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:
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:
121 comments Page 10 of 13.
Vivek said:
8 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
2y(y+3) 9x
------- =
y(y-2) 3x
Raj said:
1 decade ago
Please explain the way way of getting the value of y =12.
Kunal said:
1 decade ago
How to divide this two equation to find the value of y?
JAHNAVI said:
2 years ago
I am not understanding this. Please help me to get it.
(11)
Bhavana said:
1 decade ago
Good explanation by Amit kr Soni, Sivaraman and Amit.
Ram said:
10 years ago
Anyone find the original speed of the above question?
Damu said:
8 years ago
Just from simple trick, it is.
3 * 40 - 2 * 40 = 40.
3 * 40 - 2 * 40 = 40.
Aayush said:
1 decade ago
3X-2X = X.
40(common)40 = X.
No logic, just this.
40(common)40 = X.
No logic, just this.
Naveen said:
1 decade ago
Can anyone tell me the shortcut method to solve it?
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