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Manager  Joined: 09 May 2009
Posts: 160
Distance between A and B is 72 miles. Two men started walkin  [#permalink]

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5 00:00

Difficulty:   95% (hard)

Question Stats: 47% (03:14) correct 53% (02:58) wrong based on 144 sessions

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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 traveled uniformly with average speed 4 mph. While the other man traveled 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

i managed to solve it in my way which took 4 min , perhaps anyone can help me to solve it in 2 min

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Math Expert V
Joined: 02 Sep 2009
Posts: 58410
Re: time speed  [#permalink]

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2
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 traveled uniformly with average speed 4 mph. While the other man traveled 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

Maybe there is an easier way...

Arithmetic progression for both:

$$a_1=4+2=6$$ and common difference $$d=0.5$$, $$n$$ number of terms (hours)

$$Sum=\frac{n}{2}(2a_1+d(n-1))=\frac{n}{2}(2*6+0.5(n-1))\geq72$$

$$0.5n^2+11.5n-144\geq0$$ --> $$n^2+23n-288\geq0$$ --> $$(n+32)(n-9)\geq0$$, $$-32\geq{n}\geq9$$

Hence for n=9(hours) the above equation is exactly 72.

In 9 hours men from A covers 9*4=36, which is half of the distance.

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SVP  Joined: 29 Aug 2007
Posts: 1918
Re: time speed  [#permalink]

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xcusemeplz2009 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 traveled 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

i managed to solve it in my way which took 4 min, perhaps anyone can help me to solve it in 2 min

They meet after n hours:
4n + [2 + 2.5 + 3 + ........ + {2 + (0.5(n-1)}] + x = 72, where x has to be 0.

Lets try for n = 8:
4(8) + [2 + 2.5 + 3 + 3.5 + 4 + 4.5+ 5 + 5.5] + x = 72
x = 10 ... not correct.

Lets try for n = 9:
4(9) + [2 + 2.5 + 3 + 3.5 + 4 + 4.5+ 5 + 5.5+6] + x = 72
x = 0 .... thats possible.

When they meet after 9 hours, that is mid point.

E.
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Manager  Joined: 09 May 2009
Posts: 160
Re: time speed  [#permalink]

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thanks guys
its E
+1 kudos to bth
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Intern  Joined: 25 Dec 2009
Posts: 37
Re: time speed  [#permalink]

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3
let journey hours be=n

when they meet distance travelled is =

hence
4n=n/2*(2a+(n-1)d)
where a=2 and d=.5
8=4+(n-1)*.5
n=9

now 4n=36(midway) ='E'
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Distance between A and B is 72 miles. Two men started walkin  [#permalink]

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let t=A's and B's time to meeting
B's ratio of cumulative distance to time=(t+7)/4 mph
4t+(t)(t+7)/4=72
t^2+23t-288=0
(t+32)(t-9)=0
t=9 hours
4t=36 miles distance for A
72-36=36 miles distance for B
they meet midway between A and B

Originally posted by gracie on 29 Oct 2015, 22:03.
Last edited by gracie on 12 Nov 2015, 21:21, edited 2 times in total.
Veritas Prep GMAT Instructor V
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Re: Distance between A and B is 72 miles. Two men started walkin  [#permalink]

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xcusemeplz2009 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 traveled uniformly with average speed 4 mph. While the other man traveled 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

i managed to solve it in my way which took 4 min , perhaps anyone can help me to solve it in 2 min

In such questions, I always look for the symmetrical situation to give me a ballpark figure or at least a range. For example, in many work-rate questions, it helps if you figure out the situation in which the rate of both people is the same.
Here the average speed of A (assuming guy who starts from A is A) is 4 mph and distance is 72 miles. So this means that A would take a total of 72/4 = 18 hours to cover the distance from A to B. If B's speed were also 4, both A and B would travel for 9 hours to meet in the middle.
B has uniform speed per hour so its average will be the simple average of all speeds: 2, 2.5, 3, 3.5, 4, 4.5, ...
4 is actually right in the middle of the 9 speeds and this implies that they both have the same average speed. So they will meet midway between A and B.

Had I obtained that B's average speed is more than 4 then that would tell me that they will meet closer to point A and will take less than 9 hours and so on...

You might feel that in this question it worked because the numbers were such but you will be surprised at the number of times such simple analysis will work because GMAT gives you numbers which fit in place.
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Re: Distance between A and B is 72 miles. Two men started walkin  [#permalink]

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_________________ Re: Distance between A and B is 72 miles. Two men started walkin   [#permalink] 26 Feb 2019, 04:58
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