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 10 of 12.
Pooja Gupta said:
8 years ago
Man goes to his office at a certain speed. If he travels at a speed 4kmph more than his usual speed then he will reach his office early by 12 minutes. If he travels at a speed 8 kmph less than his usual speed he will be late by 36 minutes. Find his original speed.
How would we solve it?
How would we solve it?
Bala said:
8 years ago
Why we have to divide both equations?
Why can't do either addition or subtraction?
Let me know the reason, please.
Why can't do either addition or subtraction?
Let me know the reason, please.
Sivag said:
8 years ago
Short trick :
3 * (+40) : 2 *(-40)
120: 80 ==> 3 : 2
Then take Avg for the both speed of the Ratio
2(2*3)/1 ==> same direction so 2(XY) / X-Y.
Then we get original speed 12kmhr.
then ,12+3 = 15kmhr
12-2= 10kmhr
Diff for both time 40 - (-40) = 80mins.
The product of speed/ diff of speed *time
15*10/5 * 80/60.
Finally, we get 40 km distance.
3 * (+40) : 2 *(-40)
120: 80 ==> 3 : 2
Then take Avg for the both speed of the Ratio
2(2*3)/1 ==> same direction so 2(XY) / X-Y.
Then we get original speed 12kmhr.
then ,12+3 = 15kmhr
12-2= 10kmhr
Diff for both time 40 - (-40) = 80mins.
The product of speed/ diff of speed *time
15*10/5 * 80/60.
Finally, we get 40 km distance.
Sumit said:
8 years ago
How will you get Y=12?
my calculation is:
3x+2y-4/3y-6 = 3x-2y-6/3y+9,
(3y+9)(3x+2y-4) =( 3y-6)(3x-2y-6),
9xy+6y^2+6y+27x-36 = 9xy-6y^2-6y-18x+36,
12y^2+12y+45x-72 = 0.
please explain me.
my calculation is:
3x+2y-4/3y-6 = 3x-2y-6/3y+9,
(3y+9)(3x+2y-4) =( 3y-6)(3x-2y-6),
9xy+6y^2+6y+27x-36 = 9xy-6y^2-6y-18x+36,
12y^2+12y+45x-72 = 0.
please explain me.
Naman said:
8 years ago
What if decreasing speed and increasing speed would be same?
When a person is moving with speed x km/hr is reaching office in t time. When he increases speed by 20km/hr, he reaches 40 min early. But when he decreases speed by 20 km/hr, he reaches 40 min late. What is the distance?
Can anyone tell me?
When a person is moving with speed x km/hr is reaching office in t time. When he increases speed by 20km/hr, he reaches 40 min early. But when he decreases speed by 20 km/hr, he reaches 40 min late. What is the distance?
Can anyone tell me?
Nupur said:
8 years ago
See take distance =x and rate=y and t= time.
Now overall x/y=t.
We have one condition,
x/y+3 at t-40.
We have second condition,
x/y-y at t+40.
Now equate,
x/y=t and x/y+3=t-40 sub them,
You get x/y- x/y+3=40/60.
Now equate,
x/y=t and x/y-2=t+40 sub them,
You get x/y-2- x/y=40/60.
You will see t getting canceled.
40/60 because convert min into hr.
Now overall x/y=t.
We have one condition,
x/y+3 at t-40.
We have second condition,
x/y-y at t+40.
Now equate,
x/y=t and x/y+3=t-40 sub them,
You get x/y- x/y+3=40/60.
Now equate,
x/y=t and x/y-2=t+40 sub them,
You get x/y-2- x/y=40/60.
You will see t getting canceled.
40/60 because convert min into hr.
Sirisha said:
8 years ago
Thanks for your explanation @Akash Kishor.
Mani said:
8 years ago
Can we do like this?
(3*40)-(2*40) = D
120-80 = D
Therefore D = 40.
(3*40)-(2*40) = D
120-80 = D
Therefore D = 40.
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?
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.
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