Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 3, 2020
retagged
Nov 12, 2020
by soujanyareddy13

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Ans is option (B)

Let the milkman sells $1$L of milk originally at Rs $100$

Let him mix $’x’$ L of water in the mixture. Then, total volume of mixture is now $’x+1’$ L

Now, he would have gained Rs $100$ on selling pure milk. But now, with $’x+1’$ L of impure milk, he got Rs $125$ i.e. Rs $25$ extra.

$\therefore$ $1$ L $\rightarrow$ Rs $100$

$’x+1’$ L $\rightarrow$ Rs $125$

$\therefore$ $100(x+1)=125$ $\Rightarrow$ $x=\frac{1}{4}$ L

$\therefore$ Percentage of water in the current mixture $=$ $\frac{\frac{1}{4}}{\frac{1}{4}+1}\times100=20$%

Let the milkman sells $1$L of milk originally at Rs $100$

Let him mix $’x’$ L of water in the mixture. Then, total volume of mixture is now $’x+1’$ L

Now, he would have gained Rs $100$ on selling pure milk. But now, with $’x+1’$ L of impure milk, he got Rs $125$ i.e. Rs $25$ extra.

$\therefore$ $1$ L $\rightarrow$ Rs $100$

$’x+1’$ L $\rightarrow$ Rs $125$

$\therefore$ $100(x+1)=125$ $\Rightarrow$ $x=\frac{1}{4}$ L

$\therefore$ Percentage of water in the current mixture $=$ $\frac{\frac{1}{4}}{\frac{1}{4}+1}\times100=20$%