A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9…
20232023
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?
- A.
10
- B.
25
- C.
21
- D.
30
Attempted by 4 students.
Show answer & explanation
Correct answer: C
Key idea: use parts to represent the mixture and remove proportionally.
Steps:
Let the common multiple be x. Then total volume = 7x + 5x = 12x litres. Initially A = 7x litres and B = 5x litres.
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.
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.
Now the new ratio A : B = (7x - 21/4) : (5x + 21/4) which equals 7 : 9. Set up the equation:
(7x - 21/4) / (5x + 21/4) = 7/9.
Cross-multiply: 9(7x - 21/4) = 7(5x + 21/4). This gives 63x - 189/4 = 35x + 147/4.
Solve for x: 63x - 35x = 147/4 + 189/4 → 28x = 336/4 = 84 → x = 3.
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.