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CAT 2019 Slot 1 — QA Question 9

Basics of TSD/ProportinalityEasy

Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

Answer & solution

Correct answer: 20

  • A

    25

  • B

    10

  • C

    30

  • 20

Solution

Easy

Both cars cover the same distance and finish together; the second starts 11 hour later, so it travels 11 hour less. With equal distances, speed2speed1=tt−1\dfrac{\text{speed}_2}{\text{speed}_1}=\dfrac{t}{t-1}, which is largest when the first car's time tt is smallest. The "at least 6 hours" gives tmin⁡=6t_{\min}=6.

1

Set variables. First car: speed aa, time tt. Second car starts an hour later but arrives together, so its time is t−1t-1; let its speed be bb.

distance equal: a t=b (t−1)\begin{aligned} &\text{distance equal:}\ a\,t = b\,(t-1) \end{aligned}
2

Express the speed ratio. Rearrange the distance equation.

ba=tt−1=11−1t\begin{aligned} &\frac{b}{a} = \frac{t}{t-1} = \frac{1}{1-\frac1t} \end{aligned}
3

Maximise. ba\dfrac{b}{a} increases as tt decreases, so use the smallest allowed tt. Given the first car ran "at least 66 hours", tmin⁡=6t_{\min}=6.

(ba)max⁡=66−1=65\begin{aligned} &\left(\frac{b}{a}\right)_{\max} = \frac{6}{6-1} = \frac{6}{5} \end{aligned}
4

Convert to a percentage excess.

(ba−1)×100=(65−1)×100=20%\begin{aligned} &\left(\frac{b}{a}-1\right)\times 100 = \left(\frac{6}{5}-1\right)\times 100 = 20\% \end{aligned}
highest possible excess=20%\text{highest possible excess} = 20\%

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CAT 2019 Slot 1 QA Q9: Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach — Solution | TheCATExam