Mathematics

Question 1 – 10 marks
This question is based on your work on MU123 up to and including Unit 6.
(a) (i) Find the gradient of the straight line that passes through the points
(3, −4) and (−3, 5).
Gradient=5-(-4)-3-3 =9/-6=3/-2
The gradient will be -3/2 [1 ]
(ii) Use this gradient to find the equation of the line, clearly showing
each step of your method.
Equation of a straight line is given by
y=mx+c
y and x are the y and x-intercepts respectively
c is constant
Where m is the gradient,
Y=3/-2x+c
[3 ]
(b) Use the equation that you found in part (a) to determine whether or not
the point (−1, 2) lies on this line.
Y=3/-2x+c
2=3/-2(-1) +c
C=2-(3/2)
C=(1/2)
The point lies on the line
[2 ]
(c) (i) Use algebra to find the x-intercept of the line that you found in
part (a), clearly showing each step of your method.
Y=3/-2x+1/2
At intercept y=0
0=-3/-2x+1/2
3/2x=1/2
X=1/3
(1/3,0)
[2 ] (ii) What is the y-intercept? [1 ]
2y=-3x+1
At y intercept x=0
Hence
2y=1
Y=1/2
(0, 1/2)
(d) Plot the two points given in part (a), and draw a line through them.
Use this plot to check that your answers to parts (a), (b) and (c) are
sensible. [1 ]
Either a hand-drawn graph or a computer-generated graph is acceptable.
lefttop
Question 2 – 15 marks
This question is based on your work on MU123 up to and including Unit 6.
This question investigates the relationship between the adhesive strength of laminated wood and the time spent in the press machine. A random sample of 15 different times (minutes) and their corresponding adhesive strengths (kg per square cm) were recorded as shown in Table 1.

Table 1 Time in press machine and adhesive strength
71945533274000719455860869500
Adhesive strength (kg per square cm)
1.27
1.41
3.0 1.36
3.4 1.54
3.6 1.63
3.7 1.72
4.0 1.81
2.8 1.27
3.0 1.36
3.1 1.41
3.3 1.50
3.8 1.72
4.1 1.86
3.4 1.50
3.6 1.63
(a) Open Data plotter and enter the data, using ‘Press time’ in the first column and ‘Adhesive strength’ in the second column. Obtain a scatterplot, and display the regression line and the correlation coefficient. Include the scatterplot, with the regression line, in your answer.
You should label your scatterplot appropriately. This may be a printout from Data plotter or a scatterplot produced by hand. You can add details,
Such as the title, to a computer-generated graph by hand, if you wish. [5]
lefttop
(b) Write down the equation of the regression line, in the form y = mx + c.
What do x and y represent in the equation? [2]
A linear regression line has an equation of the form Y = a + bX
where X is the explanatory variable and Y is the dependent variable. The slope of the line is b, and a is the intercept (the value of y when x = 0).
(c) Write down the value of the correlation coefficient r. What does this tell?
You? [2]
r = sxy     (sxsy)
where sxy is the covariance of x and y,.

(d) Use the regression equation to predict the adhesive strength of laminated wood after 3.2 minutes in the press machine. How reliable do
You think your prediction is likely to be? Explain your answer. [4]
Hint: Remember that reliability and causation are different. If you are unsure about what is meant by ‘reliability,’ revisit Unit 6, Activity 25, page 106, and also read page 111 before you attempt this question.
Adhesive strength has a negative regression
(e) Can you conclude from this investigation that press time determines the adhesive strength of laminated wood? Explain your answer.
Adhesive strength has a negative correlation [2]
Question 3 – 5 marks
This question is based on your work on MU123 up to and including Unit 7.
Make k the subject of the following two equations. Show each step of your working.
132715022733000(a) 3t = 7k − 17 [2 ]
13
3t+17=7/13k
K=39/7t+221/7
135318521717000(b) 7k = 4k − 11t [3 ]
3t
Multiply 3t across we get
7kt=4k-33t2
7kt-4k=-33t2
Factoring out k
K(7t-4)=-33t2
K=-33t2/(7t-4)
Rosie is organizing a night out for the local youth group. She has decided on two possible options: going to the cinema or going bowling. For group bookings, the cinema charges £4 per person. Transport to the cinema would cost £90 for the group. The bowling alley charges £5.50 per person, and transport would cost £45. There are 50 teenagers in the youth group, but not all are expected to attend the night out.
(a) Copy and complete Table 2 by working out how much it costs to go to each venue for the numbers of teenagers shown.
1007110219075001064260249555Number of teenagers attending 26 38 50
Cinema 194 242 290
Bowling 188 254 320
00Number of teenagers attending 26 38 50
Cinema 194 242 290
Bowling 188 254 320
Table 2 Total cost in pounds for each option

[2]
(b) The linear equation y = 4x + 90 can be used to calculate the total cost y (in pounds) for x teenagers attending if they choose the cinema. Explain why this equation holds. Write down a similar equation that can be used to give the total cost y (in pounds) for x teenagers
Attending if they choose to go bowling. [2]
By replacing X with 50, we get
Y=4(50) +90
Y=200
Hence the equation holds
Equation for those attending a bowling
Y=5.5X+45
(c) (I) Using the same axes, draw graphs of the two equations from part (b). Use an x-scale from 0 to 50, and choose a y-scale that allows both lines to be seen over this range.
You may draw your graphs by hand, using squared paper, or use
Graph plotter and send a printout with the rest of your TMA.
Remember to label the lines with their equations.

[4]
(ii) What do the gradients and y-intercepts of these lines represent in
Terms of the situation being modelled? [2]
(d) The lines representing the equations in part (b) intersect. Use your graphs to write down the approximate coordinates of the intersection
Point. How can the intersection point be interpreted in this model? [2]
Y = 212
X = 29
(e) Use algebra to solve the two simultaneous equations from part (b).
Show each step of your working. [4]
I will solve your system by substitution.
y=4x+90;y=5.5x+45
Step: Solvey=4x+90for y:
Step: Substitute4x+90foryiny=5.5x+45:
y=5.5x+45
4x+90=5.5x+45
4x+90+−5.5x=5.5x+45+−5.5x(Add -5.5x to both sides)
−1.5x+90=45
−1.5x+90+−90=45+−90(Add -90 to both sides)
−1.5x=−45
−1.5x
−1.5
=
−45
−1.5
(Divide both sides by -1.5)
x=30
Step: Substitute30forxiny=4x+90:
y=4x+90
y=(4)(30)+90
y=210(Simplify both sides of the equation)
x=30 and y=210
(f) State which option gives the better deal, in terms of cost, for different
Numbers of teenagers attending. [2]
Bowling gives a better deal(a) In Figure 1, AC DF is a straight line. The straight lines AB and GDE
Are parallel, and the straight lines BC G and EF are parallel.
∠ABC = 55◦ and ∠EF D = 65◦.
B
55o E
A
1910080-109156500D C
65o
G F
Figure 1
Find ∠BC A, explaining your reasoning.
<BCA is 650
This is because ∠EF D = 65◦
Is corresponding to angle ∠BC A
[2 ] (ii) Find ∠DEF , explaining your reasoning.
55+65=120
180-120=600
<BAC=600 which is a corresponding angle to ∠DEF
Hence ∠DEF is 600
[2 ] (iii) Show that triangle ABC is similar to triangle C GD, explaining
your reasoning.
∠DEF=<BAC=600 this are corresponding angles
∠EF D=∠BC A=650 this is because they are corresponding angles
<EDF=<CBA=550
Hence triangle ABC is similar to triangle C GD, explaining
[2 ]
(b) Figure 2 shows the triangles ABE and BC D. The lengths of AE and
DB are the same. ∠ABE = ∠BC D = 38◦ . ∠EAB = 35◦ and
B
A 35o

38o
2131695-80200500E
D 107o
38o
C
Figure 2
Show that ABE and BC Dare congruent triangles. [3]
Show that ABE and BC D are congruent triangles. [3 ]
The two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent
191135060896500Center D. AC = BC and AD = BD. The length of AB is 50 cm, and the length of DB is 35 cm. ∠BEC = 90◦.
50 cm
A E B
35 cm
1586230132588000D
C
Figure 3
Give your answers to each of the following to two significant figures.
(i) Given that E is the midpoint of AB, write down the length of EB, and use Pythagoras’ Theorem in triangle EBD to find the length
of ED. [3 ]
LENGTH EB is half AB
Hence EB= 25
Pythagoras theorem is given
(EB)2+(ED)2=(BD)2
(ED)2=352-252
ED=√ (352-252
=24.49
Write down the length of C D, and hence find the length of EC. [2 ]
CD Will be 35cm since it is the radius
EC=CD+ED
Therefore EC will b given as 24.49cm+35
59.49cm
(iii) Find the area of triangle EBC. [2 }
Area=√3 *352
=√3675
60.62cm2
(iv) Find the area and circumference of the circle.
Area of the A=πr23.14*352
3846.5cm2
Circumferences is 2piD
2*3.14*70CM
439.6CM
[3 ] (v) Find the area of the circle that lies outside triangle ABC [3 ]
Area of the triangle which lies outside ABC
=Area of all circle minus that of the triangle
1/2b*h
½*50*59.49
1487.25cm2
3846.5cm2-1487.25cm2
Answer
2359.25cm2
This question is based on your work on MU123 up to and including Unit 9.
(a) Expand the brackets in the following expressions.
(i) (9x − 13)(11x + 5) [3 ]
99×2+45x-143x-65
99×2-98x-65
(ii) (7a − 4b)2
(7a-4b)(7a-4b)
49a2-28ab-28ab+16b2
49a2-56ab+16b2
[2 ]
(b) Factorise the following expressions.
(i) x2 − 22x − 48
Factor x2−22x−48
x2−22x−48
=(x+2)(x−24)
(x+2)(x−24)

[2 ] (ii) 25h2 − 64k2
(5h-8k)(5h+8k) [2 ]
(c) In parts (i), (ii) and (iii) below, you should include each step in your reasoning and check that your solution is correct.
(i) Factorise then solve the equation
x2 + 11x − 42 = 0.
Let’s solve your equation step-by-step.
X2+11x−42=0
Step 1: Factor left side of equation.
(x−3)(x+14)=0
Step 2: Set factors equal to 0.
x−3=0 or x+14=0
x=3 or x=−14
[4 ] (ii) Explain how you could use your answer to part (c)(i) to solve the
equation 4×2 + 44x − 168 = 0.
What are the solutions?
Dividing through by 4
We get the above equation which we have been given
The solution will be
x2 + 11x − 42 = 0.
Let’s solve your equation step-by-step.
X2+11x−42=0
Step 1: Factor left side of equation.
(x−3)(x+14)=0
Step 2: Set factors equal to 0.
x−3=0 or x+14=0
x=3 or x=−14
which is the same as that which is given above
[2 ]
(iii) Solve the equation
11 7
1563370958850020967709588500x − 3 = x − 7 . [5 ]
11(x-7)=7(x-3)
11x-77=7x-21
11x-7x=77-21
4x=56
X=14
Question 7 – 5 marks
This question is based on your work on MU123 up to and including Unit 9.
A student writes down the following incomplete piece of work.
Add 19x to both sides
Divide both sides by x
x2 – 19x = 0 x2 = 19x
x = 19
(a) Substitute x = 19 into the equation x2 − 19x = 0, and explain why this
shows that x = 19 is indeed a solution of the equation.
192-(19*19)=0
192-(19*19)=0
361-361=0
[2 ] (b) Write out a complete solution of the equation x2 − 19x = 0
x(x-19)=0
x=0
x=19.
[2 ] (c) Explain, as if directly to the student, why their solution is incomplete. [1 ]
Factoring out x
We get x(x-19)=0
Question 8 – 5 marks
A score out of 5 marks for good mathematical communication over the entire
TMA will be recorded under Question 8. [5 ]

Work cited
Keyfitz, Nathan. “Introduction to the mathematics of population.” (1968).