anceer wrote:
A can contains a mixture of 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 liter of liquid A was contained by the can initially?
A. 10
B. 20
C. 21
D. 25
E. 27
Using the values here might be the most straightforward way for this question.
As A:B::7:5 ---> only option C is a multiple of 7 and hence it is a good place to start. Also A:B::7:5 means that , A = (712)*Total and B = (5/12)*Total
If A = 21 , B = 15 ---> remove 9 litres ---> you remove (7/12)*9 of A ---> A remaining = 21-(7/12)*9 = 63/4
Similarly, for B, you remove (5/12)*9 ---> B remaining = 15 - (5/12)*9 = 45/4 and then add 9 more litres of B ---> 9+45/4 = 81/4
Thus A/B (final ratio) = (45/4)/(81/4) = 7:9 , the same as the final ratio mentioned in the question.
Hence C is the correct answer.
FYI, the algebraic equations will become:
A/B = 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9), where 7x and 5x are initial quantities of A and B respectively.
Thus, 7/9 = (7x-(7/12)*9)/ (5x-(5/12)*9+9) ---> giving you x=3. Thus A (original) = 7*3 = 21.
Hope this helps.