# RS Aggarwal Class 7 Math Eighth Chapter Ratio and Proportion Exercise 8A Solution

## EXERCISE 8A

**(1) Express each of the following rations in simplest form:**

(i) 24 : 40

= (24 ÷ 8) : (40 ÷ 8)

= 3 : 5

(ii) 13.5 : 15

= 135 : 150

= (135 ÷ 15) : (150 ÷ 15)

= 9 : 10

**(2) Express each of the following rations in simplest form:**

(i) 75 paise : 3 rupees

Here 3 rupees = 300 paise

= 75 : 300

= (75 ÷ 75) : (300 ÷ 75)

= 1 : 4

(ii) 1 m 5 cm : 63 cm

Here 1 m 5 cm = 105 cm

= 105 cm : 63 cm

= (105 ÷ 21) cm : (63 ÷ 21) cm

= 5 cm : 3 cm

(iii) 1 hour 5 minutes : 45 minutes

Here 1 hour 5 minutes = 65 minutes

= 65 : 45

= (65 ÷ 5) : (45 ÷ 5)

= 13 : 9

(iv) 8 months : 1 year

Here, 1 year = 12 months

= 8 : 12

= (8 ÷ 4) : (12 : 4)

= 2 : 3

(v) (2 kg 250 g) : 3 kg

Here, 2 kg 250 g = 2250 g

3 kg = 3000 g

= 2250 : 3000

= (2250 ÷ 750) : (3000 ÷ 750)

= 3 : 4

(vi) 1 km : 750 m

Here, 1 km = 1000 m

= 1000 : 750

= 100 : 75

= (100 ÷ 25) : (75 ÷ 25)

= 4 : 3

**(3) If A : B = 7 : 5 and B : C = 9 : 14, find A : C.**

Solution: A : B = 7 : 5 and B : C = 9 : 14

**(4) If A : B = 5 : 8 and B : C = 16 : 25, find A : C.**

**(5) If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.**

Solution: A : B = (3 × 2) : (5 × 2) = 6 : 10

∴ A : B : C = 6 : 10 : 13

**(6) If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.**

Solution: A : B = (5 × 2) : (6 × 2) = 10 : 12

B : C = (4 × 3) : (7 × 3) = 12 : 21

∴ A : B : C = 10 : 12 : 21

**(7) Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.**

Solution: Sum of the ratio terms = (7 + 8) = 15

**(9) Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.**

Solution: Sum of the ratio terms =(1+3+4)=8

**(10) What number must be added to each term of ratio 9 : 16 to make the ratio 2 : 3?**

Solution: Let the required number to be added be x. Then, we have,

Hence, the required number is 5.

**(11) What number must be subtracted from each term of the ratio 17 : 33 so that the ratio becomes 7 : 15?**

Solution: Let the required number is x. Then, we have,

Hence, the required number is 3.

**(12) Two numbers are in the ratio 7 : 11. If 7 is added to each term of the ratio becomes 2 : 3. Find the numbers.**

Solution: Let the required number is x. Then, we have,

Hence the numbers are (7×7) = 49 and (11×7) = 77.

**(13) Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.**

Solution: Let the required number is x. Then, we have,

Hence, the required numbers are (5×3) = 15 and (9×3) = 27.

**(14) Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.**

Solution: Let the required numbers be 3x and 4x.

Then, their LCM is 12x.

Hence, the numbers are (3×15)=45 and (4×15) = 60.

**(15) The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present ages.**

Solution: Let the present ages of A and B be 8x and 3x years respectively.

A’s age after 6 years = (8x + 6) years.

B’s age after 6 years = (3x+6) years.

Therefore, A’s present age = (8×6) = 48 years.

B’s present age = (3×6)=18 years.

**(16) The ratio of copper and zinc in an alloy is 9 : 5, If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.**

Solution: Let the weight of copper is 9x grams and zinc is 5x grams.

Therefore, the weight of the zinc is (5×5.4)=27 grams.

**(17) The ratio of boys and girls in a school is 8 : 3. If the total number of girls be 375, find the number of boys in the school.**

Solution: Let the number of boys and girls will be 8x and 3x.

Therefore, the numbers are boy is (8×125)=1000.

**(18) The ratio of monthly income to the savings of a family is 11 : 2. If the savings be Rs 2500, find the income and expenditure.**

Solution: Let the income and savings will be 11x and 2x.

Therefore, the income is (11×1250) = Rs 13750.

And expenditure is Rs (13750 – 2500) = Rs 11250.

**(19) A bag contains Rs 750 in the form of rupee, 50 P and 25 P coins n the ratio 5 : 8 : 4. Find the number of coins of each type.**

Solution: Let the number of rupee , 50 p and 25 p coins be 5x, 8x and 4x.

But the total coins = Rs 750.

∴10x = 750

or, x=75

Number of 1 rupee coins = 5 × 75 = 375

Number of 50 p coins = 8 × 75 = 600

Number of 25 p coins = 4 × 75 = 300

**(20) If (4x+5) : (3x+11) = 13 : 17, find the value of x.**

**(21) If x : y = 3 : 4, find (3x + 4y) : (5x + 6y).**

**(22) If x : y = 6 : 11, find (8x – 3y) : (3x + 2y).**

**(23) Two numbers are in the ratio 5 : 7. If the sum of the numbers is 720, find the numbers.**

Solution: Let the numbers will be 5x and 7x.

Hence, the numbers are (5 ×60) = 300 and (7 ×60) = 420.

**(24) Which ratio is greater?**

(i) (5 : 6) or (7 : 9)

(ii) (2 : 3) or (4 : 7)

Hence, (2 : 3) >(4 : 7)

(iii) (1 : 2) or (4 : 7)

Hence, (1 :2)<(4 :7).

(iv) (3 : 5) or (8 : 13)

**(25) Arrange the following ratios in ascending order:**

(i) (5 : 6), (8 : 9), (11 : 18)

Hence, (11 : 18)< (5 : 6)< (8 : 9).

(ii) (11 : 14), (17 : 21), (5 : 7) and (2 : 3)

Hence, (2 :3)<(5 :7)<(11 :14)<(17 :21).

Mera pas aap 8a ka xusara methods se baba sakta hu

Kaise bolo send karo mere ko

Hi why you don’t multiple C in Q no 6

Tell me with full explain

A:B and B:C, in where B is in both section so B is common

8a ka question no 5,6 don’t understand