Problem 1 :
A person covers a certain distance at a certain speed. If he increases his speed by 33⅓ %, he takes 15 minutes less to cover the same distance. Find the time taken by him initially to cover the distance at the original speed.
Solution :
Let the original speed be 100%.
Given : The speed is increased by 33⅓%.
Then, the speed after increment is
133⅓%
Ratio of the speeds is
100% : 133⅓%
100 : 133⅓
100 : 400/3
Divide both sides by 100.
1 : 3/4
So, ratio of the speeds is
1 : 3/4
If the ratio of the speeds is 1 : 4/3, then the ratio of time taken to cover the same distance would be
1 : 3/4
When the speed is increased by 33⅓ %, 3/4 of the original time is enough to cover the same distance.
That is, when the speed is increased by 33⅓ %, 1/4 of the original time will be decreased.
The question says that when speed is increased by 33⅓ %, time is decreased by 15 minutes.
So, we have
1/4 of the original time = 15 minutes
Multiply both sides by 4.
4 ⋅ (1/4 of the original time) = (15 minutes) ⋅ 4
Original time = 60 minutes
Original time = 1 hour
So, the time taken by the person initially is 1 hour.
Problem 2 :
A man traveled from the village to the post office at the rate of 25 k mph and walked back at the rate of 4 kmph. If the entire journey had taken 5 hours 48 minutes, find the distance of the post office from the village.
Solution :
Here, the distance covered in both the ways is same.
So, the formula to find the average speed is
= 2pq/(p + q)
Plug p = 25 and q = 4.
= (2 ⋅ 25 ⋅ 4)/(25 + 4)
= 200/29
The average speed is 200/29 km/hr.
Given :The entire journey had taken 5 hours 48 minutes
5 hour 48 min = 5 ^{48}⁄_{60 }hours
5 hours 48 min = 5 ⅘ hours
5 hours 48 min = 29/5 hours
The formula to find the distance is
= Speed ⋅ Time
Then, the distance covered in (29/5) hours at the average speed (200/29) kmph is
= (200/29) ⋅ (29/5)
= 40 km
So, the distance covered in the whole journey is 40 km.
(Whole journey : Village to post office + Post office to village)
Then the distance between the post office and village is
= 40/2
= 20
So, the distance of the post office from the village is 20 km.
Problem 3 :
If a man walks at the rate of 5 km/hr, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
Solution :
Let "x" be the distance to be covered by the person to reach the station.
The formula to find the time is
= distance/speed
When the speed is 5 km/h, time is
= x/5 hours
When the speed is 6 km/h, time is
= x/6 hours
Let "t" be the actual time required to cover the distance x.
And also,
7 minutes = 7/60 hours
5 minutes = 5/60 = 1/12 hours
Given : If the man walks at the rate of 5 km/hr, he misses the train by 7 minutes.
That is, he takes 7 minutes more than actual time.
So, we have
t = x/5 - 7/60 ----(1)
Given : If he walks at the rate of 6 km/hr, he reaches the station 5 minutes before.
That is, he takes 5 minutes less than actual time.
So, we have
t = x/6 + 1/12 ----(2)
From (1) and (2), we get
x/5 - 7/60 = x/6 + 1/12
Solving for x :
12x/60 - 7/60 = 2x/12 + 1/12
(12x - 7) / 60 = (2x + 1) / 12
L.C.M of (60, 12) is 60.
Multiply both sides by 60.
12x - 7 = 5(2x + 1)
12x - 7 = 10x + 5
2x = 12
x = 6
So,the distance covered by him to reach the station is 6 km.
Problem 4 :
A person has to cover a distance of 6 miles in 45 minutes. If he covers one-half of the distance in two-thirds of the total time. What must his speed be to cover the remaining distance in the remaining time ?
Solution :
Given : Total distance is 6 miles and total time is 45 minutes. And also, he covers one-half of the distance in two-thirds of the total time.
One-half of the total distance 6 miles is
= (1/2) ⋅ 6
= 3 km
Two-thirds of the total time 45 minutes is
= (2/3) ⋅ 45
= 30 minutes
From the above calculations, we have
Remaining distance = 6 - 3 = 3 miles
Remaining time = 45 - 30 = 15 minutes
The formula to find the speed is
= distance/time
Substitute distance = 3 and time = 15.
= 3/15
= 1/5 miles per minute
= (1/5) ⋅ 60 miles per hour
= 12 miles per hour.
So, the speed must be 12 miles per hour.
Problem 5 :
A is faster than B . A and B each walk 24 miles. The sum of their speeds is 7 miles per hour and the sum of their time taken is 14 hrs. Find A's speed and B's speed (in mph).
Solution :
Let "x" be the speed of A .
Then speed of B is
= 7 - x
The formula to find time is
= distance/speed
Then, time taken by A is
= 24/x hours
Time taken by B is
= 24/(7 - x) hours
Given : Sum of time taken is 14 hours.
So, we have
24/x + 24/(7 - x) = 14
L.C.M of x and (7 - x) is x(7 - x).
Multiply both sides by x(7 - x).
24 ⋅ (7-x) + 24 ⋅ x = 14 ⋅ x(7 - x)
168 - 24x + 24x = 98x - 14x^{2}
14x^{2 }- 98x + 168 = 0
Divide both sides by 14.
x^{2 }- 7x + 12 = 0
(x - 4)(x - 3) = 0
x = 4 or x = 3
So, A's speed can be 4 mph or 3 mph.
If A's speed is 4 mph, then B's speed
7 - x = 7 - 4
7 - x = 3 mph
If A's speed is 3 mph, then B's speed
7 - x = 7 - 3
7 - x = 4 mph
Given : A is faster than B
So, the speed of A is 4 miles per hour and B is 3 miles per hour.
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