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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.

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Solution

Let the time taken by the smaller tap to fill the tank = x hours and

time taken by larger tap = x - 9

In 1 hour, the smaller tap will fill 1x of tank

In 1 hour, the larger tap will fill 1x9 of tank.

In 1 hour both the tank will fill the tank = 1x+1x9

But its given that in 1 hour both the tank will fill 16 of the tank
16 =1x +1x9

16=x9+xx(x9) 16=2x9x29x
Solving by cross multiplication,

6(2x9)=x29x

12x54=x29x

x29x12x+54=0

x221x+54=0

Factorizing by splitting the middle term,

x218x3x+54=0

x(x18)3(x18)=0

(x18)(x3)=0
x=18,x=3

If we take x= 18

Smaller tap = (x) = 18 h

Larger tap = (x-9) = 18-9 = 9h

Hence, the time taken by the smaller tap to fill the tank = 18 hrs & the time taken by the larger tap to fill the tank = 9 h


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