WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion

Arithmetic Chapter 13 fundamental Concept of Ration And Proportion

Determine in which of the following cases a ratio can be obtained :

1. The weight of Ram and the height of Shyam.

Solution:

The weight and the height are not the same kinds of quantities. So a ratio can not be obtained in this case.

2. The amount of money that Raghab had and the amount of money that Raghab had spent.

Solution:

Here the two quantities of the same kind are comparable. So a ratio can be determined in this case.

3. The amount of water in litres that your bottle contains and the temperature of the water.

Solution:

Since the amount of water and the temperature of the water is not the same kind of quantities, any ratio can not be determined in this case.

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4. The time that you have read the book and the time that your sister have read the book.

Solution:

Here the two quantities of the same kind (i.e. the time) are comparable. Hence a ratio can be determined in this case.

WBBSE Class 6 Ratio and Proportion Solutions

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Question 2. Determine the ratio in each of the following quantities and write whether the ratio is a ratio of greater inequality or ratio of lesser inequality or a ratio of equality 

1. ₹ 30 and ₹ 22.5

Solution:

Given:

₹ 30 and ₹ 22.5

The required ratio = ₹ 30 : ₹ 22.5

= ₹ 30 / ₹ 22.5

= 30 / 22.5

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion Question 2 Q 1

= 4/3

= 4:3.

∵ 4 < 3 i.e., the antecedent > the radio is greater inequality.

₹ 30 and ₹ 22.5 = 4:3.

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion

Short Questions on Ratio and Proportion

2. 4.9 liters and 8.4 liters

Solution:

Given:

4.9 liters and 8.4 liters

The required ratio = 4-9 litres: 8.4 litres

= 4.9 / 8.4

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion Question 2 Q 2

 

= 7/12

= 7:12

∵ 7 < 12 i.e, the antecedent < the Consequent, the ratio is the ratio of lesser inequality.

4.9 liters and 8.4 liters = 7:12

3. l hour 24 minutes and 6 hours 18 minutes.

Solution:

Given:

l hour 24 minutes and 6 hours 18 minutes

1 hour 24 minutes = (1 x 60 + 24) minutes = 84 minutes.

6 hours 18 minutes = (6 x 60 + 18) minutes = 378 minutes.

∴ The required ratio = 84 minutes: 378 minutes

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion Question 2 Q 3

 

∵ 2 < 9 i.e., the antecedent < the consequent, the ratio is a ratio of lesser inequality.

l hour 24 minutes and 6 hours 18 minutes = 2:9

4. 52 m 25 cm and 522.5 cm.

Solution:

Given:

52 m 25 cm and 522.5 cm

52 m. 25 cm. = (52 x 100 + 25). cm

= 522.5 cm 522.5 dcm

= (522.5 x 10) cm = 522.5 cm.

The required ratio

= 522.5 cm: 522.5 cm

= \(\frac{5225 \mathrm{~cm}}{5225 \mathrm{~cm}}\)

= 1/1

= 1: 1

Here Antecedent = Consquent.

So the ratio is the ratio of equality.

52 m 25 cm and 522.5 cm = 1: 1

5. x2yand xy2 (x > y).

Solution:

Given:

x2yand xy2 (x > y)

The required ratio = x²y : xy²

= \(\frac{x^2 y}{x y^2}\)

= x/y

= x: y

As x > y i.e., the antecedent > the consequent, the ratio is a ratio of equality.

x2yand xy2 (x > y) = x: y

Common Questions About Ratios in Real Life

6. a2bc and ab2c (where a = b).

Solution:

Given:

a2bc and ab2c (where a = b)

The required ratio = a2bc : ab2c

= \(\frac{a^2 b c}{a b^2 c}\)

= a / b

= a:b.

As a = b i.e., the antecedent = the consequent, the ratio is a ratio of equality.

a2bc and ab2c (where a = b) = a:b

 

Question 3. What is the ratio of the 3 angles of an equilateral triangle?

Solution:

Given:

3 angles of an equilateral triangle

In an equilateral triangle, the sum of the angles = 180° and the 3 angles of the triangle are the same.

∴ The measure of each angle of the triangle = 180° / 3

= 60°

∴ The required ratio = 60° : 60° : 60°

= 1: 1: 1.

The ratio of the 3 angles of an equilateral triangle = 1: 1: 1.

Practice Problems on Ratios and Proportions

Question 4. What is the ratio of the 3 angles of an isosceles right-angled triangle?

Solution:

Given:

3 angles of an isosceles right-angled triangle

One angle of an isosceles right-angled triangle = 90° and the sum of the 3 angles of the triangle = 180°. So the sum of the angles of the remaining two angles

= 180° – 90°

= 90°

and they are equal. Hence the measure of each of the equal angles

= 180° /  90°

= 45°

The angles of the isosceles right-angled triangle

= 45°, 45°, 90°.

The required ratio of the measure of the 3 angles of the isosceles right-angled triangle

= 45°: 45°: 90° = 1:1:2.

The ratio of the 3 angles of an isosceles right-angled triangle = 45°: 45°: 90° = 1:1:2.

Important Definitions Related to Ratio and Proportion

Question 5. How much money Bidhu and Bivu will get if we divide ₹ 210 between them in a ratio of 3:4?

Solution:

Sum of the components of the ratio = 3 + 4 = 7

So Bidhy will get 3/7 of the part and Bivu will get 4/7 part of the total amount.

∴ Bidhu will et = ₹ \(\left(210 \times \frac{3}{7}\right)\)

= ₹ 90.

Bivu will get = ₹ \(\left(210 \times \frac{4}{7}\right)\)

= ₹ 120.

₹ 120 money Bidhu and Bivu will get if we divide ₹ 210 between them in a ratio of 3:4

 

Question 7. The ratio of your reading books and story books is 4 :1; if the number of your reading books is 16, then what is the number of your story books? What is the total number of your books?

Solution:

Given:

The ratio of your reading books and story books is 4 :1; if the number of your reading books is 16

It is given that the ratio of your reading books: story books = 4:1.

∴ If the number of reading books is 4, then the number of storybooks = is 1

So If the number of reading books is 1, then the number of story books = 1/4

∴ If the number of reading books is 16, then the number of story books = 1/4 x 15

= 4.

∴ Number of story books = 4

Total number of books = 16 + 4

= 20.

Total number of books = 20.

Examples of Real-Life Applications of Ratios

Question 7. In one kind of ornament, the ratio of the weight of gold to that of silver = 4: 7. In such a type of ornament, how many milligrams of gold be mixed with 357 milligrams of silver?

Solution :

Let the required weight of gold = x milligram

∵ \(\frac{the weight of gold}{the weight of silver}\) = 4/7

∴ \(\frac{x}{357}\) = 4/7

or, x x  7 = 357 x 4

or, 7x = 357 x 4

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion Question 7

= 204

∴ the required weight of gold = 204 milligrams.

 

Question 8.  The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

Solution:

1. What is the ratio of the total length of the bamboo to that of the portion painted with the orange color?

Solution :

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The length of the bamboo = 2 metres = (2 x 100) cm. = 200 cm.

The length of the bamboo in which portion is painted with the orange color

= 75 cm.

The required ratio = 200 : 75 = 200 / 75 = 8/3 =8:3.

2. What is the ratio of the total length of the bamboo to that of the portion painted with the white color?

Solution:

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The total length of the bamboo = 200 cm

The length of the bamboo in which portion is painted with the white color

= (200 – 75) cm = 125 cm.

The required ratio = 200 : 125 = 200 / 125 = 8/5 = 8 : 5

3. What is the ratio of the length of the bamboo painted with orange color to that of the portion painted with white color?

Solution:

Given:

The length of bamboo is 2 meters. 75 cm. of its length is painted with orange color and the remaining part of its length is painted with white color.

The required ratio = 75 cm ÷ 125 cm

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 13 fundamental Concept of Ration And Proportion Question 7

= 3/5

= 3:5.

3:5 ratio of the length of the bamboo painted with orange color to that of the portion painted with white color

Conceptual Questions on Equivalent Ratios

Question 9. Kamala Devi bought 6 bananas for ₹ 18. But Sarala Devi bought 2 dozen bananas for ₹ 72. Determine who gave more price for purchasing banana by expressing in ratio.

Solution:

Given:

Kamala Devi bought 6 bananas for ₹ 18. But Sarala Devi bought 2 dozen bananas for ₹ 72.

Kamala Devi bought 6 bananas in ₹ 18.

Cost of 6 bananas = ₹ 18

So the cost of 1 banana = ₹ 18/6

= ₹

Sarala Devi bought 2 dozen bananas in ₹ 72

2 dozens = 2 x 12 = 24 bananas.

Cost of 24 bananas =v₹ 72.

So the cost of 1 banana = ₹ 72/24 = ₹ 3.

The ratio of the cost of 1 banana paid by Kamala Devi to that by Sarala Devi

= ₹ 3: ₹ 3

= 1: 1.

It is a ratio of equality.

So both of them paid the same rate for bananas.

Real-Life Scenarios Involving Scale Models

Question 10. Determine which of the following ratios are equal :

1.  20: 24 and 25: 30

Solution:

20: 24

= 20 / 24

= 5/6

= 5: 6

20: 24 = 5: 6


25: 30

= 25/30

= 5/6

= 5: 6.

25: 30 = 5: 6.


2. 1.4: 0.6 and 6.3: 2.7

Solution:

1.4: 0.6 

= 1.4 / 0.6

= 14/6

= 7/3

= 7: 3

1.4: 0.6 = 7: 3


6.3: 2.7

= 6.3/2.7

= 63/27

= 7/3

= 7:3

6.3: 2.7 = 7:3

∴ The two given ratios are equal.

 

Question 11. Determine whether the following numbers are according to the given order in proportion.

1. 9, 7, 36, 28

Solution:

Here first number = 9 ; second number = 7;

Third number = 36 ; fourth number = 28.

∴ first number x fourth number = 9 x 28

= 252

second number x third number = 7 x 36

= 525.

So first number x third number = second number x third number.

∴ The given ordered numbers are in proportion.

∴ 9: 7: : 36: 28.

2. 1/2, 1, 3/5, \(1 \frac{1}{5}\)

Solution:

Here first number = 1/2 ; second number = 1 ; third number = 3/5 ;

fourth number = \(1 \frac{1}{5}\)

= 6/5

∴ The first number x fourth number =  \(\frac{1}{2} \times \frac{6}{5}\)

= 3/5

The third number x second number = \(\frac{3}{5} \times{1}\)

= 3/5

So first number x fourth number = second number x third number

∴ The given order numbers are in proportion

∴ 1/2 : 1 : : 3/5 : 6/5

or, 1/2 : 1 : : 3/5 : \(1 \frac{1}{5}\)

Class 6 Math Solutions WBBSE English Medium

Question 12. Examine whether any proportion can be formed with the quantities 12 km, 16 km, 21 kg, 28 kg. If it is possible, then how many proportions can be formed?

Solution: Here we see that 12 x 28 = 336 and 16 x 21 = 336.

∴ 12 x 28 = 16 x 21

so the given quantities form a proportion.

The proportions are :

12 km : 16 km : : 21 kg : 28 kg

 and 16 km : 12 km : : 28 kg : 21 kg

Only 2 proportions can be formed.

 

Question 13. Your height is 160 cm and your mother’s height is 170 cm. Your weight is 40 kg and your mother’s weight is 42*5 kg. Examine whether there exists a proportion between height and weight or not.

Solution:

Given:

My height is 160 cm and my mother’s height is 170 cm. MY weight is 40 kg and my mother’s weight is 42*5 kg.

The heights of you and your mother are 160 cm and 170 cm.

The ratio of the height = 160 cm : 170 cm

= 160 / 170

= 16: 17.

Again the weight of you and your mother is 40 kg and 42.5 kg.

∴ The ratio of the weights = 40 kg : 42.5 kg = 40 / 42.5

= 400 / 425

= 16 / 17

= 16: 17

∴ the height and weights of you and your mother are in proportion.

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