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 7 of 13.
Thanuj said:
9 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.
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.
Sridharan said:
9 years ago
How can I get the actual speed?
Manish said:
10 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?
Harjeet said:
10 years ago
@Pauli.
45.
.5 = 90/x - 90/x + 15.
45.
.5 = 90/x - 90/x + 15.
Dhivakar said:
10 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.
Gursewak singj said:
10 years ago
It is tough to find the solution. Can anybody solve it in a simple way?
Sidzwozz said:
10 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?
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