ACT Math Practice Test: Number, Statistics and Probability

The new mean is (400 + 140) ÷ 6 = 90, and the new median is the average of 80 and 85, which is 82.5, so the difference is 7.5. The value 10 is how much the mean rose (from 80 to 90), not the gap between mean and median.

Weighted average: 0.5(82) + 0.3(95) + 0.2(70) = 41 + 28.5 + 14 = 83.5. Averaging the three scores equally gives about 82.3, which ignores the weights.

There are 36 equally likely ordered outcomes, and 5 of them give a sum of 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). The fraction 211 treats the 11 possible sums as equally likely, which they are not.

The first draw is red with probability 58; then 4 of the remaining 7 marbles are red, so the probability is 58 × 47 = 2056 = 514. The value 2564 would be correct only if the first marble were put back.

This is a combination: ₈C₃ = (8 × 7 × 6) ÷ (3 × 2 × 1) = 56. The value 336 counts ordered selections, so it counts each committee 6 times.

Each section has probability 14, so the expected value is (0 + 2 + 5 + 9) ÷ 4 = $4.00. The value $5.33 leaves out the $0 section, and $3.50 is the median of the four amounts, not the mean.

Powers of i repeat every 4 (i, −1, −i, 1). Since 23 = 4 × 5 + 3, i²³ = i³ = −i. The answer −1 would be correct for a remainder of 2, not 3.

Multiply the numbers and add the exponents: 6 × 4 = 24 and 10^{5 + (−2)} = 10³, so the product is 24 × 10³ = 2.4 × 10⁴. The choice 2.4 × 10³ forgets to raise the exponent by 1 when 24 is rewritten as 2.4 × 10.

Take the cube root first, then square: ∛27 = 3 and 3² = 9. Multiplying 27 by 23 gives 18, which treats the exponent as a factor.

The buses leave together again after the least common multiple of 12 and 18, which is 36 minutes, so at 8:36 a.m. Multiplying 12 × 18 = 216 minutes gives a common multiple, but not the least one (11:36 a.m.).

ad − bc = 3 × 4 − 2 × 5 = 12 − 10 = 2. Adding the products instead of subtracting gives 22, and reversing the subtraction (10 − 12) gives −2.

Standard deviation measures how far the values lie from the mean. All four sets have a mean of 10, and the values 2, 6, 14, 18 lie farthest from it (4 to 8 units away), so this set has the greatest spread. The set of four 10s has a standard deviation of 0.

The choice is made only among the 45 students who play an instrument, and 27 of them are in Grade 10: 2745 = 35. The fraction 920 = 2760 divides by all Grade 10 students instead.

The probability that neither bulb is defective is 0.9 × 0.9 = 0.81, so the probability of at least one defective bulb is 1 − 0.81 = 0.19. Adding 0.1 + 0.1 = 0.20 counts the case where both are defective twice.

By the counting principle, there are 26 × 26 × 10 × 10 × 10 = 676,000 plates. The value 468,000 = 26 × 25 × 10 × 9 × 8 would be correct only if repeats were not allowed.