Central Tendency & DispersionPYQ Nov. 20Question 3002 of 473
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A fire engine rushes to a place of fire accident with a speed of 110\displaystyle 110 kmph and after the completion of operation returned to the base at a speed of 55\displaystyle 55 kmph. The average speed per hour in per-direction is obtained as ________ speeds.

Options

AAverage of
BH M of
CG M of
DHalf of HM of
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Correct Answer

✅ Option b — H M of

All Options:

  • AAverage of
  • BH M of
  • CG M of
  • DHalf of HM of

Detailed Solution & Explanation

**Step 1: Calculate average speed using the HM formula (for equal distances).** When the same distance is covered at two different speeds, the average speed is the Harmonic Mean of the two speeds: Average speed=2×v1×v2v1+v2=2×110×55110+55=12100165=2203≈73.33 km/hr\text{Average speed} = \frac{2 \times v_1 \times v_2}{v_1 + v_2} = \frac{2 \times 110 \times 55}{110 + 55} = \frac{12100}{165} = \frac{220}{3} \approx 73.33 \text{ km/hr} **Step 2: Verify the GM.** GM=110×55=6050≈77.78 km/hr\text{GM} = \sqrt{110 \times 55} = \sqrt{6050} \approx 77.78 \text{ km/hr} **Step 3: Correct answer.** The average speed over equal distances is the **HM** of the two speeds. The correct option should be **B** (HM of), not GM. Mathematically: Average speed = HM of the two speeds = 2×110×55110+55=2203\displaystyle \frac{2 \times 110 \times 55}{110 + 55} = \frac{220}{3} km/hr. Hence, **Option B** is the correct answer.

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