Question

A car leaves station $X$ for station $Y$ every $10\text{min}$. The distance between $X$ and $Y$ is $60\text{km}$. The car travels at speed $60\text{km/h}$. A man drives a car from $Y$ towards $X$ at speed $60\text{km/h}$. If he starts at the moment when first car leaves station $X$, how many cars would he meet on route?

• A. $20$
• B. $7$
• C. $10$
• D. $5$

Answer 1

The correct option is B $7$

Time interval between two consecutive cars leaving from station $X=10\text{min}=\frac{1}{6}\text{h}$
Distance between two cars leaving from station $X$ is,
$d=\text{speed}\times \text{time}=\left(\frac{1}{6}\right)\text{h}\times 60\text{km/h}=10\text{km}$
For the first time when both the cars meet, total distance travelled by both the cars $=60\text{km}$
$\Rightarrow$ Man meets the first car after time,
$t_1=\frac{\text{Total distance}}{\text{Total speed}}=\frac{60}{60+60}=\frac{1}{2}\text{h}$
He will meet the next car after time,
$t_2=\frac{10}{60+60}=\frac{1}{12}\text{h}$
In the remaining half an hour, number of cars he will meet is, $n=\frac{1/2}{1/12}=6$
$\therefore$ Total number of cars he would meet on route will be $7$. The answer is (b)