 steps of the thief

1 step of the police = 6/3 steps of the thief 

6 steps of the police = 6 x 6 / 3 steps of the thief

= 12 steps of the thief

Now, at the same time, the policeman moves 6 steps and the thief moves 5 steps.

∴ the ratio of their speeds 

= 6 steps of the police / 5 steps of the thief

= 12 steps of the thief / 5 steps of the thief

= 12/5

= 12:5

The ratio of their speeds is 12:5

Example 11

Find the missing term in place of * in the following proportion 16:20::8:*

Solution:

Given Proportion

16:20::8:*

Let, the missing term be x.

.. 16/20 = 8/x

or, 16 x x = 20 x 8

or, x = 20 x 8 /16

= 10

The missing term is 10.

Example 12

Find the fourth proportional of the following numbers: 12, 16, 18.

Solution:

Given 12, 16, 18….?

Let, the fourth proportional the x.

∴ 12/16 = 18/x

or, 12 x x = 18 x 16 

or, x = 18 x 16 / 12

= 24

The fourth proportion is 24.

Example 13

Find the fourth proportional of 2,4 1/2 and 4.

Solution:

Given 2,4 1/2 And 4……?

Let, the required fourth proportion be x.

∴ 2 / 4 ½ 

or, 2x = 9/2 x 4 = 18

or, x = 18/2 

= 9

The fourth proportion is 9.

Example 14

Find the third proportional of the following numbers: 9, 12,

Solution:

Given 9, 12,…..

Let, the third proportional be x.

∴ 9/12 = 12/x

or, 9x = 12 x 12 or, x= 12 x 12 / 9

= 16

The third proportion is 16.

Example 15

Find the mean proportional of the following numbers: 25, 81.

Solution:

Given Numbers 25, 81

Let, the mean proportion be x.

∴ 25/x = x/81

or, x2 = 25 x 81 

or, x = 5 x 9 = 45

The mean proportional is 45

Example 16

A’s money is 2/3 of B’s money and B’s money is 4/5 of C’s money. Find the ratio of A’s money to C’s money.

Solution:

Given A’s money is 2/3 of B’s money and B’s money is 4/5 of C’s money.

Since A’s money= is 2/3 of B’s money

A’s money / B’s money = 2/3

Also, B’s money = 4/5 of C’s money

B’s money/C’s money = 4/5

A’s money/B’s money X B’s money/C’s money = ⅔ x 4/5

or, A’s money / C’s money = 8:15

A’s money: C’s money 8:15

A’s money/B’s money X B’s money/C’s money = 2/3 x 4/5

or, A’s money / C’s money = 8:15

A’s money: C’s money 8:15

Example 17

If 2/3 of A = 75% of B = 0.6 of C, then find A:B: C

Solution:

Given 2/3 of A = 75% of B = 0.6 of C

2/3 of A= 75% of B = 0.6 of C

or, 2/3 × A = 75/100 x B = 6/10 x C

or, 2/3 x A = 3/4 x B = 3/5 x C = k(say)

:. A=3k/2, B=4k/3, C=5k/3

.. A: B: C= 3k/2: 4k/3: 5k/3

= 3/2:4/3:5/3

= (3/2 x 6):(4/3 x 6) : (5/3 x 6)

=9:8:10

A: B: C=9:8:10

Example 18

Divide 7000 among A, B, and C in such a way that the ratio of their shares is 8: 12 15. Find the shares of A, B, and C. 

Solution :

Given

By Dividing 7000 among A, B, and C in such a way that the ratio of their shares is 8: 12 15

Let, A will get ₹ 8x, B will get ₹ 12x and C will get ₹ 15x.

According to the question,

8x+12x+15x= 7000 or, 35x = 7000

or, x = 7000/35

= 200

∴ A will get ₹ 8×200= ₹ 1600; 

B will get ₹ 12 x 200 =  2400 

and C will get 15 x 200 = ₹ 3000.

A will get ₹ 1600, 

B will get 2400 

and C will get 3000.

Time and distance

When we travel from one place to another to cover some distance, some time is required. So to work out problems on time and distance, we have to concentrate on two main things.

1. The distance traveled and

2. the time required to travel that distance.

The relations between time, distance, and speed are as follows :

1. Distance traveled = speed x time. distance traveled

2. Speed = distance traveled/time

3. Time = distance traveled/speed

We have to remember to follow two rules:

1. When a train crosses a tree (or a telegraph post), it travels a distance equal to its own length.

2. When a train crosses a platform (or a bridge), it travels a distance equal to the sum of the lengths of the train and the platform (or bridge0)

When a boat moves on a river then we have to consider two things:

1. Speed of the boat

2. Speed of the river current

The actual speed of a boat: By actual speed of a boat we mean the distance covered by a boat in unit time in still water under the influence of oars.

Speed of a boat under the current of a river:

When the boat joint is effective along the current, then the joint effective speed = actual speed of the boat + speed of the river current.

When the boat rows against the current, then the joint effective speed = actual speed of the boat – speed of the river current.

Some problems with time and distance 

Example 1

A train travels 24 km in 4 hours. What distance will it travel in 10 minutes?

Solution: 

      Time                                            Distance 

4 x 60 minutes                                      24 km

10 minutes                                        x km (say)Since,

distance is directly proportional to the time

∴ 4 x 60 / 10 = 24/x

or, 24 x 10 / 4 x 60

= 1

1 km will it travel in 10 minutes

Alternative method:

In 4 x 60 minutes, the train travels 24 km

In 1 minute the  train travels = 24/ 4x 60 km

In 10 minutes the train travels = 24 x 10 / 4 x 60 km

= 1 km

The required distance is 1 km.

Example 2

Find the time takes to travel 600 km by train whose speed is 25 km/hour.

Solution:

Distance              Time

 25 km                  1 hour

 600 km                x hours (say)

Since distance is directly proportional to the time 

∴ 25/600 = 1/x

or, 25x = 600 or, x= 600/25

= 24

The required time is 24 hours.

Alternative method:

The train travels 25 km in 1 hour 

The train travels 25 km in 1/25 hour

The train travels 600 km in 600/25 hours 

= 25 hours.

Example 3

A train of length 210 meters crosses a telegraph post in 7 seconds. Express the speed of the train in km/hour unit. 

Solution:

Given

A train of length 210 meters crosses a telegraph post in 7 seconds.

When a train crosses a telegraph post, it travels a distance equal to its own length.

Time                                              Distance

7 seconds                                       210 meters

60 x 60 seconds                            x metres (say)

Since distance is directly proportional to the time

∴ 7/60 x 60 = 210/x

or, 7x = 210 × 60 × 60

or, x = 210 × 60 × 60 / 7

= 108000

Alternative method :

In 7 seconds the train crosses 210 meters 

In 1 second the train crosses 210/7 meters

In 60 x 60 seconds the train crosses 210/7 x 60 x 60 meters = 108000 meters 

= 108 km

The speed of the train is 108 km/hour. 

Example 4

A train crosses a platform in 30 seconds traveling at a speed of 60 km/hour. If the length of the train be 200 meters, then find the length of the platform.

Solution: When a train crosses a platform, it travels a distance equal to the sum of the lengths of the train and the platform.

Time                                      Distance

60 x 60 seconds                      60 km

30 seconds                              x km (say)

Since, the distance is directly proportional to time,

∴ 60 x 60/30 = 60/x

or, x x 60 x 60 = 30 x 60

or, x= 30 x 60/60 x 60 = 1/2

∴ Distance traveled by train in 30.

seconds = 1/2km 

= 1/2 x 1000 metres 

= 500 meters

Now, the length of the train + the length of the platform 500 meters

∴ Length of the platform = (500-200) metres = 300 metres

The length of the platform is 300 meters. 

Alternative method :

In 60×60 seconds the train travels 60×1000 meters

In 1 second the train travels 60 x 1000 / 60 x 60 meters

In 30 second the train travels 60 x 1000 x 30 / 60 × 60 meters

= 500 meters

∴ Length of the platform 

= (500-200) meters 

= 300 meters

The length of the platform is 300 meters

Example 5 

Two trains start at the same time towards each other from two stations 120 km apart. The speeds of the trains are 50 km/hour and 40 km/hour respectively. Find the distance between them after 1 hour and 15 minutes. Find also the distance between them after 2 hours. (Here assume that the two trains are traveling on parallel lines.)

Solution:

In 1 hour the distance between the two trains decreases by (50 +40) km = 90 km 

Now, 1 hour 15 minutes = 1 15/60 

hours = 1 1/4 hours

Hours = 5/4 hours

In 1 hour distance between the two trains decreases by 90 km

In 5/4  hours distance between the two trains

decreases by 90 x 5/4 km = 

= 112.5 km

∴ The distance between them after 1 hour 15 minutes will be (120 112.5) km = 7.5 km

Again, in 2 hours the two trains advance towards each other by 90 x 2 km =180 km

:. Distance between them after 2 hours will be (180-120) km = 60 km

The distance between them after 1 hr 15 min is 7.5 km and the distance between them after 2 hours is 60 km.

Example 6

The distance between A and B along a river 15 km in 1 hour is 96 km. The speed of the river current is 4 km/hour from A towards B. A boat takes 8 hours to travel from A to B. What time will it take to return from B to A? 

Solution: For the journey from A to B,

In 8 hours the boat travels 96 km

In 1 hour the boat travels 96/8 = 12 km

:. The actual speed of the boat = (12-4) km/hour = 8 km/hour

Therefore, the speed of the boat against the current (8-4) km/hour = 4 km/hour. 

So, for the journey from B to A,

To travel 4 km the boat takes 1 hour 

To travel 1 km the boat takes 1/4  hours

To travel 96 km the boat takes 96/4 hours

= 24 hours

The boat will take 24 hours to return from B to A.

Example 7

P and Q are two places on the bank of a river. The speed of the river current is from P towards Q. At the same time, a boat starts from P towards Q with the actual speed of 8 km/hour and another boat starts from Q towards P with an actual speed of 7 km/ hour. If the distance between P and Q be 150 km, when will the two boats meet? 

Solution: Let, the speed of the river current be x km/hour.

In 1 hour the first boat covers (8 + x) km. and in 1 hour the second boat covers (7-x) km

∴ In 1 hour the distance between the two boats decreases by (8+x+7-x) km = 15 km. 

The distance decreases by

15 km in 1 hour

1 km in 1/15 hour

150 km in 150/15 hours = 10 hours

The two boats will meet after 10 hours. 

Example 8

Speeds of a boat along and against the respectively. Find the actual speed of the current are 12 km/hour and 8 km/hour boat and the speed of the current.

Solution:

The actual speed of the boat + speed of the current 12 km/hr

The actual speed of the boat – speed of the current = 8 km/hr

(By addition), 2 x actual speed of the boat = 20 km/hr

∴ Actual speed of the boat= 20/2 km/hr

= 10 km/hr

Speed of the current = (12 – 10) km/hr 

= 2 km/hr

The actual speed of the boat is 10 km/hr. 

The speed of the current is 2 km/hr.

Square root of fractions

If a decimal fraction is multiplied by itself, then the product is called the square of the decimal and the first decimal fraction is called the square root of the product. 

For Example :

(0.12)2=0.12×0.12=0.0144;

hence, √0-0144

= 0.12

If a vulgar fraction is multiplied by itself, then the product is called the square of the vulgar fraction and the first vulgar fraction is called the square root of the product.

For Example :

(1/3)2 = 1/3 x 1/3

= 1/9

hence, √1/9 = 1/3

1.7 Some problems with the square root

Example 1

Find the square root of 2 up to 3 places of decimals.

Solution:

The required square root = 1.414 

Example 2

Find the square root of 52-983841 correct up to two places of decimal.

Solution: 

Hence, the square root of 52.983841 correct up to two places of decimal is 7.28. 

Example 3

The sides of the two squares are 16.5 and 22 meters respectively. Find the side of a square whose area is equal to the sum of the areas of the two squares.

Solution:

Area of the first square =(16.5)2 sq m 

= 272.25 sq m 

Area of the second square =(22)2 sq m 

= 484 sq m

∴ The area of the new square = (272.25 +484) sq m 

= 756.25 sq m

:. Each side of the new square

=√756.25 m 

= 27.5m

Each side of the new square is 27.5 meters

Example 4

Which number should be subtracted from 48.03 so that the square root of the difference is 5.7? 

Solution:

Since, the square root of the difference is 5.7 therefore, the difference = (5.7)2 

= 32.49. 

Hence, the number to be subtracted = 48.03 – 32.49 

= 15.54

The number to be subtracted is 15.54.

Example 5

A man spent 506.25. Each day he spent as many rupees as the number of days in which the money was spent. How much did he spend each day? 

Solution:

Since the number of days and the number of rupees spent each day are equal, therefore, the product of two equal numbers = 506.25

∴ Each number = √506.25

= 22.5

He spent ₹ 22.50 each day.

Example 6

A vulgar fraction multiplied by itself gives the product 841/2025. Find the fraction.

Solution:

Here the required vulgar fraction is the

the square root of 841/2025

= 27.5 m

Now, √841/2025

= √841/√2025 = 29/45

The fraction is 29/45

Example 7

The product of two numbers is 75/121 one of them is thrice the other. Find the numbers. 

Solution:

Greater number x smaller number = 75/121

or, 3 x smaller number x smaller number = 75/121

 or, (smaller number)2 = 75/25 3 x 121

= 25/121

or, smaller number = √25/121 = √25/√121

= 5/11

∴ Greater number = 3 x 5/11 

= 15/11

= 1 4/11

The numbers are 1 4/11 and 5/11.

Example 8

Find the smallest whole number by which 64/125 should be multiplied so that the product will be a fraction which is a perfect square.

Solution:

64/125 = 8×8/5x5x5

clearly, it should be multiplied at least by 5 so that the product will be a fraction which is a perfect square.

The smallest whole number by which 64/125 should be multiplied is 5.

Example 9

308 square stones are required to pave a square of area 1925 square meters. Find the length of each side of such a stone. 

Solution :

Given

308 square stones are required to pave a square of area 1925 square meters

By 308 stones 1925 sq meters are paved 

By 1 stone 1925/308 sq meters is paved 

= 25/4 sq meters is paved

Therefore, the area of each stone = 25/4 sq meters

Length of each side of a stone = //25/4

metres metres = 5/2 metres.

= 2 1/2 meters 

The length of each side of a stone is 2 1/2 meters.

Example 10

The sides of the two squares are 12/5 meters and 9/5 meters respectively. Find the side of a square whose area is equal to the sum of the areas of the two squares.

Solution:

Given:

The sides of the two squares are 12/5 meters and 9/5 meters respectively.

The area of the first square

= (12/5)2 sq meters

= 144/25 sq meters

The area of the second square= (9/2)2 sq metres

= 81/25 sq meters

∴ Area of the new square=(144/25 + 81/25) sq metres 

= 225/5 sq meters 

= 9 sq meters

∴ Each side of the new square = √9 meters 

= 3 meters

Each side of the new square is 3 meters.

1.8 Square measure

The plane area enclosed by straight lines is called a polygon. The measure of the portion enclosed by the sides of a polygon is called its area.

The sum of the lengths of the sides of a polygon is called its perimeter. 

A quadrilateral whose opposite sides are equal and the angles are right angles is called a rectangle.

The quadrilateral whose four sides are equal and the angles are right angles is called a square.

Some useful measurements:

1. Area of rectangle Length x Breadth.

2. The perimeter of rectangle = 2 x (Length + Breadth).

3. Area of square = (Side)2.

4. The perimeter of square = 4 × Side.

5. Area of the floor of a room = Length × Breadth.

6. Area of the roof of a room = Length × Breadth.

7. Area of the four walls of a room 

= 2x (Length + Breadth) x Height 

= Perimeter of the floor × Height. 

Some problems with square measure

Example 1

Solution:

Given:

Here, length = 15 m 25 cm = 1525 cm 

breadth 10 m 2 cm = 1002 cm

∴ Area of the rectangle

= length x breadth

= 1525 x 1002 sq cm

= 1528050 sq cm

= 152 sq m 80 sq dm 50 sq cm

The area of the rectangle is 152 sq m 80 sq dm 50 sq cm

Example 2

How many tiles, each 5 dm by 2 dm will be required to cover the floor of a room 15 m by 12 m?

Solution:

The area of the floor = 15×12 sqm 

= 180 sq m 

= 18000 sq dm 

The area of each tile = 5 dmx2 dm 

= 10 sq dm 

.. The number of tiles required = 18000/10 = 1800

1800 tiles will be required.

Example 3

Find the perimeter of a rectangle whose length is 15 m 5 dm and breadth is 8 m 7 dm. 

Solution:

Given:

Length is 15 m 5 dm And breadth is 8 m 7 dm.

The perimeter of the rectangle 

= 2 × (length + breadth)

= 2x (155+87) dm

= 2 × 242 dm

= 48 m 4 dm

The perimeter of the rectangle is 48 m 4 dm.

Example 4

One plot is 4 sq meters and another plot is 4 meters square. Which one is of the greater area?

Solution:

Given:

One plot is 4 sq meters and another plot is 4 meters square.

4 sq meters is meant a plot whose area is 4 sq meters.

By 4 meters square is meant a square plot 

the length of whose side is 4 meters i.e., 

its area = (4)2 sq m = 16 sq m.

The plot 4 meters square is of the greater area.

Example 5

How much fencing will be required to fence a square field of area 625 sq meters? 

Solution:

Given:

The area of the square field = 625 sq meters 

∴ The length of its side = √625 m = 25 m. 

Its perimeter = 4 x 25 m = 100 meters. 

Since the total length of the fencing is equal to the perimeter, therefore the required length of fencing = is 100 meters.

The required length of fencing is 100 meters.

Example 6

A garden is 30 m long and 20 m broad. All around outside of it, there is a path 3 m broad. What is the area of the path?

Solution:

Given

A garden is 30 m long and 20 m broad. All around outside of it, there is a path 3 m broad.

Length of the garden including the path= (30+ 3 x 2) m 

= (30+6) 

= 36 m

The breadth of the garden including the path Example 9 =(20+3×2) m 

= (20+6) m 

= 26 m Area of the garden including the path = 36 m x 26 m 

= 936 sq m

Area of the garden excluding the path = 30 m x 20 m 

= 600 sq m

Area of the path = (936-600) sq m 

= 336 sq m

The area of the path is 336 sq m. 

Example 7

The length, breadth, and height of a room are 4 m, 3 m, and 5 m respectively. What will be the cost of cementing the floor of that room at 15 per square meter? What will be the cost of whitewashing the four walls at 5 per square meter?

Solution:

Given

The length, breadth, and height of a room are 4 m, 3 m, and 5 m respectively.

Area of the floor of the room

= 4 × 3 sq m = 12 sq m.

cost of cementing 1 sq m = ₹ 15

Cost of cementing 12 sq m = 15 x 12=180

Again, area of the four walls of the room =2x(4+3) × 5 sq m 

=2x7x5 sq m 

= 70 sq m 

cost of whitewashing 1 sq m = ₹ 5

Cost of whitewashing 70 sq m = ₹ 5 x 70

= ₹ 350

The cost of cementing the floor is ₹180.

The cost of whitewashing the four walls is ₹ 350.

Example 8

Find the greatest measure of each stone of the same square shape to cover a rectangular courtyard 36 cm long and 28 m broad. Find the number of such stones.

Solution:

Given

The H.C.F. of 36 m and 28 m is 4 m.

∴ Each stone of the greatest area will be 4 meters square.

∴ Area of each stone = (4)2 sq m

= (4 × 4) sq m.

∴ The required number of stones 

= 36 x 28 / 4 x 4

= 69

 Each stone is 4 meters square and the number of stones is 63.

Example 9

What will be the cost of paper 3 m wide, at 25 paise per meter for covering the four walls of a room 15 m long 12 m broad, and 10 m high?

Solution:

Given

A paper 3 m wide, at 25 paise per meter for covering the four walls of a room 15 m long 12 m broad, and 10 m high

The area of the paper

= The area of the four walls

= 2x (length + breadth) x height

= 2 × (15 m + 12 m) x 10 m = 2 x 27 x 10 sq m

The length of the paper used 2x 27 x 10 / 3 m = 180 m

.. The required cost = 180 x 25 paise

= 4500 paise 

= ₹ 45.

The required cost is ₹45.

Example 10

The length, breadth, and height of a room are equal. If the area of the four walls is 100 sq m, then find its length, breadth, and height. 

Solution:

Given

The length, breadth, and height of a room are equal. The area of the four walls is 100 sq m,

Let the length of the room

= its breadth its height = a metres 

∴ 2x (a + a) x a = 100

or, 2 x 2a x a = 100 

or, 4a² = 100

or, a² = 100/4 

=25 

or, a = √25

=5

Each length, breadth, and height is equal to 5 m.

Example 11

The length of a rectangular field is 40 m and its breadth is 30 m.

1. Through the middle of this field a path of width 4 m parallel to length has divided the rectangular field into two equal parts. Find the area of the path. 

2. If through the middle of this field, a path of width 4 m parallel to breadth would divide the rectangular field into two equal parts what would be the area of the path?

3. If through the middle of the field two paths parallel to length and breadth of width 4 m each would divide the rectangular field into 4 equal parts what would be the area of the path?

Solution: 

1. Area of the path parallel to the length

= 40m x 4 m = 160 sq m. 

2. Area of the path parallel to the breadth 

= 30 m x 4 m = 120 sq m.

3. If there would be two paths through the middle of the field parallel to both length and breadth then its area would be

= 160 sq m + 120 sq m – 4 x 4 sq m 

=280 sq m – 16 sq m 

= 264 sq m.