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1 PLC Papers Created For: Year 11 Topic Practice Paper: Solving Quadratics (Factorisation)
2 Quadratic equations (factorisation) 1 Grade 6 Objective: Solve quadratic equations by factorising. Question 1. Solve: a) = 0 b) = 0 c) = 0 d) 2 25 = 0
3 Question 2. Solve ( + 5) = 24 Question 3. Solve by factorising = 0 TOTAL /10
4 Quadratic equations (factorisation) 2 Grade 6 Objective: Solve quadratic equations by factorising. Question 1. Solve: a) = 0 b) = 0 c) 2 4 = 0 d) = 0
5 Question 2. Solve 2 30 = Question 3. Solve by factorising = 0 TOTAL /10
6 Quadratic equations (factorisation) 3 Grade 6 Objective: Solve quadratic equations by factorising. Question 1. The diagram shows a non-regular hexagon. Diagram NOT accurately drawn All the measurements are in centimetres. The area of the hexagon is 31 cm 2. a) Show that =0 (2) b) Work out the value of. (2)
7 Question 2. A rectangular classroom is 2m longer than it is wide. Carpet tiles costs 12 per square metre and it costs 756 to carpet the whole classroom. Write down an equation to represent this problem and find out how long the classroom is. Question 3. Solve 36 =5 Total /10 (Total 2 marks)
8 Quadratics equations (factorisation) 4 Grade 6 Objective: Solve quadratic equations by factorising. Question 1. a) Show that = 4 3 is a solution to the equation =0 b) Find the other solution. (2) Question 2. A right angle triangle has a hypotenuse of 17 cm. The other two sides are ( +3) cm and ( 4) cm. Calculate the numerical area of the triangle. cm 2
9 Question 3. A cyclist rides 50km. He calculates that if he were to increase his average speed by 5km/h, he would take 30 minutes less. Work out his average speed. km/h Total /10
10 PLC Papers Created For: Year 11 Topic Practice Paper: Solving Quadratics (Factorisation)
11 Quadratic equations (factorisation) 1 Grade 6 SOLUTIONS Objective: Solve quadratic equations by factorising. Question 1. Solve: a) = 0 ( + 9)( + 6) = 0 = 9 = 6 b) = 0 ( 7)( + 5) = 0 = 7 = 5 c) = 0 ( 4)( 7) = 0 = 4 = 7 d) 2 25 = 0 ( + 5)( 5) = 0 = 5 = 5
12 Question 2. Solve ( + 5) = = 0 (M1) ( + 8)( 3) = 0 (M1) = 8 = 3 Question 3. Solve by factorising = 0 (2 + 3)( + 5) = 0 (M2) = 3 2 = 5 TOTAL /10
13 Quadratic equations (factorisation) 2 Grade 6 Objective: Solve quadratic equations by factorising. Question 1. Solve: a) = 0 ( + 6)( 1) = 0 = 6 = 1 b) = 0 ( 12)( + 4) = 0 = 12 = 4 c) 2 4 = 0 ( 2)( + 2) = 0 = 2 = 2 d) = 0 (2 4)(2 + 4) = 0 = 2 = 2
14 Question 2. Solve 2 30 = 2 30 = 0 (M1) ( + 5)( 6) = 0 (M1) = 5 = 6 Question 3. Solve by factorising = 0 (3 + 4)(2 1) = 0 (M2) = 4 3 = 1 2 TOTAL /10
15 Quadratic equations (factorisation) 3 Grade 6 SOLUTIONS Objective: Solve quadratic equations by factorising. Question 1. The diagram shows a non-regular hexagon. Diagram NOT accurately drawn All the measurements are in centimetres. The area of the hexagon is 31 cm 2. a) Show that = 0 (2 + 3)(4 + 1) + (2 3 ) = 31 (M1) ( ) = = = 0 (2) b) Work out the value of = = 0 ( 1)( + 2) = 0 (M1) = 1 ( = 2) = 1 (2)
16 Question 2. A rectangular classroom is 2m longer than it is wide. Carpet tiles costs 12 per square metre and it costs 756 to carpet the whole classroom. Write down an equation to represent this problem and find out how long the classroom is. 12( ) = = = 0 (M1) ( 7)( + 9) = 0 (M1) = 7 ( = 9) Length = (7+2) m = 9 m Question 3. Solve 36 = = = 0 (M1) ( 9)( + 4) = 0 = 9 = 4 (Total 2 marks) Total /10
17 Quadratic equations (factorisation) 4 Grade 6 SOLUTIONS Objective: Solve quadratic equations by factorising. Question 1. a) Show that = 4 is a solution to the equation = 0 (2 + 5)(3 4) = 0 (M1) If 3 4 = 0 then 3 = 4 (C1) so = 4 3 b) Find the other solution. If = 0 then 2 = 5 so = 5 2 (2) Question 2. A right angle triangle has a hypotenuse of 17 cm. The other two sides are ( + 3) cm and ( 4) cm. Calculate the numerical area of the triangle. ( + 3) 2 + ( 4) 2 = 17 2 (M1) ( ) + ( ) = = = = 0 ( 12)( + 11) = 0 = 12 ( = 11) Area = 1 (15 + 8) = 60cm2 2 (M1) cm 2
18 Question 3. A cyclist rides 50km. He calculates that if he were to increase his average speed by 5km/h, he would take 30 minutes less. Work out his average speed. Let his average speed be km/h, so the increased speed would be ( + 5) km/h. At km/h time taken = hrs, at ( + 5) km/h time taken = hrs. (M1) If the journey takes 30 mins less, then: 50 = (M1) = = = ( + 5) = (105 + ) = = 0 (M1) ( + 25)( 20) = 0 So = 20 km/h km/h Total /10
PLC Papers Created For:
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