To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

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To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

A$\frac{10m}{m-10}$

B$\frac{5m}{m-10}$

C$\frac{2m}{m-10}$

D$\frac{m}{m-10}$

Nithin ? undefined

Correct answer is: \(\frac{10m}{m-10}\)

Explanation:

Tricky approach

\(\text{ Acid = }\frac{m}{100} \times m = \frac{m^{2}}{100} \)

\(\text{New percentage =} \frac{m^{2} \times 100}{ \left(m+x\right) \times 100 }\)

\(\implies \frac{m^{2}}{m+x}=m-10 \)

\(\implies m^{2} - 10m + mx - 10x = m^{2}\)
\(\implies x \left(m-10\right)=10m\)

\(\implies x = \frac{10m}{m-10}\)

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Correct answer is: \(\frac{10m}{m-10}\) Explanation: Tricky approach \(\text{ Acid = }\frac{m}{100} \times m = \frac{m^{2}}{100} \) \(\text{New percentage =} \frac{m^{2} \times 100}{ \left(m+x\right) \times 100 }\) \(\implies \frac{m^{2}}{m+x}=m-10 \) \(\implies m^{2} - 10m + mx - 10x = m^{2}\) \(\implies x \left(m-10\right)=10m\) \(\implies x = \frac{10m}{m-10}\) Nithin ? answered undefined All Comments

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