3.3 Ratios, Proportions & Variation

Quick Answer: A ratio compares size, not raw count. Translate $a:b$ into $ax$ and $bx$, convert part-to-part into part-to-whole fractions $\frac{a}{a+b}$ and $\frac{b}{a+b}$, combine chained ratios by lining the shared term up at its LCM, and translate 'varies directly / inversely' into $y = kx$ or $xy = k$. Official Guide arithmetic review groups this with rates and percents because the same proportion engine drives all three.

Part-to-Part vs Part-to-Whole

• Part-to-part compares two subsets: 'electric cars to gas cars are $3:7$.'
• Part-to-whole compares one subset to the total. For parts $a$ and $b$, the whole is $a + b$ parts, so the electric share is $\frac{3}{3+7} = \frac{3}{10} = 30%$ and the gas share is $\frac{7}{10} = 70%$.

A three-part ratio uses the same translation. If analysts, associates, and partners are $5:3:2$, there are $5 + 3 + 2 = 10$ parts. Analysts are $\frac{5}{10} = \frac{1}{2}$ of the firm; partners are $\frac{2}{10} = 20%$. Never treat $5:3:2$ as '5 out of 3' — the denominator is the sum of all listed parts, including the ones the question is not currently asking about.

Going the other direction: if $40%$ of a class are juniors and the rest are seniors, juniors:seniors $= 40:60 = 2:3$. Percent of a whole is a part-to-whole ratio with total $100$; reducing $40:60$ recovers the part-to-part form.

If a ratio compares a part to a different part that is not the complement, you cannot read a percentage off it until you reconstruct the whole. 'The ratio of cats to dogs is $2:5$' does not mean cats are $2/5$ of the animals; cats are $2/7$ of the animals if those are the only two groups.

The Multiplier Method

A ratio encodes a scale factor, not a unique pair of numbers. For $3:5:8$, write the actual amounts as $3x$, $5x$, and $8x$. The total is $16x$. If the total is $480$, then $16x = 480 \Rightarrow x = 30$, so the amounts are $90$, $150$, and $240$.

Worked example. A scholarship pool is split among research, travel, and stipends in the ratio $4:3:1$. Research receives $$18{,}000$ more than travel. How large is the pool?

Research $= 4x$, travel $= 3x$, stipends $= x$. The given gap is $4x - 3x = 18{,}000$, so $x = 18{,}000$. The pool is $8x = $144{,}000$. The multiplier turned a ratio-plus-difference statement into one linear equation.

The multiplier $x$ is usually a positive integer on GMAT counting problems (people, cars, seats). A non-integer $x$ that produces half a person is a signal that the setup is wrong, not that the test wants a fractional employee.

Combining Successive Ratios

When two ratios share a common quantity, you cannot write $A:B:C$ by pasting the numbers until the shared term is the same number in both ratios. Scale each ratio so the bridge term equals the LCM of its two appearances.

Worked example. $A:B = 2:3$ and $B:C = 4:5$.

1. Bridge term: $B$, appearing as $3$ and as $4$. LCM$(3, 4) = 12$.
2. Scale $A:B$ by $4$: $8:12$.
3. Scale $B:C$ by $3$: $12:15$.
4. Unified ratio $A:B:C = 8:12:15$. Total parts $= 35$. If $A = 24$, then $8x = 24 \Rightarrow x = 3$, so $B = 36$ and $C = 45$.

To attach a fourth quantity, repeat the same bridge step. If also $C:D = 3:2$, the current $C$ is $15$ and the new $C$ is $3$, so scale $C:D$ by $5$ to $15:10$. Then $A:B:C:D = 8:12:15:10$. With $A = 24$, $x = 3$ and $D = 30$.

If a given ratio is inverted relative to the order you need, flip it before combining. Suppose $A:B = 2:5$ and $C:B = 3:4$. Rewrite the second as $B:C = 4:3$. Now the bridge $B$ is $5$ and $4$; LCM $= 20$; $A:B = 8:20$ and $B:C = 20:15$, so $A:B:C = 8:20:15$.

Changing Ratios When Quantities Are Added or Removed

A staple Problem Solving item states an initial ratio, adds or removes some items, and reports a new ratio. Keep the original multiplier $x$ and write the new counts as linear expressions in $x$. Cross-multiply the new proportion and solve.

Worked example. Red to blue marbles start at $4:7$. After $12$ red are added and $3$ blue are removed, the ratio is $2:3$. What was the initial total?

Initial red $= 4x$, blue $= 7x$, total $= 11x$.

$\frac{4x + 12}{7x - 3} = \frac{2}{3} \implies 3(4x + 12) = 2(7x - 3) \implies 12x + 36 = 14x - 6 \implies 2x = 42 \implies x = 21.$

Initial total $= 11 \times 21 = 231$. Always check that no count went negative: blue after the change is $7(21) - 3 = 144 > 0$.

Converting a ratio change into an added amount

Sometimes the new ratio is given and you must find how much was added. The unchanged quantity is the bridge.

Worked example. A tank holds acid and water in the ratio $2:3$. Pure acid is added until the ratio becomes $3:4$. The water (unchanged) is $60$ liters. How much acid was added?

Water is $3$ parts of the original $2:3$, so $3x = 60 \Rightarrow x = 20$ and original acid $= 40$ L. After the change, water is still $60$ L and is now $4$ parts of $3:4$, so one new part is $15$ L and acid is $3 \times 15 = 45$ L. Acid added $= 5$ L.

The same equation in multiplier form: $\frac{40 + a}{60} = \frac{3}{4} \Rightarrow 160 + 4a = 180 \Rightarrow a = 5$. When only one component changes, that unchanged component converts the old ratio into the new one without a second unknown.

If both components change, you need the full two-expression proportion (the marble example). One equation still suffices because there is still only one unknown $x$.

Direct, Inverse, and Joint Variation

Variation language is ratio language with a constant of proportionality $k$. Read the sentence, write $k$ times every direct factor, and divide by every inverse factor. Then either solve for $k$ from the first data point or equate the $k$-expressions from two states.

Direct variation

If $y$ varies directly with $x$ and $y = 15$ when $x = 6$, then $k = \frac{15}{6} = \frac{5}{2}$. When $x = 10$, $y = \frac{5}{2} \times 10 = 25$. Doubling $x$ doubles $y$; the ratio $\frac{y}{x}$ never changes. Direct variation with the square, $y = kx^2$, means quadrupling $y$ when $x$ doubles — the exponent travels with the scale factor.

Inverse variation

If $12$ workers finish a job in $15$ hours and time varies inversely with the number of workers, $TW = k$. Then $12 \times 15 = 18 \times T \Rightarrow T = 10$ hours for $18$ workers. Speed and travel time for a fixed distance behave the same way: $S_1 T_1 = S_2 T_2$. If speed rises by a factor of $\frac{5}{4}$, time falls by a factor of $\frac{4}{5}$.

Inverse variation is the statement that the product is constant. Direct variation is the statement that the quotient is constant. Mixing those two tests is the standard trap: computing $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ on an inverse problem will not match any option except by accident.

Joint variation

Electrical resistance $R$ varies directly with length $L$ and inversely with the square of diameter $d$: $R = \frac{kL}{d^2}$. A $50$ m wire of diameter $2$ mm has $R = 15$ ohms. For $120$ m and $4$ mm:

$\frac{15 \times 2^2}{50} = \frac{R_2 \times 4^2}{120} \implies \frac{60}{50} = \frac{16 R_2}{120} \implies \frac{6}{5} = \frac{2 R_2}{15} \implies R_2 = 9.$

Isolate $k$ from the first data point, then substitute the second. Squared or cubed variables mean the scale factor is squared or cubed: doubling diameter multiplies $R$ by $\frac{1}{4}$, not $\frac{1}{2}$. If a problem says $P$ varies jointly with $Q$ and $R^2$ and inversely with $S$, write $P = \frac{k Q R^2}{S}$ on the first line before touching the numbers.