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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.

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Presentation on theme: "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."— Presentation transcript:

1 GCSE - Higher Assessment 1 P2 Calculator

2 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) 7.8 12.5 – 6.8 √ 2.7³ – 15.3

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

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

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

6 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.

7 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.

8 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?

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

10 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 ≤ 403 40 < x ≤ 607 60 < x ≤ 8012 80 < x ≤ 1006 100 < x ≤ 1202

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

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

13 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

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

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

16 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

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

18 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

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

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

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

22 End of Assessment

23 Answers

24 Question 1 Using a Calculator (ii) Find the value of: (give your answer to 3 sf) 1.368421053 = 2.093561559 = (i) Find the value of: (give your answer to 1 dp) 7.8 12.5 – 6.8 √ 2.7³ – 15.3 1.4 (1 dp) 2.09 (3 sf)

25 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 2,230,000 = 200,000,000 = 2.23 × 10 6 2 × 10 8

26 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) = 45.568 too small x = 3.3 (3.3)³ + 4(3.3) = 49.137 too big x = 3.3 (1 dp) Closest 1 dp guess

27 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 × 100 165,000

28 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 = £187.50 100 Value: £5187.50 2 nd year 3.75 × £5187.50 = £194.53 100 Value: £5382.03 3 rd year 3.75 × £5382.03 = £201.83 100 Final Value: £5583.86

29 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

30 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? 1 2 3 4 5 N th ? 3n – 2 58 3(20) - 2 60 - 2

31 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 =

32 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 ≤ 403 40 < x ≤ 607 60 < x ≤ 8012 80 < x ≤ 1006 100 < x ≤ 1202 Mid M × F 20 50 70 90 110 60 350 840 540 220 30 £2010 30 = £67

33 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)² 225 + b² = 784 b² = 784 – 225 b² = 559 b = √559

34 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 28 28 × sin 50º = Opp BC = 21.4 (1dp) 21.4 cos 30º = A 21.4 21.4 × cos 30º = A BC = 18.5 (1dp)

35 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 = 8 15.6 x = tan -1 8 ÷ 15.6 x = 27.1º (1dp) 8 cm

36 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

37 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

38 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

39 Question 16 Surds Simplify: (i) (2 – √ 3 )² Rationalise the denominator (ii) 12 √3√3 4 √3 7 - 4√3

40 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 - 15 - 30 = 0 2x² - 7x - 45 = 0 Area = L × W

41 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 ± √49 -- 360 4 7 ± √409 4 7 + √409 4 7 – √409 4 = 6.81 (2 dp) = -3.31 (2 dp) 1 st Sol n 2 nd Sol n

42 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 360 12 cm = 65 × π × (12)² 360 = 81.7 cm² (1 dp)

43 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 = 716.2 r = 716.2 = 8.95 cm (2dp) 3


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