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Q. A vessel is filled with a solution of, 3 parts soda and 5 parts rum. How much of the solution must be taken out and replaced with soda so that the solution contains equal amount of sod and rum?

**(A)** 8/5

**(B)** 2/5

**(C)** 1/5

**(D)** 5/8

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(A) 8/5

Say initially the amount of solution in the vessel is 8 units.

Now say x units of the solution is taken out and replaced with soda.

Amount of soda left in the vessel = **(3 - 3/8 x + x)** units (since, 3 units soda initially, of the x units removed 3/8 part is soda, and x parts of soda added)

Amount of rum left in the vessel = **(5 - 5/8 x)** units (5 units of rum initially, and of the x units removed 5/8 part is rum)

For both to be equal proportions,

**3 - 3/8 x + x = 5 - 5/8 x
24 - 3x + 8x = 40 - 5x
10x = 16
x = 8/5**

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Vessel contains $3$ parts of soda and $5$ parts of rum.

∴ Total solution in the vessel = $(3+5) = 8$ parts

Assuming,

initially in the vessel, there are $8$ litres of solution in which $3$ litres are soda and $5$ litres is rum.

Now, some solution (assuming $S$ litre) has been taken out and replaced with soda (taken as same amount as $S$ litre) so that the solution contains an equal amount of soda and rum $(1:1)$.

As, $S$ litre of solution has been taken out-

remaining amount of soda in the $(8-S)$ litre solution = $3-\dfrac{3}{8}S$ $\left [ \text{As, S litre solution also has 3 parts of soda and 5 parts of rum } \right ]$

remaining amount of rum in the $(8-S)$ litre solution = $5- \dfrac{5}{8}S$

Now, $S \text{ litre}$ soda also been added in the vessel.

Total amount of soda in the vessel = $3-\dfrac{3}{8}S+S$

Now, as per the criteria, $\left ( 3-\dfrac{3}{8}S \right )$ & $ \left ( 5- \dfrac{5}{8}S \right )$ needs to be same.

∴ $3-\dfrac{3}{8}S+S$ = $5- \dfrac{5}{8}S$

Or, $\dfrac{24-3S+8S}{8} = \dfrac{40-5S}{8}$

Or, $24+5S = 40-5S$

Or, $10S = 40-24$

Or, $S = \dfrac{16}{10} = \dfrac{8}{5}$

∴ $\color{gold}{\dfrac{8}{5} \text{ Lt.}} \color{green}{\text{solution has been removed initially, which is equal to}}$ $\color{gold}{\dfrac{1}{5} ^{th} \text{of 8 Lt.}}$ OR $\color{gold}{\dfrac{1}{5} ^{th} \text{of initial solution}}$ $\color{green}{\text{has been taken out and replaced with soda so that }}$ $\color{green}{\text{the solution contains equal amount of soda and rum}}$

∴ Total solution in the vessel = $(3+5) = 8$ parts

Assuming,

initially in the vessel, there are $8$ litres of solution in which $3$ litres are soda and $5$ litres is rum.

Now, some solution (assuming $S$ litre) has been taken out and replaced with soda (taken as same amount as $S$ litre) so that the solution contains an equal amount of soda and rum $(1:1)$.

As, $S$ litre of solution has been taken out-

remaining amount of soda in the $(8-S)$ litre solution = $3-\dfrac{3}{8}S$ $\left [ \text{As, S litre solution also has 3 parts of soda and 5 parts of rum } \right ]$

remaining amount of rum in the $(8-S)$ litre solution = $5- \dfrac{5}{8}S$

Now, $S \text{ litre}$ soda also been added in the vessel.

Total amount of soda in the vessel = $3-\dfrac{3}{8}S+S$

Now, as per the criteria, $\left ( 3-\dfrac{3}{8}S \right )$ & $ \left ( 5- \dfrac{5}{8}S \right )$ needs to be same.

∴ $3-\dfrac{3}{8}S+S$ = $5- \dfrac{5}{8}S$

Or, $\dfrac{24-3S+8S}{8} = \dfrac{40-5S}{8}$

Or, $24+5S = 40-5S$

Or, $10S = 40-24$

Or, $S = \dfrac{16}{10} = \dfrac{8}{5}$

∴ $\color{gold}{\dfrac{8}{5} \text{ Lt.}} \color{green}{\text{solution has been removed initially, which is equal to}}$ $\color{gold}{\dfrac{1}{5} ^{th} \text{of 8 Lt.}}$ OR $\color{gold}{\dfrac{1}{5} ^{th} \text{of initial solution}}$ $\color{green}{\text{has been taken out and replaced with soda so that }}$ $\color{green}{\text{the solution contains equal amount of soda and rum}}$

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Soda in soluton = 3/(3+5) = 3/8

Rum in solution = 5/(3+5) = 5/8

Let x be the volume of solution taken out.

Then soda taken out = x * 3/8

Then rum taken out = x * 5/8

The taken out part has to be replaced with soda, means x amount of soda will be added to the new solution which makes soda and rum equal in volume/quantity. The final equation will, be :

3/8 - 3x/8 + x = 5/8 -5x/8

which gives x = 1/5

Rum in solution = 5/(3+5) = 5/8

Let x be the volume of solution taken out.

Then soda taken out = x * 3/8

Then rum taken out = x * 5/8

The taken out part has to be replaced with soda, means x amount of soda will be added to the new solution which makes soda and rum equal in volume/quantity. The final equation will, be :

3/8 - 3x/8 + x = 5/8 -5x/8

which gives x = 1/5

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