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 4 of 12.
Ruku said:
7 years ago
The distance is simply.
As if he is speed 3kmph then it takes 40 less,
And if his speed 2kmph then it takes 40 more,
(3 * 40)-(2 * 40)=120-80.
= 40km.
As if he is speed 3kmph then it takes 40 less,
And if his speed 2kmph then it takes 40 more,
(3 * 40)-(2 * 40)=120-80.
= 40km.
Jaideep said:
7 years ago
Ans-3/10+3/20+3/30+3/60=6/10 Hrs,
therefore 12/6/10=20 km/ph.
therefore 12/6/10=20 km/ph.
Vivek said:
7 years ago
This sum is not understand clearly please explain clearly.
P.abhishek said:
7 years ago
Distance= s* t.
x= actual speed,
x +3= 40 min (faster),
x-2=40 min (slower),
d= 3 * 40= 120,
d=2*40=80,
120 -80= 40.
It's an actual answer.
x= actual speed,
x +3= 40 min (faster),
x-2=40 min (slower),
d= 3 * 40= 120,
d=2*40=80,
120 -80= 40.
It's an actual answer.
(2)
Hemanth said:
7 years ago
@Amit.
Please, can you explain how to solve those 3 equations in a easy or some possible method? Please tell me.
Please, can you explain how to solve those 3 equations in a easy or some possible method? Please tell me.
Hariom saini said:
7 years ago
You can also solve it like;
If T1=T2=40 than
V=2V1V2/V1-V2
=2*3*2/3-2
=12.
Than X=V*T1(1+V/V1)
= 12*40/60(1+12/3)
= 8*5
=40KM.
T1=40/60hr V=speed V1=12km/hr.
If T1=T2=40 than
V=2V1V2/V1-V2
=2*3*2/3-2
=12.
Than X=V*T1(1+V/V1)
= 12*40/60(1+12/3)
= 8*5
=40KM.
T1=40/60hr V=speed V1=12km/hr.
Jyo said:
8 years ago
Formulae:speed=distance/time.
Case 1:
3=d/t-40
3t-120=d ---> eq(1)
Case 2:
2=d/t+40
2t+80=d ---> eq(2).
After solving eq(1) and eq(2), we get.
t=49.
Case 1:
3=d/t-40
3t-120=d ---> eq(1)
Case 2:
2=d/t+40
2t+80=d ---> eq(2).
After solving eq(1) and eq(2), we get.
t=49.
Shiva said:
8 years ago
Why they took 60 there, please can anyone give a brief explanation regarding those equations x/y - (x/(y+3)) = 40/60? And the second equation as well.
Shashank said:
8 years ago
3 km/hr Increase: 40 min less
2 km/hr Decrease: 40 min more
First, we calculate the original speed by the following trick:
=2 X increase in speed X decrease in speed / Difference of increase and decrease in speeds.
=2 x 3 x 2/(3-2).
=12 km/hr.
Now Scenario is as under:
3 km/hr Increase = 12+3 = 15 km/hr: 40 min less.
2 km/hr Decrease =12- 2 = 10 km/hr: 40 min more.
Now the distance
= products of speed / difference of speed X total time difference.
= 80/60 ( in hours ) X 15 X 10 /(15-10 ).
= 40 Km.
2 km/hr Decrease: 40 min more
First, we calculate the original speed by the following trick:
=2 X increase in speed X decrease in speed / Difference of increase and decrease in speeds.
=2 x 3 x 2/(3-2).
=12 km/hr.
Now Scenario is as under:
3 km/hr Increase = 12+3 = 15 km/hr: 40 min less.
2 km/hr Decrease =12- 2 = 10 km/hr: 40 min more.
Now the distance
= products of speed / difference of speed X total time difference.
= 80/60 ( in hours ) X 15 X 10 /(15-10 ).
= 40 Km.
Divya said:
8 years ago
@Dhivakar.
To convert 80 min into hour you divide it by 60 not multiply it. Then How did you get the answer?
To convert 80 min into hour you divide it by 60 not multiply it. Then How did you get the answer?
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