A reservoir has two pipes, A and B. A can fill the reservoir 5 hours faster than B. If both together fill the reservoir in 6 hours, the reservoir will be filled by A alone in-
Solution
Correct Answer: Option B
Let,
A alone can fill the reservoir in x hours
B can fill in x + 5 hours
Both complete in 1 hour = (1/x) + (1/ x + 5)
= (2x + 5)/(x2 + 5x)
Now,
1/{(2x + 5)/(x2 + 5x)} = 1/6
⇒ (x2 + 5x)/(2x + 5) = 6
⇒ x2 + 5x = 12x + 30
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0
⇒ x (x - 10) + 3 (x - 10) = 0
⇒ (x - 10) (x + 3) = 0
∴ x = 10 or, x = -3 , negative value not possible
A alone can fill the reservoir in 10 hours
A alone can fill the reservoir in x hours
B can fill in x + 5 hours
Both complete in 1 hour = (1/x) + (1/ x + 5)
= (2x + 5)/(x2 + 5x)
Now,
1/{(2x + 5)/(x2 + 5x)} = 1/6
⇒ (x2 + 5x)/(2x + 5) = 6
⇒ x2 + 5x = 12x + 30
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0
⇒ x (x - 10) + 3 (x - 10) = 0
⇒ (x - 10) (x + 3) = 0
∴ x = 10 or, x = -3 , negative value not possible
A alone can fill the reservoir in 10 hours
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