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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
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I think you are right the answer should be indeed C.
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
sandy wrote:
Explanation

Use the Rate Pie.

Ali is traveling 50 feet per second faster than Jeff is traveling. Therefore, that is the rate at which she is effectively gaining ground on him. Put that in the
lower-right segment of the Rate Pie. We want to know how long it will take her to gain 3,000 feet on him.

Put 3,000 in the top section of the Rate Pie. Now you can see that dividing \(\frac{3000}{50}\) will fill in the last segment of the Rate Pie, telling you how long it takes Ali to do so is 60 seconds. Be aware that choice (E) is an incorrect partial answer.

Now you need to find out how many feet Ali will travel in 60 seconds, by multiplying 60 second × 300 feet per second, which equals 18,000 feet. Divide 18,000 feet by the length of one lap, or 3,000 feet, and you’ll find that it will take Ali 6 laps to overtake Jeff.

Hence option C is correct!

what is the rate pie?
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
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Relative speed of Ali with respect to Jeff is 50 feet/second.
Since the track is 3000 ft long, it takes Ali around (3000/50 = 60) seconds to overtake Jeff.
In 60 seconds, Ali will travel 300*60 = 18000 feet.

18000 feet will translate to 6 LAPS since one lap is around 3000 ft long

Therefore, Ali takes 6 laps to overtake Jeff(Option C)
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
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sandy wrote:
Jeff and Ali race each other at the Tentleytown Speedway. Ali’s car travels at 300 feet per second, and Je?’s car travels at 250 feet per second. If one lap around the track is 3000 feet long, and each car travels at a constant rate, how many laps will it take Ali to overtake Jeff?

A. 1
B. 5
C. 6
D. 10
E. 60


sandy

Just a question here!

Since, Ali is travelling at a faster rate than Jeff
In 1 second Ali is already 50 feet ahead of Jeff

Don't you think it should be other way around?
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
sandy wrote:
Jeff and Ali race each other at the Tentleytown Speedway. Ali’s car travels at 300 feet per second, and Je?’s car travels at 250 feet per second. If one lap around the track is 3000 feet long, and each car travels at a constant rate, how many laps will it take Ali to overtake Jeff?

A. 1
B. 5
C. 6
D. 10
E. 60

Drill 4
Question: 2
Page: 531

Since the question asks for a certain number of LAPS, let's first rewrite the given information in terms of laps.
300/3000 = 1/10, which means 300 ft = 1/10 laps. So, Ali’s speed of 300 feet per second is equal to a speed of 1/10 laps per second
250/3000 = 1/12, which means 250 ft = 1/12 laps. So, Jeff’s speed of 250 feet per second is equal to a speed of 1/12 laps per second

Since each person travels for the same amount of time, we can start with the following word equation: Ali's travel time (in seconds) = Jeff's travel time (in seconds)
When Ali laps Jeff, Ali has completed 1 lap more than Jeff has.
Let x = the number of laps Ali completes (i.e., Ali's travel distance)
So, x - 1 = the number of laps Jeff completes (i.e., Jeff's travel distance)

time= distance/rate
Plug our values into our word equation to get: (x)/(1/10) = (x - 1)/(1/12)
Simplify: (10)(x) = (12/1)(x - 1)
Simplify: 10x = 12x - 12
Solve: x = 6

Answer: C
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
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Re: Je? and Ali race each other at the Tentleytown Speedway. [#permalink]
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