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CAT 2019 Slot 1 — QA Question 10

Basics of TSD/ProportinalityEasy

One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport for 1/3 of his total journey time, while Bimal took each mode of transport for 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to

Answer & solution

  • 22

  • B

    21

  • C

    20

  • D

    19

Solution

Easy

Amal splits his time equally over the three speeds, so his average speed is the simple mean. Bimal splits the distance equally, so his time is the sum of three equal-distance leg times. Pick convenient values, compute both travel times, and compare.

1

Amal (equal time). Let each segment take time tt, total time 3t3t. Distance covered == sum of (speed ×\times time).

DA=10t+20t+30t=60t(equal-time legs)\begin{aligned} &D_A = 10t + 20t + 30t = 60t \quad\text{(equal-time legs)} \end{aligned}
2

Match the distance for Bimal. Both cover the same total distance 60t60t, so each of Bimal's three equal legs is 60t3=20t\dfrac{60t}{3}=20t.

d=60t3=20t(per leg)\begin{aligned} &d = \frac{60t}{3} = 20t \quad\text{(per leg)} \end{aligned}
3

Bimal's total time (equal distance). Time == distance//speed on each leg.

TB=20t10+20t20+20t30=2t+t+2t3=11t3\begin{aligned} &T_B = \frac{20t}{10} + \frac{20t}{20} + \frac{20t}{30} = 2t + t + \frac{2t}{3} = \frac{11t}{3} \end{aligned}
4

Compare to Amal's time 3t3t.

TB−TATA×100=11t3−3t3t×100=2t33t×100=29×100≈22.2%\begin{aligned} &\frac{T_B - T_A}{T_A}\times 100 = \frac{\frac{11t}{3} - 3t}{3t}\times 100 = \frac{\frac{2t}{3}}{3t}\times 100 = \frac{2}{9}\times 100 \approx 22.2\% \end{aligned}
Bimal’s time exceeds Amal’s by≈22%\text{Bimal's time exceeds Amal's by} \approx 22\%
CAT 2019 Slot 1 QA Q10: One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, — Solution | TheCATExam