Young Fermats · Algebra 1 · Placement Diagnostic · Grades 7–9 · NYC Math Class
Algebra 1 Placement Diagnostic — Real Numbers, Order of Operations & Simplifying Expressions (30 Questions with Hidden Answers)
The pre-Algebra 1 checkpoint. Before you touch a single linear equation, you need three things absolutely automatic: the real-number system, the order of operations, and simplifying algebraic expressions with variables. This diagnostic teaches the theory for all three, then tests them with 30 questions — 10 easy, 12 medium, 8 hard (including word problems) — each with click-to-reveal step-by-step answers. Built as Class 0 of the SOMATH Young Fermats Algebra Ignite program on the Upper West Side of Manhattan.
Most students who struggle in Algebra 1 do not struggle with algebra — they struggle with the arithmetic and structure that algebra assumes. This diagnostic is designed to catch every one of those gaps in a single sitting. It is the same test every student takes on Day 1 at SOMATH before starting Algebra Ignite.
Three ideas, in this order, are the entire prerequisite:
- The real-number system — knowing what kind of number you are working with.
- The order of operations — knowing what to do first when a problem has several operations stacked on top of each other.
- Simplifying expressions with variables — distributing, combining like terms, and evaluating.
All three are practiced together in real algebra work every day. This diagnostic teaches each one, then mixes them.
- Read the three theory sections aloud (20 min). No skipping — even students who “know it” often have holes.
- Take all 30 questions with answers hidden (40–60 min). No calculator. Show all work.
- Reveal answers one at a time. Score honestly.
- Below 24/30 → work through the missed sections and retake. 24–27/30 → ready with light review. 28+/30 → ready to start Class 1.
What’s in this diagnostic
- The three prerequisite ideas
- Theory 1: the real-number system
- Theory 2: the order of operations (PEMDAS)
- Theory 3: simplifying expressions with variables
- Common mistakes (memorize these before the test)
- 30 diagnostic questions (10 easy, 12 medium, 8 hard)
- Answer key summary
- Scoring & placement guidance
- About SOMATH & Algebra Ignite
1. The three prerequisite ideas
These three skills are not “review.” They are the language algebra is written in. A student who is shaky on any one of them will feel lost by Class 3, and by Class 6 they will believe they “can’t do math.” That is a fixable problem — but only if we catch it here.
2. Theory 1: the real-number system
Every number has a name and a home
The real numbers are every number you can place on a number line. Real numbers split into an infinite nesting of sets. Each set is a subset of the next one out.
The sets nest. Every natural number is a whole number, every whole number is an integer, every integer is rational, every rational is real. Rationals and irrationals are disjoint — a number is one or the other, never both, never neither.
Two rules that catch 90% of student mistakes:
- Every whole number is an integer. Every integer is a rational number. Every rational number is a real number. The sets nest — they do not overlap in strange ways.
- Every real number is either rational or irrational — never both, never neither. There is no third category.
Rational vs. irrational — the fastest test: a number is rational if its decimal expansion terminates (like 0.75) or repeats in a fixed pattern (like 0.3333… or 0.142857142857…). If the decimal goes on forever with no repeating pattern, it is irrational.
| Number | Classification (smallest set it belongs to) |
|---|---|
7 | Natural (also whole, integer, rational, real) |
0 | Whole (also integer, rational, real) — but not natural |
−4 | Integer (also rational, real) — but not whole |
3/8 | Rational (also real) — but not integer |
0.75 | Rational (equals 3/4) |
0.3333… | Rational (equals 1/3) — repeating decimal |
√9 | Natural (equals 3) — perfect squares are rational |
√2 | Irrational (also real) — not rational |
π | Irrational (also real) |
22/7 | Rational — a rational approximation of π, but not equal to π |
The biggest trap: √9 = 3 is a natural number. √10 is irrational. The square root symbol is not what makes a number irrational — it’s whether the number under the root is a perfect square.
3. Theory 2: the order of operations (PEMDAS)
One expression, one answer — only if everyone follows the same rules
The order of operations is the convention every mathematician and calculator on Earth agrees to. It says: when an expression has more than one operation, do them in this order:
| Order | Operation | Detail |
|---|---|---|
| 1 | Parentheses (and any other grouping symbols) | Brackets, braces, the top and bottom of a fraction bar, and the inside of a square root are all “parentheses.” |
| 2 | Exponents (and roots) | Powers first, then roots at the same level. |
| 3 | Multiplication and Division | Left to right, in the order they appear. Not multiplication first. |
| 4 | Addition and Subtraction | Left to right, in the order they appear. Not addition first. |
Example 1 — the classic misread: 8 ÷ 2 × 4
Multiplication and division are the SAME level. Do them LEFT TO RIGHT. 8 ÷ 2 × 4 = 4 × 4 ← division first, because it’s on the left = 16Students who think “M before D” will do 2 × 4 = 8 first and get 1. That is wrong. PEMDAS is really P / E / M D / A S — four levels, not six.
Example 2 — exponents outrank negatives: −3² vs. (−3)²
−3² = −(3 × 3) = −9 ← exponent applies to 3 only (−3)² = (−3)(−3) = +9 ← parentheses include the negativeThe negative sign in −3² is not part of the base — it’s subtraction (or the “opposite of” operator) applied after the exponent. This distinction is worth points on every algebra exam.
Example 3 — a full stacked expression: 4 + 2 × (3 + 1)² ÷ 8 − 5
4 + 2 × (3 + 1)² ÷ 8 − 5 = 4 + 2 × (4)² ÷ 8 − 5 ← P: inside the parentheses = 4 + 2 × 16 ÷ 8 − 5 ← E: exponent = 4 + 32 ÷ 8 − 5 ← MD, left to right: multiplication first = 4 + 4 − 5 ← MD, left to right: then division = 8 − 5 ← AS, left to right: addition first = 3The fraction-bar rule: a fraction bar acts like invisible parentheses around the entire top and the entire bottom. Simplify the numerator, simplify the denominator, then divide.
3² + 7 9 + 7 16 ───────── = ─────── = ─────── = 4 10 − 6 10 − 6 44. Theory 3: simplifying expressions with variables
Distribute, combine like terms, order the result
An algebraic expression is a combination of numbers, variables, and operations — but no equals sign. 3x + 5, 2(y − 4) + 7y, and 4a²b are all expressions.
Vocabulary you need in one paragraph: a term is a piece of an expression separated by + or −. A term’s coefficient is the number multiplied by the variable part. Two terms are like terms only if their variable parts (letters and exponents) match exactly. A constant is a term with no variable.
| Expression | Terms | Coefficients | Constants |
|---|---|---|---|
3x + 5 | 3x, 5 | 3 | 5 |
7x² − 4x + 9 | 7x², −4x, 9 | 7, −4 | 9 |
2xy − y + 6 | 2xy, −y, 6 | 2, −1 | 6 |
Simplifying is always three steps in this order:
- Distribute to remove parentheses: a(b + c) = ab + ac.
- Combine like terms by adding or subtracting coefficients.
- Write the result in conventional order — usually highest exponent first, constant last.
Worked example: Simplify 3(x + 4) − 2(x − 5) + 7.
3(x + 4) − 2(x − 5) + 7 = 3x + 12 − 2x + 10 + 7 ← distribute (watch the sign of the −2) = (3x − 2x) + (12 + 10 + 7) ← group like terms = x + 29The distributive property with a minus sign is the #1 error zone. −2(x − 5) becomes −2x + 10, not −2x − 10. The minus sign flips the sign of every term inside the parentheses.
Combining like terms rules:
- 3x + 5x = 8x (same variable part)
- 7x² − 2x² = 5x² (same variable part with same exponent)
- 3x + 5x² = 3x + 5x² (DIFFERENT variable parts — cannot combine)
- 4xy + 2yx = 6xy (multiplication is commutative — xy = yx)
- 3x + 4 = 3x + 4 (a variable term and a constant are never like terms)
Evaluating an expression means substituting a specific number for each variable and using the order of operations. If x = −2, then 3x² − 5x + 1 = 3(−2)² − 5(−2) + 1 = 3(4) + 10 + 1 = 23. Always put substituted numbers in parentheses — especially negatives.
5. Common mistakes (memorize these before the test)
Nine landmines — every student steps on at least one
- √9 is rational, not irrational. √9 = 3. Only non-perfect-square roots are irrational.
- −3² is not +9. Exponents happen before the minus sign. −3² = −9. To get +9 you need (−3)².
- 8 ÷ 2 × 4 is 16, not 1. Multiplication and division are the same level; go left to right.
- 3 + 4 × 2 is 11, not 14. Multiplication before addition, always.
- −2(x − 5) = −2x + 10 (the minus flips both signs), not −2x − 10.
- 3x and 5x² are not like terms. The exponents must match.
- Evaluating with negatives requires parentheses. If x = −3 then x² = (−3)² = 9. Writing −3² loses the parentheses and returns −9.
- A fraction bar is invisible parentheses. Simplify top and bottom separately before you divide.
- Every real number is rational or irrational — never both. A number is not “both” because it’s written two ways.
30 diagnostic questions (10 easy · 12 medium · 8 hard)
No calculator. Show all work. Reveal answers one at a time.
Question 1 Easy
Classify the number −7. List every set of the real-number system it belongs to (natural, whole, integer, rational, irrational, real).
Answer: integer, rational, real
−7 is a negative counting number, so it is an integer. Every integer is rational (it can be written as −7/1). Every rational number is real. It is not natural or whole (both require the number to be zero or positive), and it is not irrational.
Question 2 Easy
Evaluate: 4 + 3 × 5
Answer: 19
4 + 3 × 5 = 4 + 15 ← multiplication first = 19Question 3 Easy
Evaluate: 20 − 6 ÷ 2
Answer: 17
20 − 6 ÷ 2 = 20 − 3 ← division before subtraction = 17Question 4 Easy
Simplify by combining like terms: 6x + 3x
Answer: 9x
Same variable part (just x), so add the coefficients: 6 + 3 = 9.
Question 5 Easy
Simplify: 8y − 5y + 2y
Answer: 5y
8y − 5y + 2y = (8 − 5 + 2)y = 5yQuestion 6 Easy
Classify each number as rational or irrational: (a) 0.75, (b) π, (c) √16.
Answer: (a) rational; (b) irrational; (c) rational
(a) 0.75 = 3/4 — a fraction of integers, so rational. (b) π is the classic irrational number — its decimal never repeats. (c) √16 = 4, which is a natural number and therefore rational. The square root symbol does not make a number irrational; only non-perfect-square roots are irrational.
Question 7 Easy
Evaluate: (6 + 2) × 3
Answer: 24
(6 + 2) × 3 = 8 × 3 ← parentheses first = 24Question 8 Easy
Simplify: 4x + 7 + 2x − 3
Answer: 6x + 4
4x + 7 + 2x − 3 = (4x + 2x) + (7 − 3) ← group like terms = 6x + 4Question 9 Easy
Evaluate 5². Then evaluate 2³.
Answer: 25 and 8
5² = 5 × 5 = 25 2³ = 2 × 2 × 2 = 8Question 10 Easy
Distribute: 3(x + 5)
Answer: 3x + 15
3(x + 5) = 3·x + 3·5 = 3x + 15Question 11 Medium
Evaluate: 12 ÷ 4 × 3 + 1
Answer: 10
12 ÷ 4 × 3 + 1 = 3 × 3 + 1 ← division first (left to right) = 9 + 1 ← then multiplication = 10Multiplication and division are the same level. Going left to right, division comes first. A student who does 4 × 3 = 12 first will get 12 ÷ 12 + 1 = 2 — wrong.
Question 12 Medium
Evaluate: −4² + (−4)²
Answer: 0
−4² = −(4 × 4) = −16 (−4)² = (−4)(−4) = +16 Sum: −16 + 16 = 0The parentheses matter. Without them, the exponent applies only to the 4, and the negative is applied after. This shows up on almost every Algebra 1 quiz.
Question 13 Medium
Simplify: 2(x + 3) + 4(x − 1)
Answer: 6x + 2
2(x + 3) + 4(x − 1) = 2x + 6 + 4x − 4 ← distribute both = (2x + 4x) + (6 − 4) ← group like terms = 6x + 2Question 14 Medium
Simplify: 5(2y − 3) − 3(y + 4)
Answer: 7y − 27
5(2y − 3) − 3(y + 4) = 10y − 15 − 3y − 12 ← distribute; watch the −3 flipping the +4 to −12 = (10y − 3y) + (−15 − 12) = 7y − 27The −3 distributes over both y and +4. It becomes −3y and −12. Missing the sign on the constant is the most common mistake in Algebra 1.
Question 15 Medium
Evaluate: 3 + 2(5 − 1)² ÷ 4
Answer: 11
3 + 2(5 − 1)² ÷ 4 = 3 + 2(4)² ÷ 4 ← P = 3 + 2·16 ÷ 4 ← E = 3 + 32 ÷ 4 ← MD, left to right: multiplication first = 3 + 8 ← then division = 11Question 16 Medium
If x = −2, evaluate 3x² − 4x + 7.
Answer: 27
Substitute x = −2 (use parentheses!): 3(−2)² − 4(−2) + 7 = 3·4 − (−8) + 7 ← (−2)² = 4; −4·(−2) = +8 = 12 + 8 + 7 = 27Always wrap the substituted value in parentheses. That way (−2)² = 4 instead of the mistaken −2² = −4.
Question 17 Medium
Simplify: 7x + 2y − 3x + 5y − y
Answer: 4x + 6y
7x + 2y − 3x + 5y − y = (7x − 3x) + (2y + 5y − y) = 4x + 6yOnly combine like terms. x terms combine with x terms; y terms combine with y terms. They cannot merge into one.
Question 18 Medium
Evaluate the fraction: (4² − 1) ÷ (2² + 1)
Answer: 3
Treat the fraction bar as parentheses on top and bottom: Top: 4² − 1 = 16 − 1 = 15 Bottom: 2² + 1 = 4 + 1 = 5 Divide: 15 ÷ 5 = 3Question 19 Medium
Which of the following is not a rational number? (a) 0.4, (b) −9, (c) √7, (d) 22/7. Explain.
Answer: (c) √7 is not rational
0.4 = 2/5 is rational. −9 is an integer and therefore rational. 22/7 is a fraction of integers, so it is rational (it is not equal to π — it is only a close approximation). √7 is irrational because 7 is not a perfect square: its decimal expansion is nonrepeating and nonterminating.
Question 20 Medium
Simplify: 4(3x − 2) − (x − 6)
Answer: 11x − 2
4(3x − 2) − (x − 6) = 12x − 8 − x + 6 ← the minus in front of (x − 6) means −1(x − 6) = −x + 6 = (12x − x) + (−8 + 6) = 11x − 2The bare minus sign in front of parentheses is really −1. It flips every sign inside.
Question 21 Medium
Evaluate: (−3)² − 2(4 − 7) + 5
Answer: 20
(−3)² − 2(4 − 7) + 5 = (−3)² − 2(−3) + 5 ← P: inside the second parentheses = 9 − 2(−3) + 5 ← E = 9 − (−6) + 5 ← M = 9 + 6 + 5 ← subtracting a negative = adding = 20Question 22 Medium
If a = 3 and b = −5, evaluate 2a − 3b + ab.
Answer: 6
2a − 3b + ab = 2(3) − 3(−5) + (3)(−5) = 6 − (−15) + (−15) = 6 + 15 − 15 = 6Question 23 Hard · Word Problem
A cell-phone plan charges a flat monthly fee of $25 plus 8 cents per minute of talk time. Write an algebraic expression for the total monthly cost in dollars if you talk for m minutes. Then evaluate the expression for m = 175 minutes.
Answer: expression = 25 + 0.08m; cost at 175 min = $39.00
Total cost = fixed fee + (rate per min × minutes) = 25 + 0.08m Evaluate at m = 175: = 25 + 0.08(175) = 25 + 14 = 39Every “flat fee plus a per-unit rate” word problem in Algebra 1 has this exact form: y = b + rx. Getting fluent at translating words into 25 + 0.08m is the single biggest predictor of success in linear equations.
Question 24 Hard · Word Problem
The perimeter of a rectangle is given by P = 2L + 2W. A rectangle has length L = 3x + 2 and width W = x − 4. Write a simplified expression for its perimeter P, then evaluate P when x = 7.
Answer: P = 8x − 4; when x = 7, P = 52
P = 2L + 2W = 2(3x + 2) + 2(x − 4) = 6x + 4 + 2x − 8 ← distribute = (6x + 2x) + (4 − 8) ← combine = 8x − 4 At x = 7: P = 8(7) − 4 = 56 − 4 = 52Two skills at once: simplifying and evaluating. Distribute before you combine like terms. Only after the expression is fully simplified should you plug in the number.
Question 25 Hard
Evaluate: [3 + (5 − 8)²] ÷ (−2)² − 3
Answer: 0
[3 + (5 − 8)²] ÷ (−2)² − 3 = [3 + (−3)²] ÷ (−2)² − 3 ← P: innermost = [3 + 9] ÷ (−2)² − 3 ← E: inner exponent = 12 ÷ 4 − 3 ← simplify brackets; (−2)² = 4 = 3 − 3 ← D = 0Nested grouping symbols work from the inside out. (−3)² = 9 because the parentheses include the negative.
Question 26 Hard · Word Problem
A student has n nickels and twice as many dimes as nickels. Write a simplified expression for the total value of the coins in cents. Then find the total in cents if n = 12.
Answer: expression = 25n cents; at n = 12, total = 300 cents ($3.00)
Nickels: n coins × 5 cents each = 5n Dimes: 2n coins × 10 cents each = 20n Total = 5n + 20n = 25n cents At n = 12: Total = 25(12) = 300 cents = $3.00Two coin quantities, both a multiple of n, so their values add cleanly. The English “twice as many dimes as nickels” translates directly to 2n. This is the template every mixture/coin word problem uses in Algebra 1.
Question 27 Hard
Simplify: 3x² − 2(x² − 4x) + 5x
Answer: x² + 13x
3x² − 2(x² − 4x) + 5x = 3x² − 2x² + 8x + 5x ← distribute −2 over both terms = (3x² − 2x²) + (8x + 5x) = x² + 13xNow both like-term families are present: x² terms combine with x² terms, and x terms combine with x terms. They do NOT combine with each other. The −2 × −4x = +8x sign flip is the step where most students slip.
Question 28 Hard · Word Problem
A rectangular garden is x feet wide and 2x + 3 feet long. It is surrounded by a walkway that is 2 feet wide on all four sides. Write a simplified expression for the outer perimeter (garden + walkway), then find it when x = 5 feet.
Answer: outer perimeter = 6x + 22; at x = 5, perimeter = 52 feet
Adding a 2-ft walkway on each side increases width AND length by 4 ft (2 ft on each of the two opposite sides). Outer width = x + 4 Outer length = (2x + 3) + 4 = 2x + 7 Outer perimeter = 2(outer length) + 2(outer width) = 2(2x + 7) + 2(x + 4) = 4x + 14 + 2x + 8 = 6x + 22 At x = 5: = 6(5) + 22 = 30 + 22 = 52 feetTwo lessons in one problem. First: adding a 2-ft border on both sides adds 4 ft total to each dimension — not 2. Second: distribute carefully, then combine like terms, then substitute. This is exactly the workflow in every geometry-based algebra problem.
Question 29 Hard
Simplify: 4(2a − 3b) − 2[3a − (a + b)] + 5b
Answer: 4a − 5b
4(2a − 3b) − 2[3a − (a + b)] + 5b Step 1: innermost parentheses first. −(a + b) = −a − b So the bracketed piece becomes 3a − a − b = 2a − b. Step 2: substitute back. = 4(2a − 3b) − 2(2a − b) + 5b Step 3: distribute. = 8a − 12b − 4a + 2b + 5b Step 4: combine like terms. = (8a − 4a) + (−12b + 2b + 5b) = 4a + (−5b) = 4a − 5bNested grouping symbols (brackets outside parentheses) are graded from the inside out. Once the inner minus is distributed, the rest is standard combine-like-terms work. This exact structure shows up in every Regents Algebra 1 exam.
Question 30 Hard · Word Problem
Challenge: A gym offers two membership plans. Plan A charges a $60 one-time enrollment fee plus $40 per month. Plan B charges no enrollment fee but $55 per month. (a) Write simplified expressions for the total cost of each plan after m months. (b) Evaluate both at m = 3 months and at m = 5 months. (c) At which of those two lengths is Plan A cheaper?
Answer: A(m) = 60 + 40m; B(m) = 55m; at m = 3 A costs $180 vs B $165 (B cheaper); at m = 5 A costs $260 vs B $275 (A cheaper).
(a) Expressions: Plan A: A(m) = 60 + 40m ← flat fee plus per-month rate Plan B: B(m) = 55m ← no fee, per-month rate only (b) Evaluate at m = 3: A(3) = 60 + 40(3) = 60 + 120 = 180 → $180 B(3) = 55(3) = 165 → $165 Evaluate at m = 5: A(5) = 60 + 40(5) = 60 + 200 = 260 → $260 B(5) = 55(5) = 275 → $275 (c) At m = 3, B is cheaper by $15. At m = 5, A is cheaper by $15. → Plan A is cheaper at 5 months.Break-even reasoning is one of the most useful ideas in Algebra 1. Plan A saves you $15 per month vs. Plan B (because $55 − $40 = $15), but starts $60 behind. So the plans cross at 60 ÷ 15 = 4 months. Below 4 months Plan B wins; above 4 months Plan A wins. This problem is the “numerical” version of the linear-systems problem you will solve algebraically in Class 6.
Answer key summary
| Q# | Answer | Q# | Answer | Q# | Answer |
|---|---|---|---|---|---|
| 1 | int, rat, real | 11 | 10 | 21 | 20 |
| 2 | 19 | 12 | 0 | 22 | 6 |
| 3 | 17 | 13 | 6x + 2 | 23 | 25 + 0.08m; $39 |
| 4 | 9x | 14 | 7y − 27 | 24 | 8x − 4; 52 |
| 5 | 5y | 15 | 11 | 25 | 0 |
| 6 | rat, irr, rat | 16 | 27 | 26 | 25n; 300¢ |
| 7 | 24 | 17 | 4x + 6y | 27 | x² + 13x |
| 8 | 6x + 4 | 18 | 3 | 28 | 6x + 22; 52 ft |
| 9 | 25 & 8 | 19 | (c) √7 | 29 | 4a − 5b |
| 10 | 3x + 15 | 20 | 11x − 2 | 30 | see solution |
Scoring & placement guidance
- 28–30 / 30 — Ready to start Algebra Ignite Class 1 (linear equations in one variable) immediately.
- 24–27 / 30 — Ready with light review. Re-read whichever theory section produced the misses (real numbers §2, order of operations §3, simplifying §4) and re-attempt those specific questions before starting.
- 18–23 / 30 — Not ready. Work through Pre-Algebra Class 1 (Integers) and Pre-Algebra Class 2 (Multiplying & Dividing Integers) first, then retake this diagnostic.
- Below 18 / 30 — Book a free evaluation. There is a gap earlier in arithmetic (fractions, negatives, or place value) that needs to be closed before Algebra 1 will feel doable.
About SOMATH & Algebra Ignite
SOMATH — School of Math is a math-focused school on the Upper West Side of Manhattan for students in grades 1–12. The Young Fermats Algebra Ignite program is our small-group Algebra 1 course — 15 classes plus this placement diagnostic (Class 0). Every student takes this diagnostic on Day 1 so we can place them in the right group and fill any gaps before Class 1.
Classes are taught by cofounder Marcelo Ambrozio (Northwestern-trained, 15+ years teaching math in NYC) and the SOMATH team.
Location: 226 W 79th St, 1st Floor, New York, NY 10024 (Upper West Side)
Phone: (646) 668-6151
Email: hello@schoolofmath.us
Book a free evaluation for your child — we’ll assess where they are and place them in the right small group.