Perpendicular Motion of Ships

CAT 2022 Slot 2 · QA · Easy · Arithmetic

All CAT 2022 Slot 2 questionsMore Arithmetic questionsAll CAT QA

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}.

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace