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Perpendicular Motion of Ships

CAT 2022 Slot 2 · Quantitative Ability · Easy · Arithmetic

This is an easy Quantitative Ability question from the CAT 2022 Slot 2 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

  1. A.

    12

  2. B.

    18

  3. C.

    20

  4. D.

    24

Answer

B

Explanation

Let the speed of the slower ship be s km/hs\text{ km/h}. Then the speed of the faster ship is s+6 km/hs + 6\text{ km/h}. In 2 hours, distance travelled by slower ship = 2s2s. Distance travelled by faster ship = 2(s+6)2(s + 6). Since south and west directions are perpendicular, the distance between them is the hypotenuse: (2s)2+(2s+12)2=602(2s)^2 + (2s + 12)^2 = 60^2. Divide by 2: s2+(s+6)2=302=900s^2 + (s + 6)^2 = 30^2 = 900. s2+s2+12s+36=900  ⟹  2s2+12s−864=0  ⟹  s2+6s−432=0s^2 + s^2 + 12s + 36 = 900 \implies 2s^2 + 12s - 864 = 0 \implies s^2 + 6s - 432 = 0. (s+24)(s−18)=0(s + 24)(s - 18) = 0. Since s>0s > 0, s=18 km/hs = 18\text{ km/h}.

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