Q.C · Quantitative comparison and number reasoning

Quantitative Comparison has four fixed relationships: A greater, B greater, equal, or cannot determine.

Check all constraints. A variable can be negative, zero or fractional unless the stem restricts it.

Use strategic substitutions to disprove a fixed relation. They can show indeterminacy but a few examples do not prove universality.

Simplify both quantities without silently dividing by a value that could be zero or negative.

Given x²=9, x can be 3 or −3. Quantity A=x; B=0. A is greater for 3 and smaller for −3, so the relationship cannot be determined. If x is positive, only 3 is allowed and A is greater.

Figures are not necessarily drawn to scale. Given labels and constraints carry the evidence.

Quantitative Comparison uses these relationships: A. Quantity A is greater. B. Quantity B is greater. C. Equal. D. Cannot be determined. Given real x with $x^2=16$, compare Quantity A: x, Quantity B: 0. Repeat with the added condition $x>0$.

Model and reasoning

Without the added condition, D: x=4 makes A greater, while x=-4 makes B greater. With $x>0$, A, because only x=4 is permitted. The principal square root of 16 is four, but the equation has both roots. One chosen example cannot prove a relationship over all allowed values.

With the same four answer categories, let $a>0$ and $b<0$. Compare Quantity A: $a/b$, Quantity B: $b/a$. Give admissible examples if the relationship can vary, and identify when equality occurs.

Check after attempting

D. For a=2,b=-1, A=-2 and B=-1, so B is greater. For $a=1$,b=-2, A=-1/2 and B=-2, so A is greater. For $a=1$,b=-1 they are equal. More generally $a/b=b/a$ means $a^2=b^2$, and the stated opposite signs give a=-b. Multiplying an inequality by the negative product ab reverses direction; ignoring this sign would reverse conclusions.