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 7 of 12.
Feroze Ahmad kumhar said:
5 years ago
Distance is same.
Therefore d = d.
3(x-40) = 2(x+40),
3x+120 = 2x+80,
3x-2x = 120-80,
X = 40km.
Therefore d = d.
3(x-40) = 2(x+40),
3x+120 = 2x+80,
3x-2x = 120-80,
X = 40km.
(134)
Lakshmi said:
1 decade ago
I solved the above two equations but I'm not getting the answer. So, please explain clearly.
Manik said:
8 years ago
Please, can anyone solve this equation in details?
On dividing (i) by (ii), we get: x = 40.
On dividing (i) by (ii), we get: x = 40.
Samir Mendhe said:
5 years ago
3times more -> Dist= 3 * 40= 120.
2tines less -> Dist= -2* 40= -80.
So, 120 - 80 = 40kmph.
2tines less -> Dist= -2* 40= -80.
So, 120 - 80 = 40kmph.
(248)
Kirti said:
1 decade ago
All I have is v+3=s/ (t-2/3).
And v-2=s/ (t+2/3). With these two eqns. How do we solve it?
And v-2=s/ (t+2/3). With these two eqns. How do we solve it?
Kavit Garg said:
1 decade ago
Please give me your calculation of how did you get y=12 by solving above two equations ?
Thanuj said:
8 years ago
3 * 40 = 120,
2 * 40 = 80,
120-80 = 40.
Does this process will come? Please tell me.
2 * 40 = 80,
120-80 = 40.
Does this process will come? Please tell me.
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.
Seema said:
1 decade ago
Hi friends please tell me how to solve this equation.
X/y-2 + 40/60 = X/y+3 -40/60.
X/y-2 + 40/60 = X/y+3 -40/60.
Mayur said:
9 years ago
Difference of speed and time = distance.
3 * 40 - 2 * 40 = x.
120 - 80 = x.
40 = x.
3 * 40 - 2 * 40 = x.
120 - 80 = x.
40 = x.
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