GCSE - Higher Assessment 1 P2 Calculator

Question 1 Using a Calculator (ii) Find the value of: (give your answer to 3 sf) (i) Find the value of: (give your answer to 1 dp) – 6.8 √ 2.7³ – 15.3

Question 2 Standard Form (i) 2.71 × 10 6 – 4.8 × 10 5 Use you calculator to work out the value of the following, giving your answers in standard form. (ii) 3.4 × 10 6 ÷ 1.7 × 10 -2

Question 3 Trial & Improvement Solve by Trial & Improvement to 1dp: x³ + 4x = 48

Question 4 % Change A house’s value decreases from £165,000 to £130,000. Find the % decrease in its value to the nearest %.

Question 5 % – Compound Interest A person invests £5,000 for 3 years at a rate of 3.75% per annum. Calculate the value of their investment after the 3 years.

Question 6 % – Finding the Original A leather settee is reduced in a sale by 30%. It now costs £595. Calculate the price of the settee before the reduction.

Question 7 Nth Term & Sequences (i) Find the Nth term in the following sequence: 1, 4, 7, 10, 13 (ii) What would the 20 th term be?

Question 8 Straight Line Graphs Find the equation of the line that passes through the points (-1, -1) and (3, 7).

Question 9 Mean from Grouped Data The table shows the amount spent per week in a supermarket by 30 families. Calculate the mean amount spent per week. Amount (£)Freq 0 < x ≤ < x ≤ < x ≤ < x ≤ < x ≤ 1202

Question 10 Pythagoras’ Theorem AB C 28 m 15 m Find the length of side AB, give your answer to 1dp.

Question 11 Trigonometry Find the length of side BD to 1 dp. A B C 28 m 50º D 30º

Question 12 3D Trigonometry In the cuboid find angle GAC to 1 dp. A B C D E F G H 12 cm 10 cm 8 cm

Question 13 Sine & Cosine Rules Find side AC to 2 dp. A B C 9 cm 55º 65º

Question 14 Expanding Brackets Expand and simplify the following: (i) 2(4x + 5) – 3(x – 2) (ii) (x + 6)(x – 7)

Question 15 Algebraic Problem – Linear The rectangle has a perimeter of 28 cm. Form an equation and solve it to find x. 3x + 4 2x

Question 16 Surds Simplify: (i) (2 – √ 3 )² Rationalise the denominator (ii) 12 √3√3

Question 17 Quadratic – Forming Expressions The rectangle has an area of 30 cm². Show that it satisfies the equation: 2x + 3 x - 5 2x² – 7x – 45 = 0

Question 18 Quadratic – Formula Solve 2x² – 7x – 45 = 0 giving your answers to 2 dp.

Question 19 Circles Calculate the area of the segment AOB to 1 dp. A B O 65º 12 cm

Question 20 Volume A sphere has a volume of 3000 cm³. Calculate its radius to 2 dp.

End of Assessment

Answers

Question 1 Using a Calculator (ii) Find the value of: (give your answer to 3 sf) = = (i) Find the value of: (give your answer to 1 dp) – 6.8 √ 2.7³ – (1 dp) 2.09 (3 sf)

Question 2 Standard Form (i) 2.71 × 10 6 – 4.8 × 10 5 Use you calculator to work out the value of the following, giving your answers in standard form. (ii) 3.4 × 10 6 ÷ 1.7 × ,230,000 = 200,000,000 = 2.23 × × 10 8

Question 3 Trial & Improvement Solve by Trial & Improvement to 1dp: x³ + 4x = 48 x = 3 (3)³ + 4(3) = 39 too small x = 4 (4)³ + 4(4) = 80 too big x = 3.2 (3.2)³ + 4(3.2) = too small x = 3.3 (3.3)³ + 4(3.3) = too big x = 3.3 (1 dp) Closest 1 dp guess

Question 4 % Change A house’s value decreases from £165,000 to £130,000. Find the % decrease in its value to the nearest %. = 21% % Change = Change × 100 Original % Change = 35,000 × ,000

Question 5 % – Compound Interest A person invests £5,000 for 3 years at a rate of 3.75% per annum. Calculate the value of their investment after the 3 years. 1 st year 3.75 × £5000 = £ Value: £ nd year 3.75 × £ = £ Value: £ rd year 3.75 × £ = £ Final Value: £

Question 6 % – Finding the Original 70% = New Price A leather settee is reduced in a sale by 30%. It now costs £595. Calculate the price of the settee before the reduction. 70% = £595 1% = £8.50 (595 ÷ 70) 100% = £850 (8.5 × 100) 100% = Original Price

Question 7 Nth Term & Sequences (i) Find the Nth term in the following sequence: 1, 4, 7, 10, 13 (ii) What would the 20 th term be? N th ? 3n – (20)

Question 8 Straight Line Graphs Find the equation of the line that passes through the points (-1, -1) and (3, 7). y = 2x + 1 y = mx + c c = m =

Question 9 Mean from Grouped Data The table shows the amount spent per week in a supermarket by 30 families. Calculate the mean amount spent per week. Amount (£)Freq 0 < x ≤ < x ≤ < x ≤ < x ≤ < x ≤ 1202 Mid M × F £ = £67

Question 10 Pythagoras’ Theorem AB C 28 m 15 m = 23.6 m (1 dp) Find the length of side AB, give your answer to 1dp. a² + b² = h² (15)² + b² = (28)² b² = 784 b² = 784 – 225 b² = 559 b = √559

BD BC Question 11 Trigonometry Find the length of side BD to 1 dp. A B C 28 m 50º D 30º sin 50º = O × sin 50º = Opp BC = 21.4 (1dp) 21.4 cos 30º = A × cos 30º = A BC = 18.5 (1dp)

Question 12 3D Trigonometry In the cuboid find angle GAC to 1 dp. A B C D Pythagoras AC = 15.6 (1dp) E F G H 12 cm 10 cm 8 cm Trigonometry tan x = x = tan -1 8 ÷ 15.6 x = 27.1º (1dp) 8 cm

Question 13 Sine & Cosine Rules Find side AC to 2 dp. AC = 9.96 (2dp) A B C 9 cm 55º 65º Sine Rule sin A a sin B b = sin 55 9 sin 65 b = sin 55 9 × sin 65 = b

Question 14 Expanding Brackets Expand and simplify the following: (i) 2(4x + 5) – 3(x – 2) 5x +16 (ii) (x + 6)(x – 7) x² – x – 42

Question 15 Algebraic Problem – Linear x = 2 10x + 8 = 28 The rectangle has a perimeter of 28 cm. Form an equation and solve it to find x. 3x + 4 2x

Question 16 Surds Simplify: (i) (2 – √ 3 )² Rationalise the denominator (ii) 12 √3√3 4 √ √3

Question 17 Quadratic – Forming Expressions (2x + 3)(x - 5) = 30 The rectangle has an area of 30 cm². Show that it satisfies the equation: 2x + 3 x - 5 2x² – 7x – 45 = 0 2x² - 10x + 3x - 15 = 30 2x² - 7x = 0 2x² - 7x - 45 = 0 Area = L × W

Question 18 Quadratic – Formula a = 2 Solve 2x² – 7x – 45 = 0 giving your answers to 2 dp. b = – 7 c = – 45 - b ± √b² - 4ac 2a -- 7 ± √(-7)² - 4(2)(-45) 2(2) + 7 ± √ ± √ √ – √409 4 = 6.81 (2 dp) = (2 dp) 1 st Sol n 2 nd Sol n

Question 19 Circles Calculate the area of the segment AOB to 1 dp. A B O 65º Area O = π × r² Area AOB = 65 × π × r² cm = 65 × π × (12)² 360 = 81.7 cm² (1 dp)

Question 20 Volume A sphere has a volume of 3000 cm³. Calculate its radius to 2 dp. Vol = 4 3 × π × r³ 3000 = 4 3 × π × r³ 3000 × 3 = r³ 4 × π r³ = r = = 8.95 cm (2dp) 3