8.4 Age Problems & Discrete Optimization

Age problems are algebraic translation, the same skill Official Guide Math Review 3.2 tests with linear equations. Discrete-optimization items — greatest possible, least possible, minimum number that guarantees — are integer cousins of the overlap bounds in Section 8.1. Both appear as Problem Solving word problems on the 21-question, 45-minute, no-calculator Quantitative Reasoning section.

Age Problems: One Clock, Constant Differences

Every living person ages at the same rate of one year per year. Two consequences follow, and both are tested:

1. The difference between two people's ages never changes.
2. The ratio of their ages does change. As years pass, any two positive ages drift toward a ratio of $1$.

The reliable setup is to pick one reference time (almost always now), assign a variable to each person at that time, and express every other epoch by adding or subtracting the same elapsed years from every person.

Do not introduce a second independent variable for the same person at a second time. "Priya in five years" is $P + 5$, not a new letter. Mixing clocks is the source of most age-algebra errors.

Translation examples:

• "In $5$ years, Arthur will be twice as old as Brenda was $3$ years ago" $\to A + 5 = 2(B - 3)$.
• "When Chris was born, Diana was already $6$" $\to D - C = 6$.
• "Four years ago, Marcus was $5$ times as old as Leo" $\to M - 4 = 5(L - 4)$.

After writing one equation per time-clause, substitute. Ages must come out positive and, on the GMAT, almost always integers. If a candidate value makes a past age zero or negative, reject it.

A useful sanity check: if the present ages are $M$ and $L$ with $M > L$, then $M - L$ must equal the difference implied by every epoch in the stem. If the past equation and the future equation produce two different differences, the algebra is wrong.

Discrete Optimization: Extreme Allocation

A typical GMAT prompt gives a sum, a mean, a median, and a distinct-positive-integer constraint, then asks for the greatest possible value of the largest term — or the least possible value of the largest term. Integers cannot take fractional values, and "distinct" forbids repeats.

Rules:

1. Maximize the largest $x_n$. Minimize every other $x_i$ to the smallest legal integer (often $1, 2, 3, \ldots$, subject to a fixed median).
2. Minimize the largest $x_n$ (minimax). Pack the list as close together as possible, usually consecutive integers around the mean $S/n$.
3. Minimize the smallest $x_1$. Maximize the other terms up to any stated cap.
4. Maximize the smallest $x_1$ (maximin). Again pack near the mean.

Median handling: for an odd count, the middle position is locked to the stated median. For an even count of $6$, the median is the average of $x_3$ and $x_4$, so $x_3 + x_4$ is twice the median. To maximize $x_6$ you still minimize $x_1, x_2, x_5$ and, subject to $x_3 < x_4$ and $x_3 + x_4$ fixed, you take the split that then allows the smallest $x_5$.

Integer Products and Sums

A second family asks for the greatest product of positive integers with a fixed sum, or the least sum of positive integers with a fixed product.

• For a fixed sum $S$ of two positive integers, the product $x(S - x)$ is maximized when $x$ is as close as possible to $S/2$. If repeats are forbidden, use the nearest distinct pair (for $S = 10$, $4 \times 6 = 24$ rather than $5 \times 5 = 25$).
• For a fixed product, the sum is minimized when the factors are as equal as possible (and usually near $3$ if unrestricted partitions are allowed). GMAT stems nearly always restrict to two or three named positive integers, so equal-as-possible is enough.
• If the product of two positive integers is $36$ and repeats are allowed, the least sum is $6 + 6 = 12$. If they must be distinct, $4 + 9 = 13$ is the next candidate to check against $3 + 12 = 15$ and $2 + 18 = 20$.

Always enforce the stated type: positive versus nonnegative, distinct versus repeats allowed, even versus odd. A beautiful fractional split is not an answer if the stem demands integers.

Worst-Case Guarantees (Pigeonhole Counting)

Prompts that ask for the minimum number that guarantees an outcome are not probability questions. They are worst-case counts:

$$\text{Guaranteed selections} = (\text{maximum unlucky selections}) + 1$$

Protocol:

1. Name the target (at least $k$ of one color; at least one matching pair).
2. Take as many items as possible while still missing the target: all of every non-target type, plus $k - 1$ of the target type, or one from each category if the target is a matching pair.
3. Add $1$. The next draw is forced.

High-yield patterns:

• At least $k$ of color $X$, with other colors present: draw every non-$X$ item, then $k$ of color $X$.
• At least $k$ of some color, with four colors available: draw $k - 1$ of each color, then one more.
• At least one matching pair from $C$ categories: draw $C + 1$.

Respect inventory caps. You cannot draw $9$ red items from a box that contains only $8$ red items; the worst case is limited by what the box actually holds.

Framework Table

Worked Example: Two-Epoch Ages

Four years ago, Marcus was $5$ times as old as Leo. In $6$ years, Marcus will be $3$ times as old as Leo. Let $M$ and $L$ be present ages. Then $M - 4 = 5(L - 4)$ so $M = 5L - 16$, and $M + 6 = 3(L + 6)$ so $M = 3L + 12$. Equating, $5L - 16 = 3L + 12$, hence $2L = 28$ and $L = 14$, $M = 54$. Check: four years ago the ages were $50$ and $10$; in six years they will be $60$ and $20$.

Worked Example: Greatest Possible Largest Integer

Five distinct positive integers have mean $40$ and median $35$. The sum is $200$. In order $x_1 < x_2 < x_3 < x_4 < x_5$ with $x_3 = 35$, minimize $1 + 2 + 35 + 36 = 74$ to maximize $x_5 = 200 - 74 = 126$.

Worked Example: Guarantee Count

A box holds $12$ red, $15$ green, $18$ blue, and $20$ yellow balls. To guarantee at least $10$ balls of one color, draw $9$ of each color first ($36$ balls, still no color at $10$), then one more ball — $37$ draws.