Question

Two taps, one big and one small, fill up ½ of the tank in 6 hours. The big tap takes

10 hours less than the small tap to fill up the full tank. How much time is required

by each of the tap to fill up the full tank?​

Answer 1

Step-by-step explanation:

Given that:Two taps, one big and one small, fill up ½ of the tank in 6 hours. The big tap takes 10 hours less than the small tap to fill up the full tank. How much time is required by each of the tap to fill up the full tank?

To find: time is required by each of the tap to fill up the full tank?

Solution:

Let big tap take x hours to fill the tank and

small tap takes (x+10) hours to fill the same tank

Part of tank full by big tap in 1 hour (1/x)th part

Part of tank full by small tap in 1 hour (1/(x+10))th part

Both taps can fill up half the tank in 6 hours,So can fill the full tank in 12 hours

Part fill by both taps in one hour

$\frac{1}{x} + \frac{1}{(x + 10)} = \frac{1}{12} \: \: \:$

Solve this equation

$\frac{x + 10 + x}{x(x + 10)} = \frac{1}{12} \\ \\ \frac{2x + 10}{ {x}^{2} + 10x } = \frac{1}{12} \\ \\ 24x + 120 = {x}^{2} + 10x \\ \\ {x}^{2} - 14x - 120 = 0 \\ \\ {x}^{2} - 20x + 6x - 120 = 0 \\ \\ x(x - 20) + 6(x - 20) = 0 \\ \\ (x - 20)(x + 6) = 0 \\ \\ x = 20 \: \: and \: x = - 6$

Time can never be a negative quantity.So,ignore x= -6

Thus,

Time taken by big tap to fill the full tank is 20 hours

and time taken by small tap to fill the full tank is 30 hours.

Hope it helps you.

Answer 2

Answer:

Time taken by big tap to fill the full tank is 20 hours

Time taken by big tap to fill the full tank is 20 hoursand time taken by small tap to fill the full tank is 30 hours.

Time taken by big tap to fill the full tank is 20 hoursand time taken by small tap to fill the full tank is 30 hours.Hope it helps you.