Quantity , 1: Value of y. A vessel contains 2 | vedclass

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(D) Initial contents: Water $= 2.5 \, 	ext{L}$,Milk $= 10 \, 	ext{L}$. Total $= 12.5 \, 	ext{L}$.
After removing $20 \%$ $(1/5)$: Remaining Water

Vedclass · Accepted Answer

Quantity $I =$ Quantity $II$

Vedclass · Answer

(D) Initial contents: Water $= 2.5 \, 	ext{L}$,Milk $= 10 \, 	ext{L}$. Total $= 12.5 \, 	ext{L}$.
After removing $20 \%$ $(1/5)$: Remaining Water $= 2.5 - (0.2 	imes 2.5) = 2 \, 	ext{L}$. Remaining Milk $= 10 - (0.2 	imes 10) = 8 \, 	ext{L}$.
Initial ratio (Water:Milk) $= 2:8 = 1:4$.
Adding $x$ liters of water to reverse the ratio to $4:1$: $\frac{2+x}{8} = \frac{4}{1} \Rightarrow 2+x = 32 \Rightarrow x = 30 \, 	ext{L}$.
Now,contents are Water $= 32 \, 	ext{L}$,Milk $= 8 \, 	ext{L}$.
Adding $y$ liters of milk to reverse the ratio back to $1:4$: $\frac{32}{8+y} = \frac{1}{4} \Rightarrow 128 = 8+y \Rightarrow y = 120 \, 	ext{L}$.
Since $y = 120$ and $Quantity \, 2 = 120$,$Quantity \, I = Quantity \, II$.

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