Year 8 - Profit, loss and discounts - Free practice

Year 8 Profit, Loss and Discounts Worksheets & Practice

Apply percentages to real money situations. Worked solutions every time, free.

Quick answer

How do you calculate profit, loss and discounts using percentages?

Profit or loss is the difference between selling price and cost price; express it as a percentage of the cost price. For a mark-up or discount, multiply the original price by the percentage and add or subtract. For example, a 20% discount on $80 gives a selling price of $64.

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Worked examples

Year 8 profit, loss and discounts: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

  1. Example 1 · Profit, loss and percentage of cost price

    Hamid buys a second-hand bicycle for $240 and sells it for $300. Find his profit in dollars and express it as a percentage of the cost price.

    Show worked solution

    Key idea: Profit = selling price − cost price. Percentage profit = profitcost price×100\dfrac{\text{profit}}{\text{cost price}} \times 100.

    1. Find the profit: 300−240=60300 - 240 = 60 So the profit is $60.
    2. Express as a percentage of cost price: 60240×100=25\frac{60}{240} \times 100 = 25

    Answer: Hamid makes a profit of $60, which is 25%25\% of the cost price.

  2. Example 2 · Finding the original price before a discount

    A pair of shoes is on sale for $126 after a 30%30\% discount. What was the original price?

    Show worked solution

    Key idea: After a 30%30\% discount, the sale price is 70%70\% of the original price. Divide the sale price by 0.700.70 to find the original.

    1. Write the equation: original price×0.70=126\text{original price} \times 0.70 = 126
    2. Solve for the original price: original price=1260.70\text{original price} = \frac{126}{0.70}
    3. Calculate: 126÷0.70=180126 \div 0.70 = 180

    Answer: The original price was $180.

  3. Example 3 · Multi-step financial problem: mark-up then discount

    A sports store buys a jacket from a supplier for $80. The store marks the price up by 60%60\% to set the retail price. During a sale, the jacket is sold at a 25%25\% discount off the retail price. Does the store make a profit or a loss on the jacket, and by how many dollars?

    Show worked solution

    Key idea: Apply the mark-up to find the retail price, then apply the discount to find the selling price, then compare to the cost price.

    1. Find the retail price after a 60%60\% mark-up: 80×1.60=12880 \times 1.60 = 128 The retail price is $128.
    2. Find the selling price after a 25%25\% discount: 128×0.75=96128 \times 0.75 = 96 The selling price is $96.
    3. Compare to the cost price: 96−80=1696 - 80 = 16 The selling price is greater than the cost price.

    Answer: The store makes a profit of $16.

Common mistakes

Where students usually lose marks

Using selling price instead of cost price for percentage profit or loss

Profit %=60300×100=20%\% = \dfrac{60}{300} \times 100 = 20\% (using selling price as the denominator)

Percentage profit and percentage loss are always calculated as a percentage of the cost price, not the selling price. Use profitcost price×100\dfrac{\text{profit}}{\text{cost price}} \times 100.

Subtracting the percentage from the price instead of multiplying

180−30%=150%180 - 30\% = 150\% (treating the percentage as if it has the same units as dollars)

Convert the percentage to a decimal multiplier first. A 30%30\% discount means you multiply by 0.700.70, so the sale price is 180×0.70=126180 \times 0.70 = 126.

Finding the original price before a discount by adding the percentage back onto the sale price

Original price =126+30%=126+37.80=163.80= 126 + 30\% = 126 + 37.80 = 163.80 (taking 30%30\% of the sale price instead of the original)

After a 30%30\% discount the sale price is 70%70\% of the original, so divide: 126÷0.70=180126 \div 0.70 = 180. Adding a percentage of the already-reduced price does not undo the discount.

Applying a mark-up and a discount by just combining the percentages

60%−25%=35%60\% - 25\% = 35\% mark-up overall, so selling price =80×1.35=108= 80 \times 1.35 = 108

The two percentages apply to different base amounts, so you cannot simply subtract them. Apply each step in order: first multiply by 1.601.60, then multiply that result by 0.750.75.

Try one yourself

Your turn

A shopkeeper buys a jacket for $150 and sells it at a 35%35\% mark-up. What is the selling price, in dollars?

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About this worksheet

Why we built it

Profit, loss and discounts bring percentages to life in Year 8 by connecting them to everyday financial situations like shopping, selling and budgeting. These worksheets guide students from calculating a simple percentage of an amount through to finding selling prices, percentage profit and discounted costs, with fully worked solutions so students can check exactly where their reasoning went right or wrong.

What's covered

Sub-skills your student will practise

  • ✓Calculating profit and loss in dollars
  • ✓Expressing profit and loss as a percentage of cost price
  • ✓Finding the selling price after a percentage mark-up
  • ✓Calculating a discounted price from a percentage discount
  • ✓Finding the original price before a discount or mark-up
  • ✓Solving multi-step financial problems using mathematical modelling

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Frequently asked

Questions parents ask about profit, loss and discounts

What is the difference between profit and percentage profit?

Profit is the dollar amount gained: selling price minus cost price. Percentage profit expresses that gain as a percentage of the cost price, so it lets you compare how good a deal was regardless of the original price.

How do students find the original price before a discount?

If a price after a 20% discount is $80, then $80 represents 80% of the original. Dividing $80 by 0.8 (or by 80 then multiplying by 100) gives the original price of $100. This reverse-percentage step is a common sticking point in Year 8.

Why do Year 8 students use mathematical modelling for these problems?

Mathematical modelling means setting up a clear representation of a real situation before calculating. For financial problems it helps students identify what is known, choose the right formula and check whether their answer makes sense in context.

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