Question

Two water taps together can fill a tank in 75/8 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:

The two taps separately can fill the tank in 25 hours and 15 hours.

Let the time taken by the larger pipe to fill the tank be x hours.
Therefore, as the smaller tap takes 10 hours more to fill the tank, time taken by it to fill the tank will be (x+10) hours.
Portion of tank filled by the larger tap in 1 hour = 1/x and portion of tank filled by the smaller tap in 1 hour = 1/(x+10).
Therefore, portion of tank filled by them together in 1 hour = 1/x + 1/(x+10)
= (x+10+x)/x(x+10)
= (2x+10)/x(x+10)
If the two taps fill this much part in 1 hour, they would fill the tank in 1 * x(x+10)/(2x+10) hours which is given to us to be 75/8 hours. Thus, we have the following equation:
x(x+10)/(2x+10) = 75/8
⇒ (x²+10x)8 = 75(2x+10)
⇒ 8x²+80x = 75*2(x+5)
⇒ 4x²+40x = 75x+375
⇒ 4x²-35x-375 = 0
⇒ x²-35x/4-375/4 = 0
⇒ x²-60x/4+25x/4-375/4 = 0
⇒ x²-15x+25x/4-375/4 = 0
⇒ (x-15)x+(x-15)*25/4 = 0
⇒ (x-15)(x+25/14) = 0
Therefore, either x-15 = 0 in which case, x=15 or x+25/14 = 0 in which case, x=-25/14.
Since the number of hours taken cannot be negative, we reject x=-25/14. Thus, we get x = 15 and x+10 = 25.


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