Multiple choice

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is: एक टैंक सामान्यतः बहाव से तीन पाइपों से भरता है। प्रथम दो पाइप एक साथ टैंक का उतने समय में भरती है जिसने समय में तीसरे पाइप अकेले भरती है। दूसरी पाइप, पहली पाइप से 5 घंटे तेज और तीसरी पाइप से 4 घंटे कम में भरती है, तो पहली पाइप से लगा समय है-

  1. 6 hours

  2. 12 hours

  3. 15 hours

  4. 30 hours
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let first pipe time = t hours. Then second pipe = t-5 hours, third pipe = t-9 hours (since second is 4 hours slower than third). Given: (1/t) + (1/(t-5)) = 1/(t-9). Solving: (t-5 + t)/[t(t-5)] = 1/(t-9), so (2t-5)(t-9) = t(t-5). This gives 2t² - 23t + 45 = t² - 5t, so t² - 18t + 45 = 0. Testing t = 15: 225 - 270 + 45 = 0. Therefore first pipe takes 15 hours.

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