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:
121 comments Page 8 of 13.
PaulI said:
10 years ago
How about this question?
Had Jake traveled 15 kph faster, the 90 km trip would have taken him 30 mins earlier. What is his speed?
Had Jake traveled 15 kph faster, the 90 km trip would have taken him 30 mins earlier. What is his speed?
Mayur said:
10 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.
Nagurum said:
10 years ago
@Zaid.
After finding the time and I cannot find the distance value, how to solve? Please help me.
After finding the time and I cannot find the distance value, how to solve? Please help me.
Ram said:
10 years ago
Anyone find the original speed of the above question?
Atchu said:
10 years ago
@Sivaram.
Is it the correct way and suitable to all this type of questions?
Is it the correct way and suitable to all this type of questions?
Zaid said:
10 years ago
Let distance = D, speed = S, time taken = T (formula, we know S*T=D).
Let's assume the man covered the distance twice(2D), one D with S+3 kmph and another D with S-2 kmph.
So, D + D = (S + 3) (T - 40) + (S - 2) (T + 40).
2D = ST + 3T - 40S - 120 + ST - 2T + 40S - 80.
=> 2D = 2ST + T - 200.
But ST = D,
=> T = 200 min.
Now with S + 3 speed, time taken = 200 - 40 = 160.
and with S - 2 speed, time was taken = 200 + 40.
= 240.
But we have a problem in our earlier format, solve it. You will get D, distance = 40 km.
Let's assume the man covered the distance twice(2D), one D with S+3 kmph and another D with S-2 kmph.
So, D + D = (S + 3) (T - 40) + (S - 2) (T + 40).
2D = ST + 3T - 40S - 120 + ST - 2T + 40S - 80.
=> 2D = 2ST + T - 200.
But ST = D,
=> T = 200 min.
Now with S + 3 speed, time taken = 200 - 40 = 160.
and with S - 2 speed, time was taken = 200 + 40.
= 240.
But we have a problem in our earlier format, solve it. You will get D, distance = 40 km.
Raj said:
1 decade ago
Please explain the way way of getting the value of y =12.
Ramakrishna said:
1 decade ago
@Sivaram,
Here 3 is kmhr, but 40 is minutes.
Our requirement is Distance in KM.
So Please explain clearly?
Here 3 is kmhr, but 40 is minutes.
Our requirement is Distance in KM.
So Please explain clearly?
Naveen said:
1 decade ago
Can anyone tell me the shortcut method to solve it?
Arjun Singh said:
1 decade ago
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.
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.
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