Aptitude - Time and Distance - Discussion

Discussion Forum : Time and Distance - General Questions (Q.No. 15)
15.
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 distance (in km) is:
35
36 2
3
37 1
2
40
Answer: Option
Explanation:

Let distance = x km and usual rate = y kmph.

Then, x - x = 40     2y(y + 3) = 9x ....(i)
y y + 3 60

And, x - x = 40     y(y - 2) = 3x ....(ii)
y -2 y 60

On dividing (i) by (ii), we get: x = 40.

Discussion:
118 comments Page 9 of 12.

Deepika said:   1 decade ago
3 kmph (faster) - 2 kmph(slower) = 1kmph

1 kmph * 40 minutes = 40.

Manzoor said:   6 years ago
Diff = 1speed.80/60time.
Add =5
Multiple = 6.
80/60 * 6 * 5/1 = 40.

Jaideep said:   7 years ago
Ans-3/10+3/20+3/30+3/60=6/10 Hrs,

therefore 12/6/10=20 km/ph.

Shakti said:   1 decade ago
Sivaram:I don't think its a correct way to solve this quesion.

Varun said:   7 years ago
d=s*t,
t=40 mins.
s=3 km/hr-2km/hr =1 km/h,
d=1*40=40km/hr.

Magnus said:   7 years ago
@ Ruchir,

For Converting Minutes to an hour you divide by 60.

Sony said:   1 decade ago
How you got y=12 from d above equations I did not understood?

Dilip said:   1 decade ago
How did you get y as 12 ? I am unable to solve the equation.

Vivek said:   7 years ago
This sum is not understand clearly please explain clearly.

Ash said:   1 decade ago
Devide both the equations
2y(y+3) 9x
------- =
y(y-2) 3x


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