Digital SAT Math — Hardest Questions (Korean Student Edition)

This is the classic trap: percent changes are multiplicative, not additive — you cannot just subtract 25% − 20% = 5%. Multiply the factors: 1.25 × 0.80 = 1.00, so the final price equals the original price exactly — a net change of 0%. Whenever a percent increase is followed by a different percent decrease, always multiply the decimal factors instead of adding/subtracting the percents.

A system of two linear equations has no solution when the lines are parallel but not identical — same slope, different intercept. Slope of line 1: −2/5. For line 2, slope = −k/15. Setting slopes equal: −k/15 = −2/5 → k = 6. Check the constants are NOT proportional: with k = 6, line 2 becomes 6x + 15y = 12, i.e. 2x + 5y = 4, which conflicts with line 1's 2x + 5y = 10 (4 ≠ 10) — confirming no solution when k = 6.

A system has infinitely many solutions only when the second equation is an exact multiple of the first — same slope AND same intercept, i.e. the two equations describe the identical line. Compare ratios of corresponding coefficients and constants: k/3 must equal −27/−9 = 3, so k = 9. Check the constant term too: 45/15 = 3, which matches the same scale factor of 3, confirming the lines are identical when k = 9. (Contrast this with the previous question: same slope + different constant → no solution; same slope + same constant ratio → infinitely many solutions.)

Because the right side must be ≥ 0 for the equation to be possible, we need x ≥ 2.5 — this alone eliminates some candidate solutions before you even finish solving. Case 1 (x − 4 ≥ 0): x − 4 = 2x − 5 → x = 1, but x = 1 fails x ≥ 2.5, so it is extraneous and must be rejected. Case 2 (x − 4 < 0): −(x − 4) = 2x − 5 → −x + 4 = 2x − 5 → 9 = 3x → x = 3, which satisfies x ≥ 2.5. Check: |3−4| = 1 and 2(3)−5 = 1. ✓ Only x = 3 works — always plug candidate solutions back into the original absolute value equation.

Multiply every term by (x − 3): x + 2 = 5 + 2(x − 3) → x + 2 = 5 + 2x − 6 → x + 2 = 2x − 1 → 3 = x. But x = 3 makes the original denominators (x − 3) equal to zero, which is undefined. This candidate is extraneous, so the equation actually has no solution. This is one of the most common traps on the digital SAT — always check that a solution to a rational equation doesn't make any denominator zero.

8^(2/3) = (8^(1/3))² = 2² = 4. And 4^(−1/2) = 1/√4 = 1/2. Multiply: 4 × (1/2) = 2. Remember: the denominator of a fractional exponent is the root, the numerator is the power, and a negative exponent means reciprocal — not a negative number.

From the table, f(x) = 2x + 1. To find g(5), first shift the input: g(5) = f(5 − 2) + 3 = f(3) + 3. From the table f(3) = 7, so g(5) = 7 + 3 = 10. The most common error is computing f(5) − 2 + 3 instead of f(5 − 2) + 3 — the horizontal shift happens inside the function first.

Read only the Science column, not the whole table: 30 out of 40 Science students are in Grade 11. 30/40 = 0.75 = 75%. A common mistake is dividing by the grand total (100) instead of the column total (40) — always match the denominator to the group named in the question ('of the students who prefer Science').

The first draw: P(red) = 4/12. Since the chip is not replaced, only 11 chips (3 red) remain for the second draw: P(red) = 3/11. Multiply: (4/12) × (3/11) = 12/132 = 1/11. Forgetting that the total and red count both shrink on the second draw is the most common error with 'without replacement' problems.

The maximum of a downward parabola occurs at the vertex, t = −b/(2a) = −64/(2(−16)) = 2 seconds. Then h(2) = −16(2)² + 64(2) + 5 = −64 + 128 + 5 = 69 feet. A common error is stopping at t = 2 and forgetting to plug it back in to find the actual height.

By the Remainder Theorem, the remainder when P(x) is divided by (x − a) equals P(a). Here, P(3) = 27 − 4(9) + 6 − 5 = 27 − 36 + 6 − 5 = −8. This avoids doing full polynomial long division.

Direct-and-inverse variation: y = k·(x/z). Solve for k using the first case: 12 = k(4/2) = 2k → k = 6. Now find y when x = 9, z = 3: y = 6 × (9/3) = 6 × 3 = 18.

Complementary angle identity: sin(θ) = cos(90 − θ). So sin(38°) = cos(90° − 38°) = cos(52°). Therefore x = 52. Students often mix this up with a supplementary-angle rule (180 − θ), which does not apply here.

Convert the central angle to radians: 120° × (π/180) = 2π/3. Arc length = r × θ(in radians) = 9 × (2π/3) = 6π. A common mistake is plugging the angle in degrees directly into the arc-length formula, which only works with radians.

FOIL the binomials: 3(1) + 3(−4i) + 2i(1) + 2i(−4i) = 3 − 12i + 2i − 8i². Since i² = −1, the last term −8i² becomes +8. Combine: (3 + 8) + (−12i + 2i) = 11 − 10i.