Question : The distance between two places A and B is 140 km. Two cars $x$ and $y$ start simultaneously from A and B, respectively. If they move in the same direction, they meet after 7 hours. If they move towards each other, they meet after one hour. What is the speed (in km/hr) of car $y$ if its speed is more than that of car $x$?
Option 1: 60
Option 2: 100
Option 3: 80
Option 4: 90
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Correct Answer: 80
Solution :
Let the speed of car $x$ as $v_x$ and the speed of car $y$ as $v_y$.
When the cars move in the same direction, they meet after 7 hours. This means that the faster car (car $y$) covers the distance of 140 km in 7 hours, at a relative speed of $v_y - v_x$.
$⇒v_y - v_x = \frac{140 \, \text{km}}{7 \, \text{hours}} = 20 \, \text{km/hr} $
When the cars move towards each other, they meet after 1 hour. This means that the cars together cover the distance of 140 km in 1 hour, at a combined speed of $v_y + v_x$.
$⇒v_y + v_x = \frac{140 \, \text{km}}{1 \, \text{hour}} = 140 \, \text{km/hr}$
Adding equations (1) and (2) gives us:
$⇒2v_y = 160 \, \text{km/hr}$
$⇒v_y = 80 \, \text{km/hr}$
Hence, the correct answer is 80.
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