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 12 of 12.
Sowmya said:
5 years ago
Thanks @Sachin Bhagat and @Dinesh.
(7)
Sukanya said:
5 years ago
Can you please explain how to slove this equation and get 2y(y+3)=9x?
(9)
Vaibhav Sharma said:
5 years ago
I have easier solution to this:
Distance is same so first consider speed x and time t:
As per the question,
Increasing speed by 3 i.e. (x+3) will decrease time by 40 min i.e (t-40/60).
Therefore,
Distance = (x+3)(t-40/60).
Now Decreasing speed by 2 i.e. (x-2) will decrease time by 40 min i.e (t+40/60).
Therefore,
Distance = (x+3)(t-40/60).
Since Distance would be same:
(x+3)(t-40/60)=(x+3)(t-40/60) -----> (1)
solving this, we get
-2x/3 + 3t -2 = 2x/3 + -2t - 4/3.
3t -2x/3 =2 -->(2)
2x/3 + -2t = 4/3 --->(3)
Solving two-equation, we get t=10/3 and x=12. Putting this value in eq. 1, we get,
Distance = 40.
Distance is same so first consider speed x and time t:
As per the question,
Increasing speed by 3 i.e. (x+3) will decrease time by 40 min i.e (t-40/60).
Therefore,
Distance = (x+3)(t-40/60).
Now Decreasing speed by 2 i.e. (x-2) will decrease time by 40 min i.e (t+40/60).
Therefore,
Distance = (x+3)(t-40/60).
Since Distance would be same:
(x+3)(t-40/60)=(x+3)(t-40/60) -----> (1)
solving this, we get
-2x/3 + 3t -2 = 2x/3 + -2t - 4/3.
3t -2x/3 =2 -->(2)
2x/3 + -2t = 4/3 --->(3)
Solving two-equation, we get t=10/3 and x=12. Putting this value in eq. 1, we get,
Distance = 40.
(60)
Jagdish said:
3 years ago
Divide both the equations.
2y(y+3) 9x.
------- =
y(y-2) 3x.
Solution:
2(y+3)÷y(y-2) = 9x÷ 3x.
2(y+3)÷(y-2) = 3,
2(y+3) = 3(y-2),
2y+6= 3y-6.
3y-2y = 6+6.
Y =12.
2y(y+3) 9x.
------- =
y(y-2) 3x.
Solution:
2(y+3)÷y(y-2) = 9x÷ 3x.
2(y+3)÷(y-2) = 3,
2(y+3) = 3(y-2),
2y+6= 3y-6.
3y-2y = 6+6.
Y =12.
(8)
Posubabu said:
3 years ago
(a(a-2)/2) * 40 = (a(a+3)/3) *40.
a=12kmph this is original speed.
Ddistance = (12(12-10)/2)*40/60.
Distance = 40km.
a=12kmph this is original speed.
Ddistance = (12(12-10)/2)*40/60.
Distance = 40km.
(12)
JAHNAVI said:
1 year ago
I am not understanding this. Please help me to get it.
(9)
Suriya said:
8 months ago
@JAHNAVI.
Initially, distance is the same so to find time and speed, we take speed x and time and now implement the distance formula.
First, speed is greater than 3 we take x+3 and if the speed is increased time is decreased so t-40/60 we convert 40 min to an hour divided by 60. Second x-2 and t+40/60 solve these equations you get the why and then x.
Initially, distance is the same so to find time and speed, we take speed x and time and now implement the distance formula.
First, speed is greater than 3 we take x+3 and if the speed is increased time is decreased so t-40/60 we convert 40 min to an hour divided by 60. Second x-2 and t+40/60 solve these equations you get the why and then x.
(2)
Mohammed shamil said:
6 months ago
Good thanks for explaining.
(1)
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