## RD Sharma Solutions For Class 7 Maths Chapter 9 Ratio And Proportion Free Online

Exercise 9.1 Page No: 9.6

**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: 42y/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

16 = 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.

Exercise 9.2 Page No: 9.10

**1. Which ratio is larger in the following pairs?**

**(i) 3: 4 or 9: 16**

**(ii) 15: 16 or 24: 25**

**(iii) 4: 7 or 5: 8**

**(iv) 9: 20 or 8: 13**

**(v) 1: 2 or 13: 27**

**Solution:**

(i) Given 3: 4 or 9: 16

LCM for 4 and 16 is 16

3: 4 can be written as = 3/4

3/4 × (4/4) = 12/16

And we have 9/16

Clearly 12 > 9

Therefore 3: 4 > 9: 16

(ii) Given 15: 16 or 24: 25

LCM for 16 and 25 is 400

15: 16 can be written as = 15/16

15/16 × (25/25) = 375/400

And we have 24/25

24/25 × (16/16) = 384/400

Clearly 384 > 375

Therefore 15: 16 < 24: 25

(iii) Given 4: 7 or 5: 8

LCM for 7 and 8 is 56

4: 7 can be written as = 4/7

4/7 × (8/8) = 32/56

And we have 5/8

5/8 × (7/7) = 35/56

Clearly 35 > 32

Therefore 4: 7 < 5: 8

(iv) Given 9: 20 or 8: 13

LCM for 20 and 13 is 260

9: 20 can be written as = 9/20

9/20 × (13/13) = 117/260

And we have 8/13

8/13 × (20/20) = 160/260

Clearly 160 > 117

Therefore 9: 20 < 8: 13

(v) Given 1: 2 or 13: 27

LCM for 2 and 27 is 54

1: 2 can be written as = 1/2

1/2 × (27/27) = 27/54

And we have 13/27

13/27 × (2/2) = 26/54

Clearly 27 > 26

Therefore 1: 2 > 13: 27

**2. Give the equivalent ratios of 6: 8.**

**Solution:**

Given 6: 8

By multiplying both numerator and denominator by 2 we equivalent ratios

6/8 × (2/2) = 12/16

And also by dividing both numerator and denominator by 2 we equivalent ratios

(6/2)/ (8/2) = 3/4

Two equivalent ratios are 3: 4 = 12: 16

**3. Fill in the following blanks:**

**12/20 = …. /5 = 9/….**

**Solution:**

12/20 = 3/5 = 9/15

**Explanation:**

Consider 12/20 = …. /5

Let unknown value be x

Therefore 12/20 = x/5

On cross multiplying

x = 60/20

x = 3

Consider 12/20 = 9/….

Let the unknown value be y

Therefore 12/20 = 9/y

On cross multiplying we get

y = 180/12

y = 15

Exercise 9.3 Page No: 9.13

**1. Find which of the following are in proportion?**

**(i) 33, 44, 66, 88**

**(ii) 46, 69, 69, 46**

**(iii) 72, 84, 186, 217**

**Solution:**

(i) Given 33, 44, 66, 88

Product of extremes = 33 × 88 = 2904

Product of means = 44 × 66 = 2904

Therefore product of extremes = product of means

Hence given numbers are in proportion.

(ii) Given 46, 69, 69, 46

Product of extremes = 46 × 46 = 2116

Product of means = 69 × 69 = 4761

Therefore product of extremes is not equal to product of means

Hence given numbers are not in proportion.

(iii) Given 72, 84, 186, 217

Product of extremes = 72 × 217 = 15624

Product of means = 84 × 186 = 15624

Therefore product of extremes = product of means

Hence given numbers are in proportion.

**2. Find x in the following proportions:**

**(i) 16: 18 = x: 96**

**(ii) x: 92 = 87: 116**

**Solution:**

(i) Given 16: 18 = x: 96

In proportion we know that product of extremes = product of means

16/18 = x/96

On cross multiplying

x = (18 × 96)/ 16

x = 256/3

(ii) Given x: 92 = 87: 116

In proportion we know that product of extremes = product of means

x/ 92 = 87/116

On cross multiplying

x = (87 × 92)/ 116

x = 69

**3. The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400.**

**Solution:**

Given that income = 1400

Given the ratio of income and expenditure = 7: 6

7x = 1400

Therefore x = 200

Expenditure = 6x = 6 × 200 = Rs.1200

Savings = Income – Expenditure

= 1400 -1200

= Rs.200

**4. The scale of a map is 1: 4000000. What is the actual distance between the two towns if they are 5cm apart on the map?**

**Solution:**

Given that the scale of map = 1: 4000000

Let us assume the actual distance between towns is x cm

1: 4000000 =5: x

x = 5 × 4000000

x = 20000000 cm

We know that 1km = 1000 m

1m = 100 cm

Therefore

x = 200 km

**5. The ratio of income of a person to his savings is 10: 1. If his savings for one year is Rs.6000, what is his income per month?**

**Solution:**

Given that the ratio of income of a person to his savings is 10: 1

Savings per year = 6000

Savings per month = 6000/12

= Rs.500

Then let income per month be x

x: 500 = 10:1

x = 500 × 10

x = 5000

Income per month is Rs. 5000

**6. An electric pole casts a shadow of length 20 meters at a time when a tree 6 meters high casts a shadow of length 8 meters. Find the height of the pole.**

**Solution:**

Given that length electric pole shadow is 20m

Height of the tree: Length of the shadow of tree

Height of the pole: Length of the shadow of pole

x: 20 = 6: 8

x = 120/8

x = 15

Therefore height of the pole is 15 meters