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:
118 comments Page 3 of 12.
Mohammed shamil said:
6 months ago
Good thanks for explaining.
(1)
Venish said:
2 decades ago
I want to know simplify method for that equations x/y - x/y+3 = 40/ 60 and x/y-2 - x/y = 40/60....
Amit Kr. Soni said:
2 decades ago
Hi, Simran & Venish,
I have a soln of this problm :-
Let Distns = X, & Spd = Y
Now fm 1st half of quesn
Spd = Y+3
Then Time = X/Y+3
2nd half of qesn
Spd = y-2
Then Time = X/Y-2
Now from quesn
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
Then, 45x+300 = 30x-300
15x = - 600
X = - 40 Km
"After neglect the minus value of X"
X = 40 Km Answer
I have a soln of this problm :-
Let Distns = X, & Spd = Y
Now fm 1st half of quesn
Spd = Y+3
Then Time = X/Y+3
2nd half of qesn
Spd = y-2
Then Time = X/Y-2
Now from quesn
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
Then, 45x+300 = 30x-300
15x = - 600
X = - 40 Km
"After neglect the minus value of X"
X = 40 Km Answer
Shaju said:
1 decade ago
After getting y=12, put the value of y in X/y - X/y+3 = 40/60
Then you will get X=40 without any minus sign.
Then you will get X=40 without any minus sign.
Amit said:
1 decade ago
vt=d
(v+3)(t-2/3)=d
(v-2)(t+2/3)=d
3 equations and three unknowns so you can find d.
(v+3)(t-2/3)=d
(v-2)(t+2/3)=d
3 equations and three unknowns so you can find d.
Mani said:
1 decade ago
Can any one explain this plz
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
I didnt get this.....
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
I didnt get this.....
Shalu said:
1 decade ago
After getting y=12, substitute y in any of the eqation to find out the x value, because it is considered as distance which we want to find.
Nikita said:
1 decade ago
Amit kr. Soni, good explanation. Great.
Sivaram said:
1 decade ago
There is simple method
3*40=120
2*40=80
Then subtract two we get 40 this simple
3*40=120
2*40=80
Then subtract two we get 40 this simple
Bhavana said:
1 decade ago
Good explanation by Amit kr Soni, Sivaraman and Amit.
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