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

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TheUltimateWinner
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The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  22 Feb 2019, 12:23
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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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varunbawa
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  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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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  23 Feb 2019, 02:27
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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
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  23 Feb 2019, 05:46
AsadAbu wrote:
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 \(= 51\).

Answer A
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Peachie
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  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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ScottTargetTestPrep
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  02 Mar 2019, 10:11
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AsadAbu wrote:
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.

Answer: A
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ShajjadH
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  02 Aug 2021, 00:06
As the fixed sink cost = 2500
CP per phone = 2500+100x
SP per phone = 150x
So the number of phone, x = 50
By selling 50 phones, profit will be 0, because CP and SP are equal.
As we know,
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.
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pratiksha1998
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink] New post  29 Aug 2021, 22:28
To Generate profit, SP - CP > 0

150x - 2500 +100x >0
Solve for X =>
x>50

Hence option A is the answer
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Re: The cost of manufacturing "X" cellphones is given below. C=2500+100x, [#permalink]
29 Aug 2021, 22:28
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