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Chapter 1 Practice Paper: Linear Graphs and Simultaneous Linear Equations — mark scheme

QuizQuipPractice Series
QQ/S2MATH/CH1

Secondary 2 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Linear Graphs and Simultaneous Linear Equations

Chapter 1 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer
1CRewriting as y = −2x + 5 shows the coefficient of x, −2, is the gradient.1
2AAt x = 1 the equation gives y = 4, so (1, 4) satisfies it.1
3DGradient 2 gives the 2x term, and the point (0, −3) fixes c = −3.1
4BOn the x-axis y = 0, so 2x = 12 and x = 6.1
5CSame gradient 2 but different intercepts: the lines never meet.1
6A3x = x + 2 gives x = 1, and then y = 3.1
7DAdding the equations gives 2x = 12, so x = 6.1
8BSubstituting x = 3: 6 + y = 8, so y = 2.1
9CThe crossing point is the one pair of coordinates that satisfies both equations.1
10ASubtracting the receipts removes the files: two notebooks cost 13 − 7 = $6, so one is $3.1

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
1(a)

y = −1, 1, 5, 7 (with the given 3 at x = 0).

  • At least two correct values [1]
  • All four values correct [1]
2
1(b)

Gradient 2; y-intercept (0, 3).

  • Gradient 2 [1]
  • y-intercept 3 [1]
2
1(c)

Yes: when x = 4, y = 8 + 3 = 11, matching the point.

  • Substitutes x = 4 and concludes yes [1]
1
Total for Question 15
2(a)

Both equations contain exactly one x, so subtracting gives x − x = 0.

  • Matching x-coefficients noted [1]
1
2(b)

(x + 2y) − (x − y) = 11 − 2 gives 3y = 9, so y = 3.

  • 3y = 9 formed [1]
  • y = 3 [1]
2
2(c)

x − 3 = 2 gives x = 5. Check: 5 + 2(3) = 11 ✓. Solution (5, 3).

  • x = 5 [1]
  • Check in x + 2y = 11 [1]
2
Total for Question 25
3(a)

x + 2(3x − 1) = 12, which expands to x + 6x − 2 = 12.

  • x + 2(3x − 1) = 12 [1]
  • Expands to 7x − 2 = 12 [1]
2
3(b)

7x = 14, so x = 2. [1]

1
3(c)

y = 3(2) − 1 = 5; the solution is (2, 5).

  • y = 5 [1]
  • Solution (2, 5) stated [1]
2
Total for Question 35
4(a)

3s + 2b = 19 and s + 2b = 9, where s is the scooter price and b the bumper-car price.

  • 3s + 2b = 19 [1]
  • s + 2b = 9 [1]
2
4(b)

Subtracting: 2s = 10, so s = 5; then 5 + 2b = 9 gives b = 2.

  • s = 5 [1]
  • b = 2 [1]
2
4(c)

4(5) + 2 = $22 [1]

1
Total for Question 45

— End of mark scheme —