SOLUTION: 1. Ramesh walks at 5/6th of his usual speed and reaches school 4 minutes late. Find the usual time taken by him to reach school. 2. Rakesh runs at 5/4 of his usual speed and reach

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Question 205963: 1. Ramesh walks at 5/6th of his usual speed and reaches school 4 minutes late. Find the usual time taken by him to reach school.
2. Rakesh runs at 5/4 of his usual speed and reaches the playground 5 minutes earlier. What is his usual time/
3.Two trains 110m and 100m long are running on parallel tracks, in the same direction with a speed of 46 km/hr and 39 km/hr respectively. How long will it take them to be clear of each other?
4. A cyclist A started his journey on cycle at 7.30 a.m. at speed 8km/hr. B another cyclist started from the same point half an hour later but with speed 10 km/hr. At what time did B overtake A?
Found 2 solutions by edjones, ankor@dixie-net.com:
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
1.
Let x=the time it takes to go to school at the usual speed
5x/6=4
x=4*6/5
=24/5 min. Time it takes to go to school at the usual speed.
.
Ed

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
1. Ramesh walks at 5/6th of his usual speed and reaches school 4 minutes late.
Find the usual time taken by him to reach school.
:
Let s = usual walking speed
then
5%2F6s = slower speed
:
Let t = usual time to walk to school
then
(t+4) = time required when walking slower
:
write a distance equation
ts = (t+4)(5%2F6s)
;
multiply both sides by 6
6ts = 5s(t+4)
6ts = 5ts + 20s
6ts - 5ts = 20s
ts = 20s
Divide for sides by s, results
:
t = 20 min; normal time to walk to school
;
:
2. Rakesh runs at 5/4 of his usual speed and reaches the playground 5 minutes earlier.
What is his usual time/
:
Let s = usual running speed
then
5%2F4s = faster running speed
:
Let t = usual time to run to school
then
(t-5) = time required when running faster
:
write a distance equation
ts = (t-5)(5%2F4s)
;
Multiply both side by 4
4ts = 5s(t-5)
4ts = 5ts - 25s
4ts - 5ts = -25s
-ts = -25s
Divide for sides by -s, results
:
t = 25 min; normal time running to school
:
:
3.Two trains 110m and 100m long are running on parallel tracks, in the same
direction with a speed of 46 km/hr and 39 km/hr respectively.
How long will it take them to be clear of each other?
:
The relative speed between the two trains: 46 - 39 = 7 km/hr
:
How long to travel 110+100 = 210m at 7 km/hr?
:
210 m = .21 km: .21%2F7 * 60 ~ 1.8 minutes
:
:
4. A cyclist A started his journey on cycle at 7.30 a.m. at speed 8km/hr.
B another cyclist started from the same point half an hour later but with
speed 10 km/hr.
At what time did B overtake A?
:
Let t = B's travel time
then
(t+.5) = A's travel time
:
When B overtakes A, they will have traveled the same distance
Write a distance equation
:
10t = 8(t+.5)
10t = 8t + 4
10t - 8t = 4
2t = 4
t = 4%2F2
t = 2 hrs is B's travel time
;
He left at 8:00, therefore he overtakes A at 10:00
:
:
Check solution by ensuring their distances are equal
2*10 = 20 km
2.5* 8 = 20 km