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Question
Ravi starts for his school at 8:20 a.m. on his bicycle. If he travels at a speed of 10 km/h, then he reaches his school late by 8 minutes but on travelling at 16 km/h he reaches the school 10 minutes early. At what time does the school start?
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Solution
Let the total distance = x km
Let the time taken by Ravi to reach the school at sharp time = t min
If the speed of the bicycle is 10 km/h, then he reach his school late by 8 min
∴ `x/10 = t + 8/60` `[∵ 1min = 1/60 h]` ...(i)
⇒ `x/10 = t + 2/15`
If the speed of the bicycle is 16 km/h, then he reach his school 10 min early.
∴ `x/16 = t - 10/60`
∴ `x/16 = t - 1/6` ...(ii)
On solving equations (i) and (ii), we get
`x/10 - x/16 = 2/15 + 1/6`
⇒ `(8x - 5x)/80 = (4 + 5)/30`
⇒ `(3x)/80 = 9/30`
⇒ `x = (9 xx 80)/(30 xx 3) = 8 km` ...[By cross-multiplication]
Now, put x = 8 in equation (i), we get
`8/10 = t + 2/15`
⇒ `t = 8/10 - 2/15 = (24 - 4)/30`
⇒ `t = 20/30 = 2/3 h`
= `2/3 xx 60` ...[∵ 1h = 60 min]
= 40 min
Hence, starting time of school is 8 : 20 + 40 min i.e. 9 : 00 am.
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