Straight-line graphs – Year 8 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Gradient = change in y ÷ change in x (1 mark for the method): \(\dfrac{11-3}{6-2}=\dfrac{8}{4}=\)2 (1 mark). For each step of 1 to the right the line goes up 2.
   2. Rewrite \(y=5-2x\) in the form \(y=mx+c\): \(y=-2x+5\). The gradient is the number multiplying x, \(-2\), and the y-intercept is 5. (1 mark)
   3. The point \((0,-4)\) is on the y-axis, so the intercept is \(-4\). With gradient 3 the equation is \(y=3x-4\). (1 mark)

   Check (a): the line \(y=2x-1\) passes through both points, since \(2\times2-1=3\) and \(2\times6-1=11\).

2. Answer to Question 2 (4 marks)

   1. Substitute each x value into \(y=2x-3\): \(x=-1\) gives \(-2-3=-5\); \(x=0\) gives \(-3\); \(x=1\) gives \(-1\); \(x=2\) gives 1; \(x=3\) gives 3. The y values are \(-5,-3,-1,1,3\). (1 mark for at least three, 2 marks for all five)
   2. Substitute \(x=5\): \(y=2\times5-3=7\). The point would need y = 7, not 8, so no, (5, 8) is not on the line; it lies just above it. (1 mark)
   3. Parallel lines have the same gradient, so the gradient is 2. The line passes through \((0,6)\), so the intercept is 6: \(y=2x+6\). (1 mark)

   The y values rise by 2 each time, which matches the gradient.

3. Answer to Question 3 (4 marks)

   1. The midpoint is the mean of the coordinates. x: \(\dfrac{-3+5}{2}=1\) (1 mark). y: \(\dfrac{2+(-6)}{2}=-2\) (1 mark). The midpoint is \((1,-2)\).
   2. Gradient \(=\dfrac{-6-2}{5-(-3)}=\dfrac{-8}{8}=\)\(-1\). The line slopes down from left to right. (1 mark)
   3. The equation has the form \(y=-x+c\). Substitute A(\(-3\), 2): \(2=-(-3)+c=3+c\), so \(c=-1\). The equation is \(y=-x-1\). (1 mark)

   Check with B(5, \(-6\)): \(-5-1=-6\). Check with the midpoint: \(-1-1=-2\). All three points satisfy the equation, so the working is consistent.

4. Answer to Question 4 (4 marks)

   1. Each cost is a fixed charge plus a daily rate times d: \(C=6d+10\) for the first shop and \(C=8d+4\) for the second. (1 mark)
   2. Set the costs equal: \(6d+10=8d+4\) (1 mark). Subtract \(6d\) from both sides: \(10=2d+4\). Subtract 4: \(6=2d\), so \(d=3\) days (1 mark).
   3. For 5 days: first shop \(6\times5+10=£40\), second shop \(8\times5+4=£44\). The first shop is cheaper, by £4. (1 mark)

   Check (b): at 3 days both cost £28. On a graph the two straight lines cross at (3, 28); after that point the steeper line, the second shop, is above, so it costs more.

5. Answer to Question 5 (4 marks)

   1. For n films the first club costs \(C=4n+15\) and the second costs \(C=7n\), a line through the origin. (1 mark)
   2. Set them equal: \(4n+15=7n\) (1 mark). Subtract \(4n\): \(15=3n\), so \(n=5\) films (1 mark). Both cost £35 at that point.
   3. For 8 films: first club \(4\times8+15=32+15=£47\); second club \(7\times8=£56\). The first club is cheaper, by \(56-47=\)£9. (1 mark)

   For fewer than 5 films the club with no joining fee is cheaper. Maya plans more than 5 films, so the joining fee is worth paying, and each extra film saves her another £3.

Total : 20 marks