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interesting LCM problem - need correct answer

Expert replies
by oops » Fri Jun 08, 2007 12:55 pm
Seems (to me) a good Q from one of the online prep sites. Am posting it only bcoz I am not sure of their answer (which I'll post a lil' later). Hope they dont come after me for copyright violation:

Jane, Chen and Boris start together to travel the same way around a circular path of 20 kms. Their speeds are 4, 5 and 1/2, and 8 kms per hour respectively. When will they meet at the starting point?

(A) 40 hours

(B) 12 hours

(C) 11 hours

(D) 44 hours

(E) 36 hours
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Source: — Problem Solving |

oops wrote:Seems (to me) a good Q from one of the online prep sites. Am posting it only bcoz I am not sure of their answer (which I'll post a lil' later). Hope they dont come after me for copyright violation:

Jane, Chen and Boris start together to travel the same way around a circular path of 20 kms. Their speeds are 4, 5 and 1/2, and 8 kms per hour respectively. When will they meet at the starting point?

(A) 40 hours

(B) 12 hours

(C) 11 hours

(D) 44 hours

(E) 36 hours

.. well the q seems to be incomplete ... it doesnt say wther it is the first, second, third time that they meet at the starting point ..

nywayz the method to solve is to take the LCM of the time it take each athlete to run around the track once .. that is the time taken by them will be 5 hrs, 40/11 hrs , 2.5 hrs .. take their LCM and u will get the time that they take to meet for the first time at the starting point ... then according to the q multiply the answer with 2,3, .. depending on whether the second , third, .. time is asked for..
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by f2001290 » Sat Jun 09, 2007 10:16 am
I will go with A. I used the answer choices to solve this problem.

Since the length of the track is 20kms, the distance covered by Jane after "x" no. of hours to be at the starting point should be a multiple of 20.

Only A(40) satisfies this condition. You can verify this using the case of Jane.
Distance covered by Jane after 40 hours = 4*40 = 160kms
Remaining options doesn't satisfy this condition.
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gabriel wrote:
oops wrote:Seems (to me) a good Q from one of the online prep sites. Am posting it only bcoz I am not sure of their answer (which I'll post a lil' later). Hope they dont come after me for copyright violation:

Jane, Chen and Boris start together to travel the same way around a circular path of 20 kms. Their speeds are 4, 5 and 1/2, and 8 kms per hour respectively. When will they meet at the starting point?

(A) 40 hours

(B) 12 hours

(C) 11 hours

(D) 44 hours

(E) 36 hours

.. well the q seems to be incomplete ... it doesnt say wther it is the first, second, third time that they meet at the starting point ..

nywayz the method to solve is to take the LCM of the time it take each athlete to run around the track once .. that is the time taken by them will be 5 hrs, 40/11 hrs , 2.5 hrs .. take their LCM and u will get the time that they take to meet for the first time at the starting point ... then according to the q multiply the answer with 2,3, .. depending on whether the second , third, .. time is asked for..

Yes, this is what I got (5, 40/11 and 2.5) and I thought the LCM was 200/11, but (as u can see) there is no such answer matching it or any integer multiple of 200/11. So am confused - suspect I am going wrong somewhere.

The given answer is A (40)
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by oops » Sat Jun 09, 2007 2:35 pm
f2001290 wrote:I will go with A. I used the answer choices to solve this problem.

Since the length of the track is 20kms, the distance covered by Jane after "x" no. of hours to be at the starting point should be a multiple of 20.

Only A(40) satisfies this condition. You can verify this using the case of Jane.
Distance covered by Jane after 40 hours = 4*40 = 160kms
Remaining options doesn't satisfy this condition.
This seems like a very good and fast way to solve the problem - and your answer agrees with the given answer. I dont use backsolving often and end up wasting a lot of time.

However, I am still very anxious to know where I am going wrong with the LCM method and why the answer is not 200/11 (plz see previous post).
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