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'.

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

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

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

Hi. Can you help point out why this method is incorrect?

Cost price is $208. So profit per camera is $42. Profit for the 54 cameras sold is 54*42=2,268. 2,268+625(refund)=2,893.

Total price to dealer is 12,480.

2893/12,480*100=23%

54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit.

suppose total cameras = 10 and each costs $10
total profit = $2*9 -$5*1 = $13 or 13% (with $100 cost)

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'.

Hello diegocml,

I hope this is what you meant by scale method (See attachment).

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

Cost to dealer for 60 cameras = \(\frac{250}{1.2}*60\) = $12,500

Revenue for 54 cameras = 54*250 = $13,500

Revenue from 6 cameras = \(\frac{250}{2.4}*6\) = $625

Total Revenue = $14,125

Profit percent = \(\frac{14125-12500}{12500}\) = 13%

Answer is D.

Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.

Kudos Please... If my post helped.

Hi MacFauz, Why did you put 2.4 in the denominator instead of 1.2 when calculating revenue from 6 cameras ? thanks!

hi,

can we substitute numbers? for example, I tried using the following:

10 is the cost price and therefore 12 is the sales price. Following the information given I get the following:

1- cost = 60*10 = 600
2- revenue = 54*12 = 648
3- refund = 6*5 = 30

therefore, I get the following:

(648 + 30) / 600 = 1.13. Now I know that I should deduct 1 from this but not sure why.

so two questions:

1. can we plug in a number for the cost?
2. why do we need to deduct 1 from the answer above (1.13)?

thanks

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

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

pushpitkc, niks18, generis, yes its me again

Friends please advise if my approach below is correct ? if yes, can you please suggest some other quick method, cause it took me almost 5 minutes to calculate all this

"60 Model X cameras to be sold for $250 each, which represents a 20 percent " i translated into following

Cost of one camera \(1.20x = 250\)

\(x\) = \(208\)

Hence Mark up = \(42\)

Now 54 were sold ---> \(Profit = 42*54 = 2,268\)

6 cameras unsold and returned at 0.50% of its cost = \(104*6 = 624\)

\(Total Profit\) = \(2268-624 = 1,644\)

Dealers Initial Cost for the 60 cameras = \(208*60 = 12,480\)

So , dealer's approximate profit s a percent of the dealer’s initial cost for the 60 cameras is \(\frac{1,644}{12,480}\) = \(0.13 %\)

Thank you

Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Answer: D.

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