Arithmetic

Time, Speed & Distance

Easy

On a long stretch of east-west road, $A$ and $B$ are two points such that $B$ is $350$ km west of $A$ . One car starts from $A$ and another from $B$ at the same time. If they move towards each other, then they meet after $1$ hour. If they both move towards east, then they meet in $7 \mathrm{hrs}$ . The difference between their speeds, in km per hour, is

Entered answer:

✅ Correct Answer: 50

We call the speeds of the cars starting from A and B as $s_A$ and $s_B$ respectively.

Distance between A and B = 350 km

B is 350 km west of A (this direction matters!)

When two objects move towards each other, their relative speed is the sum of their individual speeds.

What happens in 1 hour:

Car from A travels: $s_A \times 1 = s_A$ km

Car from B travels: $s_B \times 1 = s_B$ km

Together they cover: $s_A + s_B$ km

Since they meet after 1 hour, they must have covered the entire 350 km distance:

$s_A + s_B = 350$ km/hr ...(1)

Initial Setup:

Car from A starts moving east

Car from B also starts moving east

Since B is west of A, car from B has to "catch up" to car from A

For them to meet when both moving east:

The car from B must be faster than the car from A

In 7 hours, the faster car (from B) must travel exactly 350 km more than the slower car (from A)

Distance covered in 7 hours:

Car from A: $s_A \times 7$ km

Car from B: $s_B \times 7$ km

Extra distance covered by car from B: $(s_B - s_A) \times 7$ km

This extra distance equals the initial separation:

$(s_B - s_A) \times 7 = 350$

Therefore: $s_B - s_A = \frac{350}{7} = 50$ km/hr ...(2)

From equations (1) and (2):

$s_A + s_B = 350$

$s_B - s_A = 50$

Adding both equations:

$2s_B = 400$

$s_B = 200$ km/hr

From equation (2):

$2s_A = 300$

$s_A = 150$ km/hr

The difference between their speeds = $s_B - s_A = 200 - 150 = 50$ km/hr

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