A pump fills a water tank. If its pumping speed is increased to 9/7 of its original speed, the tank gets filled 30 minutes earlie…

Question

A pump fills a water tank. If its pumping speed is increased to 9/7 of its original speed, the tank gets filled 30 minutes earlier than usual. How much time did the pump originally take to fill the tank?

1. A. 135 minutes (Correct answer)
2. B. 140 minutes
3. C. 145 minutes
4. D. 150 minutes

Correct Answer

Option A — 135 minutes

Detailed Solution & Explanation

The correct answer is 135 minutes.

Key Points

• Speed and time are inversely proportional for a fixed volume of water.
• If the speed becomes $\frac{9}{7}$ of the original, the time becomes the reciprocal of that factor:
  • $\text{New time} = \frac{7}{9} \times \text{Original time}$
• The time saved is therefore:
  • $T - \frac{7}{9}T = \frac{2}{9}T = 30 \text{ minutes}$
  • $T = 30 \times \frac{9}{2} = \mathbf{135 \text{ minutes}}$

Shortcut Trick

• Read the fraction as a ratio directly: speed ratio $7 : 9$ means time ratio $9 : 7$. The difference of 2 parts equals 30 minutes, so 1 part = 15 minutes, and the original time of 9 parts is $9 \times 15 = 135$ minutes.

Alternate Method

• Verify by substitution. Original time 135 minutes. New time $= \frac{7}{9} \times 135 = 105$ minutes. Difference $= 135 - 105 = 30$ minutes ✓

Additional Information

• The inverse-proportion principle underlies a whole family of questions:
  • Speed and time, for a fixed distance.
  • Number of workers and time, for a fixed amount of work.
  • Rate of consumption and duration, for a fixed stock of provisions.
• The general result: if speed becomes $\frac{a}{b}$ times the original and the time saved is $t$, then
  • $\text{Original time} = t \times \frac{a}{a - b}$
• Here $a = 9$, $b = 7$ and $t = 30$: $30 \times \frac{9}{2} = 135$ minutes.
• Sanity check: an increase in speed must always reduce the time, so the original time has to be the larger figure — which rules out any answer below the new time.