Question : If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?
Option 1: 20 hours
Option 2: 18 hours
Option 3: 15 hours
Option 4: 12 hours
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Correct Answer: 20 hours
Solution :
Let the slower pipe alone can fill the tank in $x$ hours. Then, faster pipe alone can fill the tank in ($x$ – 10) hours.
Part of the tank filled by the slower pipe in 1 hour = $\frac{1}{x}$
Part of the tank filled by the faster pipe in 1 hour = $\frac{1}{(x – 10)}$
Time taken to fill the tank when both pipes are opened simultaneously = 12 hours
So, part of the tank filled by both the pipes in 1 hour = $\frac{1}{12}$
According to the question,
$\frac{1}{x}$ + $\frac{1}{(x–10)}$ = $\frac{1}{12}$
⇒ 12($x$ – 10) + 12$x$ = $x$($x$ – 10)
⇒ 12$x$ – 120 + 12$x$ = $x$
2
– 10$x$
⇒ $x$
2
– 34$x$ + 120 = 0
⇒ $x$
2
– 30$x$ – 4$x$ + 120 = 0
⇒ $x$($x$ – 30) – 4($x$ – 30) = 0
⇒ ($x$ – 4)($x$ – 30) = 0
⇒ $x$ = 4 or 30
Now, $x$ cannot be 4 as ($x$ – 10) will be negative.
⇒ $x$ = 30
Time taken by the faster pipe alone to fill the tank
= $x$ – 10 = 30 – 10 = 20 hours
Hence, the correct answer is 20 hours.
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