Question

$\huge\mathtt\red{QuEsTiOn}$

Two water taps together can fill a tank in $\dfrac{75}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time(in hrs.) in which tap of smaller diameter can separately fill the tank.​

Answer 1

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 hrs.) in which tap of smaller diameter can separately fill the tank.

Your Answer

Let the taps take x hour and x+10 hours each to fill the tank. Together they fill, in one hour,

!/x + 1/(x+10) = 8/75. The LCM of the denominators is 75x(x+10). So we have

75(x+10) +75x = 8x(x+10)

75x + 750 + 75x = 8x^2 + 80x, or

8x^2–70x-750 = 0

4x^2–35x-375 = 0

x = [35+(1225+6000)^0.5]/8

= [35+85]/8

= 120/8 = 15

The taps takes 15 hours and 25 hours to fill the same tank, working independently,

Step-by-step explanation:

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