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 8 of 12.
PaulI said:
9 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?
Sidzwozz said:
9 years ago
@Pauli.
According to my calculations, Jake has 4.5 Km/Hour. Let me know whether it's wrong or Right?
According to my calculations, Jake has 4.5 Km/Hour. Let me know whether it's wrong or Right?
Gursewak singj said:
9 years ago
It is tough to find the solution. Can anybody solve it in a simple way?
Dhivakar said:
9 years ago
If 3km/hr faster than actual speed, it takes 40 minutes less time to complete.
If 2km/hr slower, it takes 40 minutes more.
So,
If we take condition 1, it would be 2km more than the actual distance at the actual time.(3km/hr, so 2km in 40 minutes)
If we take condition 2, it would be 4/3 km less than the actual distance at the actual time.(2km/hr, so 4/3 km in 40 minutes)
Now the difference between the two condition is 80 minutes and the distance is 2 + 4/3.
Distance = speed * time.
So, we equate, 10/3 = x/80 * 60(multiplying 60 is for hour conversion).
So, x = 40.
If 2km/hr slower, it takes 40 minutes more.
So,
If we take condition 1, it would be 2km more than the actual distance at the actual time.(3km/hr, so 2km in 40 minutes)
If we take condition 2, it would be 4/3 km less than the actual distance at the actual time.(2km/hr, so 4/3 km in 40 minutes)
Now the difference between the two condition is 80 minutes and the distance is 2 + 4/3.
Distance = speed * time.
So, we equate, 10/3 = x/80 * 60(multiplying 60 is for hour conversion).
So, x = 40.
Harjeet said:
9 years ago
@Pauli.
45.
.5 = 90/x - 90/x + 15.
45.
.5 = 90/x - 90/x + 15.
Manish said:
9 years ago
WE KNOW,
DISTANCE = SPEED * TIME.
Then,
(speed+3) * (time - 40/60) = (speed - 2) (time + 40/60) = distance, I am not getting answer,
Can anyone explain by this method?
DISTANCE = SPEED * TIME.
Then,
(speed+3) * (time - 40/60) = (speed - 2) (time + 40/60) = distance, I am not getting answer,
Can anyone explain by this method?
Sridharan said:
9 years ago
How can I get the actual speed?
Singh said:
9 years ago
Simplest formulae (S + 3)/(s - 2) = 3t2/2t1,
Solving we get s = 12km/h.
Solving we get s = 12km/h.
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.
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.
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