A student walks from his house at a speed of (5/2) km per hour and reaches his school 6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house?
Solution
Correct Answer: Option B
Let, the school is at x kilometer distance , usual time is t km/hr
Now
x/(5/2) = t + (6/60)
or, 2x/5 = t + (1/10)
or, 4x = 10t + 1
or, 4x - 10t = 1
or, 28x - 70t = 7 [multiplying by 7]
Again
x/{(5/2) + 1} = t - 6/60
or, 2x/7 = t - (1/10)
or, 20x = 70t - 7 [multiplying by 70]
or, 20x - 70t = - 7
28x - 70t - 20x + 70t = 7 + 7
or, 8x = 14
or, x = 14/8
or, x = 7/4
Now
x/(5/2) = t + (6/60)
or, 2x/5 = t + (1/10)
or, 4x = 10t + 1
or, 4x - 10t = 1
or, 28x - 70t = 7 [multiplying by 7]
Again
x/{(5/2) + 1} = t - 6/60
or, 2x/7 = t - (1/10)
or, 20x = 70t - 7 [multiplying by 70]
or, 20x - 70t = - 7
28x - 70t - 20x + 70t = 7 + 7
or, 8x = 14
or, x = 14/8
or, x = 7/4
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