RD Sharma Solutions Class 7 Chapter 9 Ratio and Proportion

Question 2: If x: y = 8: 9, find the ratio (7x – 4y): 3x + 2y. 

Solution 2: 

x: y = 8: 9

the equation is written as:

=x/y = 8/9

=9x = 8y

=x = 8y/9

By replacing the value of x in the given equation 

=(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 

 

Question 3: If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. 

Solution 3: 

Two numbers are in the ratio 6: 13

Let 6x and 13x be the numbers

Take LCM of 6x and 13x is 78x

= 78x = 312

=x = 312/78

=x = 4

By putting 4 at the place of x 

So, the numbers are 6x = 6 (4) = 24

13x = 13 (4) = 52 

 

Question 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 4: 

Let 3x and 5x be the numbers 

if 8 is added to each other than ratio becomes 2: 3

That the equation 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

So, the numbers are 

=3x = 3 (8) = 24

=5x = 5 (8) = 40 

Question 5: What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 

Solution 5: 

Let x be the number to be added 

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 

 

Question 6: Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers 

Solution 6: 

The three numbers are in the ratio 2: 3: 5 and the sum of them is 800

So, sum of the values of the ratio = 2 + 3 + 5 = 10

First number is (2/10) × 800

= 2 × 80

= 160

Second number is (3/10) × 800

= 3 × 80

= 240

Third number is (5/10) × 800

= 5 × 80

= 400

The three numbers are 160, 240 and 400 

Question 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 7: 

Let 5x and 7x be the present ages of two persons 

According to the question ages of two persons are in the ratio 5: 7

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 = 10

So, the ages are 5x = 5 × 10 = 50 years

And 7x = 7 × 10 = 70 years 

 

Question 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 8: 

Let 7x and 11x  be the numbers 

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

So, the numbers are 7x = 7 × 7 =49

And 11x = 11 × 7 = 77 

Question 9: Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers. 

Solution 9: 

According to the question the two numbers are in the ratio 2: 7

And their sum = 810

So, the 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 

Question 10: Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3. 

According to the question total amount to be divided = 1350

And the sum of the terms of the ratio = 2 + 3 = 5

Share of Ravish’s money 

= 2/5 × 1350

= 2 × 270

= Rs. 540

And Share of Shikha’s money 

= 3/5 × 1350

= 3 × 270

= Rs. 810

 

Question 11: Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5. 

According to the question total amount to be divided = 2000

Sum of the terms of the ratio = 2 + 3 + 5 = 10

Share of P’s money = 2/10 × 2000

= 2 × 200

= Rs. 400

And Share of Q’s money = 3/10 × 2000

= 3 × 200

= Rs. 600

And Share of R’s money = 5/10 × 2000

= 5 × 200

= Rs. 1000

Question 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. 

According to the question that boys and the girls in a school are in the ratio 7:4

Sum of the ratio is  7 + 4 = 11

Total strength = 550

Boys strength = 7/11 × 550

= 7 × 50

= 350

Girls strength = 4/11 × 550

= 4 × 50

= 200 

 

Question 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. 

According to the question that the ratio of income and savings is 7: 2

Let be 2x is the savings 

2x = 500

So, x = 250

Income = 7x

Income = 7 × 250 = 1750

Expenditure = Income – savings

= 1750 – 500

= Rs.1250

Question 14: The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides. 

According to the question sides of a triangle are in the ratio 1: 2: 3

And the Perimeter = 36cm

Sum 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

 

Question 15: A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get? 

According to the question total amount to be divided = 5500

Sum of the ratio = 2 + 3 = 5

Share of Raman’s money = 2/5 × 5500

= 2 × 1100

= Rs. 2200

And share of Aman’s money = 3/5 × 5500

= 3 × 1100

= Rs. 3300

Question 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. 

According to the question 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

 

Question 17: In the ratio 7: 8. If the consequent is 40, what a the antecedent 

According to the question the ratio is 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

 

Question 18: Divide Rs 351 into two parts such that one may be to the other as 2: 7. 

According to the question total amount is to be divided = 351

Ratio is 2: 7

The sum of ratio = 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

Question 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. 

One score contains 20 pencils and cost per score = 16

So, the cost of pencil  = 16/20

= Rs. 0.80

Cost of one dozen ball pen = 8.40 [1 dozen = 12]

Thus, 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

 

Question 20: In a class, one out of every six students fails. If there are 42 students in the class, how many pass? 

According to the question 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.

 

(v) 1: 2 or 13: 27 

(i) 3: 4 or 9: 16

Take LCM of 4 and 16 is 16

3 : 4 we can written as = 3/4

3/4 × 4/4 = 12/16

And we have 9/16

Clearly 12/16 > 9/16

Therefore 3: 4 > 9: 16

 

(ii) 15: 16 or 24: 25

Take LCM of 16 and 25 is 400

15: 16 we written as = 15/16

15/16 × 25/25 = 375/400

And we have 24/25

24/25 × 16/16 = 384/400

Clearly 384/400 > 375/400

Therefore 15: 16 < 24: 25

 

(iii) 4: 7 or 5: 8

Take the LCM of 7 and 8 is 56

4: 7 we written as = 4/7

4/7 × 8/8 = 32/56

And we have 5/8

5/8 × 7/7 = 35/56

Clearly 35/56 > 32/56

Therefore 4: 7 < 5: 8

 

(iv) 9: 20 or 8: 13

Take the LCM of 20 and 13 is 260

9: 20 we written as = 9/20

9/20 × 13/13 = 117/260

And we have 8/13

8/13 × 20/20 = 160/260

Clearly 160/260 > 117/260

Therefore 9: 20 < 8: 13

(v) 1: 2 or 13: 27

Take the LCM of 2 and 27 is 54

1: 2 we written as = 1/2

1/2 × 27/27 = 27/54

And we have 13/27

13/23 × 2/2 = 26/54

Clearly 27/54 > 26/54

Therefore 1: 2 > 13: 27

 

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

Solution 2: 

6: 8

To make equivalent ratio we multiplying numerator and denominator by 2 

=6/8 × 2/2 = 12/16

And also to make equivalent ratio we dividing numerator and denominator by 2 

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

Two equivalent ratios are 3: 4 = 12: 16

12/20 = …. /5 = 9/…. 

12/20 = 3/5 = 9/5

Description:

Consider 12/20 = …. /5

Let x be the unknown value 

Therefore 12/20 = x/5

On cross multiplying

= 12 × 5 = 60

=x = 60/20

=x = 3

Consider 12/20 = 9/….

Let y be the unknown value

Therefore 12/20 = 9/y

On cross multiplying 

= 20 × 9 = 180

y = 180/12

y = 15

 

 

 

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

Income of a family 1400

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

 

Question 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? 

The scale of map = 1: 4000000

Let x cm assume the actual distance between towns 

1: 4000000 =5: x

x = 5 × 4000000

x = 20000000 cm [1km = 1000 m; 1m = 100 cm]

Therefore

x = 200 km

 

Question 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? 

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

 

Question 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. 

The 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