AIMO · Free Diagnostic Mock

🎁 FREE Diagnostic

15 questions to find exactly where your AIMO problem-solving is strong — and where it's not

This is a free diagnostic for the AIMO Preparation course (Australian Intermediate Mathematics Olympiad). 15 real AIMO-style problems (integer answers, 0–999) sample Number Theory, Geometry and Algebra.

It's an exam: answer all the questions first (you can change any answer freely), then submit the whole exam to see your report. The worked solutions and hints unlock only after you submit. Use the jump menu at the top to move around.

At the end you get a knowledge-point report: a bar for every topic you attempted, your weak spots flagged, and the topics to focus on first.

45–60 minutes, self-paced. No login needed. No marks deducted for a wrong answer — never leave one blank.
Self-paced, no time limit — but try to commit to an answer before you review, just like a real test.

Year 8–12 students preparing for AIMO selection.

⭐ Q1AIMOInteger answerAIMO › Number Theory

The Problem

Let $x$ denote a single digit. The tens digit in the product of the three-digit number $\overline{2x7}$ and $39$ is $9$. Find $x$.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Digit equations: name the numeral $(207 + 10x)$ and track only the digit you need.

Write $\overline{2x7}$ as $207 + 10x$. Then $(207 + 10x) \times 39 = 8073 + 390x$. Trying $x = 0\ldots9$, the tens digit is 9 only at $x = 8$: $8073 + 3120 = 11193$. Answer: 8.

✅ Answer: 8

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q2AIMOInteger answerAIMO › Number Theory

The Problem

If n is a positive integer and $n^{2}$ equals the 4-digit number $\overline{aabb}$, find $n$.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Square structure: $\overline{aabb} = 11(100a+b)$ forces $11^2$ to divide $n^2$.

$\overline{aabb} = 1100a + 11b = 11(100a + b)$, so 11 divides $n^2$ and hence $121$ divides $n^2$. Writing $n = 11m$, we need $100a + b = 11m^2$. Scanning the squares from $32^2$ to $99^2$, only $88^2 = 7744$ has the $\overline{aabb}$ shape. Answer: 88.

✅ Answer: 88

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q3AIMOInteger answerAIMO › Number Theory

The Problem

A 3-digit number $\overline{abc}$ is multiplied by 3 to give the 4-digit number $\overline{c0ba}$. Find the number $\overline{abc}$.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Reverse-digit equations: expand both numerals and reduce to one linear equation in the digits.

$3(100a + 10b + c) = 1000c + 10b + a$ reduces to $299a + 20b = 997c$. With digits, $c = 1$ forces $299a + 20b = 997$, so $a = 3$, $b = 5$. Check: $351 \times 3 = 1053$ ✓. Answer: 351.

✅ Answer: 351

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q4AIMOInteger answerAIMO › Number Theory

The Problem

The $n$th triangular number is the sum of the first $n$ positive integers. Let $T_n$ denote the sum of the first $n$ triangular numbers. Find $T_{10}$, the sum of the first 10 triangular numbers.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Summing a known sequence: stacking triangular numbers gives $T_n = n(n+1)(n+2)/6$.

$t_k = k(k+1)/2$, and stacking them gives $T_n = n(n+1)(n+2)/6$. So $T_{10} = 10 \cdot 11 \cdot 12 / 6 = 220$. Direct check: $1+3+6+10+15+21+28+36+45+55 = 220$.

✅ Answer: 220

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q5AIMOInteger answerAIMO › Number Theory

The Problem

The $n$th triangular number is $t_n = \tfrac{n(n+1)}{2}$. Notice that $t_1 + t_9 = t_4 + t_8$ — the same number is a sum of two triangular numbers in two different ways. Work out and submit this common value.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Triangular numbers: evaluate $t_n = n(n+1)/2$ at small $n$ and compare sums.

$t_1 = 1$ and $t_9 = 45$, so $t_1 + t_9 = 46$; $t_4 = 10$ and $t_8 = 36$, so $t_4 + t_8 = 46$ as well. Answer: 46.

✅ Answer: 46

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q6AIMOInteger answerAIMO › Number Theory

The Problem

Determine the number of non-negative integers x that satisfy the equation ⌊x/44⌋ = ⌊x/45⌋.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Number Theory problem (integer answer 0–999). Floor functions: turn $⌊x/44⌋ = ⌊x/45⌋ = k$ into an interval and count its integers.

$⌊x/44⌋ = ⌊x/45⌋ = k$ requires $45k \le x \le 44k + 43$, which contains $44 - k$ integers and is non-empty for $k = 0\ldots43$. Total $= \sum_{k=0}^{43}(44-k) = 44 + 43 + \cdots + 1 = 990$.

✅ Answer: 990

Revise: the AIMO Number Theory pathway in the full course.

⭐ Q7AIMOInteger answerAIMO › Geometry

The Problem

Triangles $ABC$ and $XYZ$ are congruent isosceles triangles, right-angled at $B$ and $Y$. A square is inscribed in $\triangle ABC$ with two sides along the legs and one corner at $B$; a square is inscribed in $\triangle XYZ$ with one side along the hypotenuse. If the first square has area $189$, find the area of the second square.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Inscribed squares: derive each side from similar triangles — the two orientations give an 8:9 area ratio.

Let the legs have length $a$. The legs-square has its far corner $(t, t)$ on the hypotenuse $x + y = a$, so $t = a/2$ and its area is $a^2/4 = 189$. The hypotenuse-square of side $s$ sits on a base of length $a\sqrt{2}$ under an apex of height $a\sqrt{2}/2$; the triangle’s width at height $s$ is $a\sqrt{2} - 2s = s$, so $s = a\sqrt{2}/3$ and its area is $2a^2/9$. The ratio is $(2/9) \div (1/4) = 8/9$, so the answer is $189 \times 8/9 = 168$.

✅ Answer: 168

Revise: the AIMO Geometry pathway in the full course.

⭐ Q8AIMOInteger answerAIMO › Geometry

The Problem

A rectangle has sides of length $28$ and $15$. One diagonal is divided into $7$ equal parts by $6$ points, and every division point is joined to the two opposite corners. This cuts the rectangle into $7$ quadrilaterals. Find the area of one of them.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Equal bases with a shared apex: slices along a diagonal all have equal areas.

The diagonal splits the rectangle into two triangles of area $420/2 = 210$. The six points cut each triangle into 7 slices with equal bases along the diagonal and a shared apex, so every slice has area $30$. Each of the 7 quadrilaterals is one slice from each triangle: $30 + 30 = 60$. Check: $7 \times 60 = 420 = 28 \times 15$ ✓.

✅ Answer: 60

Revise: the AIMO Geometry pathway in the full course.

⭐ Q9AIMOInteger answerAIMO › Geometry

The Problem

A triangle $ABC$ is divided into four regions by three lines parallel to $BC$. The lines divide $AB$ into four equal segments. If the second largest region has area $225$, what is the area of $\triangle ABC$?

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Similar triangles: area scales with the square of the side ratio, so parallel cuts give 1:3:5:7 strips.

The parallel cuts create similar triangles with areas in ratio $1 : 4 : 9 : 16$, so the four strips have areas $1 : 3 : 5 : 7$ (total 16). The second largest strip is $5/16$ of the whole: $5S/16 = 225$, so $S = 720$.

✅ Answer: 720

Revise: the AIMO Geometry pathway in the full course.

⭐ Q10AIMOInteger answerAIMO › Geometry

The Problem

Consider a circular sector of radius $360$ which is one-sixth of a circle. A circle is drawn inside this sector so that it is tangent to the two radii and to the circular arc. Calculate the radius of this smaller circle.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Tangency in a sector: put the centre on the bisector — $r = d\sin(θ/2)$ against the radii, $d + r = R$ against the arc.

The small circle’s centre lies on the bisector of the $60°$ sector, at distance $d$ from the vertex. Tangency to the radii gives $r = d \sin 30° = d/2$; tangency to the arc gives $d + r = 360$. So $2r + r = 360$ and $r = 120$.

✅ Answer: 120

Revise: the AIMO Geometry pathway in the full course.

⭐ Q11AIMOInteger answerAIMO › Geometry

The Problem

$ABCD$ is a trapezium for which $AB \parallel DC$, $AB = 84$ and $DC = 25$. A circle can be drawn in the trapezium so that it just touches all four sides. Find the perimeter of the trapezium.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Tangential quadrilaterals (Pitot theorem): the two pairs of opposite sides have equal sums.

A circle fits inside a quadrilateral exactly when the two pairs of opposite sides have equal sums (equal tangent lengths from each corner). So $BC + AD = AB + DC = 84 + 25 = 109$, and the perimeter is $109 + 109 = 218$.

✅ Answer: 218

Revise: the AIMO Geometry pathway in the full course.

⭐ Q12AIMOInteger answerAIMO › Geometry

The Problem

A circle is inscribed in a hexagon $ABCDEF$ so that each side of the hexagon is tangent to the circle. Find the perimeter of the hexagon if $AB = 6$, $CD = 7$, and $EF = 8$.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Geometry problem (integer answer 0–999). Equal tangent lengths: alternating sides of a tangential polygon have equal sums.

Label the tangent lengths at $A\ldots F$ as $a\ldots f$. Then $AB + CD + EF = (a+b)+(c+d)+(e+f)$ and $BC + DE + FA = (b+c)+(d+e)+(f+a)$ — both equal $a+b+c+d+e+f$. So the perimeter is $2(6 + 7 + 8) = 42$.

✅ Answer: 42

Revise: the AIMO Geometry pathway in the full course.

⭐ Q13AIMOInteger answerAIMO › Algebra

The Problem

Asha, Bree and Cala are three robots that are programmed to run athletic track races. When Asha runs a 400 m race she catches Bree at the finish line, if Bree starts 20 m ahead of Asha. Asha catches Cala at the finish line of a 1500 m race, if Cala has a 246 m start. Assuming each robot runs at constant speed, how many metres must Cala start ahead of Bree in an 800 m race, if they are to finish at the same time?

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Algebra problem (integer answer 0–999). Relative speed: chain the speed ratios, then convert the ratio into a head start.

From the 400 m race, $v_B/v_A = 380/400 = 19/20$. From the 1500 m race, $v_C/v_A = 1254/1500 = 209/250$. So $v_C/v_B = (209/250)(20/19) = 22/25$. While Bree runs 800 m, Cala covers $800 \times 22/25 = 704$ m, so Cala needs a start of $800 - 704 = 96$ m.

✅ Answer: 96

Revise: the AIMO Algebra pathway in the full course.

⭐ Q14AIMOInteger answerAIMO › Algebra

The Problem

Gaston and Jordon always misread cooking times. If the required time is “1:32” (1 hour 32 minutes), Jordon reads it as 132 minutes while Gaston reads it as 1.32 hours. For one particular recipe, the difference between Jordon's and Gaston's misread times is exactly 90 minutes. What is the actual cooking time, in minutes?

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AIMO Algebra problem (integer answer 0–999). Modelling a misreading: express both readings in the same unit and solve one linear equation.

Write the time as $h$:$mm$. Jordon reads $100h + m$ minutes; Gaston reads $h + m/100$ hours $= 60h + 0.6m$ minutes. The difference is $40h + 0.4m = 90$, so $100h + m = 225$: $h = 2$, $m = 25$. The actual time is 2 h 25 min $= 145$ minutes. Check: Jordon 225, Gaston 135, difference 90 ✓.

✅ Answer: 145

Revise: the AIMO Algebra pathway in the full course.

⭐ Q15AIMOInteger answerAIMO › Algebra

The Problem

A school adds 5 extra classrooms enabling 5 more classes and reducing the average class size by 6. Two months later another 5 classrooms are added, again enabling 5 more classes, this time reducing average class size by 4. The number of students did not change. How many students were at the school?

Your answer

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🎯 Knowledge Point — what this tests

An AIMO Algebra problem (integer answer 0–999). Average = total ÷ groups: two average-drop conditions give two equations in $c$ and $N$.

Let $c$ be the original number of classes and $N$ the students. The two average drops give $N/c - N/(c+5) = 6$ and $N/(c+5) - N/(c+10) = 4$, i.e. $5N = 6c(c+5)$ and $5N = 4(c+5)(c+10)$. Dividing: $6c = 4(c+10)$, so $c = 20$ and $N = 6 \cdot 20 \cdot 25 / 5 = 600$. Check: averages 30 → 24 → 20 ✓.

✅ Answer: 600

Revise: the AIMO Algebra pathway in the full course.

🏁 Finish

That's the end of the exam

Go back and change any answers you like. When you're ready, submit the whole exam to lock it in and see your knowledge report — then you can revisit every question to read the full worked solution.

The report stays locked until you submit — you can't open it early.

📊 Diagnostic Report

Your knowledge-point report

Here is how you performed on every knowledge area you attempted. Bars show your accuracy per topic — green is strong, amber is shaky, red needs work. Only questions you answered are scored.

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