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Rahim's Travel Time

CAT 2008 Slot 1 · Quantitative Ability · Hard · Speed Distance Time

This is a hard Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Speed Distance Time. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Rahim plans to drive from city A to station C, at the speed of 70 km/h70\text{ km/h}, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km500\text{ km} south of A, at 8:00 am and travels at a speed of 50 km/h50\text{ km/h}. It is known that C is located between west and northwest of B, with BCBC at 60∘60^\circ to ABAB. Also, C is located between south and southwest of A with ACAC at 30∘30^\circ to ABAB. The latest time by which Rahim must leave A and still catch the train is closest to

  1. A.

    6 : 15 am

  2. B.

    6 : 30 am

  3. C.

    6 : 45 am

  4. D.

    7 : 00 am

  5. E.

    7 : 15 am

Answer

B

Explanation

In △ABC\triangle ABC, ∠A=30∘,∠B=60∘  ⟹  ∠C=90∘\angle A = 30^\circ, \angle B = 60^\circ \implies \angle C = 90^\circ. Since AB=500 kmAB = 500\text{ km}: BC=500cos⁡60∘=250 kmBC = 500 \cos 60^\circ = 250\text{ km} AC=500sin⁡60∘=2503 kmAC = 500 \sin 60^\circ = 250\sqrt{3}\text{ km}

Train travel time from B to C =25050=5 hours= \frac{250}{50} = 5\text{ hours}. Train arrives at C at 8:00 am+5 hours=1:00 pm8:00\text{ am} + 5\text{ hours} = 1:00\text{ pm}. Rahim must reach C by 1:00 pm−15 min=12:45 pm1:00\text{ pm} - 15\text{ min} = 12:45\text{ pm}.

Rahim's drive time =250370=2537≈6 hours 11 mins= \frac{250\sqrt{3}}{70} = \frac{25\sqrt{3}}{7} \approx 6\text{ hours } 11\text{ mins}.

Latest departure time =12:45 pm−6 hrs 11 mins=6:34 am≈6:30 am= 12:45\text{ pm} - 6\text{ hrs } 11\text{ mins} = 6:34\text{ am} \approx 6:30\text{ am}.

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