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A man sold two cameras for £240 each. One was sold for a profit of 20%, and one for a loss of 20%. What was his overall profit or loss.

I found this question and its answer online. However I am confused by the way they did it. For example for the profit, why do they do this: 100/120, and the loss like this: 100/80? why isnt the 100 in the denominator???

Then there is this other question, which is fairly similar: A product in a shop is reduced in price by 20%. At this reduced price the shopkeeper makes only 4% profit. What percentage profit (to the nearest whole percent) does the shopkeeper make at its normal selling price?

For this question, i tried to do cross multiplication by setting up a ratio, however it does not work since there are 2 variables in 1 equation

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    It would be nice if you would give an reply. Otherwise we don´t know if the answers has helped. – 2017-01-26

2 Answers 2

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On profit -

$\frac{120}{100} × x = 240$

$ x = 240 × \frac{100}{120}$

On loss -

$\frac{80}{100} × x = 240$

$ x = 240 × \frac{100}{80}$

Hope now clears to you.

Different methods in case of Profit-

Method 1-

Let CP = Cost Price and SP = Selling Price

$\frac{(100 + \text{Gain})}{100}$ × CP = SP.

Method 2-

CP + $\frac{20}{100}$ × CP = SP

In case of loss use subtraction.

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    If any doubt please ask. – 2017-01-26
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    thank u. out of curiosuty why dont we do: 120/100x +x?? so first arent we suppose to find what 120% of the price is then add it the actual price? – 2017-01-26
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    @Kanwalijt Singh – 2017-01-26
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I go directly to your second question. You can use the definition of percentage profit:

$$\text{profit (percentage)}=\frac{S-P}{P}$$

where $S$ is the selling price and $P$ the purchase price.

Thus the equation is

$\frac{(1-0.2)S-P}{P}=\frac{0.8\cdot S-P}{P}=0.04$

If you reduce something by $20\%$ then you have to substract it from $100\%$.

$100\%-20\%=1-\frac{20}{100}=1-0.2=0.8$

$0.8S-P=0.04P$

$0.8S=1.04P$

$S=\frac{1.04}{0.8}P$

The profit for the normal price is

$\frac{ S-P}{P}$

Inserting the expression for S

$\frac{ \frac{1.04}{0.8}P-P}{P}$

P is cancelling out.

$ \frac{1.04}{0.8}-1=0.3$