Decimal Problems
Decimal Problems brings adding, subtracting, multiplying and dividing decimals together in worded questions, the last subtopic of Decimals in NSW Year 7 maths (NESA outcome MA4‑FRC‑C‑01). Choose the operation from the words, estimate, calculate, then check that the answer makes sense.
You will work with prices, petrol sold in cents per litre, gymnastics and diving scores, a bus cost shared by a class, trees spaced along a street and the profit a canteen makes, rounding to the nearest cent where needed. Each answer is then checked against the first estimate. It is part of Year 7 Decimals.
Every question with a fully worked solution.
- Decimal Problems - Video - Solving Word Problems - Decimal Operations Watch
Theory
To solve a decimal problem, choose the operation from the words, estimate, calculate, then check that the answer makes sense. This Year 7 maths page shows the key words for each operation and works through change at the canteen, fuel costs, perimeters, spacing, averages and profit.
Step 1. Read and choose the operation. Look for the key words in the table in Key facts. Some problems need two or more steps.
Step 2. Estimate with friendly numbers: \(5 \times \$2.45 \approx 5 \times \$2.50 = \$12.50\).
Step 3. Calculate. Line up the points to add or subtract; split a decimal to multiply (\(5 \times 2.45 = 10 + 2.25 = 12.25\)); make the divisor a whole number to divide.
Step 4. Check and answer in words, with units. Round money to the nearest cent.
Units. \(100\) cents \(= \$1\). Litres \(\times\) cents per litre gives cents; divide by \(100\) to change cents to dollars.
Key words and the operation they usually mean:
| Key words | Operation |
|---|---|
| total, altogether, sum | \(+\) |
| difference, how much more, change from | \(-\) |
| each, lots of, per litre times litres | \(\times\) |
| share equally, cost of one, how many fit, per minute | \(\div\) |
| average: add the values, divide by how many | \(+\) then \(\div\) |
| profit: money taken minus cost | \(-\) |
How to solve a decimal problem
- Read the question and pick the operation(s).
- Estimate with friendly numbers.
- Calculate one step at a time.
- Check against the estimate and answer in words with units.
| \(5 \times 2\) | \(=\) | \(10\) |
| \(21.5 \div 10\) | \(=\) | \(2.15\) |
There are \(10\) one-way trips. It is \(2.15\) km from home to work.
| \(2 \times 52.4 + 2 \times 18.6\) | \(=\) | \(104.8 + 37.2\) |
| \(\) | \(=\) | \(142\) |
\(142\) m of fencing.
| \(3 \times 3.45\) | \(=\) | \(9 + 1.35 = 10.35\) |
| \(10.35 + 2.80\) | \(=\) | \(13.15\) |
| \(20.00 - 13.15\) | \(=\) | \(6.85\) |
Estimate \(20 - (3 \times 3.5 + 3) = 6.5\). Sam gets \(\$6.85\) change.
| \(2.5 \times 8\) | \(=\) | \(20\) |
| \(20 \times 195.5\) | \(=\) | \(3910\) |
| \(3910 \div 100\) | \(=\) | \(39.10\) |
\(20\) L costs \(3910\) cents, which is \(\$39.10\). Check: \(20 \times 200 = 4000\) cents \(= \$40\).
| \(30.6 - 1.8 - 1.8\) | \(=\) | \(27\) |
| \(27 \div 6\) | \(=\) | \(4.5\) |
\(7\) posts have \(6\) gaps. The posts are \(4.5\) m apart, as in the first diagram.
| \(48 - 6\) | \(=\) | \(42\) |
| \(42 \times 1.50\) | \(=\) | \(63.00\) |
| \(63.00 - 38.40\) | \(=\) | \(24.60\) |
Estimate \(60 - 40 = 20\). The profit is \(\$24.60\).
Common pitfalls
Frequently asked questions
How do you know which operation to use?
Look for key words: total means add, difference means subtract, each means multiply, share means divide.
How do you change cents to dollars?
Divide by \(100\). \(3910\) cents \(= \$39.10\).
How do you find an average?
Add the values, then divide by how many there are. \((7.6 + 8.4 + 9.5) \div 3 = 8.5\).
How many gaps are there between posts?
One fewer than the number of posts. \(7\) posts have \(6\) gaps.
Why estimate in a word problem?
It shows roughly what the answer should be, so you can spot mistakes.