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Use the identity (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2. We are given x+y=8x+y=8 and x2+y2=50x^2+y^2=50. Square the sum: (x+y)2=82=64(x+y)^2 = 8^2 = 64. Substitute into the identity: 64=(x2+y2)+2xy=50+2xy64 = (x^2 + y^2) + 2xy = 50 + 2xy. Hence 2xy=64−50=142xy = 64 - 50 = 14, so xy=7xy = 7. Choice B is correct. This is a standard GRE maneuver: when a system gives you sums of first and second powers, the symmetric identity lets you avoid solving for xx and yy individually. A frequent pitfall is to try to factor or to set up a quadratic immediately; while possible, it’s slower. If you did proceed that way, you’d let tt be a root of t2−(x+y)t+xy=0t^2 - (x+y)t + xy = 0 with x+y=8x+y=8 and unknown xyxy, then use the second equation to solve for xyxy—ultimately arriving at the same result. Keep a list of core algebraic identities (sum of squares, difference of squares, binomial expansions); they streamline problems that otherwise appear to require more computation.

To find the inverse, set y=3x+2x−1y = \dfrac{3x+2}{x-1} and solve for xx in terms of yy. Cross-multiply: y(x−1)=3x+2⇒yx−y=3x+2y(x-1) = 3x + 2 \Rightarrow yx - y = 3x + 2. Collect xx terms: yx−3x=y+2⇒x(y−3)=y+2yx - 3x = y + 2 \Rightarrow x(y - 3) = y + 2. Hence x=y+2y−3x = \dfrac{y + 2}{y - 3}. Interchange variables to express f−1f^{-1}: f−1(x)=x+2x−3f^{-1}(x) = \dfrac{x+2}{x-3}. However, check carefully: our algebra gave x=(y+2)/(y−3)x = (y+2)/(y-3), so replacing yy with xx produces f−1(x)=x+2x−3f^{-1}(x)=\dfrac{x+2}{x-3}. Compare to options: that corresponds to choice C if denominator is 3−x3 - x or x−3x - 3? We found x−3x-3 in the denominator; option C is (x+2)/(3−x)(x+2)/(3-x), which equals −x+2x−3-\dfrac{x+2}{x-3} and would introduce a sign error. Re-check manipulation: yx−y=3x+2⇒yx−3x=y+2⇒x(y−3)=y+2⇒x=y+2y−3yx - y = 3x + 2 \Rightarrow yx - 3x = y + 2 \Rightarrow x(y-3)=y+2 \Rightarrow x=\dfrac{y+2}{y-3}. Thus the correct inverse is x+2x−3\boxed{\dfrac{x+2}{x-3}}, which is not option C as printed—it matches option A if A were (x+2)/(x−3)(x+2)/(x-3). Since option A is (x+2)/(x−3)(x+2)/(x-3) (verify): Option A given is x+2x−3\dfrac{x+2}{x-3}. Therefore the correct choice is A. Domain note: x≠3x\ne 3 for the inverse, corresponding to f(1)f(1) undefined; ranges and domains swap under inversion.

(Note: I selected the option matching (x+2)/(x−3)(x+2)/(x-3); if your choices display differently, pick the one with denominator x−3x-3.)

Let xx be liters of the 30% solution and yy liters of the 50% solution. We need x+y=20x+y=20 (total volume) and 0.30x+0.50y=0.40⋅20=80.30x + 0.50y = 0.40\cdot 20 = 8 (salt content). Substitute y=20−xy=20-x into the second equation:

0.30x+0.50(20−x)=8⇒0.30x+10−0.50x=8⇒−0.20x=−2⇒x=10.0.30x + 0.50(20 - x) = 8 \Rightarrow 0.30x + 10 - 0.50x = 8 \Rightarrow -0.20x = -2 \Rightarrow x = 10.

Then y=20−10=10y = 20 - 10 = 10. Thus we mix 10 L of each, choice B. A classic alternative is the “alligation” line: place 30 and 50 at the ends, target 40 in the middle. Distances from 40 are both 10, so mix in equal parts. Common pitfalls: (1) adding percentages instead of solving with weighted sums, or (2) solving the content equation but forgetting the total-volume equation. Checking reasonableness: the target concentration is exactly midway between 30% and 50%, so equal volumes make intuitive sense.

To strengthen a causal claim about output, you need direct, rigorous evidence that the new schedule caused higher output, not merely better sentiment. Option A provides randomized-pilot data tying the schedule change to a 12% increase in output per employee, with total hours constant—precisely the outcome metric and a strong design. B and D measure attitudes or preferences, not productivity. C indicates fewer sick days (a possible mechanism) but doesn’t prove output gains. GRE strengthening questions reward evidence that measures the stated outcome under controlled conditions; randomized pilots control confounds and thus most directly support the causal assertion.

The sentence contrasts honesty with avoidance: “acknowledged flaws … instead of glossing.” The blank must denote frankness. “Candid” means open and truthful—perfect. “Evasive,” “oblique,” and “circuitous” all suggest avoidance or indirection, inconsistent with acknowledging flaws. Substituting “candid” confirms coherence. GRE questions often use “instead of” to set up a direct opposition; the correct word is the opposite of avoidance.