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If two pipes function simultaneously, a tank is filled in 12 hours. One pipe fills the tank 10 hours faster than the other. How many hours does the faster pipe alone take to fill the tank?
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- 20 hrs
- 18 hrs
- 15 hrs
- 12 hrs
- 20 hrs
Correct Option: A
Let the slower pipe fills the tank in t hours, then
According to question ,
| ∴ | + | = | |||
| t | t - 10 | 12 |
| ⇒ | = | ||
| t(t - 10) | 12 |
⇒ t² – 10t = 24t – 120
⇒ t² – 34t + 120 = 0
⇒ t² – 30t – 4t + 120 = 0
⇒ t (t – 30) – 4 (t – 30) = 0
⇒ (t – 4) (t – 30) = 0
∴ t = 30 because t ≠ 4
∴ Required time = 30 – 10 = 20 hours
