The ratio of two liquids B1 and B2 in a container is 3 ∶ 2. When 10 litres of mixture is removed from the container and is replaced with B2, the ratio of B1 and B2 becomes 3 ∶ 5. How many litres of B1 was in the container initially?

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Question

This question was previously asked in

DFCCIL Executive CBT1 Electrical 25 Aug 2023 Shift 2 Official Paper

Answer 1

Given:

Initial ratio B1 : B2 = 3 : 2

Mixture removed = 10 litres

Replaced with = B2

Final ratio B1 : B2 = 3 : 5

Formula Used:

Amount of B1 removed = (Ratio of B1 / Total ratio) × Quantity removed

Amount of B2 removed = (Ratio of B2 / Total ratio) × Quantity removed

Calculations:

Let initial quantity of B1 = 3x litres

Let initial quantity of B2 = 2x litres

Amount of B1 in 10 litres removed mixture = (3 / (3 + 2)) × 10 = (3/5) × 10 = 6 litres

Amount of B2 in 10 litres removed mixture = (2 / (3 + 2)) × 10 = (2/5) × 10 = 4 litres

Remaining B1 = 3x - 6

Remaining B2 = 2x - 4

After adding 10 litres of B2, final quantity of B2 = 2x - 4 + 10 = 2x + 6

Final ratio:

⇒ (3x - 6) / (2x + 6) = 3 / 5

⇒ 5 × (3x - 6) = 3 × (2x + 6)

⇒ 15x - 30 = 6x + 18

⇒ 9x = 48

⇒ x = 48 / 9 = 16 / 3

Initial quantity of B1 = 3x = 3 × (16 / 3) = 16 litres

∴ Initially, there were 16 litres of B1 in the container.