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23 Dec 2009, 23:21
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
51% (03:13) correct
49%
(03:02)
wrong
based on 220
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
(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
24 Dec 2009, 18:52
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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
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.
Answer: E.
(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.
Answer: E.
29 Oct 2015, 22:40
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xcusemeplz2009
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.
