3.07 Decimals in the real world
Once we are comfortable performing operations with decimals, we can think about how to manipulate decimal quantities that we come across in the real world. Exchanging money, measuring lengths and weights, and recording times are all areas that make use of decimal numbers.
Worked example
Lucy pulls in to a petrol station that advertises unleaded petrol for $134.8$134.8 cents per litre. She fills up her car with $41.57$41.57 litres of petrol and picks up a $\$5.95$$5.95 bottle of engine oil before paying the cashier. How much will Lucy have to pay? Give your answer in dollars, to the nearest cent.
Think: We know the cost of petrol in cents per litre, and the amount of petrol in litres that Lucy gets. We also know the cost in dollars of the engine oil. Our goal is to combine these quantities to get the total cost in dollars.
Do: A single litre of petrol costs $134.8$134.8 cents, which is $\frac{134.8}{100}=\$1.348$134.8100=$1.348. So the cost in dollars for $41.57$41.57 litres will be given by the product $1.348\times41.57$1.348×41.57.
Next, we can add the cost of the engine oil, and round the total to two decimal places. The working out for this calculation is shown below.
| $\text{Total cost }$Total cost | $=$= | $\text{petrol price }\times\text{petrol purchased }+\text{engine oil price }$petrol price ×petrol purchased +engine oil price |
|
| $=$= | $\frac{134.8}{100}\times41.57+5.95$134.8100×41.57+5.95 |
Substitute the given information |
|
| $=$= | $1.348\times41.57+5.95$1.348×41.57+5.95 |
Simplify the fraction |
|
| $=$= | $56.03636+5.95$56.03636+5.95 |
Evaluate the multiplication |
|
| $=$= | $61.98636$61.98636 |
Evaluate the addition |
|
| $=$= | $61.99$61.99 |
Round to the nearest cent |
The total cost for the petrol and the engine oil is $\$61.98636$$61.98636, which is $\$61.99$$61.99 when rounded to the nearest cent.
Reflect: Instead of converting the price of petrol to dollars per litre, we could have found the total cost in cents and converted to dollars at the end. Since we are working in cents rather than dollars, we will round to the nearest whole number rather than to two decimal places. The working out for this method is shown below.
| $\text{Total cost }$Total cost | $=$= | $\text{petrol price }\times\text{petrol purchased }+\text{engine oil price }$petrol price ×petrol purchased +engine oil price |
|
| $=$= | $134.8\times41.57+5.95\times100$134.8×41.57+5.95×100 |
Both quantities are in cents |
|
| $=$= | $5603.636+595$5603.636+595 |
Evaluate each multiplication separately |
|
| $=$= | $6198.636$6198.636 |
Evaluate the addition |
|
| $=$= | $6199$6199 |
Round to the nearest whole number |
Our total cost is $6199$6199 cents, which is the same as $\$61.99$$61.99, as expected.
The solution to many real world problems will eventually involve some kind of calculation, but there is a lot we can do before and after this calculation that can make us more confident our answer is correct.
- What are the quantities that we are combining?
- What units do we expect the answer to have?
- What operations will combine the relevant quantities to produce the expected units?
- What magnitude do we expect the answer to have?
- Does the answer we calculate seem appropriate in the context?
Practice questions
Question 1
Harry buys an item from the school canteen for $\$3.20$$3.20. If he pays for it with a five dollar note, how much change will he get back?
Question 2
How many $0.38$0.38 L bottles can be filled from a barrel which holds $41.8$41.8 L?
Question 3
At midnight, the temperature in Darwin is $29.6$29.6 degrees Celsius.
Each hour after that the temperature decreases by $2.34$2.34 degrees until the sun comes up.
What is the temperature $4$4 hours after midnight?