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12 Oct 2022, 17:36
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Dinesh and Rahim start their journey from P and Q respectively at the same time. They meet at a point R, reverse their directions and interchange their speeds. Dinesh took t hours more to reach back P than Rahim took to reach back Q. By what percentage is the initial speed of Dinesh greater than the speed of Rahim if it is known that the ratio of the time taken by Rahim to reach R to the time t is 3:8?
A. 300
B. 150
C. 200
D. 500
E. 350
from studycopter
A. 300
B. 150
C. 200
D. 500
E. 350
from studycopter
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12 Oct 2022, 18:37
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First, going to start with some assumed variables and values.
Since the overall gap distance between the two people at the start does not matter, let’s make it a nice 100 meters.
Speed of Dinesh = D
Speed of Rahim = R
The Time it takes the two of them to cover the gap distance and meet at the meeting point we can call = a
We want to know the percentage increase of D over R:
(D - R) / (R) * [100]% = ?
(1st) They travel towards each other and Meet
Dinesh covers X of the meters:
Using (Speed)(Time) = (Distance):
(D)(a) = X
Rahim covers the remaining Gap Distance in the same Time (a):
(R)(a) = (100 - X)
(2nd) Each returns back to his or her own starting point, at the OTHER’s Speed:
Dinesh: covers the X meters at speed of (R) —- we can call this Time = (b)
(R)(b) = X
Rahim: covers the (100 - X) meters at speed of (D) —— we can call this Time = (c)
(D)(c) = (100 - X)
Dinesh’s time is (t) hours more than Rahim’s time
b = (t) + c
Finally:
(Time taken by Rahim to reach the meeting point) : (t) = (3) : (8)
Rahim and Dinesh took (a) hours to reach the meeting point
Let (a) = 3
Let (t) = 8
Now fill in the information we have:
(D) (3) = X —- speed of D: (X) / (3)
(R) (3) = (100 - X)——- speed of R: (100 - X) / (3)
(R) (b) = X —-> (b) = X / R
—and substitute R—
(b) = (X) / (100 - X / 3) = (3X) / (100 - X)
(D) (c) = (100 - X) ——> (c) = (100 - X) / (X/3) = (300 - 3X) / (3X)
Now plug in (b) and (c) into the equation assuming (t) = 8
(3X) / (100 - X) = 8 + (300 - 3X) / (X)
We can solve for X now, by multiplying both sides of the equation by * (X) (100 - X)
3(X)^2 = 8X(100 - X) + (100 - X) (300 - 3X)
3(X)^2 = 800X - 8(X)^2 + 30,000 - 600X + 3(X)^2
Subtract the 3(X)^2 from both sides and combine like terms
8(X)^2 - 200(X) - 30,000 = 0
Divide by / (8)
(X)^2 - 25X - 3,750 = 0
Factor the Quadratic:
(X - 75) (X + 50) = 0
X is a positive distance so the root must be:
X = 75 ——-> under our assumptions using the ratio values of 3 : 8
Speed of Dinesh = D = (X) / (3) = (75 / 3)
Speed of Rahim = R = (100 - X) / 3 = (25 / 3)
Lastly, find the percentage increase:
(75/3 - 25/3) / (25/3) =
(75 - 25) / (25) = 2
And
2 * [100]% = 200%
C
Posted from my mobile device
Since the overall gap distance between the two people at the start does not matter, let’s make it a nice 100 meters.
Speed of Dinesh = D
Speed of Rahim = R
The Time it takes the two of them to cover the gap distance and meet at the meeting point we can call = a
We want to know the percentage increase of D over R:
(D - R) / (R) * [100]% = ?
(1st) They travel towards each other and Meet
Dinesh covers X of the meters:
Using (Speed)(Time) = (Distance):
(D)(a) = X
Rahim covers the remaining Gap Distance in the same Time (a):
(R)(a) = (100 - X)
(2nd) Each returns back to his or her own starting point, at the OTHER’s Speed:
Dinesh: covers the X meters at speed of (R) —- we can call this Time = (b)
(R)(b) = X
Rahim: covers the (100 - X) meters at speed of (D) —— we can call this Time = (c)
(D)(c) = (100 - X)
Dinesh’s time is (t) hours more than Rahim’s time
b = (t) + c
Finally:
(Time taken by Rahim to reach the meeting point) : (t) = (3) : (8)
Rahim and Dinesh took (a) hours to reach the meeting point
Let (a) = 3
Let (t) = 8
Now fill in the information we have:
(D) (3) = X —- speed of D: (X) / (3)
(R) (3) = (100 - X)——- speed of R: (100 - X) / (3)
(R) (b) = X —-> (b) = X / R
—and substitute R—
(b) = (X) / (100 - X / 3) = (3X) / (100 - X)
(D) (c) = (100 - X) ——> (c) = (100 - X) / (X/3) = (300 - 3X) / (3X)
Now plug in (b) and (c) into the equation assuming (t) = 8
(3X) / (100 - X) = 8 + (300 - 3X) / (X)
We can solve for X now, by multiplying both sides of the equation by * (X) (100 - X)
3(X)^2 = 8X(100 - X) + (100 - X) (300 - 3X)
3(X)^2 = 800X - 8(X)^2 + 30,000 - 600X + 3(X)^2
Subtract the 3(X)^2 from both sides and combine like terms
8(X)^2 - 200(X) - 30,000 = 0
Divide by / (8)
(X)^2 - 25X - 3,750 = 0
Factor the Quadratic:
(X - 75) (X + 50) = 0
X is a positive distance so the root must be:
X = 75 ——-> under our assumptions using the ratio values of 3 : 8
Speed of Dinesh = D = (X) / (3) = (75 / 3)
Speed of Rahim = R = (100 - X) / 3 = (25 / 3)
Lastly, find the percentage increase:
(75/3 - 25/3) / (25/3) =
(75 - 25) / (25) = 2
And
2 * [100]% = 200%
C
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12 Oct 2022, 19:18
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Let David's speed D
Let Rahim's speed D*k
Distance (Q to R) = speed of Rahim * time taken by Rahim = kD*(3t/8)
Distance (P to Q) = speed of David * time taken by David = speed of David * time taken by Rahim = D * (3t/8)
On the way back, they exchanged the speeds.
Time taken by David = Distance (P-Q) / Speed of Rahim = (3tD/8)/kD = (3t/8k)
Time taken by Rahim = Distance (P-Q) / Speed of David= (3ktD/8)/D = (3tk/8)
Time taken by David = t + time taken by Rahim
3t/8k = t + (3tk/8)
3/(8k) = 1 + 3k/8 ; divide everything by t
3 = 8k + 3k^2 ;multiple everything by 8k
k=-3 or 1/3
David's speed D, Rahim's speed kd = D/3
David's speed - Rahim's speed = D - D/3 = 2D/3
Ratio of diff / Rahim's speed = (2D/3) / (D/3) = 2 or 200%
Let Rahim's speed D*k
Distance (Q to R) = speed of Rahim * time taken by Rahim = kD*(3t/8)
Distance (P to Q) = speed of David * time taken by David = speed of David * time taken by Rahim = D * (3t/8)
On the way back, they exchanged the speeds.
Time taken by David = Distance (P-Q) / Speed of Rahim = (3tD/8)/kD = (3t/8k)
Time taken by Rahim = Distance (P-Q) / Speed of David= (3ktD/8)/D = (3tk/8)
Time taken by David = t + time taken by Rahim
3t/8k = t + (3tk/8)
3/(8k) = 1 + 3k/8 ; divide everything by t
3 = 8k + 3k^2 ;multiple everything by 8k
k=-3 or 1/3
David's speed D, Rahim's speed kd = D/3
David's speed - Rahim's speed = D - D/3 = 2D/3
Ratio of diff / Rahim's speed = (2D/3) / (D/3) = 2 or 200%
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
