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Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

PAGE NO 9.14:

Question 1: Which of the following statements are true?

(i) 16 : 24 = 20 : 30

(ii) 21 : 6 = 35 : 10

(iii) 12 : 18 = 28 : 12

(iv) 51 : 58 = 85 : 102

(v) 40 men : 200 men = Rs 5 : Rs 25

(vi) 99 kg : 45 kg = Rs 44 : Rs 20

ANSWER: (i) 16 : 24 = 20 : 30

16/24 = 2/3 (Dividing numerator and denominator by 8)

20/30 = 2/3 (Dividing numerator and denominator by 10)

∴ 16/24 = 20/30

Thus, the statement is true.

(ii) 21 : 6 = 35 : 10 

21/6 = 7/2 (Dividing numerator and denominator by 3)

35/10 = 7/2 (Dividing numerator and denomenator by 5) 

∴ 21/6 = 35/10

Thus, the statement is true.

(iii) 12 : 18 = 28 : 12 

12/18 = 2/3 (Dividing numerator and denominator by 6) 

28/12 = 7/3 (Dividing numerator and denominator by 4)

∴ 16/24 = 20/30 are not equal.

Thus, the statement is not true.

(iv) 51 : 58 = 85 : 102 

51/58 

And 85/102 = 5/6 (Dividing numerator and denominator by 17) 

51/58  is not equal to 85/102.

Thus, the statement is not true.

(v) 40 men : 200 men = Rs 5 : Rs 25  

40/200 = 1/5 (Dividing numerator and denominator by 40)

5/25 = 1/5 (Dividing numerator and denominator by 5) 

∴ 40/200 = 5/25

Thus, the statement is true. 

(vi) 99 kg : 45 kg = Rs 44 : Rs 20 

99/45 = 11/5 (Dividing numerator and denominator by 9) 

44/20 = 11/5 (Dividing numerator and denominator by 4)

∴ 99/45 = 11/5

Thus, the statement is true.

Question 2: Find which of the following are in proportion:

(i) 8, 16, 6, 12
(ii) 6, 2, 4, 3
(iii) 150, 250, 200, 300

ANSWER: (i) Consider 8/16 = 1/2

And 6/12 = 1/2

∵ 8 : 16 = 6 : 12

∴ 8, 16, 6, 12 are in proportion.

(ii) Consider  6/2 = 3/1

And 4/3

∵  6 : 2 ≠≠  4 : 3
∴ 6, 2, 4, 3 are not in proportion. 

(iii) Consider  150/250 = 3/5 (Dividing numerator and denominator by 50)

And 200/300 = 2/3  (Dividing numerator and denominator by 100)

∵ 150 : 250 ≠≠  200 : 300 

∴ 150, 250, 200, 300 are not in proportion.  


Question 3: Find x in the following proportions:

(i) x : 6 = 55 : 11

(ii) 18 : x = 27 : 3

(iii) 7 : 14 = 15 : x

(iv) 16 : 18 = x : 96

ANSWER:

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(ii) 18 : x = 27 : 3 

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(iii) 7 : 14 = 15 : x1624=23 (Dividing numerator and denominator by 8)2030= 23 (Dividing numerator and denominator by 1Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

(iv) 16 : 18 = x : 96 

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics

Question 4: Set up all proportions from the numbers 9, 150, 105, 1750.

ANSWER:

All proportions are:
9 : 150 = 3 : 50
9 : 105 = 3 : 35
9 : 1750
150 : 105 = 10 : 7
150 : 1750 = 3 : 35
105 : 1750 = 3 : 50
Thus, all proportions that can be formed are:
3 : 50, 3 : 35, 10 : 7, 9 : 1750, 1750 : 9, 7 : 10, 35 : 3 and 50 : 3


Question 5:

Find the other three proportions involving terms of each of the following:
(i) 45 : 30 = 24 : 16
(ii) 12 : 18 = 14 : 21

ANSWER: (i) 45 : 30 = 24 : 16  (3 : 2 is its simplest form)
Other three proportions are:
45 : 24 = 30 : 16  (15 : 8 is its simplest form)
30 : 45 = 16 : 24  (2 : 3 is its simplest form)
16 : 30 = 24 : 45  (8 : 15 is its simplest form)
(ii) 12 : 18 = 14 : 21 (2 : 3 is its simplest form)
Other three proportions are:
12 : 14 = 18 : 21 (6 : 7 is its simplest form)
21 : 18 = 14 : 12 (7 : 6 is its simplest form)
18 : 12 = 21 : 14 (3 : 2 is its simplest form)


Question 6: If 4, x, 9 are in continued proportion, find the value x.

ANSWER: It is given that 4, x, 9 are in continued proportion; therefore, we have:
4 : x : : x : 9

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics


Question 7: If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third term.

ANSWER: In a proportion, the first, second and fourth terms are 32, 112 and 217, respectively.

Let the third term is be x. 

Then, we have:

32 : 112 : : x : 217

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics


Question 8: Show that the following numbers are in continued proportion:
(i) 36, 90, 225
(ii) 48, 60, 75
(iii) 16, 84, 441

ANSWER: (i) 36, 90, 225

Consider 36/90 = 2/5  (Dividing numerator and denominator by 18)

90/225 = 2/5   (Dividing numerator and denominator by 45)

36 : 90 : : 90 : 225
(ii) 48, 60, 75
Consider  48/60 = 4/5 (Dividing numerator and denominator by 12)

60/75 = 4/5  (Dividing numerator and denominator by 15)

⇒ 48 : 60 : : 60 : 75
(iii) 16, 84, 441
Consider 16/84 = 4/21 (Dividing numerator and denominator by 4)

84/441 = 4/21 (Dividing numerator and denominator by 4)

⇒ 16 : 84 : : 84 : 441 


Question 9: The ratio of the length of a school ground to its width is 5 : 2. Find its length if the width is 40 metres.

ANSWER: Ratio of the length to the width of a school ground = 5 : 2

∵ Width = 40 m

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics


Question 10: The ratio of the sale of eggs on  a Sunday to that of the whole week of a grocery shop was 2 : 9. If the total sale of eggs in the same week was Rs 360, find the sale of eggs on Sunday.

ANSWER: Ratio of the sale of eggs on Sunday to that of the whole week = 2 : 9
When total eggs of Rs. 9 is sold in a week, the sale of eggs on Sunday = Rs. 2
When total eggs of Rs. 1 sold in a week, th sale of eggs on Sunday = 2/9

When total eggs of Rs. 360 sold in a week, the sale of eggs on Sunday Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics


Question 11: The ratio of copper and zinc in an alloy is 9 : 7. If the weight of zinc in the alloy is 9.8 kg, find the weight of copper in the alloy.

ANSWER: The ratio of copper and zinc in an alloy is 9 : 7.
When the weight of zinc is 7 kg, the weight of copper = 9 kg

When the weight of zinc is 1 kg, the weight of copper  = 9/7 kg

When the weight of zinc is 9.8 kg, the weight of copper = 

Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics


Question 12: The ratio of the income to the expenditure of a family is 7 : 6. Find the saving if the income is Rs 1400.

ANSWER: The ratio of the income to the expenditure of a family is 7 : 6.

∴ Ratio of saving to the income = (7 − 6) : 7 = 1 : 7   (Saving = Total income −- Expenditure)
∵ Income of the family = Rs. 1400

∴ Saving = 1400 × 17 = Rs. 200


Question 13: The ratio of story books in a library to other books is 1 : 7. The total number of story books is 800. Find the total number of books in the library.

ANSWER:

The ratio of story books in a library to other books is 1 : 7.
Out of (1 + 7) = 8 books, 1 book is a story book
Therefore,
When number of story books is 1, the total number of books = 8
When number of  story books is 800, the total number of books = 8 × 800
= 6400

The document Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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FAQs on Page No.9.14, Ratio Proportion And Unitary Method, Class 6, Maths RD Sharma Solutions - RD Sharma Solutions for Class 6 Mathematics

1. What is the importance of understanding ratio and proportion in mathematics?
Ans. Understanding ratio and proportion is important in mathematics as it helps in comparing quantities and finding relationships between them. It is used in various real-life situations such as scaling, mixing solutions, calculating distances, and solving problems involving proportions.
2. How can I solve problems using the unitary method?
Ans. The unitary method is a technique used to solve problems involving proportions. To solve problems using the unitary method, first, understand the relationship between the given quantities. Then, set up a proportion equation and cross-multiply to find the unknown value. Finally, solve the equation to find the desired answer.
3. Can you give an example of a real-life situation where ratio and proportion are used?
Ans. Yes, one example of a real-life situation where ratio and proportion are used is in cooking or baking recipes. Recipes often call for ingredients in specific ratios, such as 2:1 flour to sugar. Proportions are used to scale the recipe up or down depending on the desired quantity, ensuring the right balance of ingredients.
4. How do I simplify ratios?
Ans. To simplify ratios, find the greatest common divisor (GCD) of the given numbers in the ratio. Divide both the numerator and denominator of the ratio by the GCD to obtain the simplified form. This ensures that the ratio is in its simplest form without any common factors.
5. Is it possible for two ratios to be equal even if the individual numbers are different?
Ans. Yes, it is possible for two ratios to be equal even if the individual numbers are different. Ratios are equal if their equivalent fractions or decimal representations are equal. For example, the ratios 2:4 and 1:2 are equal because both represent the same proportion of 1:2.
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