Students can view and download the PDF of RD Sharma Solutions for Class 7 Maths Exercise 9.1 of Chapter 9 Ratio and Proportion, provided here. RD Sharma Solutions for Class 7 are the best study materials for those who aim to secure excellent marks in the annual exam. The solutions are formulated by the subject expert team at BYJU’S. This exercise primarily deals with ratios of two quantities. Some of the topics covered in this exercise are mentioned below.
- Ratio – The ratio of two quantities of the same kind and in the same units is a fraction that shows how many times one quantity is of the other
- Ratio in the simplest form
RD Sharma Solutions for Class 7 Maths Chapter 9 – Ratio and Proportion Exercise 9.1
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 9 – Ratio and proportion Exercise 9.1
1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y
Solution:
Given x: y = 3: 5
We can write above equation as
x/y = 3/5
5x = 3y
x = 3y/5
By substituting the value of x in given equation 3x + 4y: 8x + 5y we get,
3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y
= (9y + 20y)/5: (24y + 25y)/5
= 29y/5: 49y/5
= 29y: 49y
= 29: 49
2. If x: y = 8: 9, find the ratio (7x – 4y): 3x + 2y.
Solution:
Given x: y = 8: 9
We can write above equation as
x/y = 8/9
9x = 8y
x = 8y/9
By substituting the value of x in the given equation (7x – 4y): 3x + 2y we get,
(7x – 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y
= (56y – 36y)/9: (24y + 18y)/9
= 20y/9: 42y/9
= 20y: 42y
= 20: 42
= 10: 21
3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers.
Solution:
Given two numbers are in the ratio 6: 13
Let the required number be 6x and 13x
The LCM of 6x and 13x is 78x
= 78x = 312
x = (312/78)
x = 4
Thus the numbers are 6x = 6 (4) = 24
13x = 13 (4) = 52
4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers.
Solution:
Let the required numbers be 3x and 5x
Given that if 8 is added to each other then ratio becomes 2: 3
That is 3x + 8: 5x + 8 = 2: 3
(3x + 8)/ (5x + 8) = 2/3
3 (3x + 8) = 2 (5x + 8)
9x + 24 = 10x + 16
By transposing
24 – 16 = 10x – 9x
x = 8
Thus the numbers are 3x = 3 (8) = 24
And 5x = 5 (8) = 40
5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3
Solution:
Let the number to be added is x
Then (7 + x)/ (13 + x) = (2/3)
(7 + x) 3 = 2 (13 + x)
21 + 3x = 26 + 2x
3x – 2x = 26 – 21
x = 5
Hence the required number is 5
6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers
Solution:
Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800
Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10
First number = (2/10) × 800
= 2 × 80
= 160
Second number = (3/10) × 800
= 3 × 80
= 240
Third number = (5/10) × 800
= 5 × 80
= 400
The three numbers are 160, 240 and 400
7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages.
Solution:
Let present ages of two persons be 5x and 7x
Given ages of two persons are in the ratio 5: 7
And also given that 18 years ago their ages were in the ratio 8: 13
Therefore (5x – 18)/ (7x – 18) = (8/13)
13 (5x – 18) = 8 (7x – 18)
65x – 234 = 56x – 144
65x – 56x = 234 – 144
9x = 90
x = 90/9
x = 10
Thus the ages are 5x = 5 (10) = 50 years
And 7x = 7 (10) = 70 years
8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers.
Solution:
Let the required numbers be 7x and 11x
If 7 is added to each of them then
(7x + 7)/ (11x + 7) = (2/3)
3 (7x + 7) = 2 (11x + 7)
21x + 21 = 22x + 14
22x – 21x = 21 – 14
x = 21 – 14 = 7
Thus the numbers are 7x = 7 (7) =49
And 11x = 11 (7) = 77
9. Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers.
Solution:
Given two numbers are in the ratio 2: 7
And their sum = 810
Sum of terms in the ratio = 2 + 7 = 9
First number = (2/9) × 810
= 2 × 90
= 180
Second number = (7/9) × 810
= 7 × 90
= 630
10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.
Solution:
Given total amount to be divided = 1350
Sum of the terms of the ratio = 2 + 3 = 5
Ravish share of money = (2/5) × 1350
= 2 × 270
= Rs. 540
And Shikha’s share of money = (3/5) × 1350
= 3 × 270
= Rs. 810
11. Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5.
Solution:
Given total amount to be divided = 2000
Sum of the terms of the ratio = 2 + 3 + 5 = 10
P’s share of money = (2/10) × 2000
= 2 × 200
= Rs. 400
And Q’s share of money = (3/10) × 2000
= 3 × 200
= Rs. 600
And R’s share of money = (5/10) × 2000
= 5 × 200
= Rs. 1000
12. The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls.
Solution:
Given that boys and the girls in a school are in the ratio 7:4
Sum of the terms of the ratio = 7 + 4 = 11
Total strength = 550
Boys strength = (7/11) × 550
= 7 × 50
= 350
Girls strength = (4/11) × 550
= 4 × 50
= 200
13. The ratio of monthly income to the savings of a family is 7: 2. If the savings be of Rs. 500, find the income and expenditure.
Solution:
Given that the ratio of income and savings is 7: 2
Let the savings be 2x
2x = 500
So, x = 250
Therefore,
Income = 7x
Income = 7 × 250 = 1750
Expenditure = Income – savings
= 1750 – 500
= Rs.1250
14. The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.
Solution:
Given sides of a triangle are in the ratio 1: 2: 3
Perimeter = 36cm
Sum of the terms of the ratio = 1 + 2 + 3 = 6
First side = (1/6) × 36
= 6cm
Second side = (2/6) × 36
= 2 × 6
= 12cm
Third side = (3/6) × 36
= 6 × 3
= 18cm
15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get?
Solution:
Given total amount to be divided = 5500
Sum of the terms of the ratio = 2 + 3 = 5
Raman’s share of money = (2/5) × 5500
= 2 × 1100
= Rs. 2200
And Aman’s share of money = (3/5) × 5500
= 3 × 1100
= Rs. 3300
16. The ratio of zinc and copper in an alloy is 7: 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy.
Solution:
Given that ratio of zinc and copper in an alloy is 7: 9
Let their ratio = 7x: 9x
Weight of copper = 11.7kg
9x = 11.7
x = 11.7/9
x = 1.3
Weight of the zinc in the alloy = 1.3 × 7
= 9.10kg
17. In the ratio 7: 8. If the consequent is 40, what a the antecedent
Solution:
Given ratio = 7: 8
Let the ratio of consequent and antecedent 7x: 8x
Consequent = 40
8x = 40
x = 40/8
x = 5
Antecedent = 7x = 7 × 5 = 35
18. Divide Rs 351 into two parts such that one may be to the other as 2: 7.
Solution:
Given total amount is to be divided = 351
Ratio 2: 7
The sum of terms = 2 + 7
= 9
First ratio of amount = (2/9) × 351
= 2 × 39
= Rs. 78
Second ratio of amount = (7/9) × 351
= 7 × 39
= Rs. 273
19. Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen.
Solution:
One score contains 20 pencils
And cost per score = 16
Therefore pencil cost = 16/20
= Rs. 0.80
Cost of one dozen ball pen = 8.40
1 dozen = 12
Therefore cost of pen = 8.40/12
= Rs 0.70
Ratio of the price of pencil to that of ball pen = 0.80/0.70
= 8/7
= 8: 7
20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass?
Solution:
Given, total number of students = 42
One out of 6 student fails
x out of 42 students
1/6 = x/42
x = 42/6
x = 7
Number of students who fail = 7 students
No of students who pass =Total students – Number of students who fail
= 42 – 7
= 35 students.
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