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:
176 comments Page 3 of 18.
Meghan Desai said:
3 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)
Pulyala ajith kumar said:
3 years ago
The difference between distances is 20km and d = s * t.
So 14x-10x = 20.
x = 5hrs
The actual distance is s*t = 10 * 5 = 50km.
So 14x-10x = 20.
x = 5hrs
The actual distance is s*t = 10 * 5 = 50km.
(56)
Aryan said:
3 years ago
@All,
According to me, the solution is;
For every 1 km/hr there is 5 km distance travelled;
So for 10km/hr there is 50km distance travelled;
which means for 4 more km/hr in question there is 20 kM distance.
So, for 10 km it's 50 km.
According to me, the solution is;
For every 1 km/hr there is 5 km distance travelled;
So for 10km/hr there is 50km distance travelled;
which means for 4 more km/hr in question there is 20 kM distance.
So, for 10 km it's 50 km.
(10)
Gaurav Raghav said:
3 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)
Habtamu dese said:
3 years ago
Good to understand. Thanks everyone for explaining the answer.
(8)
Sriya said:
4 years ago
Thanks for your explanation, it's easy to understand @Gowtham.
(13)
Salsabil said:
4 years ago
According to me, by using the simple formula we can iterate lucidly like this way,
Let the time be x hours
According to the question;
14x - 10x = 20
4x = 20
x = 5 hours.
His actual speed is 10 km/h.
So, the actual distance is 10 * 5 = 50 km.
Let the time be x hours
According to the question;
14x - 10x = 20
4x = 20
x = 5 hours.
His actual speed is 10 km/h.
So, the actual distance is 10 * 5 = 50 km.
(211)
Vineel said:
4 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.
(15)
Epaphra said:
4 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)
Abc said:
4 years ago
Thanks @Kinnari.
(1)
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