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 3 of 18.
SUKUMAR SATYEN said:
1 decade ago
Let time taken to go certain distance = t hours.
At speed of 10kmph, distance covered = 10t kms.
At speed of 14kmph, distance covered = 14t kms.
Because, 10t < 14t.
Therefore, 10t + 20 kms = 14t.
=> 14t - 10t = 20 kms.
=> 4t = 20 kms.
=> t = 5 hours.
Now, his actual speed is 10kmph.
Therefore, actual distance covered = 10t = 50kms.
At speed of 10kmph, distance covered = 10t kms.
At speed of 14kmph, distance covered = 14t kms.
Because, 10t < 14t.
Therefore, 10t + 20 kms = 14t.
=> 14t - 10t = 20 kms.
=> 4t = 20 kms.
=> t = 5 hours.
Now, his actual speed is 10kmph.
Therefore, actual distance covered = 10t = 50kms.
Gaurav Raghav said:
2 years ago
Let the actual speed be x.
Let the time be y.
Distance = Speed * time
Therefore,
X= 10y.
But, now he would have travelled 20 km more if the speed was 14km/hr.
therefore,
14x= y+20.
Subtracting both equations,
14x = y+20,
10x = y.
we get;
4x = 20,
x = 5.
Put y in the equation
10 * 5 = 50 Kms.
Verify it by putting 5 in both equations.
Let the time be y.
Distance = Speed * time
Therefore,
X= 10y.
But, now he would have travelled 20 km more if the speed was 14km/hr.
therefore,
14x= y+20.
Subtracting both equations,
14x = y+20,
10x = y.
we get;
4x = 20,
x = 5.
Put y in the equation
10 * 5 = 50 Kms.
Verify it by putting 5 in both equations.
(9)
Epaphra said:
3 years ago
@All.
According to me;
Let the actual distance = y.
At speed of 10KMPH, the distance is y.
So, 10kmph = y.
This implies, 1kmph=y/10 ------> (1).
When he travels at a speed, of 14kmph, the distance is y+20.
So, 14kmph = y+20.
This implies, 1kmph=(y+20)/14 -------> (2)
EQUATE (1)AND (2).
y/10 = (y+20)/14
14y = 10y+200
4y = 200
y = 50km.
According to me;
Let the actual distance = y.
At speed of 10KMPH, the distance is y.
So, 10kmph = y.
This implies, 1kmph=y/10 ------> (1).
When he travels at a speed, of 14kmph, the distance is y+20.
So, 14kmph = y+20.
This implies, 1kmph=(y+20)/14 -------> (2)
EQUATE (1)AND (2).
y/10 = (y+20)/14
14y = 10y+200
4y = 200
y = 50km.
(29)
Palani said:
7 years ago
Friends, we should first understand the concept.
The man walks at 10kmph covered certain distance X. Then man instead of walks at 14kmph covered certain distance X. So both are covering the distance x first. Then he will take an extra time to cover 20km.
The Time difference
D is=20km.
Formula=D = D[a/a-b].
=20(10/4)=50km.
The man walks at 10kmph covered certain distance X. Then man instead of walks at 14kmph covered certain distance X. So both are covering the distance x first. Then he will take an extra time to cover 20km.
The Time difference
D is=20km.
Formula=D = D[a/a-b].
=20(10/4)=50km.
Kalinga kumar said:
9 years ago
We know,
Time = Distance/Speed.
Let's take original distance is X.
Given,
Speed is 10km/1hr and 14km/1hr.
Now we can write,
1hr = X distance/10 speed --> (1).
Similarly,
1hr = (X + 20) distance/14 speed --> (2).
Now, put the value of (1) in (2).
We will get,
X/10 = (X + 20)/14.
14X = 10X + 200.
4X = 200.
X = 50.
Time = Distance/Speed.
Let's take original distance is X.
Given,
Speed is 10km/1hr and 14km/1hr.
Now we can write,
1hr = X distance/10 speed --> (1).
Similarly,
1hr = (X + 20) distance/14 speed --> (2).
Now, put the value of (1) in (2).
We will get,
X/10 = (X + 20)/14.
14X = 10X + 200.
4X = 200.
X = 50.
Gowtham said:
2 decades ago
For the same interval if the person travelled in 14 kmph so only.
or else you could solve like this
1st hour 10 =14
2nd hour 20 =28
3rd hour 30 =42
4th hour 40 =56
5th hour 50 =70
At speed 10 kmph he reaches 50 km at the same time if he travels in 14 kmph he would reach 70 km. So 70 - 50 = 20. I think u got it.
or else you could solve like this
1st hour 10 =14
2nd hour 20 =28
3rd hour 30 =42
4th hour 40 =56
5th hour 50 =70
At speed 10 kmph he reaches 50 km at the same time if he travels in 14 kmph he would reach 70 km. So 70 - 50 = 20. I think u got it.
(4)
Manish nair said:
1 decade ago
It came by this formula S = D/T (Speed = Distance/Time).
Here we take this formula such as:
T = D/S.
Let consider actual distance x km.
T = x/10.....(1).
T = x+20/14.....(2).
Taking both the times together we get from (1) and (2).
x/10 = x+20/14.
14x = 10x+200.
4x = 200.
x = 50km.
Hope you get it now.
Here we take this formula such as:
T = D/S.
Let consider actual distance x km.
T = x/10.....(1).
T = x+20/14.....(2).
Taking both the times together we get from (1) and (2).
x/10 = x+20/14.
14x = 10x+200.
4x = 200.
x = 50km.
Hope you get it now.
Anoop Krishna said:
8 years ago
Hey, friends.
This is the easiest way of thinking.
Let the distance travelled by walking at the speed of 10km/h be x.
Then the distance travelled by walking at the speed of 14km/h will be x+20(according to the question).
so, 14=x+20,
10=x.
Cross multiply and find the value of x.
This is the easiest way of thinking.
Let the distance travelled by walking at the speed of 10km/h be x.
Then the distance travelled by walking at the speed of 14km/h will be x+20(according to the question).
so, 14=x+20,
10=x.
Cross multiply and find the value of x.
Priya said:
4 years ago
He travels 4 km/hr extra in the other case.
So if he traveled 20 km more in that case then it means the time he took would be time= distance/speed = 20/4= 5 hours. So now that we know the time he took, we can find the actual distance he traveled.
Distance= speed* time= 10*5= 50 km.
So if he traveled 20 km more in that case then it means the time he took would be time= distance/speed = 20/4= 5 hours. So now that we know the time he took, we can find the actual distance he traveled.
Distance= speed* time= 10*5= 50 km.
Rahul said:
1 decade ago
The question is asked to find out the distance traveled by the person, and its clearly written he traveled 20km more so we equate it as (x+20) as the distance traveled by him and then when we get the value of X to be 50km. Why it is not added by 20km to get final answer as 70km ?
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