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705-805 Level|   Percent and Interest Problems|                                 
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A photography dealer ordered 60 Model X cameras to be sold 30 Sep 2020, 13:36
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Video solution with minimum math and maximum reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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A photography dealer ordered 60 Model X cameras to be sold 08 Apr 2021, 04:23
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This question can be solved using the concept of weighted average.

Good cameras = 54 (Profit of 20%)
Bad cameras = 6 (Loss of 50%)
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A photography dealer ordered 60 Model X cameras to be sold 15 Feb 2022, 22:57
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pgmat
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
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Method 1: - The long way:

SP of each camera is 120% of the cost
=> Cost of each camera = $(250/1.2) = $(1250/6)
Total cost = (1250/6) x 60 = $12500
SP from 54 cameras = 250 x 54 = $13500
The remaining 6 cameras were returned for half the cost price, i.e. (1250/6)/2 = $(625/6)
Price obtained from these 6 = 6 x 625/6 = $625
Total SP = 13500 + 625 = $14125
Profit % = (14125 - 12500)/12500 x 100 = 13%

Method 2: - The short way:

In questions that ask for percent profit, the actual values aren't required - we just need to know the ratio of the prices and the ratio of the quantities
54 cameras were sold at some profit and the remaining 6 were returned, incurring a loss => the quantity ratio is 54 : 6 = 9 : 1
The 54 cameras were sold at 20% profit and the 6 were effectively sold at 50% loss

Thus, the net percent profit or loss can be simply obtained by a weighted average (note that all cameras have the same CP):

\(\frac{[9 * 20 + 1 * (-50)]}{(9 + 1)}\) = 13%

Answer B
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A photography dealer ordered 60 Model X cameras to be sold 17 Feb 2022, 04:35
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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?

I agree with the different approaches discussed here, especially the weighted average one. So thought of discussing some time-saving tips here.

The cost price of each model X camera is the same. So let it be CP.

Note:
Whenever you solve questions based on 2 or more different transactions , it is very easy to find the overall profit/Loss % if the cost price is the same for every products involved.

Do you need to find CP here to answer this question? Not really.

Finding CP and solving this question would cost you a lot of time. Let us try a logical approach instead.

Overall profit/loss % = (Overall profit/loss)/Total_cost * 100

Overall profit /loss can be expressed in terms of CP as the profit% and loss % are given.

Since CP is common in both numerator and denominator, it will get canceled during the calculation. So, the value of CP will not be needed to get the final answer.

This will give you the freedom to completely disregard CP or plugin a suitable value of your preference for CP. Either way around, the final answer will not change.

Let's analyze in detail .

54 cameras are sold at 20 % profit and 6 cameras are sold at a loss of 50 %.

Overall profit/loss = 20 % of (54*CP) - 50 % of (6 * CP)

Profit /Loss % = (20 % of (54CP) - 50 % of (6 CP))/ (60 CP ) *100

CP will get canceled and it will not affect your final answer. Hence, we don't need to find the value of CP here.

Profit /Loss % = (20 % of 54 - 50 % of 6 )/ 60 *100 = (20*54 - 50* 6) /60 which is same as the weighted average application.

= (20*9 - 50)/10 =130/10 = 13 % profit

Option D is the answer.

Thanks,

Clifin J Francis,
GMAT QUANT SME
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A photography dealer ordered 60 Model X cameras to be sold 01 Mar 2022, 18:13
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Given: 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.

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

Dealer's initial cost of each camera = $250/1.2

Dealer's initial cost of 60 cameras = $250/1.2 * 60 = $12500

Total sales of 60 cameras = (60-6)*$250 + 6*50%*$250/1.2 = $14125

Percentage profit on 60 cameras = ($14125 - $12500) / $12500 * 100% = 13%

IMO D
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A photography dealer ordered 60 Model X cameras to be sold 12 Aug 2023, 06:51
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There is a very smooth shortcut to this problem that requires an understanding of two concepts/methods :)

1. Weighted average
2. Number putting

First, use weighted average to simplify the problem...

Essentially the question says that - He gained 20% on 54 cameras and lost 50% on 6 cameras

That means 9 parts earned 20% and 1 part lost 50%

Now, let's put numbers

Let the total number of cameras = 10

Cost of each camera = $10

Hence, total profit = $2*9 -$5*1 = $13 or 13%
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A photography dealer ordered 60 Model X cameras to be sold 29 May 2024, 21:38
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­A classic mixture wherein you use the percent revenue as your markers:

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A photography dealer ordered 60 Model X cameras to be sold 02 Nov 2024, 12:32
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Quote:
Quote:
VeritasPrepKarishma

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: https://www.gmatclub.com/forum/veritas- ... d-averages

Weighted Average Formula:

Cavg = (C1*w1 + C2*w2)/(w1 + w2)

C1 and C2 represent the quantity which we want to average so they will be profit/loss here.
C1 = 20% = .2
C2 = -50% (loss) = -.5
The weights, given by w1 and w2, are the cost prices.
Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1
So w1 =9 and w2 = 1

Avg Profit/Loss = (.2*9 + (-.5)*1)/(9 + 1) = (.2*9 + (-.5)*1)/10

Get more details on this concept from the link given in my post above.
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Hi Karishma,

I couldn't infer why we haven't taken

{1.2(9) + 0.5(1)}/ 10

Isn't the overall SP 1.2 times the cost, and the cost incurred for 1 part 0.5? (The cost divided between the manufacturer and seller)
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A photography dealer ordered 60 Model X cameras to be sold 30 Jul 2025, 08:06
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I think it can be done in this way as well

The 60 camera sold at 20% profit which means 6/5 in multiplying factor so

Cp=5
SP=6

As he sold 54 cameras at 20% profit which means that he sold
54*6 is the sp of 54 cameras
And he refund the 6 cameras to the manufacturer at 50% of it original price which is 6*2.5

So cp of 54 cameras= 54*5=270
cp of 6 cameras= 6*5 = 30

And

Sp of 54 cameras =54*6= 324
Sp of 6 cameras = 6*2.5= 15

So P% = total Sp-total Cp/cp*100

Profit=339-300= 39
P%= 39/300*100
P%= 13%

I hope it is helpful.
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A photography dealer ordered 60 Model X cameras to be sold 26 Sep 2025, 06:21
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SP = $250
CP = $\(\frac{5}{6}*250\)

PFT = \(\frac{{(250*54 = 600) - ([fraction]5}{6}*60*250}/(\frac{5}{6}*60*250 )[/fraction]\)
pgmat
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
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