IGCSE Maths 4MA1 Foundation Algebra Questions

• B1 (a) 4a
• B1 (b) 12b
• B1 (c) 4c

1. (a) Four lots of a: 4a.
2. (b) 3 × 4 = 12, so 12b.
3. (c) 5 − 2 + 1 = 4, so 4c.

Answer: (a) \(4a\) (b) \(12b\) (c) \(4c\)

• B2 3a⁴b² (B1 for two of 3, a⁴, b²)

1. 15 ÷ 5 = 3.
2. a⁶ ÷ a² = a⁴ and b³ ÷ b = b².

Answer: \(3a^4b^2\)

• B2 (a) 16x¹² (B1 for 16 or x¹²)
• B1 (b) x⁶

1. (a) (2x³)⁴ = 2⁴ × (x³)⁴ = 16x¹².
2. (b) x⁵ × x³ = x⁸, then x⁸ ÷ x² = x⁶.

Answer: (a) \(16x^{12}\) (b) \(x^6\)

• B1 4x + 12

1. Multiply each term in the bracket by 4: 4 × x = 4x and 4 × 3 = 12.

Answer: \(4x + 12\)

• M1 for (x + a)(x + b) with ab = 20 or a + b = 9
• A1 (x + 4)(x + 5)

1. Find two numbers that multiply to 20 and add to 9: 4 and 5.
2. So x² + 9x + 20 = (x + 4)(x + 5).

Answer: \((x + 4)(x + 5)\)

• B2 3xy(2x − 3y) (B1 for a correct partial factorisation, e.g. 3x(2xy − 3y²) or xy(6x − 9y))

1. The highest common factor of 6x²y and 9xy² is 3xy.
2. 6x²y ÷ 3xy = 2x and 9xy² ÷ 3xy = 3y.

Answer: \(3xy(2x - 3y)\)

• B1 (a) x + 4
• B1 (b) 2x + 4

1. (a) 4 years older: x + 4.
2. (b) x + (x + 4) = 2x + 4.

Answer: (a) \(x + 4\) (b) \(2x + 4\)

• M1 for s − 2r = 5t
• A1 t = (s − 2r)/5 oe

1. Subtract 2r from both sides: s − 2r = 5t.
2. Divide by 5: t = (s − 2r)/5.

Answer: \(t = \frac{s - 2r}{5}\)

• M1 (a) for r² = V/(πh)
• A1 (a) r = √(V/(πh)) oe
• B1 (b) 3.99

1. (a) Divide both sides by πh: r² = V/(πh). Take the square root: r = √(V/(πh)).
2. (b) r = √(500 ÷ 10π) = √15.915… = 3.989… = 3.99 (3 s.f.).

Answer: (a) \(r = \sqrt{\frac{V}{\pi h}}\) (b) \(r = 3.99\)

• B1 (a) 6
• B1 (b) 9

1. (a) Subtract 9 from both sides: x = 6.
2. (b) Divide both sides by 4: y = 9.

Answer: (a) \(x = 6\) (b) \(y = 9\)

• M1 for 2x + 3 = 35
• A1 16

1. Multiply both sides by 5: 2x + 3 = 35.
2. 2x = 32, so x = 16.

Answer: \(x = 16\)

• M1 for multiplying every term by 6, e.g. 2(x + 2) + 3(x − 1) = 24
• M1 for 5x + 1 = 24
• A1 23/5 oe, e.g. 4.6

1. Multiply every term by 6: 2(x + 2) + 3(x − 1) = 24.
2. 2x + 4 + 3x − 3 = 24, so 5x + 1 = 24.
3. 5x = 23, x = 4.6.

Answer: \(x = \frac{23}{5} = 4.6\)

• M1 for subtracting: 2x = 8
• M1 for substituting their x
• A1 x = 4, y = 3

1. Subtract: 2x = 8, so x = 4.
2. 2 × 4 + 3y = 17, so 3y = 9 and y = 3.

Answer: \(x = 4\), \(y = 3\)

• M1 for multiplying the second equation by 2: 10x − 4y = 8
• M1 for 13x = 26
• A1 x = 2, y = 3

1. Multiply the second equation by 2: 10x − 4y = 8.
2. Add to the first: 13x = 26, so x = 2.
3. 6 + 4y = 18, so y = 3.

Answer: \(x = 2\), \(y = 3\)

• M1 for subtracting: x = 5
• A1 x = 5, y = 3

1. Subtract the second equation from the first: x = 5.
2. Substitute: 5 + y = 8, so y = 3.

Answer: \(x = 5\), \(y = 3\)

• M1 for (x − 5)(x + 4)
• A1 x = 5, x = −4

1. Two numbers that multiply to −20 and add to −1: −5 and 4.
2. (x − 5)(x + 4) = 0, so x = 5 or x = −4.

Answer: \(x = 5\) or \(x = -4\)

• M1 for x² + 3x − 28 = 0
• M1 for (x + 7)(x − 4)
• A1 x = −7, x = 4

1. Rearrange to equal zero: x² + 3x − 28 = 0.
2. Factorise: (x + 7)(x − 4) = 0.
3. x = −7 or x = 4.

Answer: \(x = -7\) or \(x = 4\)

• M1 for (x − 2)(x − 3)
• A1 x = 2, x = 3

1. Factorise: (x − 2)(x − 3) = 0.
2. x = 2 or x = 3.

Answer: \(x = 2\) or \(x = 3\)

• B1 x > −2

1. An open circle means −2 itself is not included.
2. The arrow to the right means the numbers greater than −2.

Answer: \(x > -2\)

• A1 (a) −2 ≤ x < 4
• B1 (b) −2, −1, 0, 1, 2, 3

1. (a) Divide every part by 3: −2 ≤ x < 4.
2. (b) Include −2, exclude 4: −2, −1, 0, 1, 2, 3.

Answer: (a) \(-2 \le x \lt 4\) (b) \(-2, -1, 0, 1, 2, 3\)

• B1 (1, 1)
• B1 (4, 1)
• B1 (4, 4)

1. The boundaries are y = 1, x = 4 and y = x.
2. y = 1 meets y = x at (1, 1); y = 1 meets x = 4 at (4, 1); x = 4 meets y = x at (4, 4).

Answer: \((1, 1)\), \((4, 1)\), \((4, 4)\)