Question

Two vessels A and B contain spirit water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixtures are mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8:5?
a.4:8
b.3:4
c.5:6
d.7:9

Answer 1

Hint: Suppose the amount of spirit and water in the vessel A and B as 5x and 2x and 7y, 6y respectively. Now, suppose the amount taken from vessel A and vessel B as two other variables. Now, find the amounts of spirit and water in the amount taken from both vessels using ratios given in the problem. Now, equate the ratios of spirit and water in terms of variables as given in the problem. Solve it to get the ratio of amount taken from both vessels.

Complete step by step answer:
Let us suppose the amount of spirit and water in vessel A are 5x, 2x and in vessel B are 7y, 6y as per the given ratios in the question i.e. 5:2 for vessel A and 7:6 for vessel B. Now, it is given in the question that a new mixture is formed in vessel C containing spirit and water in ratio 2:5 by mixing some amount of mixture from vessel A and B both.
So, let volume ${{v}_{1}}$ is taken from vessel A and volume ${{v}_{2}}$ is taken from vessel B for getting mixture in vessel c. Hence, we need to determine the ratio of ${{v}_{1}}:{{v}_{2}}$ . So, as we know that amount $'{{v}_{1}}'$ is taken from vessel A, so it will contain an amount of spirit and water both. As we know the total amount of water and spirit in the vessel A would be 5x + 2x = 7x.
Now, we know 7x the amount of mixture contains 2s amount of water. So, 1 amount of mixture will contain $\dfrac{2x}{7x}$ amount of water i.e. $\dfrac{2}{7}$ amount of water it means ${{v}_{1}}$ amount of mixture will contain $\dfrac{2}{7}{{v}_{1}}$ amount of water.
Similarly, as 7x amount of mixture is containing 5x amount of spirit. So, 1 amount of mixture will contain $\dfrac{5x}{7x}=\dfrac{5}{7}$ amount of spirit.
Hence, ${{v}_{1}}$ amount of mixture will contain $\dfrac{5}{7}{{v}_{1}}$ amount of spirit. In the similar way, we can calculate the amount of water and spirit in vessel B as well. So, we know the total amount of mixture in vessel B is 7y + 6y = 13y and hence, the 13y amount mixture contains 7y amount of spirit. It means 1 amount of mixture contains $\dfrac{7y}{13y}=\dfrac{7}{13}$ amount of spirit.
So, ${{v}_{2}}$ amount of mixture will contain \[\dfrac{7}{13}{{v}_{2}}\] amount of mixture.
Similarly, 13y amount of mixture is contained by amount of water. So, 1 amount of mixture will contain $\dfrac{6y}{13y}=\dfrac{6}{13}$ amount of water. It means ${{v}_{2}}$ amount of mixture will contain $\dfrac{6}{13}{{v}_{2}}$ amount of water. Now, total amount of water in the mixture would be sum of water added by vessel A and B i.e. sum of amounts of water by ${{v}_{1}}$ amount and ${{v}_{2}}$ amount. So, we get
Total amount of water $=\dfrac{2}{7}{{v}_{1}}+\dfrac{6{{v}_{2}}}{13}$
And hence, total amount of spirit $=\dfrac{5}{7}{{v}_{1}}+\dfrac{7{{v}_{2}}}{13}$ .
Now, we know the ratio of spirit and water in the vessel c is 2:5 from the question. So, we get equation as
$\begin{align}
  & \dfrac{\dfrac{5}{7}{{v}_{1}}+\dfrac{7{{v}_{2}}}{13}}{\dfrac{2{{v}_{1}}}{7}+\dfrac{6{{v}_{2}}}{13}}=\dfrac{8}{5} \\
 & \Rightarrow \dfrac{\dfrac{65{{v}_{1}}+49{{v}_{2}}}{91}}{\dfrac{26{{v}_{1}}+42{{v}_{2}}}{91}}=\dfrac{8}{5} \\
 & \Rightarrow \dfrac{65{{v}_{1}}+49{{v}_{2}}}{26{{v}_{1}}+42{{v}_{2}}}=\dfrac{8}{5} \\
\end{align}$
On cross multiplying the above equation we get
$\begin{align}
  & 5\left( 65{{v}_{1}}+49{{v}_{2}} \right)=8\left( 26{{v}_{1}}+42{{v}_{2}} \right) \\
 & 325{{v}_{1}}+245{{v}_{2}}=208{{v}_{1}}+336{{v}_{2}} \\
 & 325{{v}_{1}}-208{{v}_{1}}=336{{v}_{2}}-245{{v}_{2}} \\
 & 117{{v}_{1}}=91{{v}_{2}} \\
 & \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{91}{117}=\dfrac{7}{9} \\
\end{align}$
Hence, the ratio in which these mixtures are mixed to obtain a ratio of 2:5 would be 7:9.
So, option (d) is correct.

Note: One may directly calculate the amount of water and spirit in the quantity ${{v}_{1}},{{v}_{2}}$ taken from vessel A and vessel B. As the ratio of spirit to water in vessel A is 5:2 and 7:6 in vessel 2. So, total 7 units of mixture in A will contain 5 units of spirit and 2 units of water as per ratio. It means volume ${{v}_{1}}$ , a part of the vessel A will contain $\dfrac{2}{7}{{v}_{1}}$ water and $\dfrac{5}{7}{{v}_{1}}$ amount of spirit because ratio will be maintained throughout the relation. It means no need to suppose the total amounts of spirit and water for the vessels.
Take care for writing the ratios in terms of ${{v}_{1}},{{v}_{2}}$ and the given ratio as one spirit amount and water amount in terms of ${{v}_{1}},{{v}_{2}}$ should be directly proportional to the given amount of spirit and water i.e. one may write equation as
$\dfrac{\dfrac{5}{7}{{v}_{1}}+\dfrac{7{{v}_{2}}}{13}}{\dfrac{2{{v}_{1}}}{7}+\dfrac{6{{v}_{2}}}{13}}=\dfrac{5}{8}$
Which is wrong as
$\dfrac{5}{7}{{v}_{1}}+\dfrac{7{{v}_{2}}}{13},\dfrac{2{{v}_{1}}}{7}+\dfrac{6{{v}_{2}}}{13}$
Are the amount of spirit and water respectively, but ratio 5:8 is representing the ratio of amounts from water to spirit. So, take care of this step and be careful with the position of terms with these types of questions.