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# Jeff and Ali race each other at the Tenleytown Speedway.

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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
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robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

if two cars start at the same time from the same place and they travel at a constant speed and one car is faster than the other , then how can the slower car overtake the faster car?

Unless the faster car reduces speed or the slower car speeds up, or the slower car has a head start.

in this question are we assuming that Jeff is ahead of Ali?
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
stne
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

if two cars start at the same time from the same place and they travel at a constant speed and one car is faster than the other , then how can the slower car overtake the faster car?

Unless the faster car reduces speed or the slower car speeds up, or the slower car has a head start.

in this question are we assuming that Jeff is ahead of Ali?

Assuming a circular track in a race (as it mentions lap which is possible in a circular track or any other track which starts and end at the same point.)
So, in the 1st lap, when they start Ali started getting a lead than Jeff and never looked back as Ali's speed was higher than Jeff..
But when Ali was in 2nd lap, Jeff was already in 1st lap and hence Ali got behind Jeff and then after started catching him. Since the difference between their speed was narrow 50 feet/s. It took him overall 6 laps from the start to overtake Jeff. So if you see overall Ali was ahead of Jeff in total km run.. But due to bounded nature of the road, he just got behind and tried to catch Jeff.

I hope it is clear now..
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
shashankism
stne
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

if two cars start at the same time from the same place and they travel at a constant speed and one car is faster than the other , then how can the slower car overtake the faster car?

Unless the faster car reduces speed or the slower car speeds up, or the slower car has a head start.

in this question are we assuming that Jeff is ahead of Ali?

Assuming a circular track in a race (as it mentions lap which is possible in a circular track or any other track which starts and end at the same point.)
So, in the 1st lap, when they start Ali started getting a lead than Jeff and never looked back as Ali's speed was higher than Jeff..
But when Ali was in 2nd lap, Jeff was already in 1st lap and hence Ali got behind Jeff and then after started catching him. Since the difference between their speed was narrow 50 feet/s. It took him overall 6 laps from the start to overtake Jeff. So if you see overall Ali was ahead of Jeff in total km run.. But due to bounded nature of the road, he just got behind and tried to catch Jeff.

I hope it is clear now..

Are we assuming Jeff is closer to the centre of the circle than Ali and therefore Ali has to catch up? Ali is faster than Jeff though. I am confused why we have to assume their positions here...
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
pra1785

Are we assuming Jeff is closer to the centre of the circle than Ali and therefore Ali has to catch up? Ali is faster than Jeff though. I am confused why we have to assume their positions here...

No No No.. It is like you (your speed is faster) and your friend (your friend's speed is slower) started at the same time on the same circular track...
Now at the starting point you both are at the same position..
after a few second of the start of the race, you will be ahead of your friend.. and he will be behind you..

But since the track is circular and if you will look ahead.. you will find that you can catch up your friend again.. (even though you are ahead.)

Why this happened is not because of different circular track. But because you both are in different laps.
So even if you are winning or ahead of the your friend you might be catching him again and again till the game ends as the track is circular.. Only difference will be the number of laps covered. It will be greater for you and lesser for you friend.

We can take an example :
Say you both start at 12 noon.
Now may be after 30 minutes you both meet again.. Lets say you are in 5 lap and he is in 4th lap only. So even he is behind you in total km traveled but you have catched him from behind and hence you both met.
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
Quote:
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

J--> 250 f/s
A-->300 f/s

We know that:

S_J=250*t
S_A=300*t

If we draw a circumference length = 3.000, we can see that when A overtakes J, distance traveled by A is equal to distance traveled by J plus 3.000, so we have:

250*t+3.000=300*t

And we get that t=60

If we look at the time A travels 3.000 feet (one lap) we get 3000/300= 10 seconds; 10 secs per lap; so then 6 laps.

Answer C
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Hi All,

Can you please help me to clear my doubt. For the first lap, Ali takes 10 sec to cover and Jeff takes 12 sec to cover. Hasnt Ali over taken Jeff in First Lap itself??
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
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rahul16singh28
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Hi All,

Can you please help me to clear my doubt. For the first lap, Ali takes 10 sec to cover and Jeff takes 12 sec to cover. Hasnt Ali over taken Jeff in First Lap itself??

Hi...
Both are together at starting point and he is already ahead with speed higher, so next time he overtakes is when he covers the distance between them that is one lap in the start.

Q could have been framed by mentioning that " how many laps does Ali do before taking lead of a lap over Jeff"

So here he requires to cover 3000 ft and every second he covers 300-250=50', so 3000/50=60 seconds.
Ali covers one lap in 3000/300=10 sec so will cover 60/10 in 60 secs
Ans 6
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
chetan2u
rahul16singh28
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Hi All,

Can you please help me to clear my doubt. For the first lap, Ali takes 10 sec to cover and Jeff takes 12 sec to cover. Hasnt Ali over taken Jeff in First Lap itself??

Hi...
Both are together at starting point and he is already ahead with speed higher, so next time he overtakes is when he covers the distance between them that is one lap in the start.

Q could have been framed by mentioning that " how many laps does Ali do before taking lead of a lap over Jeff"

So here he requires to cover 3000 ft and every second he covers 300-250=50', so 3000/50=60 seconds.
Ali covers one lap in 3000/300=10 sec so will cover 60/10 in 60 secs
Ans 6

Hi Bro,

Thanks for clearing the doubt. Yeah it could have been framed in a Better way..!!
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
shashankism
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Speed of ALi's car = 300 feet/s
Speed of Jeff's car = 250 feet/s

Relative speed of Ali w.r.t Jeff = 50 feet/s

So, time taken to overtake Jeff = 3000/50 = 60 sec.

Total distance covered by Ali in this time = 60 * 300 = 18000 feet

No. of laps covered = 18000/3000 = 6

Answer C

Hi

How do you know where they start??
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Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
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Ali takes (3000/300) = 10 sec to complete one lap
Jeff takes (3000/250) = 12 sec to complete one lap.

they will meet first time since start at LCM (10,12) = 60 secs
in 60 secs , Ali would have raced (60/10) = 6 laps

Answer (C)

Originally posted by hellosanthosh2k2 on 03 Nov 2017, 13:39.
Last edited by hellosanthosh2k2 on 03 Nov 2017, 13:42, edited 1 time in total.
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
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soodia
shashankism
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Speed of ALi's car = 300 feet/s
Speed of Jeff's car = 250 feet/s

Relative speed of Ali w.r.t Jeff = 50 feet/s

So, time taken to overtake Jeff = 3000/50 = 60 sec.

Total distance covered by Ali in this time = 60 * 300 = 18000 feet

No. of laps covered = 18000/3000 = 6

Answer C

Hi

How do you know where they start??

Yes true no info about where do they start, but this is not DS problem to consider the start position for sufficiency check, we have to assume they start at same point and solve the problem
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Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

when A & J start together, think of A as being exactly 1 lap behind J
J does a lap in 12 seconds
A does a lap in 10 seconds
A gains 2/12=1/6 lap on J each lap
in 6*1/6=1 lap A will overtake J
6 laps
C
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
Expert Reply
robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

Since a lap around the track is 3000 feet long, Ali’s rate in seconds per lap is 3000/300 = 10. Similarly, Jeff’s rate is 3000/250 = 12.

In order for Ali to overtake Jeff, she must travel one more lap than Jeff does. Thus, if we let p = the number of laps Jeff travels, p + 1 = the number of laps Ali travels. Since time = distance/rate, and when Ali overtakes Jeff, they will spend the same amount of time on the track, we have:

(p + 1)/12 = p/10

10(p + 1) = 12p

10p + 10 = 12p

10 = 2p

p = 5

We see that Jeff travels 5 laps. However, since Ali travels one more lap than Jeff, she travels 6 laps.

Alternate Solution:

We note that Ali overtakes Jeff precisely at the moment the distance covered by Ali is 3000 feet more than the distance covered by Jeff.

Since Ali’srate is 50 feet/sec more than Jeff’s rate, the distance between the two riders increases by 50 each second; thus, it takes 3000/50 = 60 seconds for Ali to overtake Jeff. It takes Ali 3000/300 = 10 seconds to complete a lap; therefore, it takes 60/10 = 6 laps for Ali to overtake Jeff.

Answer: C
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Re: Jeff and Ali race each other at the Tenleytown Speedway. [#permalink]
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robertops
Jeff and Ali race each other at the Tenleytown Speedway. Ali’s car travels at 300 feet per second, and Jeff’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

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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