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 5 of 12.
Anmol ratan said:
1 decade ago
X/y-2+40/60=x/y+3-40/60 is a wrong equation.
Arjun Singh said:
10 years 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.
Naveen said:
10 years ago
Can anyone tell me the shortcut method to solve it?
Ramakrishna said:
10 years 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?
Raj said:
9 years ago
Please explain the way way of getting the value of y =12.
Zaid said:
9 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.
Atchu said:
9 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?
Ram said:
9 years ago
Anyone find the original speed of the above question?
Nagurum said:
9 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.
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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