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Question

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. [CBSE 2011]

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Solution

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = Vx

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = Vx×6

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = Vx-9×6

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
V1x+1x-9×6=V1x+1x-9=16x-9+xxx-9=162x-9x2-9x=16
12x-54=x2-9xx2-21x+54=0x2-18x-3x+54=0xx-18-3x-18=0
x-18x-3=0x-18=0 or x-3=0x=18 or x=3

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

∴ x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

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