Set up and solve proportions, compute and compare unit rates, and apply percent increase, decrease, and percent change to real-world quantities.

What this topic is asking

Proportional reasoning runs through the Number and Quantity and Algebra categories. The Grade 10 MCAS asks you to set up and solve proportions, to compute and compare unit rates (best-buy and speed problems), and to handle percent increase, decrease, and percent change. These are short selected-response items and short-answer questions, and the percent-change problems are a reliable source of traps, so the reasoning matters as much as the arithmetic.

Setting up and solving proportions

A proportion is two equal ratios. The skill is building it with matching units in matching positions: if miles are on top on the left, miles must be on top on the right.

For "if 3 pounds cost \$7.50, what do 8 pounds cost?", write$\dfrac{3 \text{ lb}}{7.50} = \dfrac{8 \text{ lb}}{x}$, with pounds over dollars on both sides. Cross-multiply:$3x = 7.50 \times 8 = 60$, so$x = 20. Eight pounds cost \20.

Cross-multiplication works because multiplying both sides of $\frac{a}{b} = \frac{c}{d}$ by $bd$ clears the fractions to $ad = bc$. Always check that the answer is reasonable: more pounds should cost more, and \20 for 8 pounds is consistent with \7.50 for 3.

Unit rates and best buy

A unit rate expresses a quantity per one unit, which makes different sizes directly comparable. Divide the total by the number of units: a 20-ounce box at \$5.40 has a unit price of$\frac{5.40}{20} = \$0.27$ per ounce.

To compare buys, compute each unit price and pick the smaller one (cheaper per unit). To compare speeds, compute each as distance per unit time and pick as the question asks. The frequent error is comparing totals instead of unit rates: the bigger box usually costs more in total but can still be cheaper per ounce.

Percent increase, decrease, and percent change

A percent change is cleanest as a multiplier. A $p\%$ increase multiplies by $1 + \frac{p}{100}$; a $p\%$ decrease multiplies by $1 - \frac{p}{100}$.

• A 15% increase on \$200:$200 \times 1.15 = \$230$.
• A 30% decrease on \$200:$200 \times 0.70 = \$140$.

To find the percent change between two values, use $\dfrac{\text{new} - \text{old}}{\text{old}} \times 100$. If a price rises from \40 to \50, the change is $\frac{50 - 40}{40} \times 100 = 25\%$. The denominator is always the original value; using the new value instead is a classic error.

The subtlest trap is successive percents. A 25% markup followed by a 25% discount gives $1.25 \times 0.75 = 0.9375$ of the original, a net 6.25% decrease, not a return to the start. Each percent acts on the running base, and those bases differ.

Try this

Q1. Solve the proportion $\dfrac{x}{12} = \dfrac{15}{20}$.

• Cue. $20x = 180$, so $x = 9$.

Q2. A population grows from 1500 to 1800. What is the percent increase?

• Cue. $\frac{1800 - 1500}{1500} \times 100 = 20\%$.