A photography dealer ordered 60 Model X cameras to be sold

The weights are 54 (20% profit on 54 cameras) and 6 (50% loss on 6 cameras).
This is a ratio of 54:6 = 9:1 (in lowest terms). It is the same thing whether you use 54 and 6 as weights or 9 and 1 but 9 and 1 simplify the calculations.

Avg = (A1 * w1 + A2 * w2)/(w1 + w2) = (A1 * 9 + A2 * 1)/(9 + 1) = (A1 * 9 + A2 * 1)/10

Does this question bother anyone else in formatting terms? Perhaps I'm misinterpreting, but let me explain:

"What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?"

Profit = Revenue - Costs

Thus, the question is looking for (Revenue - Cost) / Cost

In number terms this would look like, [54*(250) + 6*(250)*(4/5)*(1/2) - 600*(250)*(4/5)]/[(60)*(250)*(4/5)]x100 ~~ 10.7%

I'd appreciate any help in clearing up this misunderstanding!

You are correct conceptually.

The error lies here:

"$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera."

Marked price is 250
This is a 20% markup on cost.

Cost * (1 + 20/100) = Marked price

Cost * (6/5) = Marked Price
Cost = (5/6) * 250

You have taken this to be (4/5)*250. Replace it by (5/6) and you should get the correct answer.

Hey all,
Why do we divide 250/2.4? Where does 2.4 come from?
Thanks!

Cost to dealer = 250/1.2 (because 250 is the selling price which 20% markup)

Revenue from the 6 cameras is half the cost price. This is = (250/1.2)*(1/2) = 250/2.4

P.S. - Saw your post later Engr2012

We are given that a dealer ordered 60 cameras to be sold for $250 each. This is a 20% markup over the initial cost to the dealer. We can use this to determine the initial cost to the dealer.

We can let C = dealer’s cost per camera. So,

1.2C = 250

12C = 2,500

C ≈ 210

Thus, we have an approximate cost of $210 per camera.

We are next given that 6 cameras were never sold and were returned to the manufacturer for a refund of 50 percent of the dealers cost. Since the approximate cost was $210 per camera, the dealer received a $105 refund per camera, or $630 total. We can now calculate the total cost of the cameras to the dealer as:

Cost = $210 x 60 – $630

Cost = $12,600 – $630

Cost = $11,970

Next, we need to determine the revenue. We know that the dealer sold 54 cameras for $250 each. So total revenue is:

Revenue = 54 x $250 = $13,500

Since profit = revenue – cost, profit = $13,500 – $11,970 = $1,530

Finally we need to determine the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras.

(Profit/initial cost) x 100

1,530/12,600 x 100

153/1260 x 100

This is roughly equal to:

150/1200 x 100

15/120 x 100

1/8 x 100 = 12.5% profit

The closest answer is answer choice D, 13% profit.

VeritasPrepKarishma, I think you method is pretty sweet - I just used another weighted average looking formula you shared in your blog:

\(\cfrac { w1 }{ w2 } :=\cfrac { \left( A2-Aavg \right) }{ \left( Aavg-A1 \right) } =\\ =\cfrac { 1 }{ 9 } =\cfrac { \left( 1.2-Aavg \right) }{ \left( Aavg-0.5 \right) } \\ =Aavg-0.5=9\left( 1.2-Aavg \right) \\ =Aavg-0.5=10.8-9Avg\\ =Aavg+9Aavg=10.8+0.5\\ =10Aavg=11.3\\ =Aavg=\cfrac { 11.3 }{ 10 } \\ Aavg=1.13\)

13% profit

I'm a big fan of the scaled method but I can't figure out how to set it up as to find the 'combined average'.

It's important to recognize that we really don't need to use the information about the cameras selling for $250 each. The question boils down to . . .

54 cameras were sold at a 20% markup, and 6 cameras were (essentially) sold at a 50% markdown. What was the approximate profit or loss as a percent of the dealer’s initial cost for all 60 cameras?

So, we can assign A NICE value of $100 to the initial cost per camera.
This means the 60 cameras cost $6000 to buy.

54 cameras were sold at a 20% markup and 6 cameras were sold at a 50% markdown.
So, 54 cameras were sold for $120, and 6 cameras were sold for $50.
(54)($120) + (6)($50) = $6780
So, the cameras were sold for $6780

This represents a profit of $780 (eliminate A and B)

If the initial cost was $6000, we must determine the percentage equivalent to $780/$6000

$780/$6000 = 78/600 = 13/100 = 13%

Answer:

Cheers,
Brent

Yes, you have used the scale method successfully to find the average sale price.
Assuming cost price of each camera is 1, 1.2 is the sale price of 54 cameras and .5 is the sale price of 6 cameras.
Hence you get the average sale price as 1.13.

Scale method is also depicted by the formula you used.

You can use the same formula with profit/loss percent and get the same answer.

w1/w2 = (A2 - Aavg)/(Aavg - A1)

1/9 = (20 - Aavg)/(Aavg - (-50))

Aavg + 50 = 9*(20 - Aavg)

10*Aavg = 130

Aavg = 13

Just that, when I have to find the average, I prefer to use the formula in the form

Aavg = (A1*w1 + A2*w2)/(w1 + w2)

It is just a realignment of the previous formula.

Hello diegocml,

I hope this is what you meant by scale method.



Based on the scale we get

\(\frac{54}{6} = \frac{(x+50)}{(20-x)}\)

\(\frac{9}{1} = \frac{(x+50)}{(20-x)}\)

\(180 - 9x = x+50\)

\(130 = 10x\)

\(x = 13\)

So, the dealer earns a profit of 13%

I learnt this approach after reading VeritasPrepKarishma's blog. Since the answer is correct, I am assuming the approach is correct

This formula is an application of the scale method:

\(\frac{w1}{w2} = \frac{A2 - Aavg}{Aavg - A1}\)

It is faster and more direct than actually using the number line.

Attached is a visual that should help.

We are told that \($250\) is equal to \(120\%\) of original price.

We can set \(X \ = \ the \ original \ price\):




We need to find profit or loss. This can be calculated by using the formula:

profit = (money earned from sales) – (money spent on purchasing)

We can approximate the values:






Profit as a percentage of initial cost \(= \frac{1500}{12500} = \frac{15}{125} = \frac{3}{25} = approximately \ 12\%\ profit\)

The final answer is .

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.

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%.

Responding to a pm - a useful tip for everyone:



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)

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