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How to solve profit and loss problems.

 This topic is part of the TCS FREE high school mathematics 'How-to Library'. It shows you how to solve profit and loss problems. (See the index page for a list of all available topics in the library.) To make best use of this topic, you need to download the Maths Helper Plus software. Click here for instructions.

### Theory:

The situations described in this topic are from the point of view of the person selling something to someone else.

The 'cost price' is how much the item cost the shop owner.
The 'selling price' is what he is selling it for.

If the shop sells a thing for more than they paid for it, then the difference is the 'profit':

profit = selling price - cost price

If the shop sells a thing for less than they paid for it, then the difference is the 'loss':

loss = cost price - selling price

Both profit and loss can be expressed either in dollars, OR as a percentage of the cost price.

Several problem types based on these formulas are discussed in the examples below.

Example 1:

A shopkeeper buys scientific calculators in bulk for \$15 each. He sells them for \$40 each.

Calculate the profit on each calculator in dollars, and as a percentage of the cost price.

Given: cost price = \$15,  selling price = \$40

profit = selling price - cost price

=   \$40 - \$15

=   \$25

Expressing the profit as a percentage of the cost price:

\$profit

profit% =  ------------------------ × 100%

\$cost price

\$25

=  -------- × 100%    =    166.7%

\$15

Example 2:

A school bookshop sells an outdated biology text book for \$49.35, making a 6% loss.

What was the cost price of the book, and what is the cash value of the loss?

Given: selling price = \$49.35,

loss = 6% of cost price

but,  (cost price) = (selling price) + loss

Expressing as percentages of the cost price:

(cost price) = (selling price)    +    loss

100%                    x%               +      6%

This means that the selling price is (100-6) = 94% of the cost price.

94

selling price  =  -------- × (cost price)

100

So:

100

cost price  =  -------- × (selling price)

94

100

=  -------- × 49.35

94

=  \$52.50

So:                     loss  =  (cost price) - (selling price)

=  \$52.50 - \$49.35

=  \$3.15

### Method:

Maths Helper Plus can solve many kinds of profit and loss problems. It will do calculations showing the working steps, as well as display a labelled diagram.

#### Step 2  Display the parameters box

Press the F5 key to display the parameters box:

You enter the given information into these edit boxes as follows:

edit box 'A' = cost price

edit box 'B' = profit ( To enter a loss, make the value negative. )

You can enter the profit or loss in 'B' as a dollar value, or as a percentage of the cost price.

If 'B' is in dollars, set 'X' to 0.

If 'B' is a percentage of the cost price, set 'X' to 1.

edit box 'C' = selling price

Set any two of A, B and C to the values you are given.

Set the unknown value to zero.

Out of the three edit boxes A, B and C, two will not be zero, and one will be zero.

NOTE: The parameters box diagram shows the correct settings for solving Example 2 from the 'Theory' section above.

'A' = 0, because we are calculating the unknown cost price.

'B' = -6, this is the percentage loss, so it is negative.

'X' = 1, which means that the 'B' value is a percentage of the cost price and not a dollar value.

'C' = 49.35, which is the selling price given.

Click the 'Update' button to refresh the diagram and calculations.