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GMAT Focus Edition · GRE & GMAT Analytical Reasoning

GRE Quantitative Comparison

Comparing two quantities and deciding which is greater or if it cannot be determined.

Quantitative Comparison is the GRE's fixed four-choice format: decide which of two quantities is larger, whether they are equal, or whether the information is not enough. Six concepts — the choices themselves, safe simplification, the unknown-sign trap, testing edge values, reading a constraint so D dies, and why two concrete numbers can never be D.

  • GMAT Focus Edition
  • Medium level
  • 6 concepts
  • 5 practice questions

1Four fixed choices on every QC question

Every Quantitative Comparison item uses the same four answers: A means Quantity A is greater, B means Quantity B is greater, C means the two quantities are equal, and D means the relationship cannot be determined from the information given.

D is a real, frequent answer — not a refuge for unfinished arithmetic. Memorise the four letters once so every later question is spent on the comparison, not on rereading the stem.

Figure. Every QC item uses the same four letters. D is a real answer when the order can still flip — not a refuge for unfinished arithmetic.

How it works

  1. Read A and BName the two quantities; any centred information above them is a shared constraint, not an optional hint.
  2. Compare, don't evaluateYou need the relationship, not the value of either quantity — stop the moment the order is fixed or clearly not fixed.
  3. Pick the letterA, B, C, or D. If a second legal value flips the order you already found, the letter is D.
The four choices, fixed for every QC item
ChoiceMeans
AQuantity A is greater
BQuantity B is greater
CThe two quantities are equal
DThe relationship cannot be determined
On a Quantitative Comparison item, choice D means
  1. You ran out of time and should guess
  2. The relationship cannot be determined from the given information
  3. Both quantities are negative

D is the official answer when the order of A and B is not fixed by the information — not a placeholder for unfinished work, and not a claim about signs.

2Simplify both quantities the same way

Because you only need the relationship, you may add or subtract the same value from both quantities, or multiply or divide both by the same positive number, and the comparison is unchanged. Subtracting a shared term or dividing out a common positive factor is usually enough — full evaluation is optional labour.

The move has to be the same on both sides and safe: a positive common factor preserves order; a negative factor would reverse it, which is the next concept.

Figure. Two vertical bars after stripping the shared 47: Quantity A at height 3 and Quantity B at height 4, with value labels, so B is visibly taller.

How it works

  1. Spot a shared pieceA common addend, a common positive factor, or an identical expression sitting in both quantities.
  2. Apply one legal moveSubtract the shared term from both, or divide both by the same positive constant.
  3. Read what remainsThe stripped comparison is the answer — A, B, or C — without ever computing the original totals.

Strip the shared 47

Quantity A: 3 + 47. Quantity B: 47 + 4. Which is greater?

  • Subtract the shared 47 from both quantitiesA → 3, B → 4
  • Compare the stripped values: 3 ? 43 < 4
  • Same order as the originals (50 vs 51)B greater
  • QC choiceB

Pro tip. You never needed 50 or 51. The moment the shared 47 is gone, the comparison is 3 against 4 — and that is the whole question.

Quantity A: 12 + 88. Quantity B: 88 + 15. After subtracting 88 from both, the comparison is
  1. 12 against 15, so B is greater
  2. 100 against 103, so you still need the totals
  3. 12 against 15, so A is greater

Subtracting the shared 88 leaves 12 against 15. B is greater. Recomputing 100 and 103 is allowed but wasted — the stripped pair already decides.

3Never cancel a factor of unknown sign

Multiplying or dividing both quantities by the same negative number reverses the inequality. If the common factor is a variable whose sign you do not know, you cannot cancel it — doing so silently assumes the factor is positive and can flip a correct A into a wrong B.

When the factor might be positive, negative, or zero, keep it and test both signs. A single convenient positive plug-in is not a proof.

Figure. Cancelling a factor of unknown sign assumes it is positive. Across x = 2 and x = −2 the order flips — keep the factor and test both signs.

How it works

  1. Name the factorIf both quantities share a multiplier or divisor involving a variable, ask whether its sign is known.
  2. Refuse the cancel when unsignedDo not divide both sides by x, or by (x − 3), unless the stem already forces that expression positive (or negative).
  3. Test both signsPlug a positive value and a negative value. If the order flips, the answer is D.

5x against 3x without knowing x

Quantity A: 5x. Quantity B: 3x. Given only that x is a real number, which is greater?

  • Try x = 2: A = 5\times 2, B = 3\times 210 vs 6, A > B
  • Try x = -2: A = 5\times(-2), B = 3\times(-2)−10 vs −6, B > A
  • Order flipped when the sign of x flippedrelationship not fixed
  • QC choiceD

Pro tip. Cancelling x from both sides would leave 5 against 3 and the false answer A. That cancel is legal only when x > 0 is given — here it is not.

Quantity A: 2x. Quantity B: x. With no information about x, cancelling x from both sides is
  1. Legal, and proves A is greater
  2. Illegal unless the sign of x is known; the answer is D
  3. Legal, and proves the quantities are equal

For x = 2, A = 4 and B = 2 so A > B. For x = −2, A = −4 and B = −2 so B > A. The order flips with the sign, so D. Cancelling x assumes x > 0.

4Test edge values until the order stabilises — or flips

When a variable is unrestricted, one convenient plug-in is never enough. Try 0, 1, −1, a proper fraction such as 1/2, and a number larger than 1. If any two legal values give different comparisons, stop — the answer is D.

The habit is not random guessing. Each edge tests a different algebraic regime: zero kills products, one leaves powers unchanged, a fraction reverses many power comparisons, and a negative flips signs.

Figure. Two clustered pairs of vertical bars. Under x = 2, Quantity A is 4 and Quantity B is 2, so A is taller. Under x = 1/2, Quantity A is 0.25 and Quantity B is 0.5, so B is taller — the order has flipped.

How it works

  1. Pick the first edgeStart with a value that is easy to compute — often 2 or 0 — and record A ? B.
  2. Pick a second regimeChange the regime: a fraction if you used an integer, or a negative if you used a positive.
  3. Stop on a flipTwo different orders mean D. Matching orders are not yet a proof — keep testing until the stem's whole domain is covered, or a flip appears.

x² against x with x unrestricted

Quantity A: x^2. Quantity B: x. Given only that x is a real number, which is greater?

  • Try x = 2: A = 2^2, B = 24 vs 2, A > B
  • Try x = \frac{1}{2}: A = \left(\frac{1}{2}\right)^2, B = \frac{1}{2}0.25 vs 0.5, B > A
  • Order flipped across the two legal valuesrelationship not fixed
  • QC choiceD

Pro tip. One case giving A > B and another giving B > A instantly forces D. You do not need a third plug-in once the flip is on the page.

Quantity A: x. Quantity B: -x, with x a nonzero real. The correct choice is
  1. A, because a number is greater than its opposite
  2. C, because the absolute values match
  3. D, because a positive x makes A greater and a negative x makes B greater

For x = 3, A = 3 and B = −3 so A > B. For x = −3, A = −3 and B = 3 so B > A. The relationship flips, so D.

5A constraint can kill D

Centred information above the two quantities is not decoration — it restricts the domain you are allowed to test. An unrestricted comparison that would be D can become A, B, or C once the stem forbids the values that caused the flip.

Re-run the edge-value habit inside the allowed set only. If every legal value keeps the same order, that order is the answer; the forbidden flips no longer count.

Figure. A plot on axes through the origin with two curves on the unit interval: the line y equals x lying above the curve y equals x squared between 0 and 1, meeting again at the origin and at (1, 1). On that open interval Quantity A is the higher graph.

How it works

  1. Read the centred givenTranslate inequalities and type restrictions into a domain — for example 0 < y < 1, or n an integer greater than 1.
  2. Test only inside itPlug values from that domain. Values outside it are irrelevant even if they would flip the order.
  3. Lock the letterA stable order on the whole allowed set is A, B, or C. A flip still inside the set is D.

y against y² on (0, 1)

Given 0 < y < 1. Quantity A: y. Quantity B: y^2. Which is greater?

  • Try y = 0.5 (inside the given): A = 0.5, B = 0.25A > B
  • Try y = 0.25: A = 0.25, B = 0.0625A > B
  • On (0,1), squaring shrinks a positive proper fractionA stays greater
  • QC choiceA

Pro tip. Without the given, y = 2 would make B greater and force D. The centred inequality deletes that case — so the unrestricted habit would over-answer D.

Given that n is an integer greater than 1. Quantity A: n. Quantity B: n². The correct choice is
  1. D, because sometimes a number exceeds its square
  2. B, because every integer n > 1 satisfies n² > n
  3. A, because n is the base

For n = 2, 4 > 2; for n = 3, 9 > 3; and so on. The fraction regime where n² < n is outside the given, so D is wrong.

6Two concrete numbers can never be D

If both Quantity A and Quantity B are fixed numbers — no unresolved variable — the relationship is determined. Cross-multiply fractions, subtract, or compare by any honest method; the answer is A, B, or C. D is impossible.

Reach for D only when a variable (or an unspecified choice inside a set) still has room to change the order. A hard-looking pair of fractions is still a concrete comparison.

Picture cross-products 27 and 28 side by side: almost the same height, one tip higher. Trust the arithmetic, not the eye — concrete pairs are A, B or C, never D.

How it works

  1. Confirm both sides are fixedNo free variable, no "some integer", no unspecified sign — just numbers.
  2. Compare by a safe methodFor positive fractions, cross-multiply: a/b ? c/d becomes a·d ? b·c.
  3. Write A, B, or C — never DArithmetic may be messy, but indeterminacy is not on the menu.

Cross-multiply 3/7 against 4/9

Quantity A: \dfrac{3}{7}. Quantity B: \dfrac{4}{9}. Which is greater?

  • Cross-multiply: compare 3\times 9 with 4\times 727 vs 28
  • 27 < 28, so \dfrac{3}{7} < \dfrac{4}{9}B greater
  • Both quantities are fixed numbersD impossible
  • QC choiceB

Pro tip. Compare positive fractions by cross-multiplication; with two concrete numbers, D is off the table before you start.

Quantity A: the number of minutes in 3 hours. Quantity B: 200. The correct choice is
  1. D, because time conversions can go either way
  2. A, because 3 hours exceeds 200 minutes
  3. B, because 3 × 60 = 180 and 180 < 200

Three hours is exactly 180 minutes. 180 < 200, so B is greater. Both sides are concrete, so D is unavailable.

Notes

  • Four Fixed Choices: Every QC question uses the same options: A (Quantity A greater), B (Quantity B greater), C (equal), or D (cannot be determined).
  • Test Multiple Cases: If variables are involved, try positive, negative, zero, and fractional values; if the relationship changes, the answer is D.
  • Simplify Both Sides: You may add, subtract, multiply, or divide both quantities by the same positive amount to compare more easily.
  • Cannot Be Determined Only With Variables: If both quantities are concrete numbers, the answer can never be D.

Formulas

  • Choices: A > B, B > A, A = B, or D (indeterminate).
  • Legal moves: add/subtract same value to both; multiply/divide by same positive value.
  • Caution: multiplying by a negative or an unknown-sign variable can flip the inequality.
  • D is impossible when both quantities are fixed numbers.

Exam traps & shortcuts

  • Plug in 0, 1, -1, and a fraction like \frac{1}{2}; if you get two different comparisons, pick D immediately.
  • Never multiply or divide both sides by a variable whose sign is unknown—it may reverse the inequality.
  • Strip away identical terms from both quantities to reduce the comparison to its essential difference.

Reference tables

Legal moves when simplifying a QC comparison
MoveSafe whenEffect on order
Add or subtract the same value from bothAlwaysUnchanged
Multiply or divide both by the same positive numberThe number is known positiveUnchanged
Multiply or divide both by a negative numberYou accept the reversalReverses
Cancel a variable factorIts sign is known from the stemUnchanged only if positive; reverses if negative

Recap

QC — keep these six pegs

Four letters
A greater, B greater, equal, or cannot be determined — the same on every item. D is a real answer.
Simplify
Add, subtract, or divide both sides by the same positive amount. Read what remains; skip the full totals.
Unknown sign
Do not cancel a factor whose sign is unknown. Test a positive and a negative; a flip means D.
Edge values
With an unrestricted variable, try 0, 1, −1, a fraction, a large number. Two different orders → D.
Constraints
Centred givens shrink the domain. Flips outside the given do not count; a stable order inside it is A, B, or C.
Concrete pair
Two fixed numbers can never be D. Cross-multiply fractions; finish with A, B, or C.

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