Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the speeds of the cars.
Given:
Places A and B are 80 km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes.
To do:
We have to find the speed of the two cars.
Solution:
We know that,
Distance$=$ Speed $\times$ Time.
Distance between the two places A and B $= 80\ km$.
Let the speed of the first car starting from A be $x\ km/hr$ and the speed of the second car starting from B be $y\ km/hr$.
Let the cars meet at point P when they are moving in the same direction and at point Q when they are moving in the opposite direction.
When they travel in the same direction, they meet in 8 hours.
Distance travelled by the first car in 8 hours $AP= 8\times x\ km=8x\ km$.
Distance travelled by the second car in 8 hours $BP= 8\times y\ km=8y\ km$.
$AP-BP=80$
$8x-8y=80$
$8(x-y)=8\times10$
$x-y=10$.....(i)
When they travel in the opposite direction, they meet after 1 hour 20 minutes$=1+\frac{20}{60}=\frac{60+20}{60}=\frac{4}{3}$ hours.
Distance travelled by the first car in 1 hour 20 minutes $AQ= \frac{4}{3}\times x\ km=\frac{4x}{3}\ km$.
Distance travelled by the second car in 1 hour 20 minutes$BQ= \frac{4}{3}\times y\ km=\frac{4y}{3}\ km$.
$AQ+BQ=AB$
$\frac{4x}{3} + \frac{4y}{3} = 80$
$\frac{4x+4y}{3}=80$
$4(x+y)=3(80)$
$x+y=60$….(ii)
Adding equations (i) and (ii), we get,
$x-y+x+y=10+60$
$2x = 70$
$x = \frac{70}{2}$
$x=35$
Substituting $x=35$ in equation (ii), we get,
$35+y=60$
$y = 60-35$
$y = 25$
Therefore, the speed of the first car is $35\ km/hr$ and the speed of the second car is $25\ km/hr$.
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