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Challenging Speed Problem

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Challenging Speed Problem

by knight247 » Fri Jul 19, 2013 2:47 pm
Two people A and B start from P and Q (Distance = D) at the same time towards each other. They meet at a point R, which is at a distance of 0.4D from P. They continue to move to and from between the two points. Find the distance from point P at which the fourth meeting takes place.

(A) 0.8D
(B) 0.6D
(C) 0.3D
(D) 0.4D
(E) 0.7D

OA is A

Detailed explanations would be appreciated. Many thanks in advance.
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by [email protected] » Fri Jul 19, 2013 5:38 pm
Hi knight247,

This question is poorly worded and the answers don't appear to be in the correct order. What is the source of this question?

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by merwin11 » Sat Jul 20, 2013 6:58 am
Knight247,

A and B are at a distance D apart from each other, and meet at a distance of 0.4 D from P, which means the point is at a distance of 0.6D from Q.
From this we know the ratio of distances covered of A and B = 4/6 = 2/3

When time taken is a constant, Distance is directly proportional to Speed and hence Ratio of speeds = Ratio of distances

Ratio of speeds of A and B = 2/3

When 2 people move towards each other from 2 opposite points on a st.line, they together cover a distance of (2n-1)D, every time they meet. (n is the nth time they meet and D is the total distance)
Individually, they will cover the distances in the ratio of ther speeds for any number of meetings.

So first time they together cover a distance of D, the next time 3D and so on. So the 4th time they would have covered a total distance of 7D.
Even without the formula you can arrive at this, since n is only 4.

We know the ratio of distance = 2/3

So distance covered by A = 2/5 * 7D = 2.8D
distance covered by B = 3/5 * 7D = 4.2D

So now you can see that the 4th meeting point will be at a distance of 0.8D from P and a 0.2D from Q

Please tell me If the explanation is ok..

Also an additional funda,

When 2 people move from the same point on a st.line in the same direction to and fro between 2 points, they together cover a distance of 2nD every time they meet. (n is the nth time they meet and D is the total distance.)

Thanks!
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by Matt@VeritasPrep » Sun Jul 21, 2013 5:57 pm
When two people are traveling toward each other, we can add their rates together. So here D = the distance between them, R = their rates added together, and T = the time that each of them has traveled when they meet. I'll call A the rate of person A and B the rate of person B.

D = (A + B)T

If R is .4 from where A starts, then .4D = AT and .6D = BT. Let's find A in terms of B, as we want the distance from A's starting point.

.6D = 1.5AT = BT, or
1.5A = B, or
3A = 2B, or
A = (2/3)B.

So A travels at 2/3 the rate of B.

For the next step, I'll assume that "between the two points" means "between points P and Q".

The first time the two people meet, they have together traveled a distance of D. But the second time they meet, they have traveled another 2D - since they must go to points P and Q, then back again to some other point on the line. Each additional time adds another 2D, so for four meetings we have a total of 7D.

If they travel 7D, and A's rate is 2/3 that of D, A travels 2/5 of the distance and B travels 3/5 of the distance. (Their combined rate is 2 + 3, so A does 2/(2+3) and B does 3/(2+3).) So A travels 2/5 of 7D, or 2.8D. 2D is a roundtrip from P back to P, so the extra .8 leaves him .8D from point P upon the fourth meeting.

Really cool problem!
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