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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. Find the time in which each tap can separately fill the tank. |
Answer» Let the smaller diameter tap fill the reservoir in ‘a’ hours Larger diameter tap fills it in ‘a – 10’ hours. Given, two water taps together can fill a tank in \(9\frac{3}{8}=\) 75/8 hours. In 1 hour, part of tank filled = 8/75 \(\Rightarrow \frac{1}{a}+\frac{1}{a\,-\,10}=\frac{8}{75}\) ⇒ 75(a + a – 10) = 8a2 – 80a ⇒ 150a – 750 = 8a2 – 80a ⇒ 8a2 – 230a + 750 = 0 ⇒ 4a2 – 115a + 375 = 0 ⇒ 4a2 – 100a – 15a + 375 = 0 ⇒ 4a(a – 25) – 15(a – 25) = 0 ⇒ (4a – 15)(a – 25) = 0 Value of a can’t be 15/4 as (a – 10) will be negative Thus a = 25 Time taken by faster tap = 25 – 10 = 15 hours |
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