Every November, Waterloo's CEMC runs two contests side by side on the same morning: the Canadian Intermediate Mathematics Contest for Grades 9 and 10 and the Canadian Senior Mathematics Contest for Grades 11, 12 and CÉGEP. They share one shape, a two-hour paper split into six boxed answers and three written solutions, and one awkward fact printed on the cover: the first written question is usually easier than the last few boxed ones. This guide is built from CEMC's contest page, the covers of the 2025 booklets and CEMC's 2025 results booklet. It sets out what the paper asks, how far students got on each question in 2025, an order of attack built on that, a complete model solution of ours, seven practice problems of ours and a four-week plan to Wednesday, November 18, 2026.
Live online since 2020 · 10,000+ students · every practice problem on this page written by our teachers
In brief
The CIMC (Grades 9 and 10) and the CSMC (Grades 11, 12 and CÉGEP) are CEMC's November math contests: nine questions in two hours, six answer-only and three full-solution, scored out of 60 and written at school on Wednesday, November 18, 2026 in the Americas. Modern Age Coders teaches both papers live online, to batches or to single students, through its course Waterloo Math Contests: Gauss to Euclid.
Courses behind the November pair
Where our CIMC and CSMC preparation is taught
The contest course runs the timed papers and returns every written solution marked. The competition course is for the Part B question nobody finishes. The high school course closes the curriculum gaps that a CSMC paper finds quickly.
Everything in this table is from CEMC's Canadian Senior and Intermediate Mathematics Contests page, read on 22 October 2026, the same day CEMC's ordering window for schools closed.
The 2026 Canadian Intermediate and Senior Mathematics Contests, from CEMC (read 22 October 2026)
Question
CEMC's answer
Contest day
"North and South America: Wednesday, November 18, 2026". Schools elsewhere write on Thursday, November 19, 2026.
School ordering deadline
Thursday, October 22, 2026. That date has now arrived, so a family that has not yet asked should check with the school today whether an order went in.
Who writes the CIMC
"Students in Grades 9 and 10 are eligible to write the CIMC."
Who writes the CSMC
"Students in Grades 11 and 12 and CÉGEP students are eligible to write the CSMC."
The paper
"9 questions; 6 are answer only and 3 are full-solution." Two hours. "Score out of 60."
How it is written
In school, individually, on paper. Some calculators are permitted.
What math
CIMC: mostly topics "commonly taught up to and including Grade 10". CSMC: mostly the curriculum "up to and including the final year of secondary school".
Start time
As close to the start of the school day as possible, in the school's own time zone, on the official date.
Results
In the Contest Supervisor Portal four weeks after the contest date.
One shipping rule, two answers
CEMC has split its shipping rule by contest for schools in Canada. CSMC question booklets are no longer posted; instead the school prints them from a CEMC download area roughly a day ahead of the contest.
CIMC question booklets are still shipped to schools in Canada as before. Either way, the student sees a printed booklet on the desk, and the finished papers travel back to CEMC together for marking.
Official participants differ too
At a school in Canada, a CSMC writer counts as official when the Student Information Form is filled in properly, the student was born in 2008 or later, and the student lives in Canada.
For the CIMC the test is grade, not birth date: a writer is official if they are in Grade 10 or below and complete the form. Unofficial writers are still marked, and CEMC says they remain eligible for certificates and medals.
Source: CEMC, Canadian Senior and Intermediate Mathematics Contests, read 22 October 2026. Modern Age Coders is an independent teaching company with no connection to CEMC or the University of Waterloo; we cannot place contest orders or mark papers. On entry, CEMC states that "all students must participate through their school."
Two parts, one cover
Part A boxes, Part B arguments, and the line that links them
The front covers of CEMC's 2025 CSMC and CIMC booklets are word for word the same apart from the contest name. They describe a paper in two halves, and one sentence on them should change how a student spends the two hours.
The 2025 CSMC and CIMC booklet covers, quoted (no problems reproduced)
Part A
Part B
Size
"six questions, each worth 5 marks"
"three questions, each worth 10 marks"
What earns marks
"full marks will be given for a correct answer which is placed in the box"
"Marks are awarded for completeness, clarity, and style of presentation."
If the answer is wrong
"Part marks will be awarded only if relevant work is shown"
Written work is marked on its own merits, part by part
Working
In the space provided in the answer booklet
"Rough work should be done separately."
Order
Roughly easier to harder
Roughly easier to harder
The sentence that matters sits just above Part A: "The early problems in Part B are likely easier than the later problems in Part A." In other words the paper is not one ladder but two, and the bottom rung of the second ladder is lower than the top rungs of the first. A student who works straight through from question A1 will often arrive at Part B with twenty minutes left, having spent forty on two boxed questions worth 5 marks each, while a Part B question worth 10 marks, whose first part is close to routine, waits unread.
Why the parts of a Part B question matter
Part B questions are usually split into (a), (b) and (c) or more, climbing as they go. The early parts are often short and checkable, and each one earns marks on its own even if the last part is never finished.
So "attempt Part B" does not mean "solve Part B". It means collect the parts that are within reach, write them properly, and leave the final part for whatever time remains.
Calculators, briefly
The cover allows calculating devices without internet access, the ability to communicate with other devices, stored notes or programs, a computer algebra system or dynamic geometry software.
On a written solution, a number from the screen is not an argument. If a calculator gave the answer, the solution still has to show where it came from.
The 2025 papers in numbers
How each 2025 CSMC and CIMC question went
CEMC's 2025 results booklet prints the average mark on every question of both contests. Part A questions are out of 5 and Part B questions out of 10, so the two halves of the table are on different scales.
Average mark per question, 2025 CSMC and CIMC, from CEMC's 2025 CSIMC Results Booklet (Problem Averages)
Question
CSMC average
CIMC average
Out of
A1
4.5
4.7
5
A2
4.4
4.6
5
A3
3.5
4.5
5
A4
2.8
1.8
5
A5
0.8
1.0
5
A6
0.2
0.1
5
B1
7.3
6.9
10
B2
4.0
4.0
10
B3
0.7
1.1
10
Headline figures for 2025, from CEMC's contest page
Measure
CSMC 2025
CIMC 2025
Average (mean), out of 60
28.6
29.1
Certificate of Distinction cutoff
35
35
School Medal Champion cutoff
28
29
Number of contestants
15 153
13 039
Our reading, and it is only ours. In both contests the first written question, B1, averaged about 7 out of 10, while the last two boxed questions, A5 and A6, averaged 1 out of 5 or less. Measured as a share of the marks available, B1 went far better than A5 or A6, which is exactly what the cover predicts. B2 averaged 4 out of 10 in both contests, so a half-finished B2 is a respectable result, not a failure. And because a correct box on A5 or A6 was rare, a student who earns one is well ahead of the field on that question; a student who spends half an hour failing to earn one has paid for it on Part B. We have not added these averages together; CEMC publishes the overall means itself, and CEMC reports 28192 students enrolled across both contests in 2025.
Sources: CEMC's 2025 CSIMC Results Booklet (Problem Averages, Key Statistics and Enrollment) and the averages and cut-offs table on the CSIMC page, read 22 October 2026. The 2025 contests were written on Wednesday, November 12, 2025 in the Americas.
Order of attack
A 120-minute plan that respects both ladders
This is the order our students practise with from their first timed paper. It is ours, not CEMC's, and it is built on the cover sentence and the 2025 averages above.
Our order of attack for a two-hour CIMC or CSMC paper (ours)
Minutes
Work on
Rule
0 to 30
A1 to A4
Box each answer; write a line of working under any answer you are not sure of
30 to 55
B1, every part
Written properly the first time, with a closing sentence
55 to 80
B2, as many parts as possible
Each part finished before the next is started
80 to 100
A5 and A6
Ten minutes each, no more; working shown in case it earns part marks
100 to 115
B3 part (a), or the unfinished B2 part
Whichever is closer to a complete answer
115 to 120
Everything
Every box filled, every written part ending in a sentence that answers it
Why A5 and A6 move to the middle
They are worth 5 marks each and, on 2025 evidence, rarely earned. B1 and B2 are worth 10 each and earn marks part by part. Doing B1 and B2 before A5 and A6 protects the larger, more reliable marks first.
The boxed questions are not abandoned. They are simply taken when a stuck student can no longer lose a written part by lingering.
When to break the plan
A student who reads A5 and sees the idea at once should take it there and then. The plan is for the student who would otherwise stare at it.
A Grade 9 student writing a first CIMC may never reach A5 at all, and that is fine: A1 to A4 plus B1 and part of B2 is a good first paper.
Students resist this at first, because finishing Part A feels tidy. We run the same past paper twice, a week apart, once straight through and once in this order, and compare the two marks. That comparison usually settles the argument better than anything a teacher says.
A written solution of ours
A Part B question with three parts, solved the way markers want
The question is our own wording of a well-known fact about sums of consecutive integers, and the solution is written by us for our CSMC students; none of it comes from a CEMC paper. It shows a feature of good Part B writing that students miss: the parts do not have to be written in the order they are asked, as long as each one is labelled.
Run numbers · our three-part question and our complete solutiontext
Part B question (ours). Call a positive integer a run number
if it equals the sum of two or more consecutive positive integers.
(a) Show that 15 is a run number in three different ways.
(b) Explain why 64 is not a run number.
(c) Prove that no power of 2 is a run number.
(a) 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.
(c) first, because (b) then follows from it.
Suppose N is the sum of k consecutive positive integers
starting at a, where k >= 2 and a >= 1. Then
N = a + (a + 1) + ... + (a + k - 1) = k(2a + k - 1) / 2,
so 2N = k(2a + k - 1).
The two factors k and 2a + k - 1 add to 2a + 2k - 1,
which is odd, so one factor is odd and the other is even.
Both factors are at least 2: k >= 2 by assumption,
and 2a + k - 1 >= 2 + 2 - 1 = 3.
So 2N has an odd factor that is at least 3.
A power of 2 doubled is still a power of 2, and its only
odd factor is 1. So N cannot be a power of 2.
(b) 64 = 2^6 is a power of 2, so by (c) it is not a run number.
Therefore 15 is a run number in the three ways shown,
and no power of 2, including 64, is a run number.
What each piece of the write-up is doing (our classroom notes)
Piece
Its job
(a) as three sums on one line
A short-answer style part needs the answer, cleanly. Three examples are all part (a) asks for
Writing (c) before (b), and saying so
The proof in (c) makes (b) a one-line consequence. A clear label lets the marker follow the order
The formula 2N = k(2a + k - 1)
Turns a sentence about runs into an equation that can be argued about
The parity step
Two numbers whose sum is odd cannot both be even: this is the idea of the whole proof, stated with its reason
"Both factors are at least 2"
Without it, the odd factor might be 1 and the proof would prove nothing. This is the line most first drafts lack
The closing sentence
Answers all three parts in words, so nothing is left for the marker to assemble
Where the marks hide
Part (b) asked only to "explain why". A student who tests every run length from 2 to 10 and finds nothing has not explained anything, because the list could go on. A student who points to (c) has.
Part (c) is where the 2025 averages say most Part B marks are lost. A proof does not need to be long; it needs every step to have a reason a stranger can check.
A follow-up for strong students
The converse is also true: every positive integer that is not a power of 2 is a run number. Proving it means choosing a and k from an odd factor, and it makes a good extension once (c) feels easy.
We set the converse as homework only after a student can write (c) cleanly from memory in under ten minutes.
Seven practice problems
Original CIMC and CSMC practice, with solutions
All seven are our practice problems, not CEMC problems, and none is adapted from a past paper. Problems 1 to 5 are boxed-answer questions at roughly Part A level; the last two are three-part written questions at roughly B1 and B2 level. Every answer was checked in Python.
Problem 1.Ours, written for this page, not from CEMC. Part A level. A rectangle has perimeter 34 and a diagonal of length 13. What is its area?
Solution. Let the sides be l and w. Then l + w = 17 and l2 + w2 = 169. Since (l + w)2 = l2 + w2 + 2lw, we get 289 = 169 + 2lw, so lw = 60. The area is 60. The sides happen to be 5 and 12, but the question never needs them.
Problem 2.Ours, written for this page, not from CEMC. Part A level. How many positive divisors does 720 have?
Solution. 720 = 24 x 32 x 5. A divisor chooses an exponent of 2 from 0 to 4, of 3 from 0 to 2 and of 5 from 0 to 1, so there are 5 x 3 x 2 = 30 divisors. Listing them one by one is how a student loses ten minutes.
Problem 3.Ours, written for this page, not from CEMC. Part A level. A box holds 3 green and 5 yellow counters. Two counters are taken out at random, one after the other, without putting the first back. What is the probability that the two match in colour?
Solution. Count unordered pairs: 8 x 7 / 2 = 28 equally likely pairs in all. Green pairs: 3 x 2 / 2 = 3. Yellow pairs: 5 x 4 / 2 = 10. The probability is (3 + 10) / 28 = 13/28.
Problem 4.Ours, written for this page, not from CEMC. Part A level. A square is drawn inside a circle of radius 5 with all four corners on the circle. What is the area of the region inside the circle but outside the square? Give an exact answer.
Solution. The diagonal of the square is a diameter, 10, so the area of the square is 102 / 2 = 50. The circle has area 25π, so the region has area 25π - 50. Writing 28.54 would lose the box: the exact form is the answer.
Problem 5.Ours, written for this page, not from CEMC. Late Part A level. How many integers from 1 to 1000 inclusive are divisible by 6 or by 10, but not by 15?
Solution. Divisible by 6: 166. By 10: 100. By both, that is by 30: 33. So 166 + 100 - 33 = 233 are divisible by 6 or 10. A number divisible by 6 and by 15 is divisible by 30, and so is a number divisible by 10 and by 15, so the ones to remove are exactly the 33 multiples of 30. The answer is 233 - 33 = 200. The trap is subtracting the multiples of 15 directly, which counts numbers like 45 that were never in the list.
Problem 6.Ours, written for this page, not from CEMC. Part B level, written solution. A triangle has side lengths 13, 14 and 15. (a) Determine its area. (b) Determine the length of the altitude to the side of length 14. (c) Determine the radius of the circle that touches all three sides.
Problem 6 · our complete written solutiontext
Part B practice 2 (ours), triangle with sides 13, 14 and 15.
Put the side of length 14 on the x-axis, from B(0, 0) to C(14, 0),
and let the third vertex be A(x, h) with h > 0.
AB = 13: x^2 + h^2 = 169
AC = 15: (x - 14)^2 + h^2 = 225
Subtract: (x - 14)^2 - x^2 = 56, so -28x + 196 = 56 and x = 5.
Then h^2 = 169 - 25 = 144, so h = 12.
(a) Area = (1/2)(14)(12) = 84.
(b) The altitude to the side of length 14 is h = 12.
(c) Join the incentre I to A, B and C. The three small triangles
have height r and bases 13, 14 and 15, so
84 = (1/2) r (13 + 14 + 15) = 21r, and r = 4.
Part (a) gives part (b) for free here, because the coordinates already found the height. Part (c) uses a different idea, splitting the triangle at the incentre, and that change of idea is typical of a final part.
Problem 7.Ours, written for this page, not from CEMC. Part B level, written solution. Let f(x) = x2 - 6x + 13. (a) Write f(x) in the form (x - p)2 + q. (b) Determine the smallest value of f(x) and the value of x where it occurs. (c) Determine all real numbers x for which f(x) + f(x + 2) = 20.
Solution. (a) x2 - 6x + 13 = (x - 3)2 + 4. (b) A square is never negative, so f(x) is at least 4, with equality exactly when x = 3. The smallest value is 4, at x = 3. (c) Using (a), f(x + 2) = (x - 1)2 + 4. The equation becomes (x - 3)2 + (x - 1)2 + 8 = 20, so 2x2 - 8x + 10 = 12, which is x2 - 4x - 1 = 0. The quadratic formula gives x = 2 + √5 or x = 2 - √5. Both are real numbers and both satisfy the equation, so both are answers.
How to use the set
Sit problems 1 to 5 in 30 minutes with the boxes only, then compare. Any wrong answer with no working shown would have earned nothing on the day.
Write problems 6 and 7 as full solutions, timed at 25 minutes each, then check every line against ours. A missing reason is a missing mark.
The habit underneath
Problems 1 and 7 both reward reusing an earlier result instead of starting again. Problem 5 rewards reading the condition exactly. Problem 4 rewards exact forms.
Those habits are worth more on a November morning than any single formula, and they carry straight into the April contests.
Choosing the contest
CIMC or CSMC: which paper a student should write
The rules below are CEMC's; the advice beside them is ours. Most families find the choice is simple once the grade is clear.
Choosing between the CIMC and the CSMC (rules from CEMC, advice ours)
Student
CEMC's rule
Our advice
Grade 9
Eligible for the CIMC
Write the CIMC; aim for A1 to A4 and B1
Grade 10
Eligible for the CIMC
Write the CIMC; a strong Grade 10 student can aim at B2 in full
Grade 11
Eligible for the CSMC
Write the CSMC; it is also the route to CEMC's Auckland workshop for Grade 11 writers
Grade 12 or CÉGEP
Eligible for the CSMC
Write the CSMC in November and treat it as the first rehearsal for the Euclid in April
Younger and keen
CEMC welcomes motivated students from lower grades
Pick the paper that stretches without crushing, and enjoy it
What the November contests lead to
CEMC's student workshops page describes the Lloyd Auckland Invitational Mathematics workshop as bringing "together high-achieving Grade 11 Canadian Senior Mathematics Contest participants from Canada for a week-long immersive experience."
Invitations are CEMC's to make. We prepare students for the paper; we cannot nominate anyone, and we never suggest a score will lead to an invitation.
What Waterloo says about the CSMC
CEMC is clear that a Waterloo Math applicant does not have to have written the CSMC. The Faculty's scholarships page adds that writing the CSMC or the Euclid "is required for some scholarship decisions."
A Grade 12 student interested in those scholarships should read the Faculty's page directly. We make no claim about how any score is used.
The contest falls in the first month of winter time. Ontario, Quebec and the Atlantic provinces moved their clocks back on 1 November 2026, so the offsets below are winter ones.
UTC offsets on Wednesday, November 18, 2026 (tz database, release 2026e)
Province or zone
Offset
Newfoundland and Labrador (island time)
UTC-3:30
The Maritimes, on Atlantic time
UTC-4
Eastern time: Ontario and Quebec
UTC-5
Manitoba, now on one offset all year
UTC-5
Prairies west of Manitoba
UTC-6
Most of BC, and Yukon
UTC-7
Each school starts close to the beginning of its own school day, so there is no single national start. One consequence for our classes: from November, Toronto and Winnipeg read the same clock, which makes a shared evening batch easier to place. We still confirm every class time with each family in their own zone.
Four weeks
A four-week plan from 22 October to 18 November 2026
For a student whose school has ordered the contest, four weeks is enough to change how the two hours are spent, even if it is not enough to learn new mathematics. Our plan, with two live classes a week:
Week 1, 22 to 28 October 2026: one past paper straight through, untimed. Mark it together. Find out which Part A questions were lost to mathematics and which to time.
Week 2, 29 October to 4 November 2026: Part B writing. Two B1 questions from past papers and our problem 6, written in full and returned marked for completeness, clarity and presentation.
Week 3, 5 to 11 November 2026: the order of attack. One full paper timed at 120 minutes using the plan above, then the same paper reviewed for which marks the order saved.
Week 4, 12 to 18 November 2026: rehearsal and rest. One last timed paper early in the week, a short review of written-solution habits, and nothing heavy the evening before the contest.
Missed this year's order?
The Fryer, Galois and Hypatia Contests on Wednesday, April 7, 2027 are CEMC's next full-solution papers for Grades 9 to 11, ordered by Thursday, March 11, 2027.
Grade 12 or CÉGEP
The Euclid on Tuesday, April 6, 2027 is the next step; CSMC practice is close to ideal preparation for its middle questions.
Starting from scratch
Begin in September 2027 with a full term of written solutions before the next November paper.
Free CEMC help
CEMC's free CIMC and CSMC resources, and how we fit them in
CEMC publishes past papers, solutions and preparation material for the November contests at no cost. We send students there rather than copying any of it, and we write our own problems alongside.
Free CEMC material for the November contests (read 22 October 2026)
Free online lessons for the curriculum the contests rest on
Targeted review when a past paper shows a curriculum gap
CEMC Visits Schools
Problem-solving workshops run by CEMC staff at schools
A school-level option worth mentioning to a head of math
Students who want the broader picture of CEMC's year, from the Gauss Contest in Grades 7 and 8 to the Euclid, will find it on our CEMC Waterloo contests page. For a Grade 9 to 11 student the February multiple-choice papers come next: Pascal, Cayley and Fermat.
Placing a student
Four stages on a two-part paper
Our stages, used in the first class to decide where a student starts.
From boxes only to both ladders (ours)
Stage
What we see
1. Box filler
Answers A1 to A3 correctly and leaves Part B nearly blank
2. Starter
Earns B1 (a) and (b) but writes them as bare numbers
3. Writer
Finishes B1 and most of B2 with reasons, and boxes A1 to A4
4. Climber
Takes a part of B3 or a box from A5 and A6 on most papers
For parents
Stage 1 is where most good school students begin, because school tests rarely ask for an argument written for a stranger. The jump to stage 3 is a writing habit, and habits take weekly practice.
A November paper is a useful checkpoint even for a student whose real goal is April. It shows, early in the year, exactly which kind of mark is being lost.
For students
Read all three Part B questions in the first five minutes. Knowing what is waiting changes how long you will tolerate a stuck box.
Label every written part clearly, and end each one with a sentence that answers it.
Further courses
Nine courses for a Grade 9 to 12 contest student
Live online classes with one teacher who follows the student all year, in a batch or one to one, joined from any province.
I
Grades 9 and 10
The CIMC years
CCCG / MID / 01
Algebra foundations
Linear and quadratic work done until it is automatic.
A weather forecasting site with live conditions for any location.
by Krish
Fees
Fees
Any contest fee CEMC charges goes through the school. Ours is for the teaching alone, billed by the month, and a student who only wants the November run can stop after it.
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Families ask
CIMC and CSMC: common questions
What are the CIMC and the CSMC?
They are the Canadian Intermediate Mathematics Contest and the Canadian Senior Mathematics Contest, run by CEMC at the University of Waterloo on the same day each November. The CIMC is for Grades 9 and 10, the CSMC for Grades 11, 12 and CEGEP.
When are the CIMC and CSMC in 2026?
CEMC lists Wednesday, November 18, 2026 in North and South America and Thursday, November 19, 2026 elsewhere. The school ordering deadline was Thursday, October 22, 2026.
What does the paper look like?
Nine questions in two hours, scored out of 60. Part A has six answer-only questions worth 5 marks each; Part B has three full-solution questions worth 10 marks each, marked for completeness, clarity and presentation.
Should my child finish Part A before starting Part B?
Not necessarily. The booklet cover itself says the early Part B problems are likely easier than the later Part A problems, and in 2025 the first Part B question averaged about 7 out of 10 in both contests. We teach an order that takes the first two Part B questions before the last two boxes.
What were the 2025 averages?
CEMC lists a 2025 mean of 28.6 out of 60 on the CSMC and 29.1 on the CIMC, with a Certificate of Distinction cutoff of 35 on both. There were 15 153 CSMC and 13 039 CIMC contestants.
Can a student register for the CIMC or CSMC alone?
No. CEMC contests are ordered by schools, and the student writes at school under supervision. If your school does not take part, CEMC suggests asking another local school whether your child can join its writers.
Is the CSMC required for University of Waterloo admission?
No. CEMC says the CSMC is not required for admission to the Faculty of Mathematics, and the Faculty says writing the CSMC or the Euclid is required for some scholarship decisions. We make no promises about admissions or scholarships.
Are calculators allowed?
Some are. The 2025 cover allows calculating devices without internet access, communication with other devices, stored notes or programs, a computer algebra system or dynamic geometry software. Written solutions still have to show the reasoning.
What if my child missed the 2026 order?
The next CEMC full-solution contests are the Fryer, Galois and Hypatia Contests on Wednesday, April 7, 2027 for Grades 9 to 11, and the Euclid on Tuesday, April 6, 2027 for the final year. Schools order both by Thursday, March 11, 2027.
When do your CIMC and CSMC classes run?
Weekdays or weekends, flexible: choose your batch, or for a very personalised plan choose 1-on-1 at your convenience, weekend mornings or weekdays. Times are set in Eastern time and confirmed in your own zone, which matters in November because the eastern provinces have just changed their clocks.
Keep reading
Related pages
Other pages for Grade 9 to 12 students in Canada following CEMC's contest year.
In the first class your child reads a Part B question of ours, writes the first two parts, and watches a teacher mark them the way a CEMC marker would. You come away knowing which of the four stages fits and what the weeks before 18 November 2026 should look like.
Prefer to look around first? How we teach walks through a real class, and the full course list has every syllabus.
Classes are taught live on video by our teachers in India, so families join from home anywhere in Canada. WhatsApp is the quickest way to reach us.
Want a class sooner? Book a priority demo
A priority demo is a full class of about 45 to 60 minutes. After the class the mentor sends a written skill report and a personal learning roadmap on WhatsApp. If you enrol, the demo fee is adjusted against your first month's fee. Share your demo Order ID with us on WhatsApp and we take it off.
