EPFO EO/AO 2023 General Ability Test · Question 74 of 120

Correct — D, (d) 28 minutes. The key idea is that the four legs are of equal length — a quarter of 40 km is 10 km each — and over equal distances the times taken are inversely proportional to the speeds. So if the speeds are in the ratio 2 : 3 : 4 : 5, the times are in the ratio 1/2 : 1/3 : 1/4 : 1/5. Clear the fractions by multiplying every term by 60: the times are in the ratio 30 : 20 : 15 : 12. Those parts add to 77, and the whole race took 3 hours, which is 180 minutes. One part is therefore 180/77 minutes, and the last quarter, being 12 parts, took 12 × 180/77 = 2160/77 = 28.05 minutes. Approximately 28 minutes, which is option (d). The same answer comes out of the algebra if you prefer it. Let the speeds be 2k, 3k, 4k and 5k km per hour. The total time is 10/2k + 10/3k + 10/4k + 10/5k = (10/k)(1/2 + 1/3 + 1/4 + 1/5) = (10/k)(77/60) = 77/6k hours. Setting that equal to 3 gives k = 77/18 ≈ 4.28 km per hour, so the last quarter was run at 5k ≈ 21.4 km per hour and took 10 ÷ 21.4 = 36/77 hours = 28.05 minutes. The reason the answer must be checked rather than estimated is that the question invites a plausible shortcut that is wrong. The runner's overall average speed is 40 km in 3 hours, or 13.33 km per hour, and it is tempting to set the ordinary average of the four speeds equal to that: (2 + 3 + 4 + 5)k ÷ 4 = 3.5k = 13.33 would give k = 3.81 and a last quarter of 31.5 minutes. Average speed over equal distances is not the arithmetic mean of the speeds — it is the harmonic mean — and that single error accounts for two of the three wrong options. A final check by substitution settles it. If the last quarter really took 28 minutes, the last-quarter speed was 21.43 km per hour, so k = 4.29, and the total time works out at 77/(6 × 4.29) = 2.994 hours — three hours, as the question says. No other option reproduces the given three hours.

• (a)32 minutes — Substitute it back and the race no longer takes three hours. A last quarter of 32 minutes means a final speed of 10 ÷ (32/60) = 18.75 km per hour, so k = 3.75, and the total time becomes 77 ÷ (6 × 3.75) = 3.42 hours — about 3 hours 25 minutes, a quarter of an hour too long. This option and option (b) bracket the 31.5 minutes produced by averaging the four speeds arithmetically instead of harmonically, which is the single most common error in speed-ratio problems.
• (b)31 minutes — The same failure, one minute closer. A last quarter of 31 minutes implies a final speed of 19.35 km per hour, so k = 3.87 and the whole race would take 77 ÷ 23.23 = 3.32 hours, about 3 hours 19 minutes rather than 3. This is the rounded-down twin of the arithmetic-mean answer of 31.5 minutes, and a candidate who has made that error will hesitate between (a) and (b) — which is precisely the hesitation the option pair is designed to create.
• (c)29 minutes — This is the near miss, offered to punish loose rounding rather than a wrong method. The exact answer is 2160/77 minutes, which is 28.05 — barely three seconds over 28 minutes and nearly a whole minute short of 29. Substituting 29 minutes gives a final speed of 20.69 km per hour, k = 4.14, and a total race time of 3.10 hours, about 3 hours 6 minutes. Since the exact value is available in two lines, there is never any need to approximate here at all.

Two facts do all the work in speed-ratio problems. First, over a fixed distance, time is inversely proportional to speed — so a ratio of speeds becomes the reciprocal ratio of times, and it is usually easier to work in the ratio of times because they add. Second, average speed is total distance divided by total time, never the average of the speeds. When a journey is split into equal distances, the average speed is the harmonic mean of the leg speeds; when it is split into equal times, it is the arithmetic mean. For two legs of equal distance at speeds u and v the average speed is 2uv/(u + v), which is always less than (u + v)/2 unless the speeds are equal. This question is the four-leg, equal-distance case, so the harmonic relationship applies and the arithmetic mean is a trap. The clean method — convert speed ratios to time ratios, clear the fractions with the LCM of the denominators, add the parts, and divide the known total time among them — avoids introducing a variable at all, and is worth practising until it is automatic, because it converts a two-minute algebra problem into a twenty-second one.

Time, speed and distance is the largest single topic in the EO/AO quantitative block, and the paper's items in it are built around a specific misconception rather than around heavy arithmetic. Here the misconception is the arithmetic mean of speeds, and the option list is constructed so that the wrong method lands between two adjacent options — a design that catches a candidate twice, first with the wrong method and then with the choice between its two roundings. The habit rewarded is checking a candidate answer by substituting it back into the given total.

• Over equal distances, times are inversely proportional to speeds, so speeds in the ratio 2 : 3 : 4 : 5 give times in the ratio 1/2 : 1/3 : 1/4 : 1/5 = 30 : 20 : 15 : 12.
• Those parts total 77; the race took 180 minutes, so one part is 180/77 minutes and the last quarter is 12 parts = 2160/77 = 28.05 minutes.
• Algebraically, with speeds 2k to 5k the total time is (10/k)(77/60) = 77/6k hours; setting this to 3 gives k = 77/18 ≈ 4.28 km per hour.
• The last quarter was therefore run at 5k ≈ 21.4 km per hour and took 36/77 of an hour.
• Average speed is total distance divided by total time — for legs of equal distance it is the harmonic mean of the speeds, not the arithmetic mean.
• For two equal legs at speeds u and v, the average speed is 2uv/(u + v).
• Averaging the four speeds arithmetically against 40/3 km per hour gives a last quarter of 31.5 minutes, which is the origin of both option (a) and option (b).

• Averaging the four speeds. Over equal distances the correct average is harmonic, and the arithmetic mean produces 31.5 minutes — two of the four options.
• Applying the speed ratio directly to the times. The ratio of times is the reciprocal ratio, 30 : 20 : 15 : 12, not 2 : 3 : 4 : 5.
• Rounding early. The exact value is 2160/77 = 28.05 minutes; approximating the sum 77/60 first can move the answer by a minute.
• Reading 'each quarter of the distance' as a quarter of the time. The legs here are 10 km each, not 45 minutes each.

EO/AO sets one speed-ratio item and one relative-speed item in most quantitative blocks, and both are answered faster by ratios than by algebra. Practise converting a speed ratio to a time ratio and dividing a known total in that proportion, and always substitute the chosen option back into the total given in the stem — it takes ten seconds and catches every method error.

No directly related past PYQ was found.

• practice — not a real PYQ

  A car covers the first half of a journey at 40 km per hour and the second half at 60 km per hour. What is its average speed for the whole journey ?

  • (a)45 km per hour
  • (b)48 km per hour
  • (c)50 km per hour
  • (d)52 km per hour

  Answer(b) 48 km per hour

• practice — not a real PYQ

  A distance is covered in three equal stages at speeds in the ratio 3 : 4 : 6. If the whole journey takes 45 minutes, the time taken for the fastest stage is :

  • (a)9 minutes
  • (b)10 minutes
  • (c)12 minutes
  • (d)15 minutes

  Answer(b) 10 minutes