CAT 2019 Slot 2 — QA Question 14

Equate the jogging times. Let track A have length $a$ and track B have length $b$ (in metres). John runs $9a$ at $6$ kmph; Mary runs $5b$ at $7.5$ kmph in the same time.

$$\begin{aligned} &\frac{9a}{6}=\frac{5b}{7.5} \quad\text{(time}=\text{distance}/\text{speed)}\\ &\Rightarrow\ \frac{3a}{2}=\frac{2b}{3} \quad\text{(simplify each side)}\\ &\Rightarrow\ a=\frac{4b}{9} \end{aligned}$$

Use the total length. The two tracks together are $325$ m.

$$\begin{aligned} &a+b=325\\ &\Rightarrow\ \frac{4b}{9}+b=325 \quad\text{(substitute }a\text{ from step 1)}\\ &\Rightarrow\ \frac{13b}{9}=325\\ &\Rightarrow\ b=225,\qquad a=\frac{4(225)}{9}=100 \end{aligned}$$

Time Mary over track A. Track A is $a=100$ m; Mary's speed is $7.5$ kmph $=7.5\times\tfrac{5}{18}$ m/s.

$$\begin{aligned} &t=\frac{100}{7.5\times\frac{5}{18}} \quad\text{(time}=\text{distance}/\text{speed)}\\ &\Rightarrow\ t=\frac{100}{\frac{37.5}{18}}=\frac{100\times18}{37.5}\\ &\Rightarrow\ t=48\ \text{seconds} \end{aligned}$$