The document RD Sharma Solutions (Part - 1) - Ex - 9.1, Ratio And Proportion, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**If x : y = 3 : 5, find the ratio 3x + 4y : 8x + 5y.**

It is given that

{dividing the numerator and the denominator by 'y'}

**If x : y = 8 : 9, find the ratio (7x âˆ’ 4y) : 3x + 2y.**

It is given that

{dividing the numerator and the denominator by 'y'}

Hence, 7x âˆ’- 4y : 3x + 2y = 10 : 21.

**If two numbers are in the ratio 6 : 13 and their l.c.m. is 312, find the numbers.**

Let the two numbers be 'x' and 'y' such that x : y = 6 : 13 â‡’ x/y = 613 .

We can assume that the HCF of 'x' and 'y' is a number 'k'.

So, x = 6k, and y = 13k.

Now, the product of any two numbers 'x' and 'y' is always equal to the product of their LCM and HCF

â‡’ xÃ—Ã—y = 312 Ã—Ã— k

â‡’ 6k Ã—Ã— 13k = 312 Ã—Ã— k

Thus, x = 6k = 6 Ã—Ã—4 = 24, and y = 13 Ã—Ã— 4 = 52.

**Two numbers are in the ratio 3 : 5. If 8 is added to each number, the ratio becomes 2 : 3. Find the numbers.**

Let the two numbers in ratio be x and y such that

x : y = 3 : 5

Now, 8 is added to each number, which means

On cross-multiplying, we get

â‡’ 3(3y + 40) = 2 Ã—Ã—5(y + 8)

â‡’ 9y + 120 = 10y + 80

â‡’ 120 âˆ’- 80 = 10y âˆ’- 9y

â‡’ y = 40

So, the numbers are 24 and 40.

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

Let the numbers that must be added to the ratio 7 : 13 be 'x'.

So,

After cross-multiplication, we get

3(7 + x) = 2(13 + x)

21 + 3x = 26 + 2x

3x - 2x = 26 - 21

x = 5

Thus, 5 must be added to each term to make the ratio = 2 : 3.

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

We have

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

Sum of the numbers = 800.

Therefore, first number

= 160

or, Second number

= 240

or, Third number

= 400

**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.**

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

Ratio of their present ages = 5 : 7.

Eighteen years ago, their ages were (5x âˆ’- 18) and (7x âˆ’- 18), respectively.

But eighteen years ago the ratio of their ages was 8 : 13.

So,

13(5x âˆ’- 18) = 8(7x âˆ’- 18)

65x âˆ’- 234 = 56x âˆ’- 144

65x âˆ’- 56x = 234 âˆ’- 144

9x = 90

So, their ages are 5x = 5Ã—Ã—10 = 50 years and 7x = 7 Ã—Ã— 10 = 70 years.

**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.**

Let the two numbers be 'x' and 'y'.

Given that x : y = 7 : 11

3 (7y + 77) = 2 Ã—Ã— 11 (y + 7)

21y + 231 = 22y + 154

22y âˆ’- 21y = 231 âˆ’- 154

Therefore, y = 77, and x

Thus, the two numbers are 49 and 77.

Two numbers are in the ratio 2 : 7. If the sum of the numbers is 810, find the numbers.

We have

Sum of the terms of the ratio = 2 + 7 = 9.

Sum of the numbers = 810.

Therefore, first number = 2/9Ã— 810 = 180

Second number = 7/9Ã— 810 = 630

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

We have

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

Therefore, Ravish's share = Rs (25Ã—1350)25Ã—1350 = Rs 540

Sikha's share = Rs (3/5Ã—1350) = Rs 810

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

We have

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

Therefore, P's share =Rs (2/10Ã— 2000) = Rs 400

Q's share = Rs (3/10Ã— 2000) = Rs 600

R's share = Rs (5/10Ã— 2000) = Rs 1000