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Munagala Janardhana Reddy

· started a discussion

· 1 Months ago

my solution (Question: 232754.0 )

d=(s1*s2)(time gap in min)/(s2 -s1)*60
d=(2.5*3.5)(12

Question:
Starting from his house one day, a student walks at a speed of 2.5 km/hr and reaches his school 6 minutes late. Next day at the same time he increases his speed by 1km/hr and reaches the school 6 minutes early. How far is the school from his house?
Options:
A) 2 km
B)


\(1\cfrac{1}{2}\) km

C) 1 km
D)


\(1\cfrac{3}{4}\) km

Solution:

Ans: (d)

Munagala Janardhana Reddy

· commented

· 1 Months ago

d=(s1*s2)(time gap in min)/(s2 -s1)*60
d=(2.5*3.5)(12)/(2.5 -3.5)*60 =7/4 km=1+3/4 km
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