From a container of milk, $5$ litres of milk is replaced with $5$ litres of water. This process is repeated again. Thus in two attempts the ratio of milk and water became $81 : 19.$ The initial amount of milk in the container was

50

litres

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45

litres

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40

litres

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25

litres

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Solution

The correct option is B $50$ litres
Milk remaining/water $= 81 / 19$
Milk remaining/milk inital $= 81 / 100$
$∴ R e m a i n i n g m i l k = i n i t i a l m i l k [1 - (q u a n t i t y t a k e n o u t / t o t a l a m o u n t)]^{n}$
$\Rightarrow 81 x = 100 x (1 - 5 / k)^{2}$
$\Rightarrow \frac{81}{100} = (1 - 5 / k)^{2}$
$\Rightarrow (\frac{9}{10})^{2} = (1 - 5 / k)^{2}$
$\Rightarrow 5 / k = 1 - \frac{9}{10}$
$\Rightarrow 5 / k = \frac{1}{10}$
$\Rightarrow k = 50 l i t r e$

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From a container of milk, 5 litres of milk is replaced with 5 litres of water. This process is repeated again. Thus in two attempts the ratio of milk and water became 81:19. The initial amount of milk in the container was

The correct option is B 50 litresMilk remaining/water=81/19Milk remaining/milk inital=81/100∴Remaining milk=initial milk [1−(quantity taken out/total amount)]n ⇒81x=100x(1−5/k)2⇒81100=(1−5/k)2⇒(910)2=(1−5/k)2⇒5/k=1−910⇒5/k=110⇒k=50 litre

From a container of milk, $5$ litres of milk is replaced with $5$ litres of water. This process is repeated again. Thus in two attempts the ratio of milk and water became $81 : 19.$ The initial amount of milk in the container was