RATE/DISTANCE QUESTION

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RATE/DISTANCE QUESTION

by Mclaughlin » Wed Jul 16, 2008 9:50 pm
warren and rubin race around an oval track. Warren runs at a uniform pace of one lap per 7 mins. and rubin at a uniform rate of one lap every 5 minutes. If they both start at the same time, how long will it take Rubin to get exactly one lap ahead of Warren.

12
13.5
15
16 1/4
17.5

OA IS E

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by sudhir3127 » Wed Jul 16, 2008 10:12 pm
My answer is E. 17.5 mints

here it goes for waren it takes 7 mints to complete 1 lap and for rubin it takes 5 mints.

we need to find how long will it take Rubin to get exactly one lap ahead of Warren.

let the total number of mints for rubin get exactly one lap ahead of Warren be X

the equation goes like this X/5 = 1+X/7

solve for X u get 17.5

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by cornell » Wed Jul 16, 2008 10:39 pm
Agree with sudhir3127, my answer is E (17.5 mins).

Warren = 1/7 laps/mins and Rubin = 1/5 laps/mins. Desired interval between Rubin & Warren = 1 lap.

We can simplify this with an equation (1/5)A – (1/7)A = 1

7A/35 – 5A/35 = 35/35

2A/35 = 35/35

2A = 35

A = 17.5
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Re: RATE/DISTANCE QUESTION

by parallel_chase » Thu Jul 17, 2008 2:47 am
Mclaughlin wrote:warren and rubin race around an oval track. Warren runs at a uniform pace of one lap per 7 mins. and rubin at a uniform rate of one lap every 5 minutes. If they both start at the same time, how long will it take Rubin to get exactly one lap ahead of Warren.

12
13.5
15
16 1/4
17.5

OA IS E

If you look at the question it tells you how many minutes it would take rubin to be exactly 1 lap ahead of warren.

Warren's rate is 7 minutes per lap

Therefore rubin will have to be 7 minutes ahead of warren.

Since the Rate between warren and rubin is increasing or decreasing (Since it is circular motion ) by 2 minutes per lap.

2 L = 7
L = 3.5

after 3.5 laps rubin will be exactly 1 lap or 7 minutes ahead of warren.

Therefore the answer is 3.5 * 5 = 17.5 minutes.

This way you can understand the logic and you dont even have to use any formulas.

Let me know if you any questions.