Key topics from JMC past papers. Quick memory tips, worked explanations, and instant feedback. Try to solve before clicking!
What is the value of $2 + 3 \times 4 - 6 \div 2$?
Apply BIDMAS: multiplication and division first. $3 \times 4 = 12$ and $6 \div 2 = 3$.
Then left-to-right: $2 + 12 - 3 = \mathbf{11}$.
Common mistake: doing addition first gives the wrong answer.
When 180 is divided by a positive integer $N$, the remainder is 5. For how many values of $N$ is this true?
If $180 \div N$ leaves remainder 5, then $N$ divides $180 - 5 = 175$ and $N > 5$.
$175 = 5^2 \times 7$. Divisors: 1, 5, 7, 25, 35, 175.
Those greater than 5: $\{7, 25, 35, 175\}$ — that is 4 values.
Answer: A
What is $\dfrac{3}{4} \div \dfrac{9}{16}$?
$\dfrac{3}{4} \div \dfrac{9}{16} = \dfrac{3}{4} \times \dfrac{16}{9} = \dfrac{48}{36} = \dfrac{4}{3}$
Answer: C
The first term of an arithmetic sequence is 7 and the common difference is 4. What is the 20th term?
$a_{20} = 7 + (20-1) \times 4 = 7 + 76 = \mathbf{83}$
Common mistake: using $n$ instead of $(n-1)$, giving $7 + 80 = 87$. Remember: 20th term needs only 19 steps from the 1st.
Answer: B
Solve: $5x - 3 = 3x + 9$
$5x - 3x = 9 + 3 \Rightarrow 2x = 12 \Rightarrow x = 6$
Answer: B
Solve simultaneously:
$2x + y = 11$
$x - y = 1$
Add the two equations: $(2x+y)+(x-y) = 11+1 \Rightarrow 3x = 12 \Rightarrow x = 4$.
Substitute: $4 - y = 1 \Rightarrow y = 3$. Check: $2(4)+3=11$ ✓
Answer: A
Idil has 12 bags of sweets. Some bags contain 3 mints, 4 toffees, and 1 fudge; some contain 4 mints, 5 toffees, and 2 fudges; the rest contain 6 mints and 3 fudges. The bags contain 31 toffees in total. How many fudges do the bags contain in total?
Let Type A bags = $a$, Type B = $b$, Type C = $c$. Then $a+b+c=12$.
Toffees: only Types A and B have toffees. $4a + 5b = 31$. Since $b = 31 - 4a - \ldots$ — try $a=4, b=3$: $16+15=31$ ✓, so $c = 5$.
Fudges: $1(4)+2(3)+3(5) = 4+6+15 = \mathbf{25}$
Answer: D
What is the size of each interior angle of a regular hexagon?
Hexagon: $n = 6$. Interior angle $= \dfrac{(6-2) \times 180}{6} = \dfrac{720}{6} = \mathbf{120°}$
Answer: B
In the diagram, $PQRS$ and $JKLM$ are squares. Each corner of $JKLM$ is one third of the way along an edge of $PQRS$. What is the ratio of the area of $JKLM$ to the area of $PQRS$?
Let outer square side $= 3$. Each corner of $JKLM$ sits 1 unit along from one vertex, 2 units from the other.
By Pythagoras, side of $JKLM = \sqrt{1^2 + 2^2} = \sqrt{5}$.
Area ratio $= \dfrac{(\sqrt{5})^2}{3^2} = \dfrac{5}{9}$, i.e. $\mathbf{5:9}$.
Answer: B
A circle has diameter 10 cm. What is its area? Give your answer in terms of $\pi$.
Diameter = 10 cm, so radius $r = 5$ cm. Area $= \pi r^2 = \pi \times 25 = \mathbf{25\pi}$ cm².
Trap: do NOT use $\pi \times 10^2$. Always halve the diameter first.
Answer: C
£540 is shared between Alice, Ben, and Clara in the ratio 2 : 3 : 4. How much does Clara receive?
Total parts: $2+3+4 = 9$. Each part $= £540 \div 9 = £60$.
Clara's share $= 4 \times £60 = \mathbf{£240}$.
Answer: C
After a 25% increase, a jacket costs £75. What was its original price?
£75 = 125% of original. Original $= \dfrac{75}{1.25} = \mathbf{£60}$.
Common trap: subtracting 25% of £75 gives £56.25, which is wrong — you'd be taking 25% of the new price, not the original.
Answer: C
$P$, $Q$, $R$, $S$ are four points in that order on a straight line. $PR = 6$ cm, $QS = 4$ cm, and $R$ is 1 cm nearer to $S$ than it is to $Q$. What is the length of $PS$?
$QS = QR + RS = 4$ and $RS = QR - 1$. So $QR + (QR-1) = 4 \Rightarrow QR = 2.5$, $RS = 1.5$.
$PR = PQ + QR = 6 \Rightarrow PQ = 3.5$.
$PS = PQ + QR + RS = 3.5 + 2.5 + 1.5 = \mathbf{7.5}$ cm.
Answer: B
A map shows three towns $P$, $Q$, $R$ and roads between them. All road lengths are whole kilometres. The route $P \to Q \to R$ is twice the direct $P \to R$. The route $P \to R \to Q$ is three times the direct $P \to Q$. Which of these could be the length of the round trip $P \to Q \to R \to P$?
Let $PQ = a$, $QR = b$, $PR = c$ (all positive integers).
Conditions: $a + b = 2c$ and $c + b = 3a$.
From these: $b = 2c - a$ and $c + (2c-a) = 3a \Rightarrow 3c = 4a \Rightarrow c = \frac{4a}{3}$. For integer solutions, $a$ must be a multiple of 3. Let $a = 3k$: then $c = 4k$, $b = 8k-3k=5k$.
Round trip $= a+b+c = 3k+5k+4k = 12k$. So the answer must be a multiple of 12: $\mathbf{108} = 12 \times 9$ ✓.
Answer: D
Given a cube, how many equilateral triangles are there whose vertices are three vertices of the cube?
Label cube vertices. Three vertices form an equilateral triangle when they are mutually connected by face diagonals of equal length.
Each such triangle uses one vertex from each of three mutually perpendicular pairs of opposite edges. For a unit cube, vertices $(0,0,0),(1,1,0),(1,0,1)$ form a triangle with all sides $\sqrt{2}$ — equilateral!
There are exactly 8 such equilateral triangles — one for each "slicing direction" through the cube, and choosing from two orientations per direction gives $4 \times 2 = 8$.
Answer: D
What is the Lowest Common Multiple (LCM) of 12 and 18?
$12 = 2^2 \times 3$, $18 = 2 \times 3^2$. LCM $= 2^2 \times 3^2 = 4 \times 9 = \mathbf{36}$.
Note: 6 is the HCF, not the LCM. A common error is to confuse the two.
Answer: C
A bag contains 3 red, 5 blue, and 2 green counters. One counter is chosen at random. What is the probability it is not blue?
Total counters: $3+5+2=10$. Blue counters: 5. P(blue) $= \frac{5}{10} = \frac{1}{2}$.
P(not blue) $= 1 - \frac{1}{2} = \mathbf{\frac{1}{2}}$. Alternatively, non-blue = $3+2=5$, so $\frac{5}{10} = \frac{1}{2}$.
Answer: A
A train travels 180 km in 2 hours 30 minutes. What is its average speed in km/h?
2 hours 30 minutes $= 2.5$ hours. Speed $= 180 \div 2.5 = \mathbf{72}$ km/h.
Trap: if you use 2 hours you get 90 km/h — always convert time to decimal hours first.
Answer: C
Expand and simplify: $(x + 3)(x - 5)$
$(x+3)(x-5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$.
Middle term: $-5x + 3x = -2x$. Don't add the numbers $3$ and $-5$ directly!
Answer: A
A right-angled triangle has legs of length 9 cm and 12 cm. What is the length of the hypotenuse?
$c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = \mathbf{15}$ cm.
This is a 3-4-5 triple scaled by 3: $(3\times3, 3\times4, 3\times5) = (9, 12, 15)$.
Answer: C