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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] New post 10 Apr 2005, 12:44
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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
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Re: PS trucks [#permalink] New post 10 Jan 2011, 18:13
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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


There are numerous ways in which you can solve this question. Brute force method if the relation between rented and non-rented trucks in not very clear:

Monday morning - 20 trucks
Saturday morning - at least 12 trucks
50% trucks rented in the week were returned.
maximum no of trucks rented out = ?

I want to maximize the no. of trucks rented so I say - If 20 trucks were rented (i.e. all of them), then we should have 50% i.e. 10 of them back. But we have more; we have at least 12.
So the no. of trucks rented out must be less than 20 (because they cannot be more than 20).
What about 18? If 18 trucks are rented out, 2 remain in the lot through the week. Out of 18, 9 are returned so total 11 are in the lot. But we need at least 12 in the lot.
Let's go further down and try 16. 4 trucks do not leave the lot. Out of 16, 8 come back so we have 12 trucks in the lot.
(As we keep reducing the number of trucks rented out, the total number of trucks in the lot of Saturday morning keeps increasing. We need to maximize the number of trucks rented out which will be at the minimum possible value of total number of trucks in the lot.)
Therefore, 16 trucks must have been rented out.

Algebraic approach:
As we increase the number of trucks rented, the total number of trucks in the lot on Saturday morning decreases since out of the rented trucks only 50% come back (while all non-rented trucks stay in the lot).
(e.g If none of the 20 trucks are rented, the lot will have 20 trucks on Saturday. If 18 trucks are not rented, the lot will have 19 (18 + 1 rented comes back) trucks on Saturday morning.)
So maximize the number of trucks rented, we should try to minimize the number of trucks in the lot on Saturday morning i.e. make it 12.
N - Not rented trucks; R - Rented trucks
N + R = 20
N + R/2 = 12
R = 16
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Re: PS trucks [#permalink] New post 09 Jan 2011, 14:16
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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 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

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.
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 [#permalink] New post 10 Apr 2005, 13:31
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B 16
I backsolved this way.
If 18 are rented out, then on Saturday morning you have 2(not rented)+50% of 18=11
If 16 are out, then on Saturday morning you have 4(not rented)+8=12
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Re: Percents question [#permalink] New post 16 Apr 2011, 19:27
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mniyer 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 percent 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

Can the pundits here please suggest the best possible way to solve this problem?



There are numerous ways in which you can solve this question. Brute force method - if you are not very clear about the relation between rented and non-rented trucks:

Monday morning - 20 trucks
Saturday morning - at least 12 trucks
50% trucks rented in the week were returned.
maximum no of trucks rented out = ?

I want to maximize the no. of trucks rented so I say - If 20 trucks were rented (i.e. all of them), then we should have 50% i.e. 10 of them back. But we have more; we have at least 12.
So the no. of trucks rented out must be less than 20 (because they cannot be more than 20).
What about 18? If 18 trucks are rented out, 2 remain in the lot through the week. Out of 18, 9 are returned so total 11 are in the lot. But we need at least 12 in the lot.
Let's go further down and try 16. 4 trucks do not leave the lot. Out of 16, 8 come back so we have 12 trucks in the lot.
(As we keep reducing the number of trucks rented out, the total number of trucks in the lot of Saturday morning keeps increasing. We need to maximize the number of trucks rented out which will be at the minimum possible value of total number of trucks in the lot.)
Therefore, 16 trucks must have been rented out.

Algebraic approach:
As we increase the number of trucks rented, the total number of trucks in the lot on Saturday morning decreases since out of the rented trucks only 50% come back (while all non-rented trucks stay in the lot i.e. 100% non rented trucks are in the lot on Saturday morning).
(If none of the 20 trucks are rented, the lot will have 20 trucks on Saturday. If 18 trucks are not rented, the lot will have 19 (18 + 1 rented comes back) trucks on Saturday morning.)
So let's make the number of trucks in the lot on Saturday morning equal to 12.
N - Not rented trucks; R - Rented trucks
N + R = 20
N + R/2 = 12
R = 16
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 13 Feb 2012, 04:14
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I found this way to be simple. Please correct me if I am wrong
Let the rental cabs be R and non rental cabs be N
R + N =20 -- 1 eq.
0.5R+N=12 -- 2 eq.
0.5R + 20-R =12
20-0.5R=12 --> R = 8 *2 = 16
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Re: PS trucks [#permalink] New post 10 Jan 2011, 17:05
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I guess the best way to solve this problem in short time is rephrase and/or back solve
1. At least 12 were back on Saturday i.e at max 8 were out on Saturday
2. Now assume 0% instead 50% were returned i.e at max 8 were rented out [we want to have at least 12 on lot i.e we can't rent more than 8 i.e upper limit is fixed]
3. 50% of those 8 were returned = 4 [We want to be at maximum and maximum we can get added to at least 12 is 50% of all rented out, i.e 50% of 8]
4. 12+4=16

Hope this helps
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 24 Aug 2014, 23:51
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russ9 wrote:

Hi karishma,

This statement is what gets me "So maximize the number of trucks rented, we should try to minimize the number of trucks in the lot on Saturday morning i.e. make it 12."

Logically speaking, since 50% of the trucks do return back, therefore to maximize the number of trucks rented, we should also maximize the number of returned trucks, at least that's what I put together. Which means that if I want to maximize the number of trucks that were rented, then i need to minimize the number of trucks that were NOT rented, but you are using a different approach?

Can you please explain why my logic is flawed?


Why do the number of trucks go down? The ones which are not rented stay there. Out of the ones which are rented, only half come back. So if more trucks are rented, the dent in the number of trucks on Sat morning will be more.
If no trucks are rented, there will be 20 trucks on Sat morning.
If all trucks are rented, there will be only 10 trucks on Sat morning.

So as you rent more and more trucks out, you will be left with fewer trucks on Sat morning...(ranging from 20 to 10)
So if we minimize the number of trucks on Sat morning, we will maximize the no of trucks rented.
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Re: PS trucks [#permalink] New post 10 Apr 2005, 18:31
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


X be unrented Y is rented

X+Y =20
X+50% of Y >= 12 ---(2)

We get y>= 14

Plug values of y>= 14 in (2) and you find 16 is the best fit
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Re: PS trucks [#permalink] New post 09 Jan 2011, 14:01
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.....
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One week a certain truck rental lot had a total of 20 trucks [#permalink] New post 16 Apr 2011, 19:08
One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent 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

Can the pundits here please suggest the best possible way to solve this problem?
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Re: Percents question [#permalink] New post 16 Apr 2011, 23:32
Let trucks rented put = x

20 -x + 0.5x >= 12

20 - 0.5x >= 12

8 >= 0.5x

16 >= x

so Max value of x = 16

Answer - B
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 13 Feb 2012, 04:19
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pavanmpv wrote:
I found this way to be simple. Please correct me if I am wrong
Let the rental cabs be R and non rental cabs be N
R + N =20 -- 1 eq.
0.5R+N=12 -- 2 eq.
0.5R + 20-R =12
20-0.5R=12 --> R = 8 *2 = 16


Welcome to Gmat Club.

Your solution is perfectly OK, (it's basically the same as shown in my previous post).
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 13 Feb 2012, 08:34
Thanks a lot for clarifying. :)
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 25 Jan 2014, 12:13
Wow, you guys make it look so easy. I was on this Q for so long. How can I not make this mistake again?
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Re: One week, a certain truck rental lot had a total of 20 [#permalink] New post 27 Jan 2014, 02:31
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Wow, you guys make it look so easy. I was on this Q for so long. How can I not make this mistake again?


Try to solve questions without using algebra. That will force you to look at the big picture. You might spend a lot of time on every question initially but the more you practice, the easier it will get.
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One week, a certain truck rental lot had a total of 20 [#permalink] New post 24 Aug 2014, 11:01
VeritasPrepKarishma wrote:
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


There are numerous ways in which you can solve this question. Brute force method if the relation between rented and non-rented trucks in not very clear:

Monday morning - 20 trucks
Saturday morning - at least 12 trucks
50% trucks rented in the week were returned.
maximum no of trucks rented out = ?

I want to maximize the no. of trucks rented so I say - If 20 trucks were rented (i.e. all of them), then we should have 50% i.e. 10 of them back. But we have more; we have at least 12.
So the no. of trucks rented out must be less than 20 (because they cannot be more than 20).
What about 18? If 18 trucks are rented out, 2 remain in the lot through the week. Out of 18, 9 are returned so total 11 are in the lot. But we need at least 12 in the lot.
Let's go further down and try 16. 4 trucks do not leave the lot. Out of 16, 8 come back so we have 12 trucks in the lot.
(As we keep reducing the number of trucks rented out, the total number of trucks in the lot of Saturday morning keeps increasing. We need to maximize the number of trucks rented out which will be at the minimum possible value of total number of trucks in the lot.)
Therefore, 16 trucks must have been rented out.

Algebraic approach:
As we increase the number of trucks rented, the total number of trucks in the lot on Saturday morning decreases since out of the rented trucks only 50% come back (while all non-rented trucks stay in the lot).
(e.g If none of the 20 trucks are rented, the lot will have 20 trucks on Saturday. If 18 trucks are not rented, the lot will have 19 (18 + 1 rented comes back) trucks on Saturday morning.)
So maximize the number of trucks rented, we should try to minimize the number of trucks in the lot on Saturday morning i.e. make it 12.
N - Not rented trucks; R - Rented trucks
N + R = 20
N + R/2 = 12
R = 16


Hi karishma,

This statement is what gets me "So maximize the number of trucks rented, we should try to minimize the number of trucks in the lot on Saturday morning i.e. make it 12."

Logically speaking, since 50% of the trucks do return back, therefore to maximize the number of trucks rented, we should also maximize the number of returned trucks, at least that's what I put together. Which means that if I want to maximize the number of trucks that were rented, then i need to minimize the number of trucks that were NOT rented, but you are using a different approach?

Can you please explain why my logic is flawed?
One week, a certain truck rental lot had a total of 20   [#permalink] 24 Aug 2014, 11:01
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