Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 7)
7.
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Answer: Option
Explanation:
| (1/2)x | + | (1/2)x | = 10 |
| 21 | 24 |
 |
x | + | x | = 20 |
| 21 | 24 |
15x = 168 x 20
| x = |
 |
168 x 20 |  |
= 224 km. |
| 15 |
Discussion:
143 comments Page 13 of 15.
Dhaval said:
7 years ago
Hear if total distance is X.
And hear fix first 1/2 at fix speed and another half at other fix speed.
So we can assume avg speed so [ speed (1) + speed (2) ] / 2 so we get avg speed for a total this x.
Then for 10 hr final answer is *10 so we get 225.
And hear fix first 1/2 at fix speed and another half at other fix speed.
So we can assume avg speed so [ speed (1) + speed (2) ] / 2 so we get avg speed for a total this x.
Then for 10 hr final answer is *10 so we get 225.
Sai kumar said:
7 years ago
I didn't understand this, can you explain this question?
GETE BASU CHAKMA said:
7 years ago
Let the total distance = X km.
The time is taken by first half journey =(X/2) * 21 hr.
The time is taken by Second half journey =(X/2) * 24 hr.
So, the total time(10 hr) =(X/2) * 21 + (X/2) * 24.
Then we get X= 224 km.
The time is taken by first half journey =(X/2) * 21 hr.
The time is taken by Second half journey =(X/2) * 24 hr.
So, the total time(10 hr) =(X/2) * 21 + (X/2) * 24.
Then we get X= 224 km.
Mani Gupta said:
7 years ago
How this is possible that a train travels with 21 km/h speed and 24 km/h speed, covers same distance?
So, we have to assume different distance for a different speed. Right?
So, we have to assume different distance for a different speed. Right?
Alpesh said:
7 years ago
Why don't you take average speed and multiply with the time?
I get average speed 72 and distance is 720. Am I right?
I get average speed 72 and distance is 720. Am I right?
Rishwanth Immaneni said:
7 years ago
Here we get 13x at the place of 15x and the remaining process is same.
Ali said:
7 years ago
10 hrs we can divide it as 5 + 5 hrs.
1st half of journey at rate of 21 km/hr,
So 21*5=105,
2nd half of journey at rate of 24 km/hr.
So 24*5 = 120.
105+120 = 225.
So option B 224 is correct.
1st half of journey at rate of 21 km/hr,
So 21*5=105,
2nd half of journey at rate of 24 km/hr.
So 24*5 = 120.
105+120 = 225.
So option B 224 is correct.
SAURAV KUMAR said:
7 years ago
We can calculate it by the average formula:
2(21*24)/21+24 = 2(504)/45.
= 1008/45,
= 22.4.
Then the distance travelled in 10hours.
= 22.4 * 10,
= 224.
2(21*24)/21+24 = 2(504)/45.
= 1008/45,
= 22.4.
Then the distance travelled in 10hours.
= 22.4 * 10,
= 224.
Rose said:
6 years ago
The average speed is 2*21*24/45 which is 336/15 km/hr.
Speed=distance/time
So 336/15 = distance/10.
Therefore, distance = 336 * 10/15 which is 224 km.
Speed=distance/time
So 336/15 = distance/10.
Therefore, distance = 336 * 10/15 which is 224 km.
Niraj said:
6 years ago
(1/2)x 21 + (1/2)x 24 = 10.
x /21 + x /24 = 10 x 2.
8x /21x8 + 7x /24x7 = 20.
8x /21x8 + 7x /24x7 = 20.
15x = 20x168.
x = 20x168/15.
x = 3360/15.
x = 224km.
x /21 + x /24 = 10 x 2.
8x /21x8 + 7x /24x7 = 20.
8x /21x8 + 7x /24x7 = 20.
15x = 20x168.
x = 20x168/15.
x = 3360/15.
x = 224km.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers