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MODEL PAPER 9 (STANDARD) — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsMODEL PAPER 9 (STANDARD)3 Marks
Question
If two pipes function simultaneously, a reservoir will be filled in $12$ hours. One pipe fills the reservoir $10$ hours faster than the other. How many hours will the second pipe take to fill the reservoir?
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Answer

Let, the slower pipe takes $x$ hours to fill the reservoir.
Hence, the faster pipe will take $(x-10)$ hours to fill the reservoir.
Since, the slower pipe takes $x$ hours to fill the reservoir.
$\therefore$ Portion of the reservoir filled by the slower pipe in $1$ hour $=\frac{1}{x}$
$\therefore$ Portion of the reservoir filled by the slower pipe in $12$ hours $=\frac{1}{x} \times 12=\frac{12}{x}$.
Now, portion of the reservoir filled by the faster pipe in $1 hr =\frac{1}{x-10}$
$\therefore$ Portion of the reservoir filled by faster pipe in $12$ hours $=\frac{1}{x-10} \times 12=\frac{12}{x-10}$
It is given that the reservoir is completely filled in $12$ hours by simultaneously operating both pipes.
$\therefore \frac{12}{x}+\frac{12}{x-10}=1$
$\Rightarrow \frac{12(x-10)+12 x}{x(x-10)}=1$
$\Rightarrow \frac{12 x-120+12 x}{x^2-10 x}=1$
$\Rightarrow x^2-10 x=24 x-120$
$\Rightarrow x^2-10 x-24 x+120=0$
$\Rightarrow x^2-34 x+120=0$
$\Rightarrow x^2-30 x-4 x+120=0$
$\Rightarrow x(x-30)-4(x-30)=0$
$\Rightarrow(x-30)(x-4)=0$
$\Rightarrow x-30=0 .$
$[\therefore x-4 \neq 0, \text { otherwise }(x-10)$ i.e time taken by faster pipe will become $ -6  hr]$
i.e. negative, which is not possible 
$\Rightarrow x=30$
Hence, the second pipe take $30$ hours to fill the reservoir.

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