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Bunuel
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Acc to given question
Assuming distance to school as d and time taken usually as t
d/5=t-4
d/4=t+2
we can write as

5t-20=4t+8
t=28.

d=5(28-4)=120

time taken when walking at 6km/h=120/6=20min Option B
Bunuel
A boy walks from his home to school at 5 kmph, reaches the school 4 minutes early. If he walks from his home to school at 4 kmph, he reaches the school 2 minutes late. What is the time taken by him to go to school if he walks at 6 kmph?

A. 15 minutes
B. 20 minutes
C. 25 minutes
D. 30 minutes
E. 35 minutes

­
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We solve using equal distance S and s = v·t.

S = 5·(t − 4/60) = 5·(t − 1/15)
S = 4·(t + 2/60) = 4·(t + 1/30)

Set equal: 5(t − 1/15) = 4(t + 1/30)
5t − 5/15 = 4t + 4/30
5t − 1/3 = 4t + 2/15
t = 1/3 + 2/15 = 5/15 + 2/15 = 7/15 hours

Now S = 5·(t − 1/15) = 5·(7/15 − 1/15) = 5·(6/15) = 5·(2/5) = 2 km

Time to complete 2 km at 6 km/h = 2/6 = 1/3 hour = 20 minutes

Answer: B
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We will assume 't' is the scheduled time to reach school.
When speed is 5 kmph, time = t-4, and distance = 5* (t-4); (distance = speed * time)
When speed is 4 kmph, time = t+2, and distance = 4* (t+2); (distance = speed * time)

Distance in both the cases would be same.
5* (t-4) = 4* (t+2),
After solving for t, t= 20
then distance = 5* (t-4) = 120

Now, when speed = 6 kmph and d= 120, speed = 20 , hence B
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