A certain car company manufactured x cars at a cost of c dollars per

Question

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)

(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

VyshakhR1995 · Accepted Answer

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

JeffTargetTestPrep · Answer

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
 
Answer: A

rohit8865 · Answer

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

LakerFan24 · Answer

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?

KarishmaB · Answer

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)}

Nunuboy1994 · Answer

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"

kdgupta · Answer

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

luisdicampo · Answer

Deconstructing the Question

The company manufactures  x  cars at a cost of  c  dollars per car, so total cost is  xc .

Let  k  be the number of cars sold at the sale price  s , and the remaining  x-k  cars are sold at the normal price  n .

Break-even means total revenue equals total cost.

Step-by-step

Total revenue:

ks + (x-k)n

Set revenue equal to cost:

ks + (x-k)n = xc

Expand:

ks + xn - kn = xc

Factor k:

k(s-n) + xn = xc

Subtract  xn  from both sides:

k(s-n) = x(c-n)

Divide by  (s-n) :

k = \frac{x(c-n)}{s-n}

Answer: A

A certain car company manufactured x cars at a cost of c dollars per

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GMAT Problem Solving (PS) Questions

A certain car company manufactured x cars at a cost of c dollars per

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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