NCEA Level 2 Mathematics 91261 Algebra free practice questions with answers

Question 1 (9 marks) [Expressions, indices and surds]

(a) Simplify (4x²y⁻⁶)^(3/2) ÷ (2y⁻¹), giving your answer with positive indices only (x and y are positive).

(a) Answer: 4x³/y⁸

• Evidence level: Merit (apply the laws of indices, including fractional indices).
• Apply the index
• 4^( • Divide by 2y⁻¹: subtract the indices of x and y, and divide the numbers.
• Write with positive indices: 4x³/y⁸.

Fractional indices: (a^m)^(p/q) means take the q-th root and raise to the power p; treat the number and each letter separately, then subtract indices when dividing.

(b) Write y = x² + 4x in the form y = a(x − p)² + q. Hence state the coordinates of the turning point, and find the values of k for which the equation x² + 4x = k has no real solutions.

(b) Answer: y = (x + 2)² − 4; turning point (−2, −4); k < −4

• Evidence level: Merit (complete the square, then link the turning point to the number of solutions).
• Halve the x coefficient (4 ÷ 2 = 2) and square it.
• y = (x + 2)² − 4.
• The turning point is (−2, −4), a minimum.
• x² + 4x = k has no real solution when the horizontal line y = k does not touch the curve: k < −4.

Completing the square gives the turning point (p, q) directly; a horizontal line y = k misses the parabola when k is beyond the minimum (or maximum) value.

(c) A car is bought for $40,000. Its value falls by 15% of its value at the start of each year. Form an equation, and solve it, to find after how many whole years the car is first worth less than 50% of its original price.

(c) Answer: 5 years

• Evidence level: Merit (model exponential decay and solve with logarithms).
• After n years the value is V = 40,000 × 0.85^n.
• We need V < 40,000 × 0.5, so 0.85^n < 0.5.
• Take logs: n × log(0.85) < log(0.5). Dividing by log(0.85), which is NEGATIVE, reverses the inequality: n > 4.27.
• The first whole number of years is n = 5.

3

2 to each factor in the bracket: 4^(3