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A container originally contains $10$ litres of pure spirit. From this container, $1$ litre of spirit replaced with $1$ litre of water. Subsequently, $1$ litre of the mixture is again replaced with $1$ litre of water and this process is repeated one more time. How much spirit is now left in the container?

1. $7.58$ litres
2. $7.84$ litres
3. $7$ litres
4. $7.29$ litres
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Quantity left after $n$ operations $=x\left(1 -\dfrac{y}{x}\right)^{n}$
where, $x=$ initial quantity
$y=$ amount of mixture withdrawn each time (this should be same every time)
$n=$ no. of times operation performed

$\quad = 10\left(1 -\dfrac{1}{10}\right)^{n}$
$\quad =10 \left(\dfrac{9}{10}\right)^{3}$
$\quad = 10\times 0.9\times 0.9\times 0.9$
$\quad = 10\times 0.729 = 7.29$ litres

Hence, option $D$ is correct.
edited by
Initial = 10L of spirit

stage 1:-

Take out 1 L of  S and add 1 L of W

9L of S + 1L of W

stage 2:-

1 L of this mixture contains 0.9 L of S and 0.1 L of W

Take out 1 L of above mixture and add 1 L of W

satge 3:-

Currently 10 L = 8.1 L of S + 1.9 L of W

1 L of this mixture contains 0.81 L of S and 0.19 L of W

Take out 1 L of above mixture and add 1 L of W

stage 4:

Currently 10 L = 7.29 L of S + 2.71 L of W

Hence the mixture contains 7.29 L of Spirit and 2.71 L of Water at final stage ie Ans is ==> D)
edited
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1 L of this mixture contains 0.9 L of S and 0.1 L of W

Take out 1 L of above mixture and add 1 L of W

Can you explain next step??