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Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. What is the time in which each tap can separately fill the tank?

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large tap = 25 h and small tap = 15 h

large tap = 30 h and small tap = 10 h

large tap = 20 h and small tap = 25 h

large tap = 35 h and small tap = 25 h

answer is A.

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Detailed Solution

Given that two water taps together can fill a tank in

9 \frac{3}{8}

hours.
Also, the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately.
Let the time taken by a smaller tap to fill the tank be n hours.
In 1 hour, the part of the tank filled is

1 n

.
Let the time taken by the larger tap to fill the tank be = n-10 hours.
In 1 hour, the part of the tank filled is

1 n − 10

.
Both the taps take together time to fill the tank

= 875

hours.
The part of the tank filled in 1 hour by both the tanks

= 875

.
Now,

1 n + 1 n − 10 = 875

Form the quadratic equation.
Take L.C.M. of

n and (n + 10)

⇒ n − 10 + n n (n − 10) = 875 ⇒ 2 n − 10 n 2 − 10 n = 875 ⇒ 75 (2 n − 10) = 8 n 2 − 10 n ⇒ 150 n − 750 = 8 n 2 − 80 n

⇒ 8 n 2 − 80 n − 150 n + 750 = 0

⇒ 8 n 2 − 230 n + 750 = 0

We know that discriminant,

D = b 2 − 4 ac

.
Here,
a=8, b=-230 and c=750.

⇒ D = (− 230) 2 − 4 (8) (750) ⇒ D = 52900 − 24000 ⇒ D = 28900

Put the values in the formula,

n = − b ± b 2 − 4 a c 2 a

⇒ n = − (− 230) ± 28900 2 (8)

⇒ n = 230 ± 170 16 {28900 = 170}

⇒ n = 230 + 170 16 ⇒ n = 40016 ⇒ n = 25

⇒ n = 230 − 170 16 ⇒ n = 6016 ⇒ n = 154

For n=

154

,
The time taken by smaller tap to fill the tank is n hours,

⇒ 154 h o u r s

⇒

3 hours 45 minutes
The time taken by larger tap to fill the tank = (n – 10) hours,

⇒ 154 − 10 ⇒ − 25 4 h o u r s

⇒

-6 hours 15 minutes.
Since time can never be negative, n=

154

is invalid.
So, n= 25.
Hence,
The time taken by a smaller tap to fill the tank is n hours = 25 hours.
The time taken by a larger tap to fill the tank,