Making calculations easier
🎯 In this topic you will
- Simplify calculations containing decimals and fractions
🧠 Key Words
- factor
- strategies
Show Definitions
- factor: A number that divides another number exactly, with no remainder.
- strategies: Different methods or approaches used to solve a mathematical problem.
When you are calculating using fractions and decimals, you can often make a calculation easier by using different strategies. This section will help you to practise the skills you need to choose the best strategy to use. You will be able to do some of the calculations mentally. For other calculations you will need to write your working. Whatever calculation you do, you must remember to use the correct order of operations.
❓ EXERCISES
1. Work out these calculations. Some working has been shown to help you.
a. $ \left(\tfrac{1}{2}+5.5\right)^{2}-1 $
b. $ \left(3\tfrac{1}{5}-0.2\right)^{3}+23 $
c. $ 6^{2}-\left(3\tfrac{3}{10}+0.7\right) $
2. Work out these calculations. Use the same strategies as in Question $1$.
a. $ 6\times\left(3.25+4\tfrac{3}{4}\right) $
b. $ \left(2.1+4\tfrac{9}{10}\right)^{2} $
c. $ 4\times 3.2-\tfrac{4}{5} $
3. Work out these calculations. Some working has been shown to help you.
a. $ 1.5\times 2.5\times 40 $
b. $ 1.25\times 3\tfrac{1}{2}\times 56 $
c. $ 2.75\times 18 $
👀 Show answer
1b. $ 3\tfrac{1}{5}=3.2 $, so $ (3.2-0.2)^{3}+23=3^{3}+23=27+23=50 $.
1c. $ 3\tfrac{3}{10}+0.7=3.3+0.7=4 $, so $ 6^{2}-4=36-4=32 $.
2a. $ 3.25+4\tfrac{3}{4}=3.25+4.75=8 $, so $ 6\times 8=48 $.
2b. $ 2.1+4\tfrac{9}{10}=2.1+4.9=7 $, so $ 7^{2}=49 $.
2c. $ 4\times 3.2=12.8 $, and $ \tfrac{4}{5}=0.8 $, so $ 12.8-0.8=12 $.
3a. $ 1.5\times 2.5\times 40=(\tfrac{3}{2}\times \tfrac{5}{2})\times 40=\tfrac{15}{4}\times 40=150 $.
3b. $ 1.25=\tfrac{5}{4},\ 3\tfrac{1}{2}=\tfrac{7}{2} $ so $ \tfrac{5}{4}\times \tfrac{7}{2}\times 56=\tfrac{35}{8}\times 56=35\times 7=245 $.
3c. $ 2.75=\tfrac{11}{4} $, so $ \tfrac{11}{4}\times 18=\tfrac{198}{4}=49.5=49\tfrac{1}{2} $.
4. Work out these calculations. Use the same strategies as in Question $3$.
a. $ 1.5 \times 3.5 \times 24 $
b. $ 2.25 \times 1 \tfrac{1}{2} \times 32 $
c. $ 3.75 \times 28 $
👀 Show answer
b. $ 2.25 \times 1 \tfrac{1}{2} \times 32 = \left(\tfrac{9}{4}\times \tfrac{3}{2}\right)\times 32 = \tfrac{27}{8}\times 32 = 27\times 4 = 108 $.
c. $ 3.75 \times 28 = \tfrac{15}{4}\times 28 = 15\times 7 = 105 $.
5. Akeno and Dae use different methods to work out the volume of this cuboid.
Akeno
$\text{Volume} = 2.5 \times 2.5 \times 12$
$= \tfrac{5}{2} \times \tfrac{5}{2} \times 12 = \tfrac{25}{4} \times 12 = 25 \times 3$
$= 75\ \text{cm}^3$
Dae
$\text{Volume} = 2.5 \times 2.5 \times 12$
$2.5 \times 12 = 2 \times 12 + \tfrac{1}{2} \times 12 = 24 + 6 = 30$
$2.5 \times 30 = 2 \times 30 + \tfrac{1}{2} \times 30 = 60 + 15 = 75\ \text{cm}^3$
a. Critique their methods.
b. Whose method do you prefer? Explain why.
c. Whose method would be easier to use when, instead of $12\ \text{cm}$, the length of the cuboid is i. $14\ \text{cm}$ ii. $15\ \text{cm}$? Explain your answers.
👀 Show answer
a. Both methods are correct. Akeno rewrites $2.5=\tfrac{5}{2}$ and uses fractions: $ \tfrac{5}{2}\times \tfrac{5}{2}\times 12 = \tfrac{25}{4}\times 12 = 25\times 3 = 75 $. Dae uses distributive thinking (breaking $2.5$ into $2+\tfrac{1}{2}$) to do easy mental products with $12$, then multiplies by $2.5$ again: $2.5\times 12=30$, then $2.5\times 30=75$. Both are efficient and show clear reasoning.
b. Prefer Dae’s method for mental arithmetic here: halving and doubling with $12$ give tidy numbers ($30$ then $75$) with minimal fraction work.
c.i. For $14\ \text{cm}$: Dae’s method is easier mentally. $2.5\times 14 = 2\times 14 + \tfrac{1}{2}\times 14 = 28 + 7 = 35$, then $2.5\times 35 = 2\times 35 + \tfrac{1}{2}\times 35 = 70 + 17.5 = 87.5\ \text{cm}^3$. Akeno’s fraction method gives $ \tfrac{25}{4}\times 14 = \tfrac{350}{4} = 87.5 $, but involves quarters that don’t cancel.
ii. For $15\ \text{cm}$: Dae’s method is again slightly easier mentally. $2.5\times 15 = 30 + 7.5 = 37.5$, then $2.5\times 37.5 = 75 + 18.75 = 93.75\ \text{cm}^3$. (Akeno: $ \tfrac{25}{4}\times 15 = \tfrac{375}{4} = 93.75 $—correct but requires fraction-to-decimal conversion.)
6. Work out these calculations. Some working has been shown to help you.
a. $ 0.28 \times 5^{2} $
b. $ 0.7 \times 4 \tfrac{2}{7} $
c. $ 1.3 \times \left( 4^{3} - 4 \right) $
👀 Show answer
b. $ 0.7=\tfrac{7}{10},\quad 4\tfrac{2}{7}=\tfrac{30}{7} \Rightarrow \tfrac{7}{10}\times \tfrac{30}{7}=\tfrac{30}{10}= \mathbf{3} $.
c. $ 4^{3}=64,\ 64-4=60,\ 1.3=\tfrac{13}{10} \Rightarrow \tfrac{13}{10}\times 60= \mathbf{78} $.
7. Work out these calculations. Use the same strategies as in Question $6$.
a. $ 1.6 \times \tfrac{5}{8} $
b. $ 0.81 \times 1 \tfrac{1}{9} $
c. $ 0.328 \times 5^{3} $
👀 Show answer
b. $ 1\tfrac{1}{9}=\tfrac{10}{9},\ 0.81=\tfrac{81}{100} \Rightarrow \tfrac{81}{100}\times \tfrac{10}{9}=\tfrac{90}{100}= \mathbf{\tfrac{9}{10}} $.
c. $ 5^{3}=125 \Rightarrow 0.328\times 125=\tfrac{328}{1000}\times 125=\tfrac{41000}{1000}= \mathbf{41} $.
8. The diagram shows a path.
The length of the path is $3.2 \, m$.
The width of the path is $ \tfrac{5}{8} \, m$.
What is the area of the path?
👀 Show answer
$ = 3.2 \times \tfrac{5}{8} $
$ = \tfrac{32}{10} \times \tfrac{5}{8} = \tfrac{160}{80} = 2 $.
✅ The area of the path is $ \mathbf{2 \, m^2} $.
🧠 Think like a Mathematician
Task: Explore strategies for working with decimals and fractions, and decide which form makes calculations easier.
Questions:
👀 show answer
- a) - $0.2^2 = 0.04$. Multiply by $1\dfrac{2}{3} = \dfrac{5}{3}$: $0.04 \times \dfrac{5}{3} = \dfrac{0.2}{3} \approx 0.0667$. - $0.75^2 = 0.5625$. Multiply by $2\dfrac{4}{9} = \dfrac{22}{9}$: $0.5625 \times \dfrac{22}{9} = \dfrac{12.375}{9} \approx 1.375$.
- b) It is often best to keep fractions when the original numbers are simple fractions, as this avoids rounding errors. Decimals may be easier when squared decimals are exact (like 0.75).
- c) Strategy: convert recurring or simple fractions to decimals only when exact (e.g., 0.25, 0.5, 0.75). Otherwise, keep them as fractions. This balances accuracy and simplicity.
- d)$0.8^2 = 0.64$. Multiply by $7\dfrac{1}{2} = \dfrac{15}{2}$: $0.64 \times \dfrac{15}{2} = \dfrac{9.6}{2} = 4.8$.
❓ EXERCISES
10. The area of this triangle is $1 \tfrac{1}{20}\, m^2$.
Work out the length of the triangle.
👀 Show answer
Given: $ A = 1 \tfrac{1}{20} = \tfrac{21}{20} \, m^2 $, height $ = 0.9 \, m $.
$ \tfrac{21}{20} = \tfrac{1}{2} \times \text{base} \times 0.9 $.
$ \tfrac{21}{20} = 0.45 \times \text{base} $.
$ \text{base} = \tfrac{21}{20} \div 0.45 = \tfrac{21}{20} \times \tfrac{100}{45} = \tfrac{2100}{900} = \tfrac{7}{3} \approx 2.33 \, m $.
✅ The length of the triangle is $ \mathbf{2.33 \, m} $.
🧠 Think like a Mathematician
Task: Explore strategies for working with square roots and fractions, and decide which form makes calculations clearer and more accurate.
Questions:
👀 show answer
- a) - $\sqrt{2.25} = 1.5$. So $1.5 + 5.5 = 7$. - $\sqrt{12.25} = 3.5$. So $10 - 3.5 = 6.5$.
- b) Both answers are exact as decimals (7 and 6.5). Fractions are not necessary here, so decimals are clearer.
- c) Strategy: simplify square roots of decimals by converting to fractions where possible (e.g., $2.25 = \dfrac{9}{4}$, so $\sqrt{2.25} = \dfrac{3}{2}$). This avoids mistakes and shows exact values. Use decimals only if the square root is a terminating decimal.
- d)$\sqrt{\dfrac{17}{9}} = \dfrac{\sqrt{17}}{3}$. So $4.25 \times \dfrac{\sqrt{17}}{3} = \dfrac{4.25\sqrt{17}}{3}$. As a decimal: $4.25 \times 1.374... \approx 5.84$.
❓ EXERCISES
12. Hiromi uses this formula in his science lesson: $ K = \tfrac{1}{2} m v^{2} $
🔎 Reasoning Tip
Remember: $\tfrac{1}{2}mv^2$ means
$\tfrac{1}{2} \times m \times v^2$
a. Use the formula to work out the value of $K$ when $ m = 2 \tfrac{1}{4}$ and $ v = 1 \tfrac{1}{3} $.
b. This is how Hiromi rearranges the formula to make $v$ the subject: $ \tfrac{1}{2} m v^{2} = K \Rightarrow m v^{2} = 2K \Rightarrow m v = \sqrt{2K} \Rightarrow v = \tfrac{\sqrt{2K}}{m} $
c. Use your values of $K, m,$ and $v$ from part a to show that Hiromi has rearranged the formula incorrectly. Rearrange the formula correctly to make $v$ the subject.
d. Use your formula in part c to work out the value of $v$ when $K=18$ and $m=25$. Check your answer is correct by substituting $m=25$ and your value for $v$ into the original formula.
👀 Show answer
a. $ m = 2 \tfrac{1}{4} = \tfrac{9}{4}, \quad v = 1 \tfrac{1}{3} = \tfrac{4}{3} $. $ K = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} \times \tfrac{9}{4} \times \left(\tfrac{4}{3}\right)^{2} = \tfrac{1}{2} \times \tfrac{9}{4} \times \tfrac{16}{9} = \tfrac{1}{2} \times \tfrac{16}{4} = \tfrac{1}{2} \times 4 = 2 $. ✅ $ K = 2 $.
b. Hiromi’s rearrangement is incorrect because he lost the square when solving for $v$. Correctly: $ K = \tfrac{1}{2} m v^{2} \Rightarrow v^{2} = \tfrac{2K}{m} \Rightarrow v = \sqrt{\tfrac{2K}{m}} $.
c. Substituting $K=2, m=\tfrac{9}{4}$: $ v = \sqrt{\tfrac{2K}{m}} = \sqrt{\tfrac{4}{9/4}} = \sqrt{\tfrac{16}{9}} = \tfrac{4}{3} $, which matches the given $v$. This shows Hiromi’s rearrangement was wrong.
d. For $K=18, m=25$: $ v = \sqrt{\tfrac{2K}{m}} = \sqrt{\tfrac{36}{25}} = \tfrac{6}{5} = 1.2 $. Check: $ K = \tfrac{1}{2} m v^{2} = \tfrac{1}{2} \times 25 \times (1.2)^{2} = 12.5 \times 1.44 = 18 $. ✅ Correct.
⚠️ Be careful! Fractions & Decimals in Calculations
- Don’t mix decimals and fractions without thinking. Choose whichever form makes the calculation cleaner. For example, $1.5=\tfrac{3}{2}$ is easier when paired with $16$.
- Follow BIDMAS/BODMAS carefully. Work out brackets and powers before multiplying or adding.
- Use fractions to avoid messy decimals. $0.32 \times 25$ is neater as $\tfrac{32}{100}\times25=\tfrac{32}{4}=8$.
- Don’t round too early. Keep exact values until the end, then simplify or convert to decimals if needed.
- Check sense with estimation. Roughly round decimals/fractions to see if your answer is reasonable (e.g. $2.75\approx3$ so $(\tfrac{1}{4}+2.75)^2+2$ should be close to $11$).
- Square and cube correctly. Apply indices to the whole bracketed expression, not to parts separately.