Solved Questions

How to Use These Questions

These 34 questions cover all six topics in mixed order. A BODMAS sum sits next to a surd, a continued fraction and a power of 3, so before you pick a method, decide what you are looking at: an "of" to handle first, a surd in the bottom to rationalise, or powers that can go to one base.

Work each one on paper with a timer, at about 36 to 45 seconds a question, before you read the answer. When one takes you over a minute, find the step that slowed you down. Usually it is a long multiplication that a pattern or an identity would have skipped. Look that pattern up in Shortcuts & Speed Methods and try the question again the next day.

Roughly a fifth of the questions are BODMAS and decimals, a fifth are surds, and the rest are split between fractions and series, roots, indices, and identities with approximation. That is close to the mix in recent papers. A question marked [Tier 2] adds one more step, such as rationalising a bottom with three terms or ordering roots of different orders. The ideas underneath are the Tier 1 ones, so try them all.

Solved Examples

Question 1

Simplify: $\frac{5}{6} \div \frac{5}{6} \text{ of } \frac{3}{5} + \frac{2}{3} \times \frac{9}{4}$

Answer. "of" goes first. $\frac{5}{6}$ of $\frac{3}{5}$ is $\frac{5}{6} \times \frac{3}{5} = \frac{1}{2}$. Then the division: $\frac{5}{6} \div \frac{1}{2} = \frac{5}{3}$. The multiplication: $\frac{2}{3} \times \frac{9}{4} = \frac{3}{2}$. Adding, $\frac{5}{3} + \frac{3}{2} = \frac{10 + 9}{6} = \mathbf{\frac{19}{6}}$, which is $3\frac{1}{6}$. If you forget the "of" rule and go left to right, you get $\frac{3}{5} + \frac{3}{2} = \frac{21}{10}$, and that is the trap.

Question 2

Simplify: $\frac{\sqrt{98} + \sqrt{50} - \sqrt{32}}{\sqrt{18}}$

Answer. I pull the square factors out of each root: $\sqrt{98} = 7\sqrt{2}$, $\sqrt{50} = 5\sqrt{2}$, $\sqrt{32} = 4\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$. The top is $7\sqrt{2} + 5\sqrt{2} - 4\sqrt{2} = 8\sqrt{2}$. Dividing by $3\sqrt{2}$, the $\sqrt{2}$ cancels and the answer is $\mathbf{\frac{8}{3}}$.

Question 3

Find $\sqrt{0.000169} + \sqrt{1.69}$.

Answer. Both come from 169 = $13^2$. A square root halves the number of decimal places. 0.000169 has 6 places, so its root has 3: 0.013. 1.69 has 2 places, so its root has 1: 1.3. The sum is 1.313. Writing 0.13 for the first root is the usual slip, because $0.13^2 = 0.0169$.

Question 4

Find $81^{\frac{3}{4}} \times 32^{-\frac{2}{5}} \times 125^{\frac{1}{3}}$.

Answer. Root first, then power. The fourth root of 81 is 3, and $3^3 = 27$. The fifth root of 32 is 2, and $2^2 = 4$, but the minus sign flips it to $\frac{1}{4}$. The cube root of 125 is 5. So the value is $27 \times \frac{1}{4} \times 5 = \mathbf{\frac{135}{4}}$, which is 33.75.

Question 5

Find $\frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \cdots + \frac{1}{380}$.

Answer. The bottoms are 2 × 3, 3 × 4, 4 × 5, and so on up to 380 = 19 × 20. Each term $\frac{1}{n(n+1)}$ splits into $\frac{1}{n} - \frac{1}{n+1}$, and all the middle pieces cancel. What's left is $\frac{1}{2} - \frac{1}{20} = \mathbf{\frac{9}{20}}$.

Question 6

Find $\frac{8.6^3 - 3.4^3}{8.6^2 + 8.6 \times 3.4 + 3.4^2}$.

Answer. The top is $a^3 - b^3$, which is $(a - b)(a^2 + ab + b^2)$. The bottom is exactly that second bracket, with a plus in the middle. So everything cancels except a − b = 8.6 − 3.4 = 5.2. No cubing needed.

Question 7

What comes in place of ? in 4.8 × 2.5 + ? ÷ 0.25 = 20.4

Answer. By BODMAS the ÷ belongs only to the ?, not to the whole left side. 4.8 × 2.5 = 12 (think 4.8 × 10 ÷ 4). So ? ÷ 0.25 has to be 20.4 − 12 = 8.4. Dividing by 0.25 is the same as multiplying by 4, so 4 × ? = 8.4 and ? = 2.1.

Question 8

If $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$, find $x^2 + y^2$.

Answer. x and y are upside-down copies of each other, so xy = 1. Rationalising x: $\frac{(\sqrt{5} + \sqrt{3})^2}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$. In the same way $y = 4 - \sqrt{15}$, so x + y = 8. Then $x^2 + y^2 = (x + y)^2 - 2xy = 64 - 2 =$ 62.

Question 9

Find $\sqrt[3]{1225043}$, given that it is a whole number.

Answer. It ends in 3, and a cube ending in 3 comes from a number ending in 7. Covering the last three digits (043) leaves 1225. $10^3 = 1000$ fits into 1225 but $11^3 = 1331$ doesn't, so the front part is 10. The root is 107. Check: $107^3$ = 12,25,043.

Question 10

Arrange in ascending order: $\frac{2}{3}$, $\frac{5}{7}$, $\frac{9}{13}$, $\frac{11}{16}$

Answer. Each one is 1 minus a gap: $\frac{1}{3}$, $\frac{2}{7}$, $\frac{4}{13}$ and $\frac{5}{16}$. A bigger gap means a smaller fraction. As rough decimals the gaps are 0.333, 0.286, 0.308 and 0.3125, so from biggest gap to smallest: $\frac{1}{3}$, $\frac{5}{16}$, $\frac{4}{13}$, $\frac{2}{7}$. So the fractions in ascending order are $\mathbf{\frac{2}{3} < \frac{11}{16} < \frac{9}{13} < \frac{5}{7}}$. The close pair is $\frac{11}{16}$ and $\frac{9}{13}$, and cross-multiplying confirms it: 11 × 13 = 143 and 9 × 16 = 144.

Question 11

If $2^{2x-1} \times 4^{x+1} = 8^{x+3}$, find x.

Answer. I put everything in base 2. $4^{x+1} = 2^{2x+2}$ and $8^{x+3} = 2^{3x+9}$. Multiplying adds the powers on the left: (2x − 1) + (2x + 2) = 4x + 1. So 4x + 1 = 3x + 9, which gives x = 8.

Question 12

Simplify $\sqrt{12 - \sqrt{140}}$.

Answer. First I need a 2 in front of the inner root: $\sqrt{140} = \sqrt{4 \times 35} = 2\sqrt{35}$. Now I want two numbers with sum 12 and product 35. That's 7 and 5. It's a minus sign, so the bigger root goes first: $\mathbf{\sqrt{7} - \sqrt{5}}$. Check: $(\sqrt{7} - \sqrt{5})^2 = 12 - 2\sqrt{35}$.

Question 13

Simplify: $9 - [6 - \{14 \div (11 - \overline{5 - 2}) \times 4\}]$

Answer. Inside out. The bar first: 5 − 2 = 3. Round bracket: 11 − 3 = 8. Curly bracket: 14 ÷ 8 × 4, left to right, is 1.75 × 4 = 7. Square bracket: 6 − 7 = −1. Last step: 9 − (−1) = 10. Doing 8 × 4 before the division gives 55/16, a typical wrong option.

Question 14

Find $\frac{2.5^3 - 1.5^3 - 1}{3 \times 2.5 \times 1.5}$.

Answer. The trick is to see the 1 as $1^3$. Then the top is $a^3 + b^3 + c^3$ with a = 2.5, b = −1.5 and c = −1, and 2.5 − 1.5 − 1 = 0. So the top equals 3abc = 3 × 2.5 × (−1.5) × (−1) = 3 × 2.5 × 1.5. That is exactly the bottom, so the answer is 1.

Question 15 [Tier 2]

Solve for x:

$$\cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{x}}} = \frac{10}{23}$$

Answer. I peel it from the outside. Flip both sides: $2 + \cfrac{1}{3 + \frac{1}{x}} = \frac{23}{10}$. Take away 2: $\cfrac{1}{3 + \frac{1}{x}} = \frac{3}{10}$. Flip: $3 + \frac{1}{x} = \frac{10}{3}$. Take away 3: $\frac{1}{x} = \frac{1}{3}$. So x = 3.

Question 16

Find $\sqrt{17 + \sqrt{57 + \sqrt{33 + \sqrt{256}}}}$.

Answer. Start from the innermost root and work out. $\sqrt{256} = 16$. Then 33 + 16 = 49, and $\sqrt{49} = 7$. Then 57 + 7 = 64, and $\sqrt{64} = 8$. Then 17 + 8 = 25, and $\sqrt{25}$ = 5. With these questions every stage is built to be a perfect square, so if one isn't, recheck the step before.

Question 17

Arrange in descending order: $\sqrt{8} + \sqrt{3}$, $\sqrt{7} + \sqrt{4}$, $\sqrt{6} + \sqrt{5}$, $\sqrt{9} + \sqrt{2}$

Answer. The numbers under the roots add to 11 every time. When I square each one I get 11 + $2\sqrt{ab}$, so only the products ab matter: 24, 28, 30 and 18. Bigger product, bigger value. So the order is $\mathbf{\sqrt{6} + \sqrt{5} > \sqrt{7} + \sqrt{4} > \sqrt{8} + \sqrt{3} > \sqrt{9} + \sqrt{2}}$. The pair whose numbers are closest together comes first.

Question 18

Find $\frac{(0.0016)^{\frac{1}{4}} \times (0.000027)^{\frac{1}{3}}}{(0.0036)^{\frac{1}{2}}}$.

Answer. Each root splits the decimal places evenly. 0.0016 has 4 places and 16 = $2^4$, so its fourth root is 0.2. 0.000027 has 6 places and 27 = $3^3$, so its cube root is 0.03. 0.0036 has 4 places and 36 = $6^2$, so its square root is 0.06. The value is 0.2 × 0.03 ÷ 0.06 = 0.006 ÷ 0.06 = 0.1.

Question 19

Find $\frac{0.0016 \times 0.25}{0.08 \times 0.002}$.

Answer. I ignore the points first. Top: 16 × 25 = 400. Bottom: 8 × 2 = 16. So the digits give 400 ÷ 16 = 25. Now the decimal places: the top has 4 + 2 = 6 and the bottom has 2 + 3 = 5. That leaves 6 − 5 = 1 place on top, so the answer is 25 with one decimal place: 2.5.

Question 20

$\frac{29.98\% \text{ of } 801 + 15.03^2}{\sqrt{48.9} + 2.97}$ is closest to: (a) 39 (b) 47 (c) 55 (d) 63

Answer. Round everything. 30% of 800 = 240. $15^2 = 225$. So the top is about 465. At the bottom, $\sqrt{49} = 7$ and 2.97 is about 3, so the bottom is about 10. The value is about 46.5, so the answer is (b) 47. (The exact value is about 46.8.)

Question 21 [Tier 2]

Rationalise $\frac{1}{2 + \sqrt{3} + \sqrt{7}}$.

Answer. Three terms is too many to handle at once, so I treat $2 + \sqrt{3}$ as one block and multiply top and bottom by $(2 + \sqrt{3}) - \sqrt{7}$. The bottom becomes $(2 + \sqrt{3})^2 - 7 = 4 + 4\sqrt{3} + 3 - 7 = 4\sqrt{3}$. So the fraction is $\frac{2 + \sqrt{3} - \sqrt{7}}{4\sqrt{3}}$. One more step to clear the $\sqrt{3}$: multiply top and bottom by $\sqrt{3}$. The answer is $\mathbf{\frac{2\sqrt{3} + 3 - \sqrt{21}}{12}}$.

Question 22 [Tier 2]

Find the value of $\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\left(1 - \frac{1}{25}\right) \cdots \left(1 - \frac{1}{225}\right)$.

Answer. The bottoms are $3^2, 4^2, 5^2, \ldots, 15^2$, so this chain starts at 3, not 2. Each bracket is $1 - \frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$, so I split the whole thing into two chains:

$$\left(\frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{14}{15}\right) \times \left(\frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{16}{15}\right) = \frac{2}{15} \times \frac{16}{3}$$

So the answer is $\mathbf{\frac{32}{45}}$. Careful with the shortcut here: $\frac{n+1}{2n}$ only works when the chain starts at $1 - \frac{1}{4}$. With n = 15 it would give $\frac{16}{30}$, because it also multiplies in the missing first bracket: $\frac{3}{4} \times \frac{32}{45} = \frac{16}{30}$.

Question 23

If $\sqrt{5} = 2.236$, find $\frac{\sqrt{5}}{2} + \frac{1}{\sqrt{5}}$.

Answer. Dividing by $\sqrt{5}$ is messy, so I flip it into a multiple of $\sqrt{5}$: $\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} = 0.2\sqrt{5}$. And $\frac{\sqrt{5}}{2} = 0.5\sqrt{5}$. Together that is $0.7\sqrt{5}$ = 0.7 × 2.236 = 1.5652. One multiplication instead of a long division by 2.236.

Question 24 [Tier 2]

Simplify:

$$\frac{3\frac{1}{3} \div 2\frac{1}{2} \text{ of } 1\frac{1}{3}}{\frac{1}{2} + \frac{1}{3} \text{ of } \frac{3}{4}}$$

Answer. I change the mixed numbers first: $3\frac{1}{3} = \frac{10}{3}$, $2\frac{1}{2} = \frac{5}{2}$ and $1\frac{1}{3} = \frac{4}{3}$. On top, "of" goes first: $\frac{5}{2} \times \frac{4}{3} = \frac{10}{3}$, and then $\frac{10}{3} \div \frac{10}{3} = 1$. At the bottom, $\frac{1}{3}$ of $\frac{3}{4}$ is $\frac{1}{4}$, and $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. So the answer is $1 \div \frac{3}{4} = \mathbf{\frac{4}{3}}$.

Question 25

Find $\frac{3^{n+3} - 4 \times 3^{n+1}}{7 \times 3^n - 3^{n+1}}$.

Answer. I take $3^n$ out of every term. Top: $3^n(27 - 4 \times 3) = 3^n \times 15$. Bottom: $3^n(7 - 3) = 3^n \times 4$. The $3^n$ cancels, and the answer is $\mathbf{\frac{15}{4}}$. It doesn't matter what n is.

Question 26

Find $\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \cdots + \frac{1}{\sqrt{35} + \sqrt{36}}$.

Answer. In every term the two numbers differ by 1, so rationalising gives a bottom of 1. That makes $\frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}$. The sum becomes $(\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + \cdots + (\sqrt{36} - \sqrt{35})$, and everything cancels except $\sqrt{36} - \sqrt{4} = 6 - 2 =$ 4.

Question 27 [Tier 2]

Write $\sqrt{3\sqrt{3\sqrt{3\sqrt{3}}}}$ as a power of 3.

Answer. I work from the inside. $\sqrt{3} = 3^{\frac{1}{2}}$. Next, $3 \times 3^{\frac{1}{2}} = 3^{\frac{3}{2}}$, and its root is $3^{\frac{3}{4}}$. Next, $3 \times 3^{\frac{3}{4}} = 3^{\frac{7}{4}}$, and its root is $3^{\frac{7}{8}}$. Last, $3 \times 3^{\frac{7}{8}} = 3^{\frac{15}{8}}$, and its root is $\mathbf{3^{\frac{15}{16}}}$. The pattern for four roots is $1 - \frac{1}{2^4} = \frac{15}{16}$, which saves the working next time.

Question 28

Find $\frac{(8.73 + 1.27)^2 - (8.73 - 1.27)^2}{8.73 \times 1.27}$.

Answer. $(a + b)^2 - (a - b)^2 = 4ab$. The top is 4 × 8.73 × 1.27, and the bottom is 8.73 × 1.27, so the answer is 4. Working out $10^2 - 7.46^2$ would get there too, but only after a lot of decimal work.

Question 29

Find 3 + 33 + 333 + 3333 + 33333.

Answer. I add column by column. Units: five 3s make 15, so write 5 and carry 1. Tens: four 3s make 12, plus 1 is 13, so write 3 and carry 1. Hundreds: three 3s make 9, plus 1 is 10, so write 0 and carry 1. Thousands: two 3s make 6, plus 1 is 7. Ten-thousands: one 3. Reading upwards, the answer is 37,035.

Question 30 [Tier 2]

Arrange in descending order: $3^{\frac{1}{3}}$, $4^{\frac{1}{4}}$, $5^{\frac{1}{5}}$, $6^{\frac{1}{6}}$

Answer. All four at once would need the LCM of 3, 4, 5 and 6, which is 60. That's too big, so I compare neighbours.

• $3^{\frac{1}{3}}$ and $4^{\frac{1}{4}}$: raise both to the power 12. $3^4 = 81$ and $4^3 = 64$, so $3^{\frac{1}{3}}$ is bigger.
• $4^{\frac{1}{4}}$ and $5^{\frac{1}{5}}$: power 20. $4^5 = 1024$ and $5^4 = 625$, so $4^{\frac{1}{4}}$ is bigger.
• $5^{\frac{1}{5}}$ and $6^{\frac{1}{6}}$: power 30. $5^6 = 15625$ and $6^5 = 7776$, so $5^{\frac{1}{5}}$ is bigger.

So the order is $\mathbf{3^{\frac{1}{3}} > 4^{\frac{1}{4}} > 5^{\frac{1}{5}} > 6^{\frac{1}{6}}}$. From 3 onwards, $n^{\frac{1}{n}}$ keeps getting smaller.

Question 31 [Tier 2]

Solve: $\frac{\sqrt{x + 7} + \sqrt{x - 7}}{\sqrt{x + 7} - \sqrt{x - 7}} = 4$

Answer. Componendo–dividendo on both sides. On the left, (top + bottom) over (top − bottom) leaves $\frac{\sqrt{x + 7}}{\sqrt{x - 7}}$. On the right it gives $\frac{4 + 1}{4 - 1} = \frac{5}{3}$. Squaring: $\frac{x + 7}{x - 7} = \frac{25}{9}$. Using the rule again: $\frac{2x}{14} = \frac{34}{16}$, so $\frac{x}{7} = \frac{17}{8}$ and $x = \mathbf{\frac{119}{8}}$.

Question 32

A tank is $\frac{2}{5}$ full. After 21 litres are poured in, it is $\frac{3}{4}$ full. How much does the tank hold, and how much water was in it at first?

Answer. The 21 litres filled the gap between $\frac{2}{5}$ and $\frac{3}{4}$. That gap is $\frac{15}{20} - \frac{8}{20} = \frac{7}{20}$ of the tank. So $\frac{7}{20}$ of the tank is 21 litres, $\frac{1}{20}$ is 3 litres, and the whole tank holds 60 litres. At first it had $\frac{2}{5}$ of 60 = 24 litres.

Question 33

Simplify: 25% of 480 ÷ 6 of 4 + 3 × 7

Answer. Both "of" parts go first. 25% of 480 is 120, and 6 of 4 is 24. Then the division: 120 ÷ 24 = 5. Then 3 × 7 = 21. So the answer is 5 + 21 = 26. Doing 120 ÷ 6 × 4 instead would give 80 + 21 = 101.

Question 34 [Tier 2]

If a = 101, b = 102 and c = 103, find $a^3 + b^3 + c^3 - 3abc$.

Answer. Cubing 101, 102 and 103 is the slow way. Instead I use $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. The second bracket is half the sum of the squared gaps: $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2] = \frac{1}{2}(1 + 1 + 4) = 3$. And a + b + c = 306. So the answer is 306 × 3 = 918.