3.1 Fraction Arithmetic & Operations

Key Takeaways

  • Convert mixed numbers to improper fractions first: a b/c = (ac + b)/c, and a leading minus sign applies to the entire mixed number.
  • Add or subtract unlike fractions with the LCD (the LCM of the denominators), not with the product of the denominators.
  • Cross-cancel prime factors before multiplying; divide by multiplying by the reciprocal.
  • Clear a complex fraction in one step by multiplying numerator and denominator by the LCM of every interior denominator.
  • Never multiply an inequality by a variable denominator unless its sign is known; an unknown sign requires casework or an always-positive multiplier such as x^2.
Last updated: August 2026

Quick Answer: GMAT Quantitative Reasoning is 21 Problem Solving questions in 45 minutes with no calculator. Fraction arithmetic — LCD addition, cross-cancelling products, complex-fraction clearing, and signed mixed-number conversion — is the operational core. Convert mixed numbers to improper fractions first, cancel prime factors before multiplying, and clear nested fractions by multiplying through by the LCM of every interior denominator.

Proper, Improper, and Mixed Numbers

A fraction $\frac{a}{b}$ is the quotient of integer numerator $a$ by nonzero integer denominator $b$.

  • Proper fraction: $|a| < |b|$ (for example $\frac{3}{7}$ or $-\frac{5}{8}$). The value lies strictly between $-1$ and $1$.
  • Improper fraction: $|a| \ge |b|$ (for example $\frac{9}{4}$ or $-\frac{11}{3}$). The absolute value is at least $1$.
  • Mixed number: an integer plus a proper fraction, written $3\frac{1}{4} = 3 + \frac{1}{4} = \frac{13}{4}$.

The mixed-number sign trap

Convert mixed numbers to improper fractions before any other operation:

abc=ac+bc,abc=(a+bc)=ac+bc.a\frac{b}{c} = \frac{ac + b}{c}, \qquad -a\frac{b}{c} = -\left(a + \frac{b}{c}\right) = -\frac{ac + b}{c}.

So $-3\frac{2}{5} = -\frac{17}{5}$, not $-3 + \frac{2}{5} = -\frac{13}{5}$. The minus sign applies to the entire mixed number, not only to the integer part. On a timed, no-calculator section this conversion is the first line of work, not an optional rewrite.

Addition and Subtraction: Finding the LCD

Unlike denominators require a common denominator. The least common denominator (LCD) is the least common multiple of the denominators — not their product, which is usually larger than necessary and invites arithmetic errors.

LCD algorithm

  1. Prime-factor each denominator.
  2. For every prime that appears, take the highest power.
  3. Multiply those prime powers to obtain the LCD.
  4. Scale each fraction so that every denominator equals the LCD, then add or subtract numerators and reduce.

Worked example. Compute $\frac{5}{18} - \frac{7}{24} + \frac{1}{30}$.

  • $18 = 2 \times 3^2$
  • $24 = 2^3 \times 3$
  • $30 = 2 \times 3 \times 5$
  • LCD $= 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$

5×203607×15360+1×12360=100105+12360=7360.\frac{5 \times 20}{360} - \frac{7 \times 15}{360} + \frac{1 \times 12}{360} = \frac{100 - 105 + 12}{360} = \frac{7}{360}.

Using $18 \times 24 \times 30 = 12{,}960$ as a common denominator would be valid and slower. Always prefer the LCD. If a result such as $\frac{14}{360}$ appears, reduce immediately: $\frac{7}{180}$.

Multiplication and Division

Multiplication and cross-cancelling

Multiply numerators together and denominators together. Cancel common prime factors before multiplying so the integers stay small:

1425×1528×1021=145×5×3×52×14×2×53×7=235.\frac{14}{25} \times \frac{15}{28} \times \frac{10}{21} = \frac{14}{5 \times 5} \times \frac{3 \times 5}{2 \times 14} \times \frac{2 \times 5}{3 \times 7} = \frac{2}{35}.

Leaving the product uncancelled produces $2{,}100 / 14{,}700$, which then has to be reduced by hand — a waste of the 45-minute clock.

Division (multiply by the reciprocal)

ab÷cd=ab×dc=adbc(b,c,d0).\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \quad (b, c, d \ne 0).

Worked example. $\frac{9}{16} \div \frac{15}{8} = \frac{9}{16} \times \frac{8}{15} = \frac{9}{2 \times 15} = \frac{3}{10}$.

A mixed-number division such as $2\frac{1}{3} \div 1\frac{1}{6}$ must be rewritten first: $\frac{7}{3} \div \frac{7}{6} = \frac{7}{3} \times \frac{6}{7} = 2$. Skipping that conversion is the most common arithmetic miss on mixed-number items.

Complex and Nested Fractions

A complex fraction has a fraction in the numerator, the denominator, or both. Do not simplify the top and bottom as separate mini-problems if you can clear every interior denominator in one stroke.

Clearing method: multiply the entire numerator and the entire denominator by the LCM of all interior denominators.

Worked example. Simplify $\dfrac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x^2} - \frac{1}{y^2}}$ for $x \ne y$ and $xy \ne 0$.

Interior denominators: $x$, $y$, $x^2$, $y^2$. LCM $= x^2 y^2$.

x2y2(1x1y)x2y2(1x21y2)=xy2x2yy2x2=xy(yx)(yx)(y+x)=xyx+y.\frac{x^2 y^2 \left(\frac{1}{x} - \frac{1}{y}\right)}{x^2 y^2 \left(\frac{1}{x^2} - \frac{1}{y^2}\right)} = \frac{xy^2 - x^2 y}{y^2 - x^2} = \frac{xy(y - x)}{(y - x)(y + x)} = \frac{xy}{x + y}.

The factor $y - x$ cancels because $x \ne y$. This is faster than adding $\frac{1}{x} - \frac{1}{y}$ and $\frac{1}{x^2} - \frac{1}{y^2}$ as separate steps, and it keeps the algebra factored.

A purely numeric nested fraction is the same idea, evaluated inside-out: $1 - \frac{1}{3} = \frac{2}{3}$, then $\frac{1}{2/3} = \frac{3}{2}$, and so on. Either inside-out evaluation or one-shot LCD clearing is acceptable; mixing the two midstream is what produces sign errors.

Comparing Fractions Without a Calculator

TechniqueWhen to use itRule / example
Cross-multiplicationTwo positive fractions $\frac{a}{b}$ and $\frac{c}{d}$$\frac{a}{b} > \frac{c}{d}$ if and only if $ad > bc$. Compare $\frac{7}{11}$ vs $\frac{9}{14}$: $7 \times 14 = 98$ and $11 \times 9 = 99$, so $\frac{9}{14}$ is larger.
Same numeratorNumerators matchLarger denominator is smaller: $\frac{5}{17} > \frac{5}{19}$.
Same denominatorDenominators matchLarger numerator is larger: $\frac{11}{23} < \frac{14}{23}$.
BenchmarkNearby halves or quarters$\frac{13}{27} < \frac{1}{2}$ because $13 < 13.5$, while $\frac{17}{32} > \frac{1}{2}$, so $\frac{17}{32} > \frac{13}{27}$.
Distance to 1Proper fractions just below 1$\frac{14}{15} = 1 - \frac{1}{15}$ and $\frac{16}{17} = 1 - \frac{1}{17}$. Since $\frac{1}{17} < \frac{1}{15}$, $\frac{16}{17}$ is closer to 1 and therefore larger.
Added-constant ruleAdding $k > 0$ to numerator and denominatorIf $\frac{a}{b} < 1$, then $\frac{a+k}{b+k} > \frac{a}{b}$. If $\frac{a}{b} > 1$, then $\frac{a+k}{b+k} < \frac{a}{b}$. In both cases the fraction moves toward 1.

The added-constant rule is why $\frac{123}{127} < \frac{489}{493}$: both equal $1 - \frac{4}{\text{denominator}}$, and the larger denominator subtracts a smaller piece from 1. Cross-multiplication still works, but rewriting as $1 - \frac{k}{n}$ is faster when every option has the same numerator-denominator gap.

Algebraic Fraction Inequalities

Never multiply an inequality by a variable expression unless you know the sign. If the denominator can be negative, multiplying reverses the inequality for those values and you will drop an entire solution set.

Example. Solve $\frac{1}{x} < 3$.

  • Wrong: multiply by $x$ to get $1 < 3x$, hence $x > \frac{1}{3}$. This drops every negative $x$.
  • Two cases. If $x > 0$, the inequality direction is preserved: $1 < 3x \Rightarrow x > \frac{1}{3}$. If $x < 0$, the direction reverses: $1 > 3x \Rightarrow x < \frac{1}{3}$, which together with $x < 0$ is simply $x < 0$. Combined: $x < 0$ or $x > \frac{1}{3}$.
  • Always-positive multiplier. Multiply both sides by $x^2 > 0$ (valid for $x \ne 0$): $x < 3x^2 \Rightarrow 3x^2 - x > 0 \Rightarrow x(3x - 1) > 0$. A sign chart on the roots $0$ and $\frac{1}{3}$ recovers $x < 0$ or $x > \frac{1}{3}$.

On GMAT Problem Solving, if a denominator can change sign, casework or an always-positive multiplier is mandatory. The excluded value $x = 0$ must stay excluded.

Test Your Knowledge

If x = 1 / (1 - 1 / (1 - 1/5)), what is the value of (x^2 - 1) / (x^2 + 1)?

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Test Your Knowledge

Which of the following fractions has the greatest value?

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Test Your Knowledge

A grant spends 1/3 of its budget on research, then 1/4 of the remaining budget on travel, then 2/5 of what is left on supplies. If $18,000 remains after those three allocations, what was the original budget?

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