A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9…

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A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

  1. A.

    10

  2. B.

    25

  3. C.

    21

  4. D.

    30

Attempted by 4 students.

Show answer & explanation

Correct answer: C

Key idea: use parts to represent the mixture and remove proportionally.

Steps:

  1. Let the common multiple be x. Then total volume = 7x + 5x = 12x litres. Initially A = 7x litres and B = 5x litres.

  2. When 9 litres of the mixture is removed, it is removed in the same 7:5 proportion. Amount of A removed = 9 * (7/12) = 21/4 litres. So A left = 7x - 21/4 litres.

  3. Amount of B left after removal = 5x - 9 * (5/12) = 5x - 15/4. Then we add 9 litres of B, so B becomes 5x - 15/4 + 9 = 5x + 21/4 litres.

  4. Now the new ratio A : B = (7x - 21/4) : (5x + 21/4) which equals 7 : 9. Set up the equation:

  5. (7x - 21/4) / (5x + 21/4) = 7/9.

  6. Cross-multiply: 9(7x - 21/4) = 7(5x + 21/4). This gives 63x - 189/4 = 35x + 147/4.

  7. Solve for x: 63x - 35x = 147/4 + 189/4 → 28x = 336/4 = 84 → x = 3.

  8. Initial amount of A = 7x = 7 * 3 = 21 litres.

Answer: 21 litres.

Quick check: Initially A = 21, B = 15 (total 36). After removing 9 L, remaining total = 27 L; A left = 21 - 9*(7/12) = 63/4 = 15.75; B after adding 9 = 45/4 + 9 = 81/4 = 20.25. Ratio 15.75 : 20.25 = 63 : 81 = 7 : 9, which matches.

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