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# John.s car dealership contains m cars, 20% of which are

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John.s car dealership contains m cars, 20% of which are [#permalink]

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26 Feb 2012, 15:17
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79% (02:44) correct 21% (03:06) wrong based on 39 sessions

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John.s car dealership contains m cars, 20% of which are minivans and 80% are sedans. Kevin's car dealership contains n cars, 40% of which are minivans are 60% are trucks. Larry's car dealership contains p cars, 50% of which are minivans and 50% of which are convertibles. If 25% of the m + n + p cars contained at the three dealerships are minivans, what is m in terms of n and p?

(A) n + 3p
(B) 3n + 5p
(C) 4n + 5p
(D) (n+5p)/2
(E) (4n+5p)/3

Ok I tried doing this this question in 2 ways. 1----> By picking numbers for m, n and p and I struggled to get the right answer and I need help. 2-----------> Algebraically and I got the correct answer.

So Method 1 (Number picking)

John Dealership
Let say m = 200
Minivans = 20 ----------------------> I will only talk about Minivan's as the question is about minivans.

Kevin dealership
Let say n = 300
Minivans = 40

Larry dealership
let say n = 300
Minivan's = 50

Total cars at the dealership = 600
Minivans = 150 -------------------------------(25% of 600)

MY algebraic approach

0.2m + 0.4n + 0.5p = 0.25m + 0.25n + 0.25p

Solving for m will give m = 3n + 5p which is the right answer.
[Reveal] Spoiler: OA

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Re: Number of cars in a dealership [#permalink]

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26 Feb 2012, 16:54
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John.s car dealership contains m cars, 20% of which are minivans and 80% are sedans. Kevin's car dealership contains n cars, 40% of which are minivans are 60% are trucks. Larry's car dealership contains p cars, 50% of which are minivans and 50% of which are convertibles. If 25% of the m + n + p cars contained at the three dealerships are minivans, what is m in terms of n and p?
(A) n + 3p
(B) 3n + 5p
(C) 4n + 5p
(D) (n+5p)/2
(E) (4n+5p)/3

For this question algebraic ways is the easiest: $$0.2m+0.4n+0.5p=0.25(m + n + p)$$ --> $$m=3n+5p$$.

enigma123 wrote:
John Dealership
Let say m = 200
Minivans = 20 ----------------------> I will only talk about Minivan's as the question is about minivans.

Kevin dealership
Let say n = 300
Minivans = 40

Larry dealership
let say n = 300
Minivan's = 50

Total cars at the dealership = 600
Minivans = 150 -------------------------------(25% of 600)

First of all you have miscalculated # of minivans and total cars:
20% of 200 = 40;
40% of 300 = 120;
50% of 300 = 150;
Total minivans = 310;
Total cars = 800.

But this is not the main problem: you can not arbitrary assign numbers to m, n and p here. Because 0.2m+0.4n+0.5p must equal to 0.25(m + n + p). For your numbers 25% of (m + n + p)=800 is 200 and not 310.

To summarize this is not a good problem for number plugging.
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Re: John.s car dealership contains m cars, 20% of which are [#permalink]

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10 Oct 2014, 18:57
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Re: John.s car dealership contains m cars, 20% of which are [#permalink]

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21 Oct 2014, 23:04
$$\frac{25}{100} (m+n+p) = \frac{20m}{100} + \frac{40n}{100} + \frac{50p}{100}$$

5m = 15n + 25p

m = 3n + 5p

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Re: John.s car dealership contains m cars, 20% of which are   [#permalink] 21 Oct 2014, 23:04
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