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 6 of 12.
Puneet negi said:
8 years ago
How can we get the answer in km when the time is given in minutes? @Shivraman.
Sonu kumar said:
8 years ago
@Akash.
I can't understand why have you divided by e3 and 2 on both side of equations. Can you explain please?
I can't understand why have you divided by e3 and 2 on both side of equations. Can you explain please?
Uroosa said:
8 years ago
@Sivaram.
How can you multiply km/hr with minutes without changing it to hours?
How can you multiply km/hr with minutes without changing it to hours?
Manik said:
8 years ago
Please, can anyone solve this equation in details?
On dividing (i) by (ii), we get: x = 40.
On dividing (i) by (ii), we get: x = 40.
Neeraj said:
8 years ago
3/x+3/2/x-2= 40/40.
On solving we get x = 12.
On solving we get x = 12.
Akash Kishor said:
8 years ago
LET'S SOLVE IN VERY EASY MANNER.
Easiest formula :- DISTANCE = (s1*s2)/(speed difference) * time difference
So, now
D= {s*(s+3)/3} * 40/60 = {s*(s-2)/2} * 40/60.
=> (s+3)/3 = (s-2)/2 [ s and 40/60 will cancel out each other at both side],
=> 2s+6 = 3s-6,
=> s = 12 km/hr.
And now,
D= (12*15)/3 * 40/60 = 40 km Ans. [ according to formula].
Easiest formula :- DISTANCE = (s1*s2)/(speed difference) * time difference
So, now
D= {s*(s+3)/3} * 40/60 = {s*(s-2)/2} * 40/60.
=> (s+3)/3 = (s-2)/2 [ s and 40/60 will cancel out each other at both side],
=> 2s+6 = 3s-6,
=> s = 12 km/hr.
And now,
D= (12*15)/3 * 40/60 = 40 km Ans. [ according to formula].
Amit said:
8 years ago
Hi,
3*40 =120
2*40 = 80
120-80 = 40.
In this one quantity is in hr and second one is in minutes how you cancel the different units.
km/hr*min not relevant.
3*40 =120
2*40 = 80
120-80 = 40.
In this one quantity is in hr and second one is in minutes how you cancel the different units.
km/hr*min not relevant.
Thanuj said:
8 years ago
3 * 40 = 120,
2 * 40 = 80,
120-80 = 40.
Does this process will come? Please tell me.
2 * 40 = 80,
120-80 = 40.
Does this process will come? Please tell me.
Pravinya said:
9 years ago
Hi, I'm giving a practical example on it.
Suppose initial time is 5minute and after increasing speed is 3 minute so 5-3 = 2minute that 2 minute is time taken to cover the distance at 3kmph faster speed.
Now make an equation for 5 i.e. as per question and subtract equation 3 i.e as per question and 2 minute is 40/60.
Suppose initial time is 5minute and after increasing speed is 3 minute so 5-3 = 2minute that 2 minute is time taken to cover the distance at 3kmph faster speed.
Now make an equation for 5 i.e. as per question and subtract equation 3 i.e as per question and 2 minute is 40/60.
Jagan said:
9 years ago
Let distance be x speed be y and time taken by him in actual speed is p, ie., x/y=p (distance/speed=time) (km/kmphr=hr).
Then go for 1st condition 3 km faster hence 40 min less, x/(y+3)=p-40/60 (converting 40 min to hr).
Already we know p=x/y sub it we get x/(y+3)= x/y-40/60, that's how we get x/y -x/(y+3)= 40/60.
Then go for 1st condition 3 km faster hence 40 min less, x/(y+3)=p-40/60 (converting 40 min to hr).
Already we know p=x/y sub it we get x/(y+3)= x/y-40/60, that's how we get x/y -x/(y+3)= 40/60.
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