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Factor quadratic: x2−7x+12=(x−3)(x−4)=0x^2 - 7x + 12 = (x-3)(x-4)=0. Solutions: x=3x=3 or x=4x=4. Choice A. Verify: plug 3 → 9−21+12=09-21+12=0. Plug 4 → 16−28+12=016-28+12=0. Works. Pitfalls: mispairing factors; remember factors must multiply to +12 and sum to -7. On GRE quadratics, factoring is usually fastest, but quadratic formula (7±(−7)2−4⋅12)/2=(7±49−48)/2=(7±1)/2=3,4(7 \pm \sqrt{(-7)^2-4\cdot 12})/2 = (7\pm \sqrt{49-48})/2 = (7\pm1)/2 = 3,4 also confirms.

The argument presumes that removing access in schools reduces total caloric intake of sugary beverages for students. For that to translate into lower obesity rates, students must not “make up” the missing sugar by buying sodas after school or substituting high-calorie snacks. Option B articulates this displacement (or “compensatory behavior”) assumption. If it is false, the intervention may have negligible net impact on obesity despite reducing on-campus purchases. Option A (parental support) affects implementation politics, not the causal mechanism. Option C is logistical/financial, not logically necessary for the health effect. Option D introduces an unrelated variable that might influence obesity but isn’t assumed in the argument’s chain. On GRE assumptions, identify where the proposed intervention could be offset by behavior change; the necessity is with preventing substitution that nullifies the intended effect.

Cross multiply: 3⋅12=9x⇒36=9x⇒x=36/9=4.3 \cdot 12 = 9x \Rightarrow 36 = 9x \Rightarrow x=36/9=4. Choice C. Quick error check: sometimes test takers mistakenly invert instead of cross-multiplying. Confirm: plug back: LHS = 3/4 = 0.75; RHS = 9/12 = 0.75. Correct. On GRE Quant, always verify by substitution to avoid arithmetic slips.

Use the complement rule. “At least one defective” = 1 − P(no defective). “No defective” means all 3 drawn from the 16 good ones. The number of 3-component samples with no defective is (163)=560\binom{16}{3} = 560. The total number of 3-component samples from 20 is (203)=1140\binom{20}{3} = 1140. Thus

P(no defective)=(163)(203)=5601140=56114=2857.P(\text{no defective})=\frac{\binom{16}{3}}{\binom{20}{3}}=\frac{560}{1140}=\frac{56}{114}=\frac{28}{57}.

Therefore

P(at least one defective)=1−2857=57−2857=2957.P(\text{at least one defective}) = 1 - \frac{28}{57} = \frac{57-28}{57} = \frac{29}{57}.

Choice C. This structure—“at least one”—nearly always favors complementary counting because the direct approach would require summing multiple cases (exactly 1 defective + exactly 2 + exactly 3). With combinations, ensure you use the same denominator for all outcomes and simplify fractions at the end to avoid arithmetic clutter. A quick plausibility check: since 20% of the lot is defective, drawing three without replacement should yield a probability modestly above 1−0.83=0.4881 - 0.8^3 = 0.488. Our exact result 29/57≈0.50929/57 \approx 0.509 is consistent, because without replacement slightly increases the chance of hitting a defective compared with independent draws.

The contrast is clear: “fiery speeches” versus “calm, measured tones.” The blank must signal tranquility. “Placid” means calm and peaceful, exactly fitting. “Vehement,” “vociferous,” and “ardent” all indicate passion or intensity, contradicting the description. On GRE completions, note contrast words like “although … surprisingly.” They require an antonymic relation between first and second clauses. Substituting “placid” yields: “fiery … but surprisingly placid” — fully coherent. Wrong options misalign with the tone markers.