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Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.
- a)25 km/h
- b)10 km/h
- c)6 km/h
- d)15 km/h
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Sheldon had to cover a distance of 60 km. However, he started 6 minute...
Amount of work P can do in 1 day = 1/20
Amount of work Q can do in 1 day = 1/30
Amount of work R can do in 1 day = 1/60
P is working alone and every third day Q and R is helping him
Work completed in every three days = 2 * (1/20) + (1/20 + 1/30 + 1/60) = 1/5
So work completed in 15 days = 5 * 1/5 = 1
Ie, the work will be done in 15 days
Most Upvoted Answer
Sheldon had to cover a distance of 60 km. However, he started 6 minute...
Let the total work be LCM of (20,30,60) = 60 units
P's 1 day work = 3 units. but P works for all 3 days so units done by P = 3 x 3 = 9
Q's 1 day work = 2 units.
R's 1 day work = 1 unit.
units done by P+R+Q at the end of the 3rd day = 9+2+1 = 12 units
required days = 3/12 x 60 = 15 days
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Community Answer
Sheldon had to cover a distance of 60 km. However, he started 6 minute...
Given:
Distance to be covered, d = 60 km
Delay in starting, t = 6 minutes = 1/10 hour
Let the originally planned speed be x km/h.
Sheldon's speed in the actual race = (x+1) km/h
To find:
The speed at which Sheldon travelled during the journey described.
Solution:
Let's assume that Sheldon took 't' hours to complete the race at the speed of 'x' km/h.
So, the time taken to cover the distance of 60 km at the speed of 'x' km/h would be:
t = d/x
According to the given condition, Sheldon started 6 minutes (1/10 hour) late and completed the race at the same time as he would have completed if he had started on time and raced at the planned speed. This means that he took less time than 't' hours to complete the race.
Let's say he took 'h' hours to complete the race at the speed of (x+1) km/h.
So, the time taken to cover the distance of 60 km at the speed of (x+1) km/h would be:
h = d/(x+1)
According to the question, h = t - 1/10
d/(x+1) = d/x - 1/10
Solving the above equation, we get:
x = 24 km/h
So, the speed at which Sheldon travelled during the journey described = x+1 = 24+1 = 25 km/h
Therefore, option 'A' is the correct answer.
Distance to be covered, d = 60 km
Delay in starting, t = 6 minutes = 1/10 hour
Let the originally planned speed be x km/h.
Sheldon's speed in the actual race = (x+1) km/h
To find:
The speed at which Sheldon travelled during the journey described.
Solution:
Let's assume that Sheldon took 't' hours to complete the race at the speed of 'x' km/h.
So, the time taken to cover the distance of 60 km at the speed of 'x' km/h would be:
t = d/x
According to the given condition, Sheldon started 6 minutes (1/10 hour) late and completed the race at the same time as he would have completed if he had started on time and raced at the planned speed. This means that he took less time than 't' hours to complete the race.
Let's say he took 'h' hours to complete the race at the speed of (x+1) km/h.
So, the time taken to cover the distance of 60 km at the speed of (x+1) km/h would be:
h = d/(x+1)
According to the question, h = t - 1/10
d/(x+1) = d/x - 1/10
Solving the above equation, we get:
x = 24 km/h
So, the speed at which Sheldon travelled during the journey described = x+1 = 24+1 = 25 km/h
Therefore, option 'A' is the correct answer.
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