Question

There are three containers of equal capacity.The ratio of sulphuric acid to water in the first container is
\[3:2\] that in the second container is \[7:3\]and in the third container it is \[11:4\].If all the liquids are mixed together
 then the ratio of sulphuric acid to water in the mixture will be?
\[\begin{array}{*{35}{l}}
   A)\text{ }61\text{ }:\text{ }29 \\
   B)\text{ }61\text{ }:\text{ }28 \\
   C)\text{ }60\text{ }:\text{ }29 \\
   D)\text{ }59\text{ }:\text{ }29 \\
\end{array}\]

Answer 1

Hint: Equate the quantity of mixture in each container to any unknown variable like .For equal capacity when they are mixed, the sum of the quantities of sulphuric acid in each container is divided by the sum of quantities water in each container. And by simplifying those ratios we will get the actual ratio of the mixture when they are mixed.

Complete step-by-step answer:
Let us consider,
The quantity of mixture in each container is “$x$”
From the problem we have given that,
⟹The ratio of sulphuric acid to water in the first container = \[3:2\]
From this, the quantity of sulphuric acid in first container =$\dfrac{3}{3+2}x$= $\dfrac{3}{5}x$
Similarly, the quantity of water in first container =$\dfrac{2}{3+2}x$= $\dfrac{2}{5}x$
⟹The ratio of sulphuric acid to water in second container = \[7:3\]
From this , the quantity of sulphuric acid in second container = $\dfrac{7}{7+3}x=\dfrac{7}{10}x$
 Similarly, the quantity of water in second container = $\dfrac{3}{7+3}x=\dfrac{3}{10}x$
⟹ The ratio of sulphuric acid to water in third container = \[11:4\]
From this , the quantity of sulphuric acid in third container = $\dfrac{11}{11+4}x=\dfrac{11}{15}x$
Similarly, the quantity of water in third container = $\dfrac{4}{11+4}x=\dfrac{4}{15}x$
If the all the liquids are mixed then,
The required mixture ratio when they mixed up = (Sum of sulphuric acid quantities in three containers) :(Sum of water quantities in three containers)
 Therefore,
The required ratio = $\left( \dfrac{3x}{5}+\dfrac{7x}{10}+\dfrac{11x}{15} \right):\left( \dfrac{2x}{5}+\dfrac{3x}{10}+\dfrac{4x}{15} \right)$
Now take ‘$x$’as common and cancel it , then we get as follows
The required ratio = $\left( \dfrac{3}{5}+\dfrac{7}{10}+\dfrac{11}{15} \right):\left( \dfrac{2}{5}+\dfrac{3}{10}+\dfrac{4}{15} \right)$
On taking L.C.M for first two terms we get
⇒$\left( \dfrac{30+35}{50}+\dfrac{11}{15} \right):\left( \dfrac{20+15}{50}+\dfrac{4}{15} \right)$
⇒$\left( \dfrac{65}{50}+\dfrac{11}{15} \right):\left( \dfrac{35}{50}+\dfrac{4}{15} \right)$
⇒$\left( \dfrac{13}{10}+\dfrac{11}{15} \right):\left( \dfrac{7}{10}+\dfrac{4}{15} \right)$
Now let us take 150 as L.C.M we get
⇒$\left( \dfrac{(13\times 15)+(11\times 10)}{150} \right):\left( \dfrac{(7\times 15)+(4\times 10)}{150} \right)$
⇒$\left( \dfrac{305}{150} \right):\left( \dfrac{145}{150} \right)$
Here ‘150’ on both sides gets cancelled out
⇒$305:145$
Now the simplified ratio is ⇒$61:29$
Hence,
 If all the liquids are mixed together then the ratio of sulphuric acid to water in the mixture will be “$61:29$”
The answer is option (A)

Note: a)We must be careful that with respect to which liquid the ratio is asked for and according to that we must proceed.
 b)For the ratios ,the quantities must be in the same unit.If they are not equal,they should be expressed in the same unit before the ratio is to be taken.And the ratio should be in the simplified form. The simplified form will get only when we multiply with any related factor or with the same number both in numerator and denominator.