A Level

A Level Paper 5: Mixed Applications

Covers sequences and series with the binomial expansion, exponentials and logarithms, vectors, sampling and data presentation, kinematics, the normal distribution and hypothesis testing, and moments.

14 questions - 60 marks - calculator allowed

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Questions

Question 1 [2 marks]

Exponentials and Logarithms

Without using a calculator, find the exact value of log_2(32).

Question 2 [2 marks]

Moments

A uniform beam of weight 60 N and length 4 m is suspended horizontally in equilibrium by two vertical strings, one at each end.

Find the tension in each string.

Question 3 [3 marks]

Sampling and Data Presentation

The table shows the age, in years, of 40 people attending a clinic: 0-20 (6 people), 20-40 (14 people), 40-60 (12 people), 60-80 (8 people).

Using the midpoint of each class, estimate the mean age.

Question 4 [3 marks]

Sequences, Series and the Binomial Expansion

The first three terms of a geometric sequence are 8, 12, 18.

Find the common ratio and the 6th term.

Question 5 [4 marks]

Vectors

Find the magnitude of the vector v = 3i - 4j, and find the vector of magnitude 15 in the same direction as v.

Question 6 [4 marks]

Sampling and Data Presentation

The box plot summarising the times, in minutes, taken by 200 runners to complete a race gives: minimum 28, lower quartile 35, median 41, upper quartile 50, maximum 68.

Find the interquartile range, and state, with a reason, whether a time of 66 minutes would be classed as an outlier using the rule 'more than 1.5 x IQR above the upper quartile'.

Question 7 [5 marks]

Kinematics

A train accelerates uniformly from rest at 0.5 m/s^2 for 20 seconds, then travels at constant speed for a further 90 seconds.

Find the total distance travelled by the train.

Question 8 [5 marks]

The Normal Distribution and Hypothesis Testing

A machine fills bags of flour with a mean mass of 1000 g. A sample of 60 bags is taken and found to have a mean mass of 1015 g, with population standard deviation known to be 40 g.

Test, at the 5% significance level, whether there is evidence that the mean mass has increased. State your hypotheses, test statistic and conclusion.

Question 9 [5 marks]

Vectors

Vectors a and b are such that |a| = 5, |b| = 3 and the angle between a and b is 60 degrees.

Find a.b, and find |a + b|.

Question 10 [5 marks]

Moments

A non-uniform plank AB has length 5 m and weight 120 N, and rests horizontally on two supports at A and B. When a block of weight 60 N is placed at a point 2 m from A, the reaction at A is 100 N.

Find the reaction at B in this situation, and find the distance of the centre of mass of the plank from A.

Question 11 [4 marks]

Sampling and Data Presentation

A sample of 12 pairs of data gives a product moment correlation coefficient of r = 0.62. The critical value for a sample of size 12 at the 5% significance level (one-tail) is 0.497.

Test, at the 5% significance level, whether this provides evidence of positive correlation between the variables in the population.

Question 12 [6 marks]

Moments

A non-uniform ladder AB of length 5 m and weight 240 N rests with end A on rough horizontal ground and end B against a smooth vertical wall. The centre of mass of the ladder is 2 m from A. The coefficient of friction between the ladder and the ground is 0.4.

Find the least angle the ladder can make with the ground without slipping.

Question 13 [6 marks]

Kinematics

A particle P moves in a straight line so that its displacement s metres from a fixed point O at time t seconds (t >= 0) is given by s = t^3 - 9t^2 + 24t.

Find the velocity and acceleration of P as functions of t, find the times at which P is instantaneously at rest, and find the total distance travelled by P in the first 5 seconds.

Question 14 [6 marks]

Moments

A uniform ladder AB of length 6 m and weight 180 N rests with end A on rough horizontal ground and end B against a smooth vertical wall, making an angle of 60 degrees with the ground. A person of weight 700 N stands on the ladder at a point 4 m from A.

Given that the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
recognising 32 = 2^5M1
log_2(32) = 5A1
Final answer: 5
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
using symmetry, since the beam is uniform and each string is the same distance from the centre of massM1
a tension of 30 N in each stringA1
Final answer: 30 N in each string
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
using the midpoints 10, 30, 50 and 70M1
computing the sum of (frequency x midpoint) = 1640M1
the mean = 1640/40 = 41 yearsA1
Final answer: Estimated mean = 41 years
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
common ratio r = 12/8 = 1.5M1
using the 6th term = a*r^5M1
6th term = 60.75A1
Final answer: r = 1.5, 6th term = 60.75
Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or workingMarks
|v| = sqrt(3^2 + 4^2)M1
|v| = 5A1
the scale factor 15/5 = 3M1
the vector 9i - 12jA1
Final answer: |v| = 5; the vector of magnitude 15 is 9i - 12j
Mark scheme for Question 6 [4 marks]
Question 6[4 marks]
Answer or workingMarks
IQR = upper quartile - lower quartileM1
IQR = 15A1
the outlier boundary = upper quartile + 1.5 x IQR = 72.5M1
the conclusion: since 66 < 72.5, 66 minutes is not an outlierA1
Final answer: IQR = 15; 66 minutes is not an outlier since the boundary is 72.5
Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or workingMarks
finding the velocity at the end of the acceleration phase v = 0 + 0.5 x 20 = 10 m/sM1
finding the distance in stage 1 using s = ut + 0.5at^2M1
distance 1 = 100 mA1
finding the distance in stage 2 = 10 x 90M1
the total distance = 1000 mA1
Final answer: Total distance = 1000 m
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
stating hypotheses H0: mu = 1000, H1: mu > 1000M1
computing the standard error 40/sqrt(60) = 5.16 (3 sf)M1
computing the test statistic z = 15/5.16 = 2.90 (3 sf)M1
comparing with the critical value z = 1.645 (5% one-tail)A1
the conclusion: reject H0, there is evidence the mean mass has increasedA1
Final answer: z = 2.90 (3 sf), exceeds the critical value 1.645, so reject H0: evidence the mean mass has increased
Mark scheme for Question 9 [5 marks]
Question 9[5 marks]
Answer or workingMarks
using a.b = |a||b|cos(theta)M1
a.b = 7.5A1
using |a + b|^2 = |a|^2 + 2(a.b) + |b|^2M1
|a + b|^2 = 25 + 15 + 9 = 49A1
|a + b| = 7A1
Final answer: a.b = 7.5, |a + b| = 7
Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or workingMarks
vertical equilibrium to find R_B = 180 - 100 = 80 NM1
taking moments about A: R_B x 5 = 120x + 60 x 2, using x for the distance of the centre of mass from AM1
forming the equation 400 = 120x + 120A1
solving for xM1
x = 2.33 m (3 sf)A1
Final answer: R_B = 80 N; centre of mass is 2.33 m from A (3 sf)
Mark scheme for Question 11 [4 marks]
Question 11[4 marks]
Answer or workingMarks
stating hypotheses H0: rho = 0, H1: rho > 0, where rho is the population correlation coefficientB1
comparing the sample value r = 0.62 with the critical value 0.497M1
noting 0.62 > 0.497A1
the conclusion: reject H0, there is evidence of positive correlation between the variables in the populationA1
Final answer: 0.62 > 0.497, so reject H0: evidence of positive correlation in the population
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
vertical equilibrium: R = 240 NM1
horizontal equilibrium: F = SM1
taking moments about A: S x (5 sin(theta)) = 240 x (2 cos(theta))M1
S = 96/tan(theta)A1
using F = mu R = 0.4 x 240 = 96 N at limiting equilibrium, and setting S = 96M1
tan(theta) = 1, so theta = 45 degreesA1
Final answer: theta = 45 degrees
Mark scheme for Question 13 [6 marks]
Question 13[6 marks]
Answer or workingMarks
differentiating to find v = 3t^2 - 18t + 24M1
differentiating to find a = 6t - 18A1
setting v = 0 and solving to get t^2 - 6t + 8 = 0M1
t = 2 and t = 4A1
evaluating s at t = 0, 2, 4, 5 (s = 0, 20, 16, 20) and using these to find the distance travelled in each intervalM1
total distance = 28 mA1
Final answer: v = 3t^2 - 18t + 24, a = 6t - 18; at rest at t = 2 s and t = 4 s; total distance in first 5 s = 28 m
Mark scheme for Question 14 [6 marks]
Question 14[6 marks]
Answer or workingMarks
vertical equilibrium: R = 180 + 700 = 880 NM1
horizontal equilibrium F = SM1
taking moments about A: S x (6 sin(60)) = 180 x (3 cos(60)) + 700 x (4 cos(60))M1
S = 321 N (3 sf)A1
using F = S at limiting equilibrium with mu = F/RM1
mu = 0.365 (3 sf)A1