A solution has 18% of A, 40% of B and the rest is C. If 50% of B is wi

1.	A solution has 18% of A, 40% of B and the rest is C. If 50% of B is withdrawn and the same amount of solution A is put back into the mixture, the quantity of C would be 160 litres more than the final quantity of A. Find the initial quantity of B in the mixture?1. 15602. 15803. 16804. 1600
Answer» Correct Answer - Option 4 : 1600 Given: Quantity of A = 18% of total solution Quantity of B = 40% of total solution Quantity of C = Total solution – (Total quantity of A and B) 50% of B is withdrawn and the same amount of solution A is put back into the mixture Initial quantity of C = 160 Calculations: Let the total solution be 100x Quantity of A = 100x × 18/100 = 18x Quantity of B = 100x × 40/100 = 40x Quantity of C = Total solution – (Total quantity of A and B) ⇒ 100x – (18x + 40x) ⇒ 100x – 58x = 42x It is given that 50% of B is withdrawn and the same amount of solution A is put back into the mixture = 50% of B = (50/100) × 40x = 20x A will be = 18x + 20x = 38x The quantity of C would be 160 litres more than the final quantity of A ⇒ 42x – 38x = 160 ⇒ 4x = 160 ⇒ x = 40 Initial quantity of B in the mixture = 40x = 40 × 40 = 1600 litres ∴ Initial quantity of B in the mixture is 1600 litres

A solution has 18% of A, 40% of B and the rest is C. If 50% of B is withdrawn and the same amount of solution A is put back into the mixture, the quantity of C would be 160 litres more than the final quantity of A. Find the initial quantity of B in the mixture?1. 15602. 15803. 16804. 1600

Answer» Correct Answer - Option 4 : 1600

Given:

Quantity of A = 18% of total solution

Quantity of B = 40% of total solution

Quantity of C = Total solution – (Total quantity of A and B)

50% of B is withdrawn and the same amount of solution A is put back into the mixture

Initial quantity of C = 160

Calculations:

Let the total solution be 100x

Quantity of A = 100x × 18/100 = 18x

Quantity of B = 100x × 40/100 = 40x

Quantity of C = Total solution – (Total quantity of A and B)

⇒ 100x – (18x + 40x)

⇒ 100x – 58x = 42x

It is given that 50% of B is withdrawn and the same amount of solution A is put back into the

mixture = 50% of B = (50/100) × 40x = 20x

A will be = 18x + 20x = 38x

The quantity of C would be 160 litres more than the final quantity of A

⇒ 42x – 38x = 160

⇒ 4x = 160

⇒ x = 40

Initial quantity of B in the mixture = 40x = 40 × 40 = 1600 litres

∴ Initial quantity of B in the mixture is 1600 litres