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2n ESO · Fractions Revision 29 Sep 2026

14. Revision exercises for the exam

Homework to prepare for the exam: equivalence, ordering, adding and subtracting, multiplying and dividing, powers and combined operations.

Exercises to prepare for the exam, one for each part of the exam topics. Try them before opening the solution.

1. Find $R$ so that they are equivalent

f) $\displaystyle \dfrac{R}{18} = \dfrac{25}{45}$

Simplify the known fraction: $\dfrac{25}{45} = \dfrac{5}{9}$. From $9$ to $18$ we multiply by $2$, so the numerator too:

$$\dfrac{R}{18} = \dfrac{5}{9} = \dfrac{5 \cdot 2}{9 \cdot 2} \;\Longrightarrow\; R = 10$$

Cross-multiplying gives the same:

$$45 \cdot R = 25 \cdot 18 \;\Longrightarrow\; R = \dfrac{25 \cdot 18}{45} = \dfrac{5^2 \cdot 2 \cdot 3^2}{3^2 \cdot 5} = 10$$
$$R = 10$$
g) $\displaystyle \dfrac{120}{R} = \dfrac{40}{24}$

Simplify the known fraction by dividing by $8$:

$$\dfrac{40}{24} = \dfrac{5}{3}$$

Cross-multiply:

$$\dfrac{120}{R} = \dfrac{5}{3} \;\Longrightarrow\; 5 \cdot R = 3 \cdot 120 \;\Longrightarrow\; R = \dfrac{360}{5} = 72$$

Check: $\dfrac{120}{72} = \dfrac{5}{3}$.

Correction to the board. In class $\dfrac{40}{24}$ was simplified to $\dfrac{8}{3}$, giving $R = 45$. But $\dfrac{40}{24} = \dfrac{5}{3}$ (dividing top and bottom by $8$), and $R = 45$ would give $\dfrac{120}{45} = \dfrac{8}{3}$, which is not equal to $\dfrac{40}{24}$.

$$R = 72$$

2. Order from smallest to largest

e) $\displaystyle \dfrac{5}{18}, \quad -\dfrac{2}{15}, \quad -\dfrac{1}{9}, \quad \dfrac{5}{36}$

$\text{LCM}(18, 15, 9, 36) = 180$. Scale up:

$$\dfrac{50}{180}, \quad -\dfrac{24}{180}, \quad -\dfrac{20}{180}, \quad \dfrac{25}{180}$$

With the same denominator the numerators decide: $-24 < -20 < 25 < 50$. Among negatives, the one further from zero is smaller.

$$-\dfrac{2}{15} < -\dfrac{1}{9} < \dfrac{5}{36} < \dfrac{5}{18}$$
f) $\displaystyle \dfrac{7}{18}, \quad \dfrac{7}{4}, \quad \dfrac{23}{15}, \quad \dfrac{7}{18}$

$\text{LCM}(18, 4, 15) = 180$. Scale up:

$$\dfrac{70}{180}, \quad \dfrac{315}{180}, \quad \dfrac{276}{180}, \quad \dfrac{70}{180}$$

The first and last are the same fraction, so they get an equals sign.

$$\dfrac{7}{18} = \dfrac{7}{18} < \dfrac{23}{15} < \dfrac{7}{4}$$

3. Work out

i) $\displaystyle \dfrac{6}{7} - \left(\dfrac{3}{4} + \dfrac{5}{6}\right) - \left(-\dfrac{7}{5}\right)$

Remove the brackets: the minus in front changes the sign of everything inside.

$$\dfrac{6}{7} - \dfrac{3}{4} - \dfrac{5}{6} + \dfrac{7}{5}$$

$\text{LCM}(7, 4, 6, 5) = 420$:

$$\dfrac{360 - 315 - 350 + 588}{420} = \dfrac{283}{420}$$
$$\dfrac{283}{420}$$
j) $\displaystyle \left(5 - \dfrac{3}{8}\right) - \left(-\dfrac{8}{5} - \dfrac{3}{20}\right)$

Remove the brackets. The second one changes sign:

$$\dfrac{5}{1} - \dfrac{3}{8} + \dfrac{8}{5} + \dfrac{3}{20}$$

$\text{LCM}(8, 5, 20) = 40$:

$$\dfrac{200}{40} - \dfrac{15}{40} + \dfrac{64}{40} + \dfrac{6}{40} = \dfrac{255}{40} = \dfrac{51}{8}$$
$$\dfrac{51}{8}$$

4. Work out by factorising

d) $\displaystyle -\dfrac{300}{75} : \dfrac{115}{95} : \dfrac{6}{10}$

The first stays and the other two are flipped:

$$-\dfrac{300}{75} \cdot \dfrac{95}{115} \cdot \dfrac{10}{6}$$

Factorise and cancel:

$$-\dfrac{(2^2 \cdot 3 \cdot 5^2) \cdot (5 \cdot 19) \cdot (2 \cdot 5)}{(3 \cdot 5^2) \cdot (5 \cdot 23) \cdot (2 \cdot 3)} = -\dfrac{2^2 \cdot 5 \cdot 19}{3 \cdot 23} = -\dfrac{380}{69}$$
$$-\dfrac{380}{69}$$
e) $\displaystyle -\dfrac{147}{20} : \dfrac{70}{9} \cdot \dfrac{10}{7}$

The division becomes a multiplication by the reciprocal:

$$-\dfrac{147}{20} \cdot \dfrac{9}{70} \cdot \dfrac{10}{7}$$

Factorise and cancel:

$$-\dfrac{(3 \cdot 7^2) \cdot 3^2 \cdot (2 \cdot 5)}{(2^2 \cdot 5) \cdot (2 \cdot 5 \cdot 7) \cdot 7} = -\dfrac{3^3}{2^2 \cdot 5} = -\dfrac{27}{20}$$
$$-\dfrac{27}{20}$$

5. Work out the powers

f) $\displaystyle \left(\dfrac{8}{11}\right)^3 \cdot \left(\dfrac{8}{11}\right)^2 : \left(\dfrac{8}{11}\right)^7$

Same base: add and subtract the exponents.

$$\left(\dfrac{8}{11}\right)^{3+2-7} = \left(\dfrac{8}{11}\right)^{-2}$$

Negative exponent: flip the fraction.

$$\left(\dfrac{11}{8}\right)^2 = \dfrac{121}{64}$$
$$\dfrac{121}{64}$$
g) $\displaystyle \left[\left(\dfrac{5}{3}\right)^2\right]^{-3} : \left(\dfrac{10}{9}\right)^{-6}$

Power of a power: $2 \cdot (-3) = -6$. That leaves two powers with the same exponent:

$$\left(\dfrac{5}{3}\right)^{-6} : \left(\dfrac{10}{9}\right)^{-6} = \left(\dfrac{5}{3} : \dfrac{10}{9}\right)^{-6} = \left(\dfrac{45}{30}\right)^{-6} = \left(\dfrac{3}{2}\right)^{-6}$$

Negative exponent: flip the fraction.

$$\left(\dfrac{2}{3}\right)^6 = \dfrac{2^6}{3^6} = \dfrac{64}{729}$$
$$\left(\dfrac{2}{3}\right)^6 = \dfrac{64}{729}$$

6. Combined operations

e) $\displaystyle \dfrac{28}{60} - \dfrac{5}{36} \cdot \left(-\dfrac{27}{75}\right) + \sqrt{\dfrac{256}{625}} + 20 : (-8)$

Each term on its own. The multiplication has two minus signs, which give plus:

$$\dfrac{28}{60} = \dfrac{7}{15} \qquad -\dfrac{5}{36} \cdot \left(-\dfrac{27}{75}\right) = +\dfrac{5 \cdot 3^3}{(2^2 \cdot 3^2) \cdot (3 \cdot 5^2)} = \dfrac{1}{20}$$
$$\sqrt{\dfrac{256}{625}} = \dfrac{16}{25} \qquad 20 : (-8) = -\dfrac{20}{8} = -\dfrac{5}{2}$$

$\text{LCM}(15, 20, 25, 2) = 300$:

$$\dfrac{7}{15} + \dfrac{1}{20} + \dfrac{16}{25} - \dfrac{5}{2} = \dfrac{140 + 15 + 192 - 750}{300} = -\dfrac{403}{300}$$
$$-\dfrac{403}{300}$$
f) $\displaystyle \left(\dfrac{6}{7}\right)^3 : \left(\dfrac{6}{7}\right)^2 - \dfrac{65}{18} \cdot \dfrac{12}{39} + \left(-\dfrac{9}{5}\right)^3 \cdot \left(-\dfrac{9}{5}\right)^{-5}$

The powers with the same base, and the multiplication by factorising:

$$\left(\dfrac{6}{7}\right)^{3-2} = \dfrac{6}{7} \qquad \dfrac{65}{18} \cdot \dfrac{12}{39} = \dfrac{(5 \cdot 13) \cdot (2^2 \cdot 3)}{(2 \cdot 3^2) \cdot (3 \cdot 13)} = \dfrac{10}{9}$$
$$\left(-\dfrac{9}{5}\right)^{3-5} = \left(-\dfrac{9}{5}\right)^{-2} = \left(-\dfrac{5}{9}\right)^2 = \dfrac{25}{81}$$

$\text{LCM}(7, 9, 81) = 567$:

$$\dfrac{6}{7} - \dfrac{10}{9} + \dfrac{25}{81} = \dfrac{486 - 630 + 175}{567} = \dfrac{31}{567}$$
$$\dfrac{31}{567}$$