ACT Math: Number and Quantity (Drill 3)

Question 1. Given i = √−1, what is (5 + 3i) + (2 − 7i)?

• A) 7 + 4i
• B) 7 − 4i ✓
• C) 3 + 10i
• D) 3 − 10i

Explanation: Add real parts and imaginary parts separately. Real: 5 + 2 = 7. Imaginary: 3i + (−7i) = −4i. The result is 7 − 4i. Choice A has the wrong sign on the imaginary part, a common error when students drop the negative sign from the 7i term. Choices C and D result from subtracting the real parts (5 − 2 = 3) instead of adding them.

Question 2. Given i = √−1, what is (3 + 2i)(1 − 4i)?

• A) 3 − 8i²
• B) −5 − 10i
• C) 11 − 10i ✓
• D) 11 + 10i

Explanation: Use FOIL: (3)(1) + (3)(−4i) + (2i)(1) + (2i)(−4i) = 3 − 12i + 2i − 8i². Since i² = −1, the term −8i² = −8(−1) = 8. Combine real parts: 3 + 8 = 11. Combine imaginary parts: −12i + 2i = −10i. The answer is 11 − 10i. Choice A leaves i² unsimplified. Choice B results from treating i² as 0 and losing the +8 that comes from simplifying −8i². Choice D has the correct real part but the wrong sign on the imaginary part.

Question 3. Vector u has components ⟨3, 4⟩ and vector v has components ⟨−1, 2⟩. What is the magnitude of u + v?

• A) √20
• B) √40 ✓
• C) √52
• D) √61

Explanation: First add the vectors component by component: u + v = ⟨3 + (−1), 4 + 2⟩ = ⟨2, 6⟩. Then find the magnitude: |u + v| = √2² + 6² = √4 + 36 = √40. Choice A results from computing √2² + 4² = √20, using the original y-component of u (4) instead of the summed y-component (6). Choice C results from computing √4² + 6² = √52, squaring the wrong x-component. Choice D results from computing the sum of the individual magnitudes of u and v before squaring: (5 + √5)² ≈ 61... actually from adding magnitudes of u (= 5) and v (= √5) and then squaring.

Question 4. Given i = √−1, what is the value of i⁴³?

• A) 1
• B) −1
• C) i
• D) −i ✓

Explanation: Powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, then the pattern repeats. Divide the exponent by 4: 43 ÷ 4 = 10 remainder 3. So i⁴³ = i³ = −i. Choice A corresponds to a remainder of 0 (e.g. i⁴⁴ = 1). Choice B corresponds to a remainder of 2 (e.g. i⁴² = −1). Choice C corresponds to a remainder of 1 (e.g. i⁴¹ = i). Since the remainder is 3, the answer is −i.

Question 5. Vector w has components ⟨−2, 5⟩. What are the components of the vector 3w − ⟨1, 1⟩?

• A) ⟨−7, 14⟩ ✓
• B) ⟨−5, 16⟩
• C) ⟨−5, 14⟩
• D) ⟨−7, 16⟩

Explanation: First multiply w by the scalar 3: 3w = 3⟨−2, 5⟩ = ⟨−6, 15⟩. Then subtract ⟨1, 1⟩ component by component: ⟨−6 − 1, 15 − 1⟩ = ⟨−7, 14⟩. Choice B results from adding ⟨1, 1⟩ instead of subtracting: ⟨−6 + 1, 15 + 1⟩ = ⟨−5, 16⟩. Choice C adds in the first component but correctly subtracts in the second: ⟨−5, 14⟩. Choice D correctly subtracts in the first component but adds in the second: ⟨−7, 16⟩.