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:
173 comments Page 2 of 18.
Liashee said:
3 years ago
D= (S1*T ) - (S2*T)
20= (14*T) - (10*T)
20= 14T - 10T
T = 20/4
T= 5 hrs.
Therefore Actual distance where actual speed was 10km/ hr.
D= S*T
= 10km/hr * 5hr.
= 50km.
I hope this was helpful.
20= (14*T) - (10*T)
20= 14T - 10T
T = 20/4
T= 5 hrs.
Therefore Actual distance where actual speed was 10km/ hr.
D= S*T
= 10km/hr * 5hr.
= 50km.
I hope this was helpful.
(32)
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)
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)
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)
Chahat said:
2 years ago
Thanks, everyone for explaining.
(20)
Anomiee said:
1 year ago
Thanks for explaining.
(18)
Vineel said:
3 years ago
Distance = a/x-y *y.
20/14 - 10 * 10,
20/4 * 10 = 5 * 10 = 50km.
20/14 - 10 * 10,
20/4 * 10 = 5 * 10 = 50km.
(14)
Sriya said:
3 years ago
Thanks for your explanation, it's easy to understand @Gowtham.
(13)
Vishal said:
2 years ago
Why t in both the equation is same?
If a person is walking at 14km/hr then it will take less time then 10km/hr, which means both t are different.
If a person is walking at 14km/hr then it will take less time then 10km/hr, which means both t are different.
(12)
Rishika V said:
4 months ago
In both speed, the time period is same right?
(11)
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