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# The cost of manufacturing "X" cellphones is given below. C=2500+100x,

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Director
Joined: 23 Feb 2015
Posts: 756
GMAT 1: 720 Q49 V40
The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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22 Feb 2019, 12:23
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25% (medium)

Question Stats:

81% (00:57) correct 19% (00:48) wrong based on 27 sessions

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The cost of manufacturing "X" cellphones is given below.
C=2500+100x, and the revenue from selling these is given by R=150x. What is the least number of cellphones that must be produced and sold to realize some profit?

A) 51
B) 50
C) 49
D) 48
E) 41

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Intern
Joined: 22 May 2017
Posts: 17
Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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22 Feb 2019, 13:39
The Answer for this is A

As the fixed sink cost = 2500
CP per phone = 100
SP per phone = 150
So the profit per phone = 50

As the question asks, for what value of X i.e. number of phones sold should be > 2500 which is the fixed sink cost.
50x > 2500 for min of x
For x=50 => 50x = 2500. So the minimum value for 50x to be greater than 2500, X = 51

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Senior Manager
Joined: 15 Feb 2018
Posts: 255
Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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23 Feb 2019, 02:27
1
Profit occurs when revenue is greater than cost.
Revenue>cost
150x>2500+100x
50x>2500
x>50
A phone is an integer (can't sell a fraction of a phone) so the next number larger than 50 is 51.

A
Director
Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
Schools: ISB '19, IIMA , IIMB, XLRI
WE: Information Technology (Consulting)
Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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23 Feb 2019, 05:46
The cost of manufacturing "X" cellphones is given below.
C=2500+100x, and the revenue from selling these is given by R=150x. What is the least number of cellphones that must be produced and sold to realize some profit?

A) 51
B) 50
C) 49
D) 48
E) 41

Total Cost; $$C = 2500 + 100x$$

Total Revenue; $$R = 150x$$

To get Profit, Revenue must be greater than Cost.

Therefore; $$Revenue > Cost$$

$$150x > 2500 + 110x$$

$$150x - 110x > 2500$$

$$50x > 2500$$

$$x > \frac{2500}{50}$$

$$x > 50$$

Hence $$x$$ has to be greater than $$50$$ to get profit. Least number of cell phone to be produced to get least profit $$= 51$$.

Intern
Joined: 15 Feb 2019
Posts: 7
Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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25 Feb 2019, 10:25
Profit=Revenue-cost
Revenue=2500+100X
Cost=150X
Profit= 2500+100X-150X
Profit=2500-50X
If X=50, Profit=0
i.e 2500-50(50)=0
So you want the least possible value for X greater than 50, which is 51. And is A

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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,  [#permalink]

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02 Mar 2019, 10:11
The cost of manufacturing "X" cellphones is given below.
C=2500+100x, and the revenue from selling these is given by R=150x. What is the least number of cellphones that must be produced and sold to realize some profit?

A) 51
B) 50
C) 49
D) 48
E) 41

To realize some profit, we need the revenue to be greater than the cost; that is,

150x > 2500 + 100x

50x > 2500

x > 50

Thus, at least 51 cell phones must be sold to realize a profit.

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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x,   [#permalink] 02 Mar 2019, 10:11
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