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A photography dealer ordered 60 Model X cameras to be sold

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A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 02 Nov 2019, 13:38
My thought process with very similar with akang. After reading the problem, I just saw "percent", "percent" "What percent...?" So I didn't even bother with the marked-up "$250"

In my head, I saw:

(6 out of 60) = 10% of stuff (never sold, returned, = loss) at 50% loss

(rest of stuff, not 6, minus 10%) = 90% of stuff (gives him 20% markup) at 20% profit

And his T (total) whatever percentage is Profit - Loss

So, my scratch paper for this problem literally looks like:

90% of 20% = 18%
10% of 50% = 5%
P - L = ___________

18% - 5% = 13%

Then I thought, "profit" since the gain was bigger than the loss.

The other clue to ditch the $250 and any other non-percentages was when I tried to figure out the actual cost per camera. When I tried to divvy up 250 by 1.2 and started to get something funky, I thought, "Ok, they don't want me to mess with that."

Right? So is my thought process applicable to other problems like this? We don't actually have to mess with actual total profit and dividing by huge numbers to get the answer, huh? Or am I narrowing my train of thought down too much that it will only work for this problem? And I totally feel you on the weighted averages equation, but is it a good habit to use weighted averages for all problems like this? Do we need the full meal deal, or can I get away with just the burger?

I just think it's funny when I see contributors online write out work like: "1,644/12,480 = ~13%. See?" As if were all supposed to be like, "Oh, of course! Divided by twelve thousand four hundred and eighty IS about thirteen percent. Makes total sense." Like us normal people could just come up with that answer that easily and that quickly, without a calculator, during the actual test. No offense to anyone, I love this site, but I already have the answers in the back of the book. What helps, at least, helps me, is when I can be shown how to break the work down into small digestible pieces that can be used on other questions as well. Like eating cookies in bed. The trick is to break them up inside the bag, then put the small cookie pieces in your mouth. No crumbs all over the sheets.

But maybe that's just me.

Thanks everyone.
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 02 Nov 2019, 14:56
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit



Approach 1:
54 cameras yield a 20% profit, while 6 cameras suffer a 50% loss.
Average for all 60 cameras \(= \frac{(54*20 - (6*50)}{60} = \frac{780}{60} = 13\)



Approach 2:
Since the question stem asks for a PERCENTAGE, ignore the $250 selling price.
Plug in values that satisfy the given conditions and make the math easy.
Fraction of cameras not sold \(= \frac{6}{60} = \frac{1}{10}\)

Let the number of cameras ordered \(= 10\)
Let the cost per camera \(= 10\)
Total cost \(= 10*10 = 100\)

Number not sold \(= \frac{1}{10}*10 = 1\)
For this one camera, the refund received \(= 5*10 =5\)

Number of cameras sold \(= 9\)
With a markup of 20%, the selling price per camera \(= 12\)
Total revenue \(= 9*12 = 108\)

Refund + revenue = 5+108 = 113, a profit of 13%.


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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 26 Dec 2019, 01:30
1
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit


Responding to a pm - a useful tip for everyone:

Quote:
In this question, the sold price of each camera is 250$ after 20% markup price. So i tried to find the original cost using the method 250 * .80 which calculates to 200 and 20% is just 240 but not 250. I have also tried with other numbers as well such as 250*.75, 90 but i am sure, am missing something :( .

Could you please help me understand, how should I do the reverse percentage calculation?


Don't try to calculate the reverse percentage lest you make a mistake. Use the straight forward percentage manipulation even if that needs a variable. Even if you do not want to actually write it down, make a straight forward equation in your mind like this:

Cost * (6/5) = 250

So
Cost = 250 * (5/6)
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 26 Dec 2019, 03:47
VeritasKarishma wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit


Responding to a pm - a useful tip for everyone:

Quote:
In this question, the sold price of each camera is 250$ after 20% markup price. So i tried to find the original cost using the method 250 * .80 which calculates to 200 and 20% is just 240 but not 250. I have also tried with other numbers as well such as 250*.75, 90 but i am sure, am missing something :( .

Could you please help me understand, how should I do the reverse percentage calculation?


Don't try to calculate the reverse percentage lest you make a mistake. Use the straight forward percentage manipulation even if that needs a variable. Even if you do not want to actually write it down, make a straight forward equation in your mind like this:

Cost * (6/5) = 250

So
Cost = 250 * (5/6)


Nice! Thank You VeritasKarishma!
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 31 Dec 2019, 09:07
I got the question wrong with this approach.

Sales Price (SP) = 250
Initial Cost (IC) =x

IC + .20 Sp = 1SP
IC = .80 SP
Thus dealer had an initial cost of $200 per camera. I know that is wrong I should have done the following

IC + .20 IC = SP
1.20 IC = SP = 1.20 IC = 250

I just don't know why I should have done this approach. What clues in the question stem should I look for? I have made this mistake on similar problems, any solutions or tips on how to fix my interpretation VeritasKarishma Bunuel JeffTargetTestPrep
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 02 Jan 2020, 04:39
whollymoses wrote:
I got the question wrong with this approach.

Sales Price (SP) = 250
Initial Cost (IC) =x

IC + .20 Sp = 1SP
IC = .80 SP
Thus dealer had an initial cost of $200 per camera. I know that is wrong I should have done the following

IC + .20 IC = SP
1.20 IC = SP = 1.20 IC = 250

I just don't know why I should have done this approach. What clues in the question stem should I look for? I have made this mistake on similar problems, any solutions or tips on how to fix my interpretation VeritasKarishma Bunuel JeffTargetTestPrep


Mark up is on cost and discount is on marked price.
Anyway the question gives you "a 20 percent markup over the dealer’s initial cost for each camera"

So you know that the mark up is over cost so 20% will be the 20% of cost.

"B is 20% more than A" means
B = A + 20% of A
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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New post 28 Jan 2020, 15:28
[(20*54)-(50*6)]/60 = 13%
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Re: A photography dealer ordered 60 Model X cameras to be sold   [#permalink] 28 Jan 2020, 15:28

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