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time speed

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time speed [#permalink]  23 Dec 2009, 23:21
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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 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

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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Re: time speed [#permalink]  24 Dec 2009, 18:31
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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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Re: time speed [#permalink]  24 Dec 2009, 18:52
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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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Re: time speed [#permalink]  25 Dec 2009, 18:58
thanks guys
its E
+1 kudos to bth
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Re: time speed [#permalink]  28 Dec 2009, 04:02
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'
Re: time speed   [#permalink] 28 Dec 2009, 04:02
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