Peter usually travels from town P to town Q in eight hours. One day, he increased his average speed by 5km per hour so that he arrived 20 minutes earlier. Find his usual average speed, in km per hour.
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Answered by
1
Usual time taken by Peter = 8 hours
20 minutes earlier than usual time = 8 - 20/60
= 8 - 1/3
= 7 2/3
His speed = 5 kmph
In 7 2/3 hours he will cover = 7 2/3 × 5
= 23/3 × 5
=115/3
≈ 38.3
Therefore, distance covered by Peter = 38.3
His speed = Distance travelled / time taken = 38.3/8
= 4.79 kmph
20 minutes earlier than usual time = 8 - 20/60
= 8 - 1/3
= 7 2/3
His speed = 5 kmph
In 7 2/3 hours he will cover = 7 2/3 × 5
= 23/3 × 5
=115/3
≈ 38.3
Therefore, distance covered by Peter = 38.3
His speed = Distance travelled / time taken = 38.3/8
= 4.79 kmph
Answered by
0
4.79 kmph is the answer.
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