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 in which each tap can separately fill the tank.
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Let the time taken by the smaller tap to fill the tank = x hours
time taken by larger tap = x - 9
In 1 hour, the smaller tap will fill 1/x of tank
In 1 hour, the larger tap will fill 1/(x-9) of tank.
In 1 hour both the tank will the tank = ⅙
In 1 hour both the tank will fill the tank= 1/x + 1/x-9
⅙ = (x-9)+ (x)/(x) (x-9)
⅙ = 2x-9/x²-9x
6(2x-9) = x²-9x
12x-54= x²-9x
x²- 9x-22x +54= 0
x²- 21x+54= 0
x²-18x-3x +54=0
x(x-18)-3(x-18)=0
(x-18) (x-3)=0
x= 18, x=3
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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Hope this will help you....
time taken by larger tap = x - 9
In 1 hour, the smaller tap will fill 1/x of tank
In 1 hour, the larger tap will fill 1/(x-9) of tank.
In 1 hour both the tank will the tank = ⅙
In 1 hour both the tank will fill the tank= 1/x + 1/x-9
⅙ = (x-9)+ (x)/(x) (x-9)
⅙ = 2x-9/x²-9x
6(2x-9) = x²-9x
12x-54= x²-9x
x²- 9x-22x +54= 0
x²- 21x+54= 0
x²-18x-3x +54=0
x(x-18)-3(x-18)=0
(x-18) (x-3)=0
x= 18, x=3
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
==================================================================
Hope this will help you....
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