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man travelled with varying speeds

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man travelled with varying speeds

by sanju09 » Wed Dec 23, 2009 4:17 am
Distance between A and B is 72 miles. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed 4 mph. While the other man travelled with varying speeds as follows: In first hour his speed was 2 mph, in the second hour it was 2.5 mph, in the third hour it was 3 mph, and so on. When will they meet each other?

(A) 35 miles away from A
(B) 32 miles away from B
(C) after 32 hours from start
(D) after 10 hours from start
(E) midway between A and B
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by rohan_vus » Wed Dec 23, 2009 5:09 am
IMO E...

Did backsolving, one choice had 10 hours , so started with that , but found A + B exceeding the total distance , then went to cross check with 9 hours , which turns out to satisfy the conditions.

In 9 hrs , A travels 36 miles
in 9 hrs , B travels - 2 + 2.5 + 3.5...... ( till the 9th term)..this is AP with common diff of .5 and is equal to 36
Last edited by rohan_vus on Wed Dec 23, 2009 5:42 am, edited 1 time in total.
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by sanju09 » Wed Dec 23, 2009 5:27 am
Great going rohan, you got it; in fact backsolving is a great strategy on GMAT Quantitative section. Another classical approach, anyone?
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by linkinpark » Wed Dec 23, 2009 8:28 am
interesting question

here time will be same for both A and B say its T


so for A it will be 72/4 and for B it will be 72/(sum of speeds)/n, where n = number of different speeds

after solving I get 4n = (sum of diff speeds). I'm stuck here :( how do I find n and sum
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by Stuart@KaplanGMAT » Wed Dec 23, 2009 6:31 pm
linkinpark wrote:interesting question

here time will be same for both A and B say its T


so for A it will be 72/4 and for B it will be 72/(sum of speeds)/n, where n = number of different speeds

after solving I get 4n = (sum of diff speeds). I'm stuck here :( how do I find n and sum
The algebra is incredibly complicated - we need to use arithmetic series to find the distance travelled by B; backsolving is definitely the way to go.

As an aside, the total distance travelled is 72; that's not how far each of them travels alone. So, our final equation would be:

72 = 4n + (sum of arithmetic series for B)
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by sanju09 » Wed Dec 23, 2009 11:58 pm
Stuart Kovinsky wrote:
linkinpark wrote:interesting question

here time will be same for both A and B say its T


so for A it will be 72/4 and for B it will be 72/(sum of speeds)/n, where n = number of different speeds

after solving I get 4n = (sum of diff speeds). I'm stuck here :( how do I find n and sum
The algebra is incredibly complicated - we need to use arithmetic series to find the distance travelled by B; backsolving is definitely the way to go.

As an aside, the total distance travelled is 72; that's not how far each of them travels alone. So, our final equation would be:

72 = 4n + (sum of arithmetic series for B)
Let me finish Stuart's take please...

Actually, he took n to be the number of hours after what they met, ?if? it's an integer, then we can talk about forming an AP for (person from B)'s run, as under:

sum of arithmetic series for B = n/2 {2*6 + (n - 1) (0.5)}

I think that to make sure that n is a positive integer, we should ideally write it as an inequality here, I mean...

4 n + n/2 {2*6 + (n - 1) (0.5)} ≤ 72

this ends up in...

(n - 9) (n + 32) ≤ 0

therefore n ≤ 9 permits us to go with n = 9; and in 9 hours person from A will travel 36 miles and will meet person from B too. This point must be the midway between A and B. Go with E.
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by pink_08 » Thu Dec 24, 2009 2:53 pm
where did u get that 6 in

sum of arithmetic series for B = n/2 {2*6 + (n - 1) (0.5)}


Is the sum of AP = (a1+an)/2 ??
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by Stuart@KaplanGMAT » Thu Dec 24, 2009 6:11 pm
pink_08 wrote:where did u get that 6 in

sum of arithmetic series for B = n/2 {2*6 + (n - 1) (0.5)}


Is the sum of AP = (a1+an)/2 ??
It's actually n(a1 + an)/2.
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by maihuna » Fri Dec 25, 2009 5:35 am
sanju09 wrote:Distance between A and B is 72 miles. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed 4 mph. While the other man travelled with varying speeds as follows: In first hour his speed was 2 mph, in the second hour it was 2.5 mph, in the third hour it was 3 mph, and so on. When will they meet each other?

(A) 35 miles away from A
(B) 32 miles away from B
(C) after 32 hours from start
(D) after 10 hours from start
(E) midway between A and B
Assuming that other man moved in AP: 2, 2.5, 3, 3.5, 5 etc.

On an average one is @4mph, another is minimum @2mph, so a min of @6mph a max of 12 hrs are needed.

A mid of 4 hrs
==Option C is out. Here is a table covering hr and distance covered:

Hour One's speed Two's speed Distance covered
==========================================
1 4 2 6
2 4 2.5 6.5
3 4 3 7
4 4 3.5 7.5
4 4 4 8
4 4 4.5 8.5
======================================

Basically it turns out to be an AP which S = 72 with terms 6 6.5 7 7.5
72 = n/2(2*6 + (n-1)/2) = n/4(24+n-1) = n/4(23+n)
72*4 = n(23+n)
9*(9+23) = n(23+n)

So they will meet in 9th hours. In 9 hrs they might have gone from A by 9*4 = 36 miles. which is the mid point, I will choose E.
Charged up again to beat the beast :)
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