walking 3/4 of his usual speed,a student is 10 mintues late to his school.find the usual time to cover the distance
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Let the usual speed be 'S'
Time = T hrs
D = S* T
When speed is 3/4 usual, he reaches 10 mins late
S = 3*S/4
T = T + 10 = T + 10/60 = T + 1/6 hrs
D = 3*S/4 *(T + 1/6)
Since, D is same, equate
S*T = 3*S/4 * (T + 1/6)
T = 3*T/4 + 3/24
T - 3T/4 = 1/8
4T-3T/4 = 1/8
T/4 = 1/8
T = 1/2 hours or 30 minutes.
Hence usual time = 1/2 hours or 30 minutes
Hope it helps.
Time = T hrs
D = S* T
When speed is 3/4 usual, he reaches 10 mins late
S = 3*S/4
T = T + 10 = T + 10/60 = T + 1/6 hrs
D = 3*S/4 *(T + 1/6)
Since, D is same, equate
S*T = 3*S/4 * (T + 1/6)
T = 3*T/4 + 3/24
T - 3T/4 = 1/8
4T-3T/4 = 1/8
T/4 = 1/8
T = 1/2 hours or 30 minutes.
Hence usual time = 1/2 hours or 30 minutes
Hope it helps.
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