Problem Solving

A photography dealer ordered 60 model X cameras to be sold for $250 each, which represent a 20% markup over the dealer's initial cost for each camera. Of the cameras ordered , 6 were never sold and were returned to manufacturer for a refund of 50% of the dealer's initial cost.what was the dealer's aprrox profit or loss as a percent of dealers initial cost of 60 cameras.
a) 7%loss
b) 13%loss
c) 7% profit
D) 13%profit
e) 15 % profit

Was able to solve it but took me around 4 minutes. Is there any shortcut??

This is a pretty long one to work on. The key is to work quickly between individual prices and group prices. What was your method of solving?

This was my way of doing it. I took just more than 2 minutes but this was due to the final step which required lengthy calculations.

The dealer plans to cell the cameras for USD 250 which is 20% more than dealer's intial cost.

Let the initial cost be "a"

Therefore, a + 0.2a = 250

Solve for "a" , a = 208.

Therefore initial cost = 208 * 60 = $ 12480.

The dealer sold 54 cameras at $ 250 = $ 13500

The dealer sold 6 cameras at $ 104 (50% refund on initial cost) = $ 624

Total selling price = $ 14124.

Profit = (14124 - 12480)/ 12480 = 13 % profit.

Thanks all

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

Since the question asks for a PERCENTAGE, ignore the numbers given.
Plug in values that satisfy the given conditions and make the math easy.
The fraction of cameras not sold = 6/60 = 1/10.

Let the number of cameras ordered = 10.
Let the cost per camera = 10.
Total cost = 10*10 = 100.

The number not sold = (1/10)10 = 1.
For this one camera, the refund received = .5(10) = 5.

The number of cameras sold = 9.
With a markup of 20%, the selling price = 12.
Total revenue = 9*12 = 108.

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

The correct answer is D.

By plugging in our own values that satisfy all of the given conditions, the correct answer here can be determined MUCH more quickly.
Essentially, the problem above is no different from the following:

A photography dealer ordered x cameras to be sold for y dollars each, which represents a 20 percent markup over the dealer's initial cost for each camera. Of the cameras ordered, 1/10 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 cameras?

Seeing this problem I just had a doubt whether there will be any on screen calculator that can be used in the exam?? Please let me know.

On screen calculator will be available for the IR section but not for the Quant section.

Hi,

According to me answer is E. Below is the explanation

Let x be the C.P

x + 0.2x = 250

Initial Cost of 60 Cameras = 60*250/1.2
Revenue = 60*250
Loss encountered for 6 cameras = 6*250*0.5/1.2

Profit = 60*250 - 60*250/1.2 - 6*250*0.5/1.2

Divide the profit by initial cost of 60 cameras and the answer will be E.

Can u please tell me where i am wrong with respect to you guys as your answer is coming as D

// Quote//

Profit = 60*250 - 60*250/1.2 - 6*250*0.5/1.2

//Unquote//

This equation should be as follows :

Profit = 54*250 + 6*250*0.5/1.2 - 60*250/1.2

See, out of 60 cameras, the dealer sold 54 at $ 250. He returned the remaining cameras for 50% of the intial cost. So basically he " sold" it back to the manufacturer at half the initial cost. So to calculate the profit, you add these two selling prices and then subtract the initial price from this sum.

prat_agl wrote:A photography dealer ordered 60 model X cameras to be sold for $250 each, which represent a 20% markup over the dealer's initial cost for each camera. Of the cameras ordered , 6 were never sold and were returned to manufacturer for a refund of 50% of the dealer's initial cost.what was the dealer's aprrox profit or loss as a percent of dealers initial cost of 60 cameras.
a) 7%loss
b) 13%loss
c) 7% profit
D) 13%profit
e) 15 % profit

Was able to solve it but took me around 4 minutes. Is there any shortcut??

One more shortcut:

Of the 60 cameras, 54 earn 20% and 6 lose 50%.
Average earnings = (54*20 - 6*50)/60 = 780/60 = 13.

The correct answer is D.

Hi,

For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula:
Income: 54*1.2C + 6*0.5C
Cost: 60*C
Then percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.