Free GRE Exam Questions That Helps Candidates Pass

Divisible by both 2 and 3 = divisible by LCM(2,3)=6. Count multiples of 6 under 100: ⌊99/6⌋=16.\lfloor 99/6 \rfloor=16. These are {6,12,18,…,96}. Exclude those divisible by 4. Numbers divisible by 6 and 4 are divisible by LCM(6,4)=12. Count multiples of 12 under 100: ⌊99/12⌋=8.\lfloor 99/12 \rfloor=8. So desired count = 16–8=8. Correct choice: A. Re-check: among 16 multiples of 6, half (the even multiples of 12) also divisible by 4. So indeed 8 remain. Thus correct answer is 8. (Note: if original options differ, the mathematically correct result is 8.)

The second clause—“leave room for adaptation”—signals openness rather than rigidity. “Flexible” directly expresses that. “Rigid,” “prescriptive,” and “doctrinaire” all suggest inflexible strictness, inconsistent with adaptation. On GRE Text Completion, attend to positive signals like “leave room” that require words connoting openness or adjustability. Substituting “flexible” creates coherence: “recommendations are flexible … leave room for adaptation.”

Percent change from an initial to a final value depends only on those two endpoints. From Q1 to Q3, revenue increases from $120,000 to $165,000, a rise of $45,000. Compute 45,000/120,000=0.375=37.5%45{,}000 / 120{,}000 = 0.375 = 37.5\%. Option C is correct. The intermediate step (Q2) is irrelevant when asked for overall change from Q1 to Q3; avoid compounding sequential percent changes unless specifically requested. A common error is to average Q1→Q2 and Q2→Q3 percentage increases; that would not equal the net change. Another mistake is to use the wrong base (e.g., dividing by 165,000), which yields a different metric (fraction of final). Quick mental check: a 25% increase from 120k would be +30k (to 150k), and an additional 15k is half of 30k, i.e., another 12.5% of 120k; 25% + 12.5% = 37.5%. Structuring changes relative to the original baseline helps ensure you divide by the correct quantity.

Compute the area via Heron’s formula. Semiperimeter s=(13+14+15)/2=21s = (13+14+15)/2 = 21. Then

A=s(s−a)(s−b)(s−c)=21⋅8⋅7⋅6=7056=84.A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\cdot 8\cdot 7\cdot 6}=\sqrt{7056}=84.

The altitude hh to side b=14b=14 satisfies A=12bhA = \frac{1}{2} b h. Thus

84=12⋅14⋅h⇒84=7h⇒h=12.84 = \tfrac{1}{2}\cdot 14 \cdot h \Rightarrow 84 = 7h \Rightarrow h = 12.

Choice B. (Note that option C simplifies to 6, which is not correct.) A quick check: the area can also be found using base 15 and its altitude; the result must be consistent. Recognizing the classic 13–14–15 triangle is helpful: it has integer area 84, and the altitude to the middle side (14) is an integer (12), which is a common GRE-friendly feature. Avoid arithmetic slips when taking the square root: 7056=8427056=84^2, not ±84 \pm 84; area is positive.

Solve the inequality: ∣x+1∣≥4  ⟺  x+1≤−4 or x+1≥4|x+1|\ge 4 \iff x+1\le -4 \text{ or } x+1\ge 4, i.e., x≤−5x\le -5 or x≥3x\ge 3. Count within [−10,10][-10,10].

Left block: −10,−9,−8,−7,−6,−5-10,-9,-8,-7,-6,-5 → 6 integers.

Right block: 3,4,5,6,7,8,9,103,4,5,6,7,8,9,10 → 8 integers.

Total =6+8=14=6+8=14. Hence choice C. Check endpoints: “≥\ge” includes −5-5 and 33. A quick sanity test: middle values like −1,0,1,2-1,0,1,2 should not satisfy (distance from −1-1 less than 4); indeed they aren’t counted. On absolute-value inequalities, convert to a compound inequality and mind inclusivity; counting errors usually come from missing endpoints or off-by-one at the interval edges.