Diagnostic Test

This is a linear equation in one variable. Distribute, collect the x-terms, and isolate x. Distributing the 3 opens the left side to 6x − 12. Both sides are now linear in x, so gather them: subtracting 5x leaves x − 12 = −6, and adding 12 gives x = 6. The one place to move slowly is that first distribution: the 3 multiplies both the 2x and the −4, not the 2x alone.

Slope is a single idea, how fast y changes as x moves, and it is read off any two points as the change in y over the change in x. Going from (−2, 5) to (4, −7), y falls by 12 while x rises by 6, so the slope is −12/6 = −2. Keep the order of the points the same top and bottom; subtracting the y-values one way and the x-values the other is the usual source of a sign error.

A discount hides a shortcut: taking 35 percent off is the same as keeping 65 percent, so one multiplication settles it instead of two steps. Sixty-five percent of 80 is 0.65 × 80 = 52 dollars. Stopping at the 28-dollar discount itself is the easy error; that is choice A, the part removed rather than the part that remains.

Evaluating a function is pure substitution: put −2 everywhere x appears and let the arithmetic finish. The main risk is sign handling: (−2)² is +4, so 2(−2)² = 8, and −3(−2) = +6. Adding, 8 + 6 + 1 = 15. The safe habit is to wrap −2 in parentheses before squaring; writing −2² drops the sign and points the wrong way.

Two equations, two unknowns, and these two are built for elimination, because the y-terms are already equal and opposite, +y and −y. Adding the equations straight down cancels y in one stroke: 3x = 12, so x = 4. There is no need to find y at all; the question asks only for x, and the structure hands it over directly.

When a transversal crosses parallel lines, every angle it makes is one of just two sizes, and any two are either equal or supplementary. The 70° angle and x sit on the same side of the transversal, between the lines, same-side interior angles, and that pairing is supplementary, summing to 180. So x = 180 − 70 = 110. The figure rewards a moment of labeling: decide which relationship the two marked angles share before reaching for a number.

A quadratic set to zero is asking which inputs make a product vanish, so factor it into the two numbers whose product is −15 and whose sum is −2: those are −5 and +3, giving (x − 5)(x + 3) = 0. A product is zero only when one factor is, so the solutions are 5 and −3. The answer list offers −3 (choice C). If factoring feels slow, testing each choice back in the equation reaches the same place.

An absolute-value inequality is a statement about distance: |x − 3| < 4 says x lies within 4 units of 3. Unfolding it gives −4 < x − 3 < 4, and adding 3 throughout opens it to −1 < x < 7. Now count the integers strictly inside: 0, 1, 2, 3, 4, 5, 6, seven of them. The strict < is what keeps −1 and 7 out; counting those endpoints is the easy way to arrive at 8 by accident.

A mean is a total in disguise, the sum shared out equally, so a question about an average is really a question about a sum. The five numbers add to 30. For six numbers to average 7, they must total 6 × 7 = 42. The appended value is whatever closes that gap: 42 − 30 = 12. Working through the totals is steadier than trying to average the numbers and adjust by feel.

Doubling is repeated multiplication, so the only quantity that matters is how many doublings occur. Twelve hours holds 12 ÷ 3 = 4 doubling periods, and four doublings multiply the population by 2⁴ = 16. Starting from 500, that is 500 × 16 = 8,000. The common wrong turn is to grow additively, multiplying by 2 and then by the hours, but the exponent counts periods, not hours.

Two ticket types, two facts, a count and a total, so name the unknowns and let the equations pin them. With a + c = 200 and 12a + 8c = 2,040, substitute a = 200 − c into the money equation: 2,400 − 4c = 2,040. That collapses to 4c = 360, so c = 90 child tickets. The question asks for children; solving for adults and forgetting to switch back is the easy way to hand over the wrong number.

A sector is a slice of the whole circle, and its central angle says exactly what fraction: 120 out of 360 degrees is one third. So its area is one third of πr², set equal to the given 12π. Then (1/3)πr² = 12π, the π cancels, and r² = 36, giving r = 6. The 120° central angle represents one third of the circle, so the sector equation follows directly.

Vertex form lays the parabola bare: f(x) = (x − 3)² − 4 sits lowest at x = 3. Setting f(x) = 0 gives (x − 3)² = 4, so x − 3 = ±2 and the roots are 1 and 5; their sum is 6. There is an even quicker read: a parabola's two roots are mirror images across the vertex line x = 3, so they average 3 and sum to 2 × 3 = 6. Seeing that symmetry turns a two-step solve into a single line.

“Constant rate” is the signal for a linear model: a steady slope plus a starting height. The slope is the change in height over the change in time, (12 − 18)/(5 − 2) = −2 centimeters per hour, negative because the candle shrinks. Write the height as −2t + b and use the t = 2 reading: 18 = −2(2) + b, so b = 22. That b is precisely what the question wants: the height at t = 0, before any burning. The starting value is the intercept, not one of the measured points.

A rational expression simplifies only when the denominator hides inside the numerator as a factor, so factor first and look for the shared piece. The numerator 2x² + 7x + 3 factors as (2x + 1)(x + 3), and the denominator is (x + 3). Canceling the common (x + 3) leaves 2x + 1. The note x ≠ −3 is what makes the cancellation legal, it keeps that denominator from being zero.

Sine ties an angle to the side across from it and to the hypotenuse, opposite over hypotenuse, so the figure's job is to tell you which side is opposite. Angle A looks across the triangle to BC, so sin A = BC/AB = 3/5. With the hypotenuse AB = 20, that gives BC = (3/5)(20) = 12. The resulting 12-16-20 triangle is consistent with a scaled 3-4-5 triangle.

Two fractions of x add most easily once they share a denominator, and the smallest common one for 4 and 6 is 12. Rewritten, x/4 = 3x/12 and x/6 = 2x/12, so the left side becomes 5x/12 = 5. Multiplying both sides by 12 gives 5x = 60, and x = 12. Clearing the denominators at the start, multiplying the whole equation by 12, reaches the same place if you would rather skip the rewriting.

“Exactly one real solution” is the discriminant speaking: a quadratic has a single repeated root precisely when b² − 4ac = 0. Here that is k² − 4(1)(9) = 0, so k² = 36 and k = ±6. The question asks for a positive value, so k = 6. The condition "exactly one real solution" requires the discriminant to equal zero.

Percent change always measures the change against where you started, never where you ended. The increase is 320 − 250 = 70, and as a fraction of the original 250 that is 70/250 = 0.28, or 28 percent. Dividing by 320 instead is the most common wrong turn, it quietly answers a different question. The grid takes 28, with the percent sign left off as instructed.

An equation with the unknown in the exponent wants the power isolated, then the bases matched. Dividing 3 · 2^x = 96 by 3 gives 2^x = 32. Since 32 is 2⁵, the exponents must agree: x = 5.

A mixture problem is bookkeeping on one ingredient: here, the acid. The starting mixture holds 0.30 × 20 = 6 liters of acid in 20 liters of liquid. Pouring in y liters of pure acid adds to both the acid and the total, so the new concentration is (6 + y)/(20 + y), set equal to 0.50. Clearing the fraction: 6 + y = 10 + 0.5y, which gives 0.5y = 4 and y = 8 liters. Tracking the acid and the total separately, rather than juggling percentages, keeps the equation honest.

A circle hides inside a scrambled equation, and completing the square pulls its center and radius back into view. Group the terms: (x² − 6x) + (y² + 8y) = −9, then add the number that completes each square (9 for the x-group, 16 for the y-group) to both sides. That gives (x − 3)² + (y + 4)² = 16. The right side is r², so r = √16 = 4. The amount you add is always the square of half the middle coefficient: 9 from half of −6, 16 from half of 8.

▸ Use the skills map to choose what to review next

Tally the questions you answered wrong by domain. A cluster in one domain (three wrong in Advanced Math, say) points to where a few hours of focused review will move your score the most. A scatter of single careless errors across domains points instead to the accuracy habits of Part V. Either way, the diagnostic has done its work: it has turned a score into a short, specific list of what to do next. Work that list, then take the first full-length test.