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# A certain vehicle is undergoing performance adjustments during a set

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Math Expert
Joined: 02 Sep 2009
Posts: 56304

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28 Aug 2015, 02:14
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14
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Difficulty:

65% (hard)

Question Stats:

68% (03:27) correct 32% (02:57) wrong based on 91 sessions

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A certain vehicle is undergoing performance adjustments during a set of three trials on an 8-mile track. The first time it travels around the track at a constant rate of p miles per minute, the second time at a constant rate of p^2 miles per minute, and the third time at a constant rate of p^3 miles per minute. If the vehicle takes the same time to travel the first 18 miles of these trials as it does to travel the last 8 miles, how many minutes does it take to complete all three trials (not counting any time between trials)?

(A) 18
(B) 36
(C) 60
(D) 96
(E) 112

Kudos for a correct solution.

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Joined: 23 Jun 2015
Posts: 1
Re: A certain vehicle is undergoing performance adjustments during a set  [#permalink]

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28 Aug 2015, 09:55
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Bunuel wrote:
A certain vehicle is undergoing performance adjustments during a set of three trials on an 8-mile track. The first time it travels around the track at a constant rate of p miles per minute, the second time at a constant rate of p^2 miles per minute, and the third time at a constant rate of p^3 miles per minute. If the vehicle takes the same time to travel the first 18 miles of these trials as it does to travel the last 8 miles, how many minutes does it take to complete all three trials (not counting any time between trials)?

(A) 18
(B) 36
(C) 60
(D) 96
(E) 112

Kudos for a correct solution.

1. Time taken for first 18 miles = $$\frac{8}{p} + \frac{8}{p^2}+\frac{2}{p^3}$$
2. For the last 8 miles =$$\frac{8}{p^3}$$
As per the problem,
$$\frac{8}{p} + \frac{8}{p^2}+\frac{2}{p^3}$$= $$\frac{8}{p^3}$$
Solving for p,
p=$$\frac{1}{2}$$

Therefore, time taken for all the 3 trials =$$\frac{8}{(1/2)} +\frac{8}{(1/4)} +\frac{8}{(1/8)}$$ = 16 +32 + 64 = 112
Ans: E
##### General Discussion
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Joined: 24 Aug 2014
Posts: 7

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29 Aug 2015, 21:47
1
This is a rate-time-distance problem. We would be better off setting a rate-time-distance table.

The first journey takes 18 miles whereas the last journey takes 8 miles.

We are told that:
r1= p
r2=p^2
and r3= p^3 so we set up a table like

r x t = d
p x t1 = 8
p^2 x t2= 8
p^3 x t3 = 2 (because 18-16)

so the time taken to complete each leg individually amounts to t1 + t2 + t3 = (8/p) + (8/p^2) + (2/p^3). ----> (equation 1)

On the other hand, the time taken to complete the last leg of miles is (8/p^3) ---> (equation 2) because p^3 is the speed on the last leg. Since both the times are equal, equate them to get:

(8/p) + (8/p^2) + (2/p^3) = (8/p^3)

Taking the LCM of the Left hand equation and solving both these equations gives us the value of p=1/2 and p= (-3/2) Since p can not be negative, we have p=1/2 as our value.

Now put this value in the time taken to finish each trial which is the sume of (8/p) + (8/p^2) + (8/p^3) and it will become 16+32+64 = 112 minutes.
Option E is the correct answer.

p.s: if you like the explanation, please give a kudos
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Joined: 13 May 2017
Posts: 1
Re: A certain vehicle is undergoing performance adjustments during a set  [#permalink]

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20 Sep 2017, 04:40
Hi, I had a doubt on this question. As per my understanding, it is said "If the vehicle takes the same time to travel the first 18 miles of these trials" and then "the last 8 miles" Adding the above two cases(18+8=26) but the total length is (8*3=24)miles. How is this possible. Please let me know
Math Expert
Joined: 02 Sep 2009
Posts: 56304
Re: A certain vehicle is undergoing performance adjustments during a set  [#permalink]

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20 Sep 2017, 04:50
Tapabrata18 wrote:
Hi, I had a doubt on this question. As per my understanding, it is said "If the vehicle takes the same time to travel the first 18 miles of these trials" and then "the last 8 miles" Adding the above two cases(18+8=26) but the total length is (8*3=24)miles. How is this possible. Please let me know

The first 18 miles consists of 8 miles of the first trial + 8 miles of the second trial + the first 2 miles of the third trial. The last 8 miles consists of 8 miles of the third trial. So, there is an overlap of 2 miles.
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Re: A certain vehicle is undergoing performance adjustments during a set  [#permalink]

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25 Jun 2019, 06:05
As per the problem,
8/p+8/p^2+2/p^3= 8/p^3
=> 8/p+8/p^2=6/p^3
Reqd time is 8/p + 8/p^2 + 8/p^3 =>Substituting above value =>6/p^3 + 8/p^3 i.e 14/p^3.
Only 112 is a multiple of 14.
Manager
Joined: 07 May 2018
Posts: 68
Re: A certain vehicle is undergoing performance adjustments during a set  [#permalink]

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25 Jun 2019, 12:46
umair4 wrote:
This is a rate-time-distance problem. We would be better off setting a rate-time-distance table.

The first journey takes 18 miles whereas the last journey takes 8 miles.

We are told that:
r1= p
r2=p^2
and r3= p^3 so we set up a table like

r x t = d
p x t1 = 8
p^2 x t2= 8
p^3 x t3 = 2 (because 18-16)

so the time taken to complete each leg individually amounts to t1 + t2 + t3 = (8/p) + (8/p^2) + (2/p^3). ----> (equation 1)

On the other hand, the time taken to complete the last leg of miles is (8/p^3) ---> (equation 2) because p^3 is the speed on the last leg. Since both the times are equal, equate them to get:

(8/p) + (8/p^2) + (2/p^3) = (8/p^3)

Taking the LCM of the Left hand equation and solving both these equations gives us the value of p=1/2 and p= (-3/2) Since p can not be negative, we have p=1/2 as our value.

Now put this value in the time taken to finish each trial which is the sume of (8/p) + (8/p^2) + (8/p^3) and it will become 16+32+64 = 112 minutes.
Option E is the correct answer.

p.s: if you like the explanation, please give a kudos

Hi,

Could you please show the working out of how you got p =1/2
Re: A certain vehicle is undergoing performance adjustments during a set   [#permalink] 25 Jun 2019, 12:46
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