Free GRE Exam Questions That Helps Candidates Pass

The argument draws a causal conclusion (make the module mandatory → boost completion) from an observed correlation (those who choose it finish more often). A key threat is selection bias: students who voluntarily complete an optional module may be more motivated or better prepared. If higher completion is due to those preexisting traits, mandating the module will not replicate the effect. Option A precisely articulates the required assumption that the module itself drives the improvement, not self-selection. Option B concerns implementation efficiency, not the logic of the claim. Option C introduces an equity issue that could matter operationally but is not logically necessary for the causal inference. Option D is peripheral; instructor training may be useful but is not the linchpin assumption. On GRE Argument tasks, always scrutinize whether an intervention’s observed benefit among volunteers truly generalizes; ruling out selection effects is often the necessary bridge from correlation to causation.

The structure contrasts “initially seemed …” with “subsequent experiments provided strong support.” The blank must describe doubt or skepticism, later overturned by evidence. “Implausible” means not seeming reasonable or probable, fitting perfectly. “Compelling,” “obvious,” and “incontrovertible” suggest strong acceptance, contradicting the idea of initial doubt. GRE Text Completion questions often hinge on contrast markers like “though” or “initially seemed … but later.” Substituting “implausible” keeps the narrative coherent: theory appeared implausible, later supported.

Possible sums range from 2 to 12. Prime sums in this range are 2, 3, 5, 7, and 11. Count ordered pairs:

Sum 2: (1,1) → 1 way

Sum 3: (1,2), (2,1) → 2 ways

Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 ways

Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways

Sum 11: (5,6), (6,5) → 2 ways

Total favorable outcomes = 1+2+4+6+2=151 + 2 + 4 + 6 + 2 = 15. Total outcomes = 6×6=366 \times 6 = 36. Probability =15/36=5/12= 15/36 = 5/12. Option B is correct. This item tests discrete probability, counting, and recognition of primes—core GRE Quant skills. A common pitfall is missing symmetric pairs or including composite sums like 9 or 10. Another is forgetting that dice outcomes are ordered; (2,3) and (3,2) are distinct. When time is short, note that the distribution of sums is triangular: counts increase up to 7 and then decrease symmetrically. Knowing that 7 has 6 outcomes and using symmetry speeds enumeration. Reduce fractions to simplest terms; 15/36 simplifies by dividing numerator and denominator by 3, yielding 5/12.

The sentence praises the memo for resolving a dispute “in barely half a page” with “crisp precision.” The blank must convey brevity without any negative implication. “Laconic” means concise, terse, or using few words—exactly what the sentence celebrates. “Garrulous” means excessively talkative, which contradicts “barely half a page.” “Equivocal” indicates ambiguity or the use of language open to multiple interpretations; the memo is described as precise, not ambiguous. “Abstruse” means difficult to understand, again clashing with “crisp precision.” On GRE Text Completion, look for multiple cues that converge: (1) length cue (“barely half a page”), (2) outcome cue (“resolved … dispute”), (3) style cue (“crisp precision”). Together they signal brevity paired with clarity—hallmarks of “laconic.” A good tactic is to plug choices back into the sentence; with “laconic,” the rhetoric remains coherent and complimentary, whereas the other options either invert the tone or introduce a meaning the author clearly rejects.

Compute inside-out. First evaluate g(5)=2⋅5+3=13g(5)=2\cdot 5 + 3 = 13. Then apply ff to that result:

f(13)=132−4(13)=169−52=117.f(13) = 13^2 - 4(13) = 169 - 52 = 117.

Thus choice C. This problem checks function composition, a recurring GRE Quant theme. The order matters: f(g(5))f(g(5)) differs from g(f(5))g(f(5)). For instance, f(5)=25−20=5f(5)=25 - 20 = 5 and g(f(5))=g(5)=13g(f(5))=g(5)=13, not 117. A helpful tip is to box the inner evaluation (here, 13) before substitution, reducing the likelihood of plugging 5 into both functions accidentally. You can also rewrite f(g(x))f(g(x)) symbolically to verify: f(2x+3)=(2x+3)2−4(2x+3)=4x2+12x+9−8x−12=4x2+4x−3f(2x+3)=(2x+3)^2 - 4(2x+3) = 4x^2 + 12x + 9 - 8x - 12 = 4x^2 + 4x - 3. Then f(g(5))=4(25)+4(5)−3=100+20−3=117f(g(5))=4(25) + 4(5) - 3 = 100 + 20 - 3 = 117, acting as a consistency check. When time is tight, numeric substitution is faster; symbolic composition is useful for verifying and for items that ask for expressions rather than values.