AMC Y7-8 · Free Diagnostic Mock

🎁 FREE Diagnostic

15 questions to find exactly where your maths is strong — and where it's not

This is a free diagnostic for AMC Y7-8 (Australian Mathematics Competition) preparation. 12 multiple-choice plus 3 short-answer (Q26-28 level) questions sample Number Theory, Geometry, Combinatorics 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 exact lessons to revise.

~25–35 minutes (12 quick multiple-choice, then 3 challenge questions). 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 7–8 students preparing for the AMC (Australian Mathematics Competition).

⭐ Q1AMC Y7-8★★★★AMC › Number Theory

The Problem

A three-digit number n has digit sum s. It is known that n = 15s + 27. Given that n is the largest such number, find n.

Your answer

❓ Pick the best answer:

A 162
B 297
C 324
D 432

🎯 Knowledge Point — what this tests

An AMC Number Theory problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

Any number is congruent to its digit sum mod 9, so 15s + 27 ≡ s (mod 9), which forces s to be a multiple of 9: s = 9, 18 or 27. s = 9 → n = 162 and 1+6+2 = 9 ✓. s = 18 → n = 297 and 2+9+7 = 18 ✓. s = 27 → n = 432 but 4+3+2 = 9 ≠ 27 ✗. Two numbers satisfy the equation (162 is option A — the trap), and the question asks for the largest: n = 297.

✅ Answer: B — 297

Revise: Number Theory lessons.

⭐ Q2AMC Y7-8★★★AMC › Number Theory

The Problem

A two-digit number has digit sum 11. Reversing its digits gives a number 27 larger than the original. What is the original number?

Your answer

❓ Pick the best answer:

A 38
B 47
C 58
D 74

🎯 Knowledge Point — what this tests

An AMC Number Theory problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

Reversed − original = 9×(units − tens) = 27, so units − tens = 3. With tens + units = 11, the tens digit is 4 and units 7, giving 47. Check: 74 − 47 = 27 ✓.

✅ Answer: B — 47

Revise: Number Theory lessons.

⭐ Q3AMC Y7-8★★AMC › Geometry

The Problem

A square has side length 6. A smaller similar square sits inside it with side length 4. The remaining "frame" (large square minus small square) is then cut into 4 congruent rectangles. What is the total area of two of these rectangles?

Your answer

❓ Pick the best answer:

A 10
B 14
C 16
D 20

🎯 Knowledge Point — what this tests

An AMC Geometry problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

The frame's area is 6² − 4² = 36 − 16 = 20. Cut into 4 congruent rectangles, each is 20 ÷ 4 = 5, so two of them total 10.

✅ Answer: A — 10

Revise: Geometry lessons.

⭐ Q4AMC Y7-8★★★AMC › Geometry

The Problem

A rectangle measures 10 by 6. The midpoints of its four sides are joined in order to form a quadrilateral inside it. What is the area of that quadrilateral?

Your answer

❓ Pick the best answer:

A 15
B 24
C 30
D 36

🎯 Knowledge Point — what this tests

An AMC Geometry problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

Each corner triangle has perpendicular legs 5 and 3, so area 5×3/2 = 7.5. The four corners total 30, leaving 10×6−30 = 30 for the midpoint rhombus.

✅ Answer: C — 30

Revise: Geometry lessons.

⭐ Q5AMC Y7-8★★★AMC › Combinatorics

The Problem

A box contains red, blue and green balls. The numbers of red, blue and green balls are three different positive integers that sum to 10. How many unordered sets of three different positive integer counts {red, blue, green} are possible?

Your answer

❓ Pick the best answer:

A 4
B 6
C 8
D 10

🎯 Knowledge Point — what this tests

An AMC Combinatorics problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

The sets of three different positive integers summing to 10 are {1,2,7}, {1,3,6}, {1,4,5} and {2,3,5} — four in all.

✅ Answer: A — 4

Revise: Combinatorics lessons.

⭐ Q6AMC Y7-8★★★AMC › Combinatorics

The Problem

A 4-digit PIN uses the digits 1, 2, 3, 4 and 5 only. No digit may repeat, and the PIN must start with an odd digit. How many such PINs are possible?

Your answer

❓ Pick the best answer:

A 60
B 72
C 120
D 144

🎯 Knowledge Point — what this tests

An AMC Combinatorics problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

The first digit must be odd: 3 choices (1, 3, 5). The other three places use 3 of the 4 remaining digits: 4×3×2 = 24. Total = 3 × 24 = 72.

✅ Answer: B — 72

Revise: Combinatorics lessons.

⭐ Q7AMC Y7-8★★★AMC › Algebra

The Problem

Anna and Ben paint a wall. Working alone, Anna would finish in 6 hours. Working alone, Ben would finish in 4 hours. They start together but after 1 hour Ben leaves and Anna continues alone. How many additional hours, after Ben leaves, does Anna need to finish the wall?

Your answer

❓ Pick the best answer:

A 2.5
B 3
C 3.5
D 4

🎯 Knowledge Point — what this tests

An AMC Algebra problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

In the first hour together they paint $\frac{1}{6}$ + $\frac{1}{4}$ = $\frac{5}{12}$, leaving $\frac{7}{12}$. Anna alone paints $\frac{1}{6}$ per hour, so she needs ($\frac{7}{12}$) ÷ ($\frac{1}{6}$) = 3.5 more hours.

✅ Answer: C — 3.5

Revise: Algebra lessons.

⭐ Q8AMC Y7-8★★★★AMC › Algebra

The Problem

A shop sells stamps in two denominations: 5 cents and 8 cents. Customers pay in exact change with no change given back. What is the largest postage amount (in cents, integer ≥ 1) that cannot be paid using a combination of these stamps?

Your answer

❓ Pick the best answer:

A 13
B 27
C 35
D 40

🎯 Knowledge Point — what this tests

An AMC Algebra problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

27 cannot be formed: the possible numbers of 8-cent stamps are 0,1,2,3, leaving 27,19,11,3 cents, none divisible by 5. The next five amounts are possible: 28=8+4×5, 29=3×8+5, 30=6×5, 31=2×8+3×5, 32=4×8. Adding 5 repeatedly to these covers every amount above 27, so 27 is the largest impossible amount.

✅ Answer: B — 27

Revise: Algebra lessons.

⭐ Q9AMC Y7-8★★★AMC › Number Theory

The Problem

How many of these numbers are divisible by 3? 1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789

Your answer

❓ Pick the best answer:

A 2
B 3
C 4
D 5
E 6

🎯 Knowledge Point — what this tests

An AMC Number Theory problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

A number is divisible by 3 when its digit sum is. The digit sums are 1, 3, 6, 10, 15, 21, 28, 36, 45; six of them (3, 6, 15, 21, 36, 45) are multiples of 3.

✅ Answer: E — 6

Revise: Number Theory lessons.

⭐ Q10AMC Y7-8★★AMC › Combinatorics

The Problem

James chooses two electives — one from each group:
Group A: Mandarin, Japanese, Indonesian
Group B: French, German, Arabic, Italian
How many different pairs are possible?

Your answer

❓ Pick the best answer:

A 7
B 8
C 12
D 15
E 16

🎯 Knowledge Point — what this tests

An AMC Combinatorics problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

One choice from Group A (3 ways) and one from Group B (4 ways): 3 × 4 = 12 different pairs.

✅ Answer: C — 12

Revise: Combinatorics lessons.

⭐ Q11AMC Y7-8★★★AMC › Algebra

The Problem

I have four consecutive odd numbers. The largest is one less than twice the smallest. Which of the following is the largest of the four numbers?

Your answer

❓ Pick the best answer:

A 9
B 11
C 13
D 15
E 21

🎯 Knowledge Point — what this tests

An AMC Algebra problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

Let the smallest be x; the four odds are x, x+2, x+4, x+6. The largest is one less than twice the smallest: x + 6 = 2x − 1, so x = 7 and the largest is 13.

✅ Answer: C — 13

Revise: Algebra lessons.

⭐ Q12AMC Y7-8★★★AMC › Algebra

The Problem

A tank is $\frac{7}{12}$ full. After 30 litres are added it is $\frac{3}{4}$ full. What is the tank’s capacity, in litres?

Your answer

❓ Pick the best answer:

A 120
B 150
C 160
D 180

🎯 Knowledge Point — what this tests

An AMC Algebra problem. Read the conditions, set up the relationship and work to the answer — method beats guessing.

The 30 litres raised the level by $\frac{3}{4}$ − $\frac{7}{12}$ = $\frac{9}{12}$ − $\frac{7}{12}$ = $\frac{1}{6}$ of the tank. So $\frac{1}{6}$ of the capacity is 30 litres, giving a capacity of 180 litres.

✅ Answer: D — 180

Revise: Algebra lessons.

⭐ Q13AMC Y7-8★★★★AMC › Number Theory

The Problem

I start with a positive integer, multiply it by 10, and then subtract a positive multiple of 9 less than 100. My answer is 5347. What number did I start with?

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AMC Number Theory problem (a harder Q26-28 short-answer). Read the conditions, set up the relationship and work to the answer — method beats guessing.

10×(start) − 9k = 5347 with 9k < 100. For 10×start to end in 0, 5347 + 9k must end in 0, so 9k ends in 3: k = 7 (9×7 = 63). Then 10×start = 5410, so the start is 541.

✅ Answer: 541

Revise: Number Theory lessons.

⭐ Q14AMC Y7-8★★★★AMC › Combinatorics

The Problem

Amelia noticed that the names of three friends, Mei, Emma and Liam, were all made from letters of 'Amelia'. She chose different values from 0 to 9 for each of A, E, I, L and M so that M + E + I = E + M + M + A = L + I + A + M. What is the largest possible value of A + M + E + L + I + A?

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AMC Combinatorics problem (a harder Q26-28 short-answer). Read the conditions, set up the relationship and work to the answer — method beats guessing.

The two equalities give I = M + A and E = L + A, so A+M+E+L+I+A = 4A + 2L + 2M. Maximising over distinct digits gives A=3, M=6, L=5 (so I=9, E=8): A+M+E+L+I+A = 3+6+8+5+9+3 = 34. (Equivalently the target equals 2(E+I), and E+I ≤ 9+8 = 17.)

✅ Answer: 34

Revise: Combinatorics lessons.

⭐ Q15AMC Y7-8★★★★AMC › Geometry

The Problem

There are three sets of three parallel lines in a plane. Lines in different sets are not parallel, and every pair of non-parallel lines intersect. Find the largest possible number of intersection points and the smallest possible number of intersection points, then multiply these two numbers.

Your answer

❓ Enter your answer (a whole number):

🎯 Knowledge Point — what this tests

An AMC Number Theory problem (a harder Q26-28 short-answer). Read the conditions, set up the relationship and work to the answer — method beats guessing.

Parallel lines never meet, so points come only from different sets: at most 3 set-pairs × 3 × 3 = 27 points. Every point carries at most one line per family, so each intersection point uses 2 or 3 lines, and T triple points give 27 − 2T points. At most one line of the third family can pass through three of the nine grid points made by the first two families (two such lines would be forced to coincide), so T ≤ 3 + 2 + 2 = 7, giving a minimum of 27 − 14 = 13 — achieved by a 3×3 grid with diagonals y = x−1, x, x+1. Product = 27 × 13 = 351.

✅ Answer: 351

Revise: Geometry lessons.

🏁 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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