VT · Fractions: Four Operations

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In each set, one thing changes and everything else stays the same. Work them out in order and look for the pattern — the last line tells you what to notice.

①Multiplying – the first fraction stays $\tfrac{3}{4}$, the second changeschanging: the second fraction
Work out each product, giving your answer in its lowest terms.
$\dfrac{3}{4}\times\dfrac{1}{2}$
=
$\dfrac{3}{4}\times\dfrac{2}{3}$
=
$\dfrac{3}{4}\times\dfrac{4}{5}$
=
$\dfrac{3}{4}\times\dfrac{8}{9}$
=
Every answer is smaller than $\tfrac{3}{4}$. Why does multiplying by a fraction less than 1 make it smaller?
②Dividing – the first fraction stays $\tfrac{1}{2}$, the divisor changeschanging: the divisor
Work out each division, giving your answer in its lowest terms.
$\dfrac{1}{2}\div\dfrac{1}{4}$
=
$\dfrac{1}{2}\div\dfrac{1}{2}$
=
$\dfrac{1}{2}\div\dfrac{3}{4}$
=
$\dfrac{1}{2}\div 2$
=
Dividing by a number less than 1 makes the answer bigger; dividing by a number greater than 1 makes it smaller. Where does the turning point happen?
③Adding – the first fraction stays $\tfrac{1}{2}$, the second changeschanging: the second fraction
Work out each sum, giving your answer in its lowest terms.
$\dfrac{1}{2}+\dfrac{1}{3}$
=
$\dfrac{1}{2}+\dfrac{1}{4}$
=
$\dfrac{1}{2}+\dfrac{1}{6}$
=
$\dfrac{1}{2}+\dfrac{1}{5}$
=
You need a common denominator each time. How do you choose it from the two denominators?
④Same two fractions $\tfrac{3}{4}$ and $\tfrac{1}{2}$ – the operation changeschanging: the operation
Work out each one, giving your answer in its lowest terms.
$\dfrac{3}{4}+\dfrac{1}{2}$
=
$\dfrac{3}{4}-\dfrac{1}{2}$
=
$\dfrac{3}{4}\times\dfrac{1}{2}$
=
$\dfrac{3}{4}\div\dfrac{1}{2}$
=
Same numbers, four operations. Which gives the biggest answer? Why does $\div$ give more than $\times$ here?

Answers · Fractions: Four Operations

Variation practice
① Multiplying – the first fraction stays $\tfrac{3}{4}$, the second changes
$\dfrac{3}{4}\times\dfrac{1}{2}$: $\dfrac{3}{8}$$\dfrac{3}{4}\times\dfrac{2}{3}$: $\dfrac{1}{2}$$\dfrac{3}{4}\times\dfrac{4}{5}$: $\dfrac{3}{5}$$\dfrac{3}{4}\times\dfrac{8}{9}$: $\dfrac{2}{3}$
② Dividing – the first fraction stays $\tfrac{1}{2}$, the divisor changes
$\dfrac{1}{2}\div\dfrac{1}{4}$: 2$\dfrac{1}{2}\div\dfrac{1}{2}$: 1$\dfrac{1}{2}\div\dfrac{3}{4}$: $\dfrac{2}{3}$$\dfrac{1}{2}\div 2$: $\dfrac{1}{4}$
③ Adding – the first fraction stays $\tfrac{1}{2}$, the second changes
$\dfrac{1}{2}+\dfrac{1}{3}$: $\dfrac{5}{6}$$\dfrac{1}{2}+\dfrac{1}{4}$: $\dfrac{3}{4}$$\dfrac{1}{2}+\dfrac{1}{6}$: $\dfrac{2}{3}$$\dfrac{1}{2}+\dfrac{1}{5}$: $\dfrac{7}{10}$
④ Same two fractions $\tfrac{3}{4}$ and $\tfrac{1}{2}$ – the operation changes
$\dfrac{3}{4}+\dfrac{1}{2}$: $\dfrac{5}{4}$$\dfrac{3}{4}-\dfrac{1}{2}$: $\dfrac{1}{4}$$\dfrac{3}{4}\times\dfrac{1}{2}$: $\dfrac{3}{8}$$\dfrac{3}{4}\div\dfrac{1}{2}$: $\dfrac{3}{2}$
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