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A and B started together to cover a certain distance, at a speed of 8 km/h and 10 km/h, respectively. If A arrives 15 minutes after B, then find the distance.

Solution

✅ Correct Option: 4

A walks at 8 km/h and B walks at 10 km/h. They start together at the same time and travel to the same destination. A arrives 15 minutes after B.

Let the distance be dd km.

Time taken by A = d8\frac{d}{8} hours

Time taken by B = d10\frac{d}{10} hours


Since A arrives 15 minutes after B, the time difference is 15 minutes.

Converting to hours: 15 minutes = 1560=14\frac{15}{60} = \frac{1}{4} hour

The equation becomes:

d8−d10=14\frac{d}{8} - \frac{d}{10} = \frac{1}{4}


Finding LCM of 8 and 10, which is 40:

5d−4d40=14\frac{5d - 4d}{40} = \frac{1}{4}

d40=14\frac{d}{40} = \frac{1}{4}

d=404d = \frac{40}{4}

d=10d = 10


Therefore, the distance is 10 km.

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