# Moderate Alligations or Mixtures Solved QuestionAptitude Discussion

 Q. A can contains a mixture of two liquids A and B in 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. 28 litres ✔ B. 21 litres ✖ C. 45 litres ✖ D. 36 litres

Solution:
Option(B) is correct

Suppose the can initially contains $7x$ and $5x$ litres of mixtures A and B respectively.

When 9 litres of mixture are drawn off, quantity of A in mixture left:

$7x-\left(\dfrac{7}{12}\right)\times 9 = 7x -\dfrac{21}{4}$ litres
Similarly quantity of B in mixture left:
$5x-\left(\dfrac{5}{12}\right)\times 9 = 5x -\dfrac{15}{4}$ litres

Therefore ratio becomes:
\begin{align*}
\dfrac{7x-\dfrac{21}{4}}{5x-\dfrac{15}{4}+9}&=\dfrac{7}{9}\\
\Rightarrow \dfrac{28x-21}{20x+21}&=\dfrac{7}{9}\\
\Rightarrow 252x-189&=140x+147\\
\Rightarrow 112x &=336\\
\Rightarrow x&=3
\end{align*}
So the can contained:
$7\times x=7\times 3= 21$ litres of A initially.

## (7) Comment(s)

AYUSH
()

A:B

7:5

REPLACING 9 LITRES OF SOLN BY B

A:B

7:9

VALUE OF B INCREASES BY 9-5 = 4 UNITS

4 UNITS = 9 LITRS

1 UNIT = 9/4 LTRS

SINCE A:B

7:9

THEREFORE 16 UNITS = 16 *(9/4) = 36

INITIAL VOLUME OF A 7*36/12 =21 LITRES

Bing
()

a much simpler solution is:

initial quantity of A = 7/12 (x)

final quantity of A = 7/16 (x)

quantity of A removed = 21/4

assuming that the total quantity of solution remains after addition of 9 litres of B.

the equation becomes = 7/12 (x) - 21/4 = 7/16 (x)

solving for x we get = 36 litres, which is the total quantity of solution.

since initially A= 7/12 (X) , therefore the initial quantity of A turns out to be 21.

Akanksha
()

how did you get 21/4?

Elamurugan
()

I think initial mixture should be, $7+5=12\times 3=36$

Sreethi Reddy
()

what abt addition of b??

Mangesh Kumar
()

Deepak
()

Mangesh, I think the when the ratio becomes $7:9$, there is a typo in the solution.

the $3^{rd}$ equation given should be modified from:
$\dfrac{7x-(21/4)}{5x-(15/4)}=\dfrac{7}{9}$ to
$\dfrac{7x-(21/4)}{5x-(15/4)+ \textbf{9}}=\dfrac{7}{9}$

Which will take us to the correct next step as given in the existing solution.

I guess they will take it into consideration and and update the solution.