Race Across Three Stretches

CAT 1997 Slot 1 · DILR · Hard · Data Interpretation

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Passage / data set

A certain race is made up of three stretches: A, B and C, each 2 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in km/hr) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is 10 min.

StretchModeMin Speed (km/hr)Max Speed (km/hr)
ACar4060
BMotorcycle3050
CBicycle1020

Question 1 of 3

Anshuman travels at minimum speed by car over A and completes stretch B at the fastest speed. At what speed should he cover stretch C in order to break the previous record?

  1. A.

    Maximum speed for C

  2. B.

    Minimum speed for C

  3. C.

    This is not possible

  4. D.

    None of these

Answer

C

Explanation

Time taken on stretch A at 40 km/hr = 240\frac{2}{40} hr = 3 min. Time taken on stretch B at 50 km/hr = 250\frac{2}{50} hr = 2.4 min. Total time for first two stretches = 3+2.4=5.43 + 2.4 = 5.4 min.

To break the 10 min record, total time must be <10< 10 min, so time on C must be <4.6< 4.6 min. Speed needed on stretch C = 2 km4.6 min=24.6/60≈26.08\frac{2 \text{ km}}{4.6 \text{ min}} = \frac{2}{4.6/60} \approx 26.08 km/hr.

However, the maximum speed possible for bicycle on C is 20 km/hr. Thus, it is not possible to break the record.

Question 2 of 3

Mr Hare completes the first stretch at the minimum speed and takes the same time for stretch B. He takes 50% more time than the previous record to complete the race. What is Mr Hare's speed for stretch C?

  1. A.

    10.9 km/hr

  2. B.

    13.3 km/hr

  3. C.

    17.1 km/hr

  4. D.

    None of these

Answer

B

Explanation

Time for stretch A = 240\frac{2}{40} hr = 3 min. Time for stretch B = 3 min. Total time taken for race = 50% more than 10 min = 15 min.

Time taken for stretch C = 15−3−3=915 - 3 - 3 = 9 min. Speed on stretch C = 2 km9/60 hr=1209=13.33\frac{2 \text{ km}}{9/60 \text{ hr}} = \frac{120}{9} = 13.33 km/hr.

Question 3 of 3

Mr Tortoise completes the race at an average speed of 20 km/hr. His average speed for the first two stretches is four times that for the last stretch. Find the speed over stretch C.

  1. A.

    15 km/hr

  2. B.

    12 km/hr

  3. C.

    10 km/hr

  4. D.

    This is not possible

Answer

C

Explanation

Total distance = 2+2+2=62 + 2 + 2 = 6 km. Average speed = 20 km/hr   ⟹  \implies Total time = 620=0.3\frac{6}{20} = 0.3 hr = 18 min.

Let speed on stretch C be vv. Average speed for first two stretches (distance 4 km) = 4v4v.

Time for first two stretches = 44v=1v\frac{4}{4v} = \frac{1}{v}. Time for stretch C = 2v\frac{2}{v}. Total time = 1v+2v=3v\frac{1}{v} + \frac{2}{v} = \frac{3}{v} hr.

Given total time = 0.3 hr: 3v=0.3  ⟹  v=10 km/hr\frac{3}{v} = 0.3 \implies v = 10 \text{ km/hr}

This speed is within the permissible range for stretch C (10 to 20 km/hr). So speed over stretch C is 10 km/hr.

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