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.
Kamu said:
7 years ago
Why subtracting the equation?
Ruchir said:
7 years ago
How come 40/60? Please explain in detail.
Magnus said:
7 years ago
@ Ruchir,
For Converting Minutes to an hour you divide by 60.
For Converting Minutes to an hour you divide by 60.
Manzoor said:
6 years ago
Diff = 1speed.80/60time.
Add =5
Multiple = 6.
80/60 * 6 * 5/1 = 40.
Add =5
Multiple = 6.
80/60 * 6 * 5/1 = 40.
Durgesh Singh said:
6 years ago
S + s1/s1*t1=S+s2/s2*t2=T.
So S+3/3*40=S-2/2*40,
S= 12.
Then T= 12+3/3*40/60 = 10/3.
ST = D.
So 10/3*12 = 40.
So S+3/3*40=S-2/2*40,
S= 12.
Then T= 12+3/3*40/60 = 10/3.
ST = D.
So 10/3*12 = 40.
Md Parvez Islam said:
6 years ago
Thanks @Sachin Bhagat.
Sandjive said:
6 years ago
@Pauli.
For Jake sum.
S*t=90.....1.
(S+15)(t-30/60)=90.......2.
Solving 2 eqn, answer: 45km/hr.
For Jake sum.
S*t=90.....1.
(S+15)(t-30/60)=90.......2.
Solving 2 eqn, answer: 45km/hr.
Shweta said:
6 years ago
Thank you @Sachin bhagat.
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