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Manager  G
Joined: 26 Dec 2015
Posts: 236
Location: United States (CA)
Concentration: Finance, Strategy
WE: Investment Banking (Venture Capital)
A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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12 00:00

Difficulty:   65% (hard)

Question Stats: 64% (02:45) correct 36% (03:25) wrong based on 152 sessions

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A certain car company manufactured x cars at a cost of c dollars per car. If a certain number of cars were sold below cost at a sale price of s dollars per car, while the rest of the cars were sold for the normal retail price of n dollars per car, how many cars could the company afford to sell at the sale price in order to break even (no profit and no loss)?

(A) $$\frac{x(c-n)}{(s-n)}$$

(B) $$\frac{x(n-c)}{(s-n)}$$

(C) $$\frac{x(c-n)}{(s-c)}$$

(D) $$\frac{x(s-n)}{(c-n)}$$

(E) (x-n)(x-s)

P.S. we definitely need more of these question types on this forum. they definitely pop up on the practice CATs i've taken from all the different major test prep companies

Originally posted by LakerFan24 on 21 Apr 2017, 16:51.
Last edited by Bunuel on 21 Apr 2017, 16:55, edited 1 time in total.
Renamed the topic.
VP  D
Joined: 05 Mar 2015
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Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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LakerFan24 wrote:
A certain car company manufactured x cars at a cost of c dollars per car. If a certain number of cars were sold below cost at a sale price of s dollars per car, while the rest of the cars were sold for the normal retail price of n dollars per car, how many cars could the company afford to sell at the sale price in order to break even (no profit and no loss)?

(A) $$\frac{x(c-n)}{(s-n)}$$

(B) $$\frac{x(n-c)}{(s-n)}$$

(C) $$\frac{x(c-n)}{(s-c)}$$

(D) $$\frac{x(s-n)}{(c-n)}$$

(E) (x-n)(x-s)

P.S. we definitely need more of these question types on this forum. they definitely pop up on the practice CATs i've taken from all the different major test prep companies

Assuming nos.are quiet faster method for solving this one

let
x=10
c=50
n=55 for 5cars out of 10
then @ retail price ,its income be = 55*5 =275
but his CP= 50*10 = 500
again let s=45 (discounted)
thus he must earn after discount = 500-275 = 225 for no loss no gain
no of cars sold @ discount = 225/45 =5
now put individual assumption , which yeilds 5 as our answer

clearly option 1 does

Ans A
Senior Manager  G
Joined: 02 Mar 2017
Posts: 256
Location: India
Concentration: Finance, Marketing
Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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3
Let K be the value of Number of cars sold at retail price
For break even profit = 0

KS + (X-K)N - XC = 0

Rearranging K = X(N-C)/N-S
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Manager  G
Joined: 26 Dec 2015
Posts: 236
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Concentration: Finance, Strategy
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Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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Thank you rohit8865 and VyshakhR1995.

I'm looking for other ways of using numbers to determine the answer (as you did, rohit). Anyone else care to solve?
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Joined: 16 Oct 2010
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Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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1
LakerFan24 wrote:
A certain car company manufactured x cars at a cost of c dollars per car. If a certain number of cars were sold below cost at a sale price of s dollars per car, while the rest of the cars were sold for the normal retail price of n dollars per car, how many cars could the company afford to sell at the sale price in order to break even (no profit and no loss)?

(A) $$\frac{x(c-n)}{(s-n)}$$

(B) $$\frac{x(n-c)}{(s-n)}$$

(C) $$\frac{x(c-n)}{(s-c)}$$

(D) $$\frac{x(s-n)}{(c-n)}$$

(E) (x-n)(x-s)

P.S. we definitely need more of these question types on this forum. they definitely pop up on the practice CATs i've taken from all the different major test prep companies

It may not be advisable to use number plugging with 5-6 variables. Chances of silly mistakes are quite high in such cases.

You want no profit no loss so profit made by selling at retail should be equal to loss borne by selling at sale price. Say y cars were sold at same price.

$$(x - y)*(n - c) = y*(c - s)$$

We need to separate out y from this:
$$x*(n - c) - y*(n - c) = y*(c - s)$$

$$x*(n - c) = y(c - s +n - c)$$

$$y = x*\frac{(n - c)}{(n - s)} = x*\frac{(c - n)}{(s - n)}$$
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Director  S
Joined: 12 Nov 2016
Posts: 699
Location: United States
Schools: Yale '18
GMAT 1: 650 Q43 V37 GRE 1: Q157 V158 GPA: 2.66
A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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LakerFan24 wrote:
A certain car company manufactured x cars at a cost of c dollars per car. If a certain number of cars were sold below cost at a sale price of s dollars per car, while the rest of the cars were sold for the normal retail price of n dollars per car, how many cars could the company afford to sell at the sale price in order to break even (no profit and no loss)?

(A) $$\frac{x(c-n)}{(s-n)}$$

(B) $$\frac{x(n-c)}{(s-n)}$$

(C) $$\frac{x(c-n)}{(s-c)}$$

(D) $$\frac{x(s-n)}{(c-n)}$$

(E) (x-n)(x-s)

P.S. we definitely need more of these question types on this forum. they definitely pop up on the practice CATs i've taken from all the different major test prep companies

This is indeed a very realistic GMAT question- other examples include (see Kaplan) express x y z etc- basically the ability to express three variables. Anyways, what this question is simply stating is that with a specific amount of cars produced, how many cars could be sold at a lower cost than the cost it took to produce them- so if took $30,000.00 to make each Tesla car out of a batch of 100 then how many Tesla's could be sold out of that batch of 100 for a price less than$30,000- assuming, remember, that the only other price offered by Tesla will be sold at a price higher higher than $30,000. That is to say if the total cost of producing all these Tesla cars is$3,000,000 then what amount of Tesla's sold at a price below $30,000 would add up to$3,000,000 if the other Tesla cars must be sold at a constant price above the production cost. Basically $30,000 (S) +$30,0000(N)= X (C)- but we can use even simpler numbers

x= 100
c= 2.00
s= 1.00
n=3.00

Only A satisfies the condition

Thus
"A"
Target Test Prep Representative G
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Joined: 04 Mar 2011
Posts: 2815
Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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LakerFan24 wrote:
A certain car company manufactured x cars at a cost of c dollars per car. If a certain number of cars were sold below cost at a sale price of s dollars per car, while the rest of the cars were sold for the normal retail price of n dollars per car, how many cars could the company afford to sell at the sale price in order to break even (no profit and no loss)?

(A) $$\frac{x(c-n)}{(s-n)}$$

(B) $$\frac{x(n-c)}{(s-n)}$$

(C) $$\frac{x(c-n)}{(s-c)}$$

(D) $$\frac{x(s-n)}{(c-n)}$$

(E) (x-n)(x-s)

The total cost of all the cars was cx dollars.

Since x total cars were sold, we can let a = the total number of cars sold at s dollars per car and x - a = the number of cars sold for n dollars per car. Since we need to break even, we can create the following equation:

cx = as + (x - a)n

cx = as + xn - an

cx - xn = as - an

cx - xn = a(s - n)

cx - xn/(s - n) = a

x(c - n)/(s - n) = a

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Intern  B
Joined: 02 Sep 2019
Posts: 9
Re: A certain car company manufactured x cars at a cost of c dollars per  [#permalink]

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1
Cost of Manufacturing = X cars * C $= XC Loss Sales = Lets assume Y Cars * S$ = YS
Profit Sales = (X-Y) Cars * N \$ = (X-Y)N
Question is What is Y in terms of X,C,S,N

Break Even (no profit or No Loss) = Profit + Loss = Manufacturing Cost

Therefore, YS + (X-Y)N = XC
=> Y(S-N) = X(C-N)
=> Y = X(C-N)/ (S-N). Answer A Re: A certain car company manufactured x cars at a cost of c dollars per   [#permalink] 18 Sep 2019, 06:56
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