SSC CHSL Model Paper 8

Question : 68 of 100

Marks: +1, -0

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The approx distance (in km) is:

Validate

Solution:

Let distance = x km and usual rate = y kmph
Then,

x/y-x/{y+3}=(40/60)

⇒

2 y( y+3)=9x

…(i)
And,

x/{y-2}-x/y=40/60

⇒

y( y-2)=3x

…(ii)
On dividing Eq. (i) by (ii),
we get:

x = 40

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