Mixed Topic Worksheet Grades 6/7: Day 065
- 1
Question 1. A straight line passes through (1, 5) and (4, 14).
Does the point (6, 20) lie on the line?
You must show your working.Equations of a Straight Line Using Two Points▶ Show worked answerHide worked answer
- M1: gradient dfrac14 - 54 - 1 = 3 and c = 2
- M1: 3 × 6 + 2 = 20
- A1: yes, 20 matches
Answer: Yes. The line is y = 3x + 2 and when x = 6, y = 20.[3 marks] - 2
Question 2. Shape A is drawn on the grid.
Enlarge shape A by scale factor -2 with centre of enlargement (-1, 1).
Label the image B.Enlargements with Negative Integer Scale Factors▶ Show worked answerHide worked answer
- B2: image B drawn with vertices (−3, −1), (−7, −1), (−7, −3) (one mark for the correct image on the wrong side of the centre, or for two correct vertices)
Answer: Image B has vertices (−3, −1), (−7, −1), (−7, −3)[2 marks] - 3
Question 3. (2n + a)² - (2n - a)² is a multiple of 40 for all integer values of n.
a is a positive integer.
Work out the smallest possible value of a.Proof with Multiples▶ Show worked answerHide worked answer
- M1: 4n² + 4an + a² - (4n² - 4an + a²)
- A1: 8an
- M1: 8an is a multiple of 40 for every n when a is a multiple of 5
- A1: a = 5
Answer: a = 5[4 marks] - 4
Question 4. Line mathbfL_1 has equation y = 4x - 3
Line mathbfL_2 has equation 4y + x = 37
Show that mathbfL_1 and mathbfL_2 are perpendicular.Properties of Perpendicular Lines▶ Show worked answerHide worked answer
- M1: gradient of mathbfL_1 is 4; mathbfL_2 rearranged to y = -dfrac14x + dfrac374, gradient -dfrac14
- C1: 4 × ≤ft(-dfrac14) = -1, so the lines are perpendicular
Answer: gradients 4 and -dfrac14, product -1[2 marks] - 5
Question 5. Solve 2x² + 3x - 4 = 0
Give your solutions in the form dfraca ± √bc where a, b and c are integers.Solving Quadratics with the Quadratic Formula▶ Show worked answerHide worked answer
- M1: x = dfrac-3 ± √3² - 4 × 2 × (-4)4
- M1: √41 seen
- A2: dfrac-3 ± √414 (A1 for one solution)
Answer: x = dfrac-3 ± √414[4 marks]
Mixed Topic Worksheet Grades 6/7: Day 065
- 1
Question 1. A straight line passes through (1, 5) and (4, 14).
[3 marks]
Does the point (6, 20) lie on the line?
You must show your working.Equations of a Straight Line Using Two Points▶ Show worked answerHide worked answer
- M1: gradient dfrac14 - 54 - 1 = 3 and c = 2
- M1: 3 × 6 + 2 = 20
- A1: yes, 20 matches
Answer: Yes. The line is y = 3x + 2 and when x = 6, y = 20. - 2
Question 2. Shape A is drawn on the grid.
[2 marks]
Enlarge shape A by scale factor -2 with centre of enlargement (-1, 1).
Label the image B.Enlargements with Negative Integer Scale Factors▶ Show worked answerHide worked answer
- B2: image B drawn with vertices (−3, −1), (−7, −1), (−7, −3) (one mark for the correct image on the wrong side of the centre, or for two correct vertices)
Answer: Image B has vertices (−3, −1), (−7, −1), (−7, −3) - 3
Question 3. (2n + a)² - (2n - a)² is a multiple of 40 for all integer values of n.
[4 marks]
a is a positive integer.
Work out the smallest possible value of a.Proof with Multiples▶ Show worked answerHide worked answer
- M1: 4n² + 4an + a² - (4n² - 4an + a²)
- A1: 8an
- M1: 8an is a multiple of 40 for every n when a is a multiple of 5
- A1: a = 5
Answer: a = 5 - 4
Question 4. Line mathbfL_1 has equation y = 4x - 3
[2 marks]
Line mathbfL_2 has equation 4y + x = 37
Show that mathbfL_1 and mathbfL_2 are perpendicular.Properties of Perpendicular Lines▶ Show worked answerHide worked answer
- M1: gradient of mathbfL_1 is 4; mathbfL_2 rearranged to y = -dfrac14x + dfrac374, gradient -dfrac14
- C1: 4 × ≤ft(-dfrac14) = -1, so the lines are perpendicular
Answer: gradients 4 and -dfrac14, product -1 - 5
Question 5. Solve 2x² + 3x - 4 = 0
[4 marks]
Give your solutions in the form dfraca ± √bc where a, b and c are integers.Solving Quadratics with the Quadratic Formula▶ Show worked answerHide worked answer
- M1: x = dfrac-3 ± √3² - 4 × 2 × (-4)4
- M1: √41 seen
- A2: dfrac-3 ± √414 (A1 for one solution)
Answer: x = dfrac-3 ± √414
Mixed Topic Worksheet Grades 6/7: Day 065
- 1
Question 1. A straight line passes through (1, 5) and (4, 14).
Does the point (6, 20) lie on the line?
You must show your working.Equations of a Straight Line Using Two Points▶ Show worked answerHide worked answer
- M1: gradient dfrac14 - 54 - 1 = 3 and c = 2
- M1: 3 × 6 + 2 = 20
- A1: yes, 20 matches
Answer: Yes. The line is y = 3x + 2 and when x = 6, y = 20.[3 marks] - 2
Question 2. Shape A is drawn on the grid.
Enlarge shape A by scale factor -2 with centre of enlargement (-1, 1).
Label the image B.Enlargements with Negative Integer Scale Factors▶ Show worked answerHide worked answer
- B2: image B drawn with vertices (−3, −1), (−7, −1), (−7, −3) (one mark for the correct image on the wrong side of the centre, or for two correct vertices)
Answer: Image B has vertices (−3, −1), (−7, −1), (−7, −3)[2 marks] - 3
Question 3. (2n + a)² - (2n - a)² is a multiple of 40 for all integer values of n.
a is a positive integer.
Work out the smallest possible value of a.Proof with Multiples▶ Show worked answerHide worked answer
- M1: 4n² + 4an + a² - (4n² - 4an + a²)
- A1: 8an
- M1: 8an is a multiple of 40 for every n when a is a multiple of 5
- A1: a = 5
Answer: a = 5[4 marks] - 4
Question 4. Line mathbfL_1 has equation y = 4x - 3
Line mathbfL_2 has equation 4y + x = 37
Show that mathbfL_1 and mathbfL_2 are perpendicular.Properties of Perpendicular Lines▶ Show worked answerHide worked answer
- M1: gradient of mathbfL_1 is 4; mathbfL_2 rearranged to y = -dfrac14x + dfrac374, gradient -dfrac14
- C1: 4 × ≤ft(-dfrac14) = -1, so the lines are perpendicular
Answer: gradients 4 and -dfrac14, product -1[2 marks] - 5
Question 5. Solve 2x² + 3x - 4 = 0
Give your solutions in the form dfraca ± √bc where a, b and c are integers.Solving Quadratics with the Quadratic Formula▶ Show worked answerHide worked answer
- M1: x = dfrac-3 ± √3² - 4 × 2 × (-4)4
- M1: √41 seen
- A2: dfrac-3 ± √414 (A1 for one solution)
Answer: x = dfrac-3 ± √414[4 marks]