# Concept of cost price

### Concept of cost price

Most of us are quite comfortable with the concepts of percentages, cost price and sale price, but when we come across a toughie from these topics, we feel lost. Then we go back to the theory but there seems to be nothing new there – nothing new that could potentially help us tackle such questions with ease in the future. The point is, the basic theory of these topics is quite simple – there isn’t anything else to it – but it’s application to questions is an altogether different deal. There are small but critical things that you need to keep in mind, one of which we will discuss today: what is the cost price?

Let’s take a look at this with an official question:

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) 14% profit

Solution:

Here are the various data points:

• 60 cameras bought at 20% markup.
• Selling Price = \$250
• 6 not sold and 50% of initial cost refunded
• Profit/Loss = ?

Now look at the solution:

The cost price per camera = 250/1.2 = 1250/6

The total cost price = (1250/6)*60 = \$12,500

50% of the cost of 6 cameras was returned.

The cost price of 6 cameras = (1250/6)*6 = \$1250

50% of this = 1250/2 = \$625

This means the effective cost price = 12,500 – 625 = \$11,875

If the selling price per camera = \$250, the total selling price = 54 * 250 = \$13500 (only 54 cameras were sold)

Hence, the profit % = [(13500 – 11875) / 11875] x 100 = (1625/11875) x 100 = 13.684%

This gives us approximately 14% as the answer (rounding up). But that is not correct. Before you move ahead, try to figure out the problem with this solution. If you are able to, it means you do understand this topic very well.

Here is the problem with the solution:

The cost price is the total initial cost price. You cannot subtract the refund out of it. The refund is effectively the price at which the 6 cameras were sold. You cannot cancel off your cost price with your sale price and have a smaller cost price. Your initial investment in the transaction is your cost price. When you reduce it by cancelling off some sale price (or refund), you are artificially increasing your profit percentage.

Say, we buy a few thing for \$100. While selling them off, we get \$50 for half of them. We reduce our cost price by \$50 and get \$50 as cost price. For the other half, we sell them for \$60. We say that \$50 is out cost price and \$60 is our selling price. The profit we made is \$10, which is fine. The issue is that our profit percentage is not (10/50) * 100 = 20%. Rather, our profit percentage will be (10/100) * 100 = 10% only, so \$100 would be our actual cost price.

Keeping this in mind, here is the correct algebra solution:

The total cost price = (1250/6)*60 = \$12,500

The total selling price = 54 * 250 + \$625 = \$13,500 + \$625 = \$14,125 (60 cameras were sold, 54 at \$250 each and 6 at 50% of cost price)

The profit = 14,125 – 12,500 = \$1625 (same as before)

The profit percentage = (1625/12,500) * 100 = 13%

Therefore, the answer is (D).

Obviously, we can always use our trusted weighted averages formula here for a quick and efficient solution:

Weighted Averages

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

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