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One week, a certain truck rental lot had a total of 20

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One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 13 Jun 2015, 08:46
I used the same approach as Bunuel, but I would just like to show the notes I made to reach this solution:

Monday: 20 trucks (1)
Week: x rented out (2)
From (1) and (2) we conclude that 20-x were left on the lot.
Finally, out of those that were rented out (x), 50% (0.50x) were returned.

From the above we gather that the trucks that were left on the lot are those that were returned (0.50x) plus those that were left on the lot (20-x). As we know that there were at least 12 trucks on the lot on Saturday, these 12 are those that were returned plus those that were left:

(20-x) + (0.50x) = 12
20-x+0.50x = 12
8 = 0.50x
x = 16. So, ANS B

ps1: do we know about where on the 600-700 level this question falls? So, is it 650+ or 650-?
ps2: i might be asking for too much, because you are already doing an amazing job, but could we have a 650 to 700 range of questions? I guess there should be enough people looking to score on that range, and even though 600-700 is a good indication of your level, 650-700 is an even safer one...
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 02 Aug 2015, 06:32
Spent a lot of time just to understand the question. Simply put, there were 12 trucks at the lot on Saturday. Hence, there should be 8 trucks that were rented and had not been returned to the lot. If 50% of rented trucks were returned before or on Saturday, then another 50% should be those 8 trucks that had not been returned. Therefore, the number of trucks that was rented must be 8/50%, 16.
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 23 Jun 2016, 23:28
the idea is to keep your focus on converting the language into mathematical equation...total number of trucks is 20.....let x be the number of trucks rented out during the week....if x/2 were rented out returned...and 20-x stationed at lot already.....set up the equation...x/2+(20-x) greater than equal to 12...to maximize one quantity...minimize the other quantity..solve for x
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 13 Jul 2016, 08:05
That's an easy question to solve, if you understand that only 50% of the trucks were returned by Saturday moring.
That means the other half was not returned by that time.
We are trying to minimize the number of cars the lot had by Saturday so that we can maximize the number of trucks missing.

Monday- Total trucks=20

Saturday- Trucks in the lot >= 12
So, at most 8 trucks are not returned yet.
That will have to be the 50% of those rented throughout the week.

Therefore 8=50%*y => y=<16

So, the maximum number of trucks rented is 16.
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 02 Jun 2017, 08:42
cloaked_vessel wrote:
One week, a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

A 18
B 16
C 12
D 8
E 4


We can solve this problem via backsolving or algebra.

Backsolving:

Rented cars = 18.

A) 18 cars were rented, and 9 were returned by Saturday. Two cars were not rented. So that means 9+2 = 11 cars should be in the lot, but we know there are 12 cars, so we can eliminate this choice.

B) 16 cars were rented, and 8 were returned by Saturday. Four cars were not rented. So that means 8+4 = 12 cars should be in the lot, and 12 cars are in fact in the lot. Since this problem involves maximization, we don't have to check the other choices because we only care about the highest value, and the answer choices are in descending order.

Algebra:

R + U = 20

R/2 + U = 12

U = 12 - R/2

R + 12 - R/2 = 20

R/2 = 8

R = 16.
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 05 Jun 2017, 15:57
cloaked_vessel wrote:
One week, a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

A 18
B 16
C 12
D 8
E 4


We are given that a truck rental lot had a total of 20 trucks, that 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and that at least 12 trucks were on the lot that Saturday morning.

We can let t = the total number of trucks rented. Then, (20 - t) trucks were not rented at all. Since 0.5t of the trucks were returned, there were (20 - t) + 0.5t = 20 - 0.5t trucks on the lot Saturday morning. Thus, we can create the following inequality:

20 - 0.5t ≥ 12

8 ≥ 0.5t

16 ≥ t

We see that the greatest possible value of t is 16.

Answer: B
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Re: One week, a certain truck rental lot had a total of 20  [#permalink]

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New post 08 Apr 2018, 12:27
Bunuel wrote:
katealpha wrote:
This seems like the easiest question ever and I am not comprehending it. I have wasted more than one hour on that thing. Could please someone explain it one more time to obviously slow person.....


One week, a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were [b]at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?[/b]
A.18
B.16
C.12
D.8
E.4

First how to deal with "at least" and "greatest number" part of the question.

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.


So to maximize the # of trucks rented we should minimize # of trucks at the lot on Saturday. We are told that # of trucks at the lot on Saturday was at least 12, so to minimize it, we should consider this number to be 12 (minimum possible).

Next, the # of trucks at the lot on Saturday, 12, equals to {the # of trucks not rented} plus {half of the # of trucks rented} --> \((20-R)+\frac{1}{2}R=12\) --> \(R=16\).

Or: as "50% of the trucks that were rented out during the week were returned" then another 50% of the trucks that were rented were not returned --> not returned = 20-12=8 trucks, which is 50% of the trucks that were rented --> # of truck were rented = 2*8 = 16.

Answer: B.

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Hope it's clear.


hello Bunuel :) how are you ?:)

you know i dont understand the wording of the question ....

it says total are 20 trucks. 50 % of trucks were rented.

now my reasoning turns on :) so if 50 % were rented and there were20 trucks. it means that total number of truck is 40 and 20 were rented which is 50 % no ?

another question "If 50% of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were [b]at least 12 trucks on the lot that Saturday morning" [/b]

are we talking about next week, or the week before :?
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Re: One week, a certain truck rental lot had a total of 20 &nbs [#permalink] 08 Apr 2018, 12:27

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