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A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Answer & Solution
Correct Answer:
Option
C
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = $$\left( {3 - \frac{{3x}}{8} + x} \right)$$ litres
Quantity of syrup in new mixture = $$\left( {5 - \frac{{5x}}{8}} \right)$$ litres
$$\eqalign{ & \therefore {3 - \frac{{3x}}{8} + x} = {5 - \frac{{5x}}{8}} \cr & \Rightarrow 5x + 24 = 40 - 5x \cr & \Rightarrow 10x = 16 \cr & \Rightarrow x = \frac{8}{5} \cr} $$
So, part of the mixture replaced
$$\eqalign{ & = {\frac{8}{5} \times \frac{1}{8}} \cr & = \frac{1}{5} \cr} $$
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = $$\left( {3 - \frac{{3x}}{8} + x} \right)$$ litres
Quantity of syrup in new mixture = $$\left( {5 - \frac{{5x}}{8}} \right)$$ litres
$$\eqalign{ & \therefore {3 - \frac{{3x}}{8} + x} = {5 - \frac{{5x}}{8}} \cr & \Rightarrow 5x + 24 = 40 - 5x \cr & \Rightarrow 10x = 16 \cr & \Rightarrow x = \frac{8}{5} \cr} $$
So, part of the mixture replaced
$$\eqalign{ & = {\frac{8}{5} \times \frac{1}{8}} \cr & = \frac{1}{5} \cr} $$
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Login5/8-5x/8=1/2
Solving x=1/5
Let the volume of the vessel = x liters.
Out of this x liters 3/8(x) is water & 5/8(x) is syrup.
Let y liters of liquid is taken out. So out of this y liters 3/8(y) is water and 5/8(y) is syrup.
The water in the remaining liquid is 3/8(x-y) .
The syrup in the remaining liquid is 5/8(x-y).
Now y liters of water is added to the remaining liquid so that the final liquid contains half water and half syrup.
=> 3/8(x-y)+y=5/8(x-y)
=>y=1/5(x)
or y = 1/5(x).
Answer: The amount of liquid drawn out = The amount of water added = 1/5 of x.