Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 3)
3.
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
Answer: Option
Explanation:
Let the actual distance travelled be x km.
| Then, | x | = | x + 20 |
| 10 | 14 |
14x = 10x + 200
4x = 200
x = 50 km.
Discussion:
174 comments Page 2 of 18.
Ravi said:
1 decade ago
As time is same in both the case. Therefore distance is directly proportional to speed. Let distance is x.
x directly proportional to 10km/hr.
So, x/10 directly proportional to 1 km/hr.
And (x/10) *14 directly proportional to 14 km/hr.
And (x/10) *14 is the total distance covered in 2nd case which x+20.
So just compare both the distance i.e.
(x/10) *14 = x+20. You will get x=50. Answer.
x directly proportional to 10km/hr.
So, x/10 directly proportional to 1 km/hr.
And (x/10) *14 directly proportional to 14 km/hr.
And (x/10) *14 is the total distance covered in 2nd case which x+20.
So just compare both the distance i.e.
(x/10) *14 = x+20. You will get x=50. Answer.
Vijeth BK said:
1 year ago
The speed of man is =10km/hr.
In question that given if he walked 14km/hr instead of 10km/hr,he will reach
20km more(but the time is same).
If he walked 10km/hr, we take that distance covered is Xkm.
If he walked 14km/hr, we take that distance covered X+20km.
But for a both cases time is the same.
(Time = distance/speed).
Since X/10 = (X+20)/14.
14X = 10X+200
4X = 200
X = 50km.
In question that given if he walked 14km/hr instead of 10km/hr,he will reach
20km more(but the time is same).
If he walked 10km/hr, we take that distance covered is Xkm.
If he walked 14km/hr, we take that distance covered X+20km.
But for a both cases time is the same.
(Time = distance/speed).
Since X/10 = (X+20)/14.
14X = 10X+200
4X = 200
X = 50km.
(87)
Anwaar said:
1 year ago
At usual time Period, a man Covers a Distance of 10 KM/hr ,But he has increased the Speed to 14 km/hr.
That means he will going 4 Km/hr additionally.
To Cover the Distance of 20 km with the Same of 14 km/hr he has to travel additional 5 hours.
4km * 5hrs = 20 km.
After 5 hours of travel if the person is travelling at the speed of 10 km/hr covers;
D = S * T/
D = 1 0 *5.
D = 50 KM.
That means he will going 4 Km/hr additionally.
To Cover the Distance of 20 km with the Same of 14 km/hr he has to travel additional 5 hours.
4km * 5hrs = 20 km.
After 5 hours of travel if the person is travelling at the speed of 10 km/hr covers;
D = S * T/
D = 1 0 *5.
D = 50 KM.
(37)
Sohail said:
1 decade ago
We can do this in simple unitary method; like,
If the man travels at the speed of 14 km/hr he covers 20 extra km than if he travels at 10 km/hr.
Therefore we can see this extra 4 km/hr speed helps him to cover 20 km.
Now we can write in:
4 km/hr speed he covers 20 km;
1 km/hr speed he cover 20/4 km;
Now in (original speed) 10 km/hr speed he covers 20/4 x 10 = 50 km.
If the man travels at the speed of 14 km/hr he covers 20 extra km than if he travels at 10 km/hr.
Therefore we can see this extra 4 km/hr speed helps him to cover 20 km.
Now we can write in:
4 km/hr speed he covers 20 km;
1 km/hr speed he cover 20/4 km;
Now in (original speed) 10 km/hr speed he covers 20/4 x 10 = 50 km.
Krish said:
1 decade ago
Guys if two boys are given a constant time (10min). One boy who walks little slower covered small distance 500m only at 10min. Another boy who walks faster covers more distance 800m at that constant time (10min). If time is kept constant means only distance covered varies.
Likewise If a constant distance given means time taken by two boys to reach that distance varies.
Likewise If a constant distance given means time taken by two boys to reach that distance varies.
Bdboy Hasan said:
9 years ago
Speed increase 14 - 10 = 4 kmph and so that distance increase 20 km.
When distance increases 4 km then time is 1 hour.
When distance increases 1 km then time is 1/4 hour.
When distance increases 20 km then time is 20/4 hour or 5 hours.
So, he journeyed 5 hours in 10 kmph.
So, distance is 5 * 10 = 50 hour.
Note : This technique is more useful for quick calculation.
When distance increases 4 km then time is 1 hour.
When distance increases 1 km then time is 1/4 hour.
When distance increases 20 km then time is 20/4 hour or 5 hours.
So, he journeyed 5 hours in 10 kmph.
So, distance is 5 * 10 = 50 hour.
Note : This technique is more useful for quick calculation.
Meghan Desai said:
2 years ago
Let
Actual distance is 'X' - (at 10kmph),
Distance afterwards is 'X+20' - (at 14kmph).
Also, the time be 'Y'.
FORMULA: Distance = Speed x Time.
1. X = 10Y ---> (a)
2. X+20 = 14Y
X = 14Y-20 ---> (b).
Comparing a&b we have;
10Y = 14Y-20
So, 14Y-10Y = 20;
and, Y = 5hrs.
Actual distance (at actual speed):-
10kmphx 5hrs = 50km.
That's the answer.
Actual distance is 'X' - (at 10kmph),
Distance afterwards is 'X+20' - (at 14kmph).
Also, the time be 'Y'.
FORMULA: Distance = Speed x Time.
1. X = 10Y ---> (a)
2. X+20 = 14Y
X = 14Y-20 ---> (b).
Comparing a&b we have;
10Y = 14Y-20
So, 14Y-10Y = 20;
and, Y = 5hrs.
Actual distance (at actual speed):-
10kmphx 5hrs = 50km.
That's the answer.
(30)
Kaviprema k said:
2 years ago
We don't know the actual distance so let we consider it as x.
The actual speed is 10 km/ hr ->x.
Another speed is 14 km/hr ->x + 20 km.
( because we have considered distance is x and if he had walked 14 km/ hr then he would have walked 20 km more)
So,
X/ 10 = x+20/14,
14x = 10x+200,
14x-10x = 200,
4x = 200,
x = 200/4,
x = 50.
So,the actual distance is 50 km.
The actual speed is 10 km/ hr ->x.
Another speed is 14 km/hr ->x + 20 km.
( because we have considered distance is x and if he had walked 14 km/ hr then he would have walked 20 km more)
So,
X/ 10 = x+20/14,
14x = 10x+200,
14x-10x = 200,
4x = 200,
x = 200/4,
x = 50.
So,the actual distance is 50 km.
(29)
Tariq said:
2 years ago
Given:
S1= 14km/h
S2= 10km/h
D1= 20km
D2=?
Solution:
We know that; (Distance= Speed * Time)
We first need to know the time taken,
By putting values:
20km = 14km/h (s1) - 10km/h (s2)
20km = 4 km/h,
20/4 = 5.
So the time (T) is 5.
Similarly, putting values we get;
D2 (actual distance)= 10km/h (actual speed) * 5 (time).
Distance= 10*5km.
Actual distance= 50km.
S1= 14km/h
S2= 10km/h
D1= 20km
D2=?
Solution:
We know that; (Distance= Speed * Time)
We first need to know the time taken,
By putting values:
20km = 14km/h (s1) - 10km/h (s2)
20km = 4 km/h,
20/4 = 5.
So the time (T) is 5.
Similarly, putting values we get;
D2 (actual distance)= 10km/h (actual speed) * 5 (time).
Distance= 10*5km.
Actual distance= 50km.
(131)
Santoooosh said:
1 decade ago
It is very clear. The time in both the cases is same that is 5 hrs.
T = 5;
Now the distance travelled by him at a speed of 10 kmph is x/10 in 5 hrs.
Similarly the distance travelled by him at a speed of 14 kmph is (x+20)/14 in 5 hrs.
Form that we can say that x/10 = 5, and (x+20)/14 = 5;
From the above x/10 = (x+20)/14.
Solve this you will get the answer.
T = 5;
Now the distance travelled by him at a speed of 10 kmph is x/10 in 5 hrs.
Similarly the distance travelled by him at a speed of 14 kmph is (x+20)/14 in 5 hrs.
Form that we can say that x/10 = 5, and (x+20)/14 = 5;
From the above x/10 = (x+20)/14.
Solve this you will get the answer.
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