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A merchant paid $300 for a shipment of x identical calculators. The

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A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 21 Apr 2012, 21:18
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E

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A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

PS45502.01

Is this question correct? Shipping cost given but cost of each calculator missing.....

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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 21 Apr 2012, 23:03
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boomtangboy wrote:
A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

Is this question correct? Shipping cost given but cost of each calculator missing.....


"A merchant paid $300 for a shipment of x identical calculators" means that there was a shipment of x calculators which cost merchant $300.

The average cost of each calculator is \(\frac{300}{x}\);
The average price of each calculator is \(\frac{300}{x}+5\);
Number of calculators sold is \(x-2\);
The total revenue is \(300+120=420\);

So, \((x-2)*(\frac{300}{x}+5)=420\). At this point it's better to substitute values. Answer choice E fits: \((30-2)*(\frac{300}{30}+5)=28*15=420\).

Answer: E.

Hope it's clear.
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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 04 Oct 2013, 13:21
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1
Try plugging in the answer:-
1st Try selecting options which are divisible by 300
If you select 30 as x.Therefore avg cost of each calculator 300/30 i.e 10$.
Set aside 2 .So we have 28 calc.
Each calc sold at 5$ more i.e 10+5=15$ & 15x28=$420
Difference is $420-$300=120
E) is the answer hope it helps :)
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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 11 Jan 2014, 08:52
boomtangboy wrote:
A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?

A. 24
B. 25
C. 26
D. 28
E. 30

Is this question correct? Shipping cost given but cost of each calculator missing.....


I'd go with backsolving on this one though I must admit that originally I tried to solve algebraically and was stuck with variables and quadratics

We will have (300/x + 5)(x-2) = 420

Then begin replacing, I started with 25 cause 26 would not yield an integer when divided by 300 but it didn't work, so I realized I needed a bigger number and the only factor of 30 that could work was E

So E

Hope it helps
Cheers!
J :)
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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 11 Jan 2014, 11:29
Cost of calculator=300/x
Given,(x-2)(300/x +5)=420
We can check by options that only x=30 satisfies this equation.
Hence,
Ans. E

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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 20 Mar 2014, 19:56
2
Answer = E

Total Calculators = x

Cost per calculator = \(\frac{300}{x}\)

Calculators sold = x-2

Selling price per calculator = \(\frac{300}{x} +5\)

Total amount received = \((\frac{300}{x} + 5) * (x-2) = 420\)

Solving,

\(x^2 - 26x - 120 = 0\)

x = 30
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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 06 Dec 2017, 14:25
2
Hi All,

The information in the prompt include a variety of nice 'round' numbers, so it's likely that the number of calculators in the shipment is also a round number AND divides evenly into 300. Only two of the answers 'fit' that idea - 25 and 30, so let's TEST THE ANSWERS and see if either of those values fits all of the information given in the prompt.

IF.... X = 30 and there are 30 calculators in the shipment....

The merchant paid $300/30 = $10 per calculator
2 calculators were used for demonstrations, so 30 - 2 = 28 calculators were sold
Each of the 'sold' calculators brought in $10+$5 = $15
The total revenue was ($15)(28) = $420
The total profit was $420 - $300 = $120

This is an exact match for what we were told, so this MUST be the answer.

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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 27 Dec 2017, 16:36
5 is gained every calculator sold. Also, if 300 calculators cost x dollars, then 2 cost 600/x.

5(x-2) -600/x=120

(X-2)-120/x=24
X^2-2x-120=24x
x^2-26x-120=0

x^2=26x+120

X=13+- sqroot(169+120)

X=13+-17

x= 30

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Re: A merchant paid $300 for a shipment of x identical calculators. The  [#permalink]

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New post 28 Dec 2018, 21:27
Cost of each calculator = (300/x)
selling price of each calculator is = 420/(x-2)
Since, Selling price = Cost Price + Profit

Selling Price - Cost Price = 5 (per unit)

(420/(x-2)) - (300/x) = 5
(84/(x-2) ) - (60/x) = 1
Solving the above equation,
x^2 - 26x - 120 = 0
x^2 - 30x + 6 - 120 = 0
x = 30,-6

Number of calculators can't be negative = 30
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Re: A merchant paid $300 for a shipment of x identical  [#permalink]

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Re: A merchant paid $300 for a shipment of x identical   [#permalink] 21 Sep 2019, 06:09
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