The weight of a container alone is 25 % of th | vedclass

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Question

(NONE) Let the total weight of the container filled with fluid be $W$.
Weight of the container $= 0.25W$.
Weight of the fluid initially $= W - 0.25W

Vedclass · Accepted Answer

$\frac{8}{15}$

Vedclass · Answer

(NONE) Let the total weight of the container filled with fluid be $W$.
Weight of the container $= 0.25W$.
Weight of the fluid initially $= W - 0.25W = 0.75W$.
After removing some fluid,the new weight is $60 \%$ of $W$,which is $0.60W$.
Weight of the remaining fluid $= 0.60W - 0.25W = 0.35W$.
Weight of the removed fluid $= 0.75W - 0.35W = 0.40W$.
Fraction of fluid removed $= \frac{	ext{Weight of removed fluid}}{	ext{Initial weight of fluid}} = \frac{0.40W}{0.75W} = \frac{40}{75} = \frac{8}{15}$.

The weight of a container alone is $25 \%$ of the container filled with a certain fluid. When some fluid is removed,the weight of the container and remaining fluid is $60 \%$ of the original total weight. What fractional part of the liquid has been removed?

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