Andrew, twelve weeks on the clock: eight and a half to the first sitting, the rest held for the retake. One page, three jobs: work out where you stand, commit to a rhythm, and taste the book that waits at the top of the hill.
DiagnosticTue 13 Oct 2026
First sittingThu 10 Dec 2026
Hedge retakeTue 5 Jan 2027
Weekly budget15 to 20 hours
Tonight, the night before the diagnostic: do not study. Sit Official Practice Exam 1 tomorrow timed, in one 2h15m sitting, sections in the order you plan to use on test day. The moment it ends, write down four numbers: total, Quant, Verbal, Data Insights. Question 2 below is where they go.
Part A. Where you stand
The book on your desk, Manhattan Prep GMAT Advanced Quant, has an unusually blunt front door. It says, in effect: come in with a Quant scaled score of 80 or better out of 90, having already worked through the core Quant syllabus (algebra; fractions, decimals, percents and ratios; number properties; word problems), aiming to push to 83+. If you do not match that, it says, you will find it too difficult at this stage, and you should go build the foundations first and come back. That is not a discouragement, it is a map: this book belongs to the middle and back of your twelve weeks, not the front. Part A works out exactly where the door is for you.
Part B. Commitments
Plans fail in the gap between intention and calendar. This part is not about enthusiasm; it is about slots. Answer with days and hours you believe on a bad week, not a good one. The statistics course eats about 16 hours a week, so honesty here is the whole game.
Part C. First taste: eight problems from the book
These are the book's own problems, quoted for your private study. Give each one an honest two to three minutes with pen and paper; the book itself says of its hardest ones that you should not expect to solve them in two minutes, and that getting stuck is the moment to switch gears: eliminate what you can, guess, move on. Being stuck here is data, not failure. Every card names its chapter, so a problem that bites you tells you exactly where to go in December's edition of yourself. Answers hide behind each Show answer link.
How to read your score at the end: 0 to 2 correct is the normal cold start after nineteen years off maths, and it is precisely what the foundations fortnight (19 Oct to 1 Nov) is for. 3 to 5 means the door is nearer than the book expects. 6 to 8 means start Chapter 1 alongside foundations in week 2 and we talk about accelerating the plan.
Introduction: Try-It #0-2hard, the book's own door test
The Introduction tests you with three problems before it will let you in, and warns they are at least as hard as anything on the real GMAT. This is one of them.
When Mira exercises, she likes to do pushups in short sets as follows: 1 pushup, then 2 pushups, then 3 pushups, and so on, with the number of pushups in each set ascending consecutively until a set of n pushups, where n is a positive integer, after which the number of pushups in each set descends consecutively until a final set of just 1 pushup. If Mira did less than 100 total pushups, what is the maximum number of pushups she could have done following this pattern?
(A) 64
(B) 72
(C) 81
(D) 90
(E) 97
Show answer and where to go next
Answer: (C) 81.
The whole pyramid 1 + 2 + ... + n + ... + 2 + 1 equals twice the triangle sum minus n: 2 × n(n+1)/2 − n = n². Every valid total is a perfect square, and the largest square under 100 is 81 = 9². (Going to n = 10 gives exactly 100, which is not less than 100.)
Where to go next: Introduction, then Chapter 3 (Pattern Recognition). The hidden perfect square here is the sum-of-consecutive-integers idea the chapter teaches explicitly.
Chapter 1: Problem Set #3easy
A baker makes a combination of chocolate chip cookies and peanut butter cookies for a school bake sale. His recipes only allow him to make chocolate chip cookies in batches of 7 and peanut butter cookies in batches of 6. If he makes exactly 95 cookies for the bake sale, what is the minimum number of chocolate chip cookies that he could make?
(A) 7
(B) 14
(C) 21
(D) 28
(E) 35
Show answer and where to go next
Answer: (E) 35.
Write 7C + 6P = 95 and, because the choices are small multiples of 7, test them from the bottom up: 95 − 7 = 88, 95 − 14 = 81, 95 − 21 = 74 and 95 − 28 = 67 are none divisible by 6, but 95 − 35 = 60 = 6 × 10 works. Five batches of chocolate chip, ten of peanut butter.
Where to go next: Chapter 1, whose four principles (Understand, Plan, Solve, Review) are its whole message: pen to paper, test, check.
Chapter 2: Try-It #2-4medium
If n is a positive integer, what must be true of n³ − n?
(A) It is divisible by 4.
(B) It is odd.
(C) It is a multiple of 6.
(D) It is a prime number.
(E) It has, at most, two distinct prime factors.
Show answer and where to go next
Answer: (C) It is a multiple of 6.
Test cases to kill the others: n = 1 gives 0, which is neither odd nor prime; n = 2 gives 6, not divisible by 4; n = 5 gives 120 = 2 × 2 × 2 × 3 × 5, three distinct prime factors. Why (C) always holds: n³ − n = (n − 1) × n × (n + 1), three consecutive integers, so the product always contains an even number and a multiple of 3.
Where to go next: Chapter 2, the Test Cases tactic and its siblings (Work Backwards, Choose Smart Numbers), plus the guessing tactics for when nothing else is available.
Chapter 3: Problem Set #3medium
What is the remainder when 13¹⁷ + 17¹³ is divided by 10?
Show answer and where to go next
Answer: 0.
A remainder on division by 10 is just the units digit, so only the last digits of the bases matter: 3 and 7. Units digits of powers of 3 cycle 3, 9, 7, 1 with period 4, and 17 is 1 past a full cycle, so 13¹⁷ ends in 3; likewise 7's cycle 7, 9, 3, 1 makes 17¹³ end in 7. The sum's last digit is 3 + 7 = 10, so the remainder is 0.
Where to go next: Chapter 3, the units-digit cycle section: the chapter's law that big powers are never computed, only cycled.
Chapter 4: Try-It #4-6easy, ten seconds if you see it
What is 198 × 202?
Show answer and where to go next
Answer: 39,996.
Rewrite the factors as (200 − 2)(200 + 2), a difference of squares in disguise: 198 × 202 = 200² − 2² = 40,000 − 4 = 39,996. No long multiplication ever happens.
Where to go next: Chapter 4, Quadratic Templates in Disguise: the algebra identities you knew in 2005, wearing costumes.
Chapter 5: Try-It #5-6medium
A convenience store stocks soda in 12-ounce and 24-ounce bottles. If the average capacity of all the bottles in the store is 22 ounces, then what fraction of the bottles in the store are 12 ounces?
Show answer and where to go next
Answer: 1/6.
Picture a teeter-totter balanced at the average, 22. The 12 end sits 10 units away, the 24 end 2 units away; each side's share of the bottles equals the distance to the opposite end. So the 24-ounce bottles are 10/12 = 5/6 of the stock and the 12-ounce bottles are 2/12 = 1/6. Check: (1/6)(12) + (5/6)(24) = 2 + 20 = 22.
Where to go next: Chapter 5, the number line and teeter-totter pictures for averages and statistics: the chapter that turns mean and median questions into drawings.
Chapter 6: Try-It #6-4medium
A carnival card game gives the contestant a one in three probability of successfully choosing the correct card and thereby winning the game. If a contestant plays the game repeatedly, what is the minimum number of times that he must play the game so that the probability that he never loses is less than 1%?
Show answer and where to go next
Answer: 5.
Never losing means winning every time: probability (1/3)ⁿ after n plays. You need (1/3)ⁿ < 1/100, that is 3ⁿ > 100. Since 3⁴ = 81 is not enough and 3⁵ = 243 clears it, the minimum is 5. One probability idea, one exponent idea, one inequality: a hybrid, sequenced.
Where to go next: Chapter 6, Hybrid Problems: the skill of naming the two or three skills inside one question and doing them in the right order.
Introduction: the Illustrationdessert, harder than the GMAT and proud of it
Last one, for fun. The book warns this one is not exactly GMAT-like, because no rule is mathematically definite, but says: you will know when you have solved it, because the answer will be elegant.
What should the next number in this sequence be?
1 2 9 64 ___
Show answer and where to go next
Answer: 625.
The standard probes dead-end: the differences 1, 7, 55 mean nothing, and reading 9 as 3² and 64 as 8² breaks on the 2. The breakthrough is writing every term as a power: 1 = 1⁰, 2 = 2¹, 9 = 3², 64 = 4³. The base climbs 1, 2, 3, 4 while the exponent stays one behind it, so the next term is 5⁴ = 625, which is even 25², an elegance the book says announces the answer. Its lesson: the reading of 64 as 8² had to be actively abandoned, and letting go of a good-looking dead end is itself a top-scorer skill.
Where to go next: the Introduction's An Illustration section, where this problem is worked slowly as a demonstration of how to think, not what to know.