EQN-02-08

Proportional Division & Age Problems

Split a quantity three ways and solve age/ratio word problems with the "parts" method and a single variable.

~5 min read•Difficulty 3/5

The parts method extends directly. Add all ratio terms, divide the total by the number of parts, then scale each share.

Divide 360 in 2 : 3 : 4 \to parts = 9, one part = 40

shares = 80, 120, 160

A difference equals (difference of ratio terms $) \times$ one part.

In 2 : 3 : 4 of 360 : largest - smallest = (4 - 2) \times 40 = 80.

Let the common part be x. Write each age as (ratio term $) \cdot x,$ then use the extra condition.

Ages are in ratio 3 : 5 and differ by 8 years .

5 x - 3 x = 8 \to 2 x = 8 \to x = 4 \to ages 12 and 20.

In t years, add t to every person's age — the difference stays the same, but the ratio changes. Set up the new ratio at the new ages.

Worked examples

Example 1 (difficulty 2/5)

An amount of 250 is shared in the ratio 1 : 2 : 2. What is the largest share?

◎ Predict the answer first:

Example 2 (difficulty 2/5)

Two brothers’ ages are in the ratio 4 : 7 and differ by 9 years. How old is the younger?

◎ Predict the answer first:

Add all the ratio terms to get the number of parts, then find one part from the total.
A difference of shares $= ($ difference of ratio terms $) \times$ one part.
For age problems, write each age as (term $) \cdot x$ and use the extra condition to find x.

Changing the ratio terms (not the ages) when years pass

Why: The ratio is the visible quantity.

Fix: Add the years to the actual ages, then rebuild the ratio.

Forgetting that the age difference stays constant over time

Why: Both the ratio and ages change, so the difference seems to as well.

Fix: Everyone ages equally, so the gap never changes — only the ratio does.