Question

6. Two water taps together can fill a tank in 9 hours. The tap of larger diameter takes 10
hours less than the smaller one to fill the tank separately. Find the time in which each tap
can separately fill the tank.​

Answer 1

$\bf\large\underline\pink{Solution:-}$

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.

Answer 2

$\huge\purple{Answer:-}$

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