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ZIMSEC A Level · 9164/2 · J2012

Pure Mathematics Paper 2 June 2012

Questions
51
Total marks
120
Syllabus code
9164/2

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Questions
51
Pass mark
31
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Section a

Section a, Question 1

[2 marks]Numerical methods, Newton-Raphson
Use the Newton-Raphson method to find, correct to 3 decimal places, the root of the equation sin⁡x+cos⁡x=ex−1\sin x+\cos x=e^{x}-1 that lies between −4-4 and −3-3.

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[2 marks]Numerical methods, Newton-Raphson
The equation sin⁡x+cos⁡x=ex−1\sin x+\cos x=e^{x}-1 is to be solved by the Newton-Raphson method, writing f(x)=sin⁡x+cos⁡x−ex+1f(x)=\sin x+\cos x-e^{x}+1. Which of these is the correct iterative formula?
  1. Axn+1=xn−sin⁡xn+cos⁡xn−exn+1cos⁡xn+sin⁡xn−exnx_{n+1}=x_n-\dfrac{\sin x_n+\cos x_n-e^{x_n}+1}{\cos x_n+\sin x_n-e^{x_n}}
  2. Bxn+1=xn+sin⁡xn+cos⁡xn−exn+1cos⁡xn−sin⁡xn−exnx_{n+1}=x_n+\dfrac{\sin x_n+\cos x_n-e^{x_n}+1}{\cos x_n-\sin x_n-e^{x_n}}
  3. Cxn+1=xn−sin⁡xn+cos⁡xn−exn+1cos⁡xn−sin⁡xn−exnx_{n+1}=x_n-\dfrac{\sin x_n+\cos x_n-e^{x_n}+1}{\cos x_n-\sin x_n-e^{x_n}}
  4. Dxn+1=xn−cos⁡xn−sin⁡xn−exnsin⁡xn+cos⁡xn−exn+1x_{n+1}=x_n-\dfrac{\cos x_n-\sin x_n-e^{x_n}}{\sin x_n+\cos x_n-e^{x_n}+1}
[2 marks]Numerical methods, Newton-Raphson
For f(x)=sin⁡x+cos⁡x−ex+1f(x)=\sin x+\cos x-e^{x}+1 it is found that f(−4)=1,085f(-4)=1,085 and f(−3)=−0,181f(-3)=-0,181. Why does this show that the equation sin⁡x+cos⁡x=ex−1\sin x+\cos x=e^{x}-1 has a root between −4-4 and −3-3?
  1. ABecause ff is continuous on [−4,−3][-4,-3] and changes sign across it, so it takes the value zero inside.
  2. BBecause f(−4)f(-4) is larger than f(−3)f(-3), and a decreasing function is guaranteed to have a root somewhere.
  3. CBecause the two values differ by more than 1, which is what guarantees that a root lies between them.
  4. DBecause −4-4 and −3-3 are consecutive integers, and every interval of length 1 contains at least one root.

Section a, Question 2

[2 marks]Mathematical induction and series
Given that ∑r=1n(r+3)(2r+1)=n6(4n2+27n+41)\displaystyle\sum_{r=1}^{n}(r+3)(2r+1)=\frac{n}{6}\left(4n^{2}+27n+41\right), evaluate ∑r=110(r+3)(2r+1)\displaystyle\sum_{r=1}^{10}(r+3)(2r+1).

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[2 marks]Mathematical induction and series
Expand and simplify the product (r+3)(2r+1)(r+3)(2r+1).

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[2 marks]Mathematical induction and series
In proving by induction that ∑r=1n(r+3)(2r+1)=n6(4n2+27n+41)\displaystyle\sum_{r=1}^{n}(r+3)(2r+1)=\frac{n}{6}\left(4n^{2}+27n+41\right), what must be checked in the base case n=1n=1?
  1. AThat the formula is a quadratic in nn multiplied by nn, which it plainly is for every value of nn.
  2. BThat n6(4n2+27n+41)\dfrac{n}{6}\left(4n^{2}+27n+41\right) is a whole number when n=1n=1, since a sum must be a whole number.
  3. CThat 16(4+27+41)\dfrac{1}{6}(4+27+41) equals 1, which is the number of terms in the sum at that stage.
  4. DThat both sides equal 12, since the sum is (1+3)(2+1)(1+3)(2+1) and the formula gives 16(4+27+41)\dfrac{1}{6}(4+27+41).

Section a, Question 3

[2 marks]Differentiation, stationary values
An isosceles triangle has a constant perimeter of 2k2k and its two equal sides are each of length xx cm. Find, in terms of xx and kk, the length of the base of the triangle.

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[2 marks]Differentiation, stationary values
An isosceles triangle has a constant perimeter of 2k2k, its two equal sides are each of length xx cm and its base is therefore 2k−2x2k-2x. Find, in terms of xx and kk, the perpendicular height of the triangle.

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[2 marks]Differentiation, stationary values
An isosceles triangle has a constant perimeter of 2k2k, equal sides of length xx cm each, base 2k−2x2k-2x and perpendicular height 2kx−k2\sqrt{2kx-k^{2}}. Which expression gives its area AA in terms of xx and kk?
  1. AA=(2k−2x)2kx−k2A=(2k-2x)\sqrt{2kx-k^{2}}
  2. BA=12x2kx−k2A=\dfrac{1}{2}x\sqrt{2kx-k^{2}}
  3. CA=(k−x)(2kx−k2)A=(k-x)\left(2kx-k^{2}\right)
  4. DA=(k−x)2kx−k2A=(k-x)\sqrt{2kx-k^{2}}
[2 marks]Differentiation, stationary values
The area of an isosceles triangle of constant perimeter 2k2k, whose equal sides are each xx cm, is A=(k−x)2kx−k2A=(k-x)\sqrt{2kx-k^{2}}. Find, in terms of kk, the value of xx at which the area is stationary.

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Section a, Question 4

[2 marks]Differential equations, exponential decay
A car is bought for $4 000\$4\,000. A model assumes that the value $V\$V of the car, tt months after purchase, decreases at a rate which is proportional to VV. Which differential equation states this model?
  1. AdVdt=−kV\dfrac{dV}{dt}=-kV, with k>0k>0
  2. BdVdt=kV\dfrac{dV}{dt}=kV, with k>0k>0
  3. CdVdt=−kt\dfrac{dV}{dt}=-kt, with k>0k>0
  4. DdVdt=−kV\dfrac{dV}{dt}=-\dfrac{k}{V}, with k>0k>0
[3 marks]Differential equations, exponential decay
A car bought for $4 000\$4\,000 has value $V\$V after tt months, where V=4 000e−ktV=4\,000e^{-kt}. After 3 years the car is valued at $2 000\$2\,000. Find the value of kk, correct to 4 decimal places.

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[3 marks]Differential equations, exponential decay
A car bought for $4 000\$4\,000 has value $V\$V after tt months, where V=4 000e−ktV=4\,000e^{-kt} and k=ln⁡236k=\dfrac{\ln 2}{36}. Calculate the value of the car when it is 15 months old, giving your answer to the nearest dollar.

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[3 marks]Differential equations, exponential decay
A car bought for $4 000\$4\,000 has value $V\$V after tt months, where V=4 000e−ktV=4\,000e^{-kt} and k=ln⁡236k=\dfrac{\ln 2}{36}. Calculate the age of the car when its value is $1 600\$1\,600, giving your answer to the nearest month.

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Section a, Question 5

[3 marks]Vectors, lines and planes
The lines L1L_1 and L2L_2 have vector equations r=(2+λ)i+(−3+2λ)j+(1+2λ)k\mathbf{r}=(2+\lambda)\mathbf{i}+(-3+2\lambda)\mathbf{j}+(1+2\lambda)\mathbf{k} and r=(8+3μ)i+(5+2μ)j+(13+6μ)k\mathbf{r}=(8+3\mu)\mathbf{i}+(5+2\mu)\mathbf{j}+(13+6\mu)\mathbf{k} respectively. Find the coordinates of their point of intersection.

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[2 marks]Vectors, lines and planes
The line L1L_1 has vector equation r=(2+λ)i+(−3+2λ)j+(1+2λ)k\mathbf{r}=(2+\lambda)\mathbf{i}+(-3+2\lambda)\mathbf{j}+(1+2\lambda)\mathbf{k} and the plane π\pi is 2x−z=32x-z=3. Which statement verifies that L1L_1 lies wholly in π\pi?
  1. AThe direction of L1L_1 is perpendicular to the normal of π\pi, which on its own settles the matter entirely.
  2. BThe point (2,−3,1)(2,-3,1) satisfies the equation 2x−z=32x-z=3, which on its own settles the matter entirely.
  3. C2(2+λ)−(1+2λ)=32(2+\lambda)-(1+2\lambda)=3 holds when λ=0\lambda=0, so the whole of the line L1L_1 must lie inside π\pi.
  4. D2(2+λ)−(1+2λ)=32(2+\lambda)-(1+2\lambda)=3 for every λ\lambda, so every point of L1L_1 satisfies the plane's equation.
[3 marks]Vectors, lines and planes
The plane π\pi is 2x−z=32x-z=3 and A is the point with position vector 12i+5j+6k12\mathbf{i}+5\mathbf{j}+6\mathbf{k}. B is the foot of the perpendicular from A to π\pi. Which is a vector equation of the line AB?
  1. Ar=12i+5j+6k+t(2i+j−k)\mathbf{r}=12\mathbf{i}+5\mathbf{j}+6\mathbf{k}+t\left(2\mathbf{i}+\mathbf{j}-\mathbf{k}\right)
  2. Br=12i+5j+6k+t(2i−k)\mathbf{r}=12\mathbf{i}+5\mathbf{j}+6\mathbf{k}+t\left(2\mathbf{i}-\mathbf{k}\right)
  3. Cr=2i−k+t(12i+5j+6k)\mathbf{r}=2\mathbf{i}-\mathbf{k}+t\left(12\mathbf{i}+5\mathbf{j}+6\mathbf{k}\right)
  4. Dr=12i+5j+6k+t(3i)\mathbf{r}=12\mathbf{i}+5\mathbf{j}+6\mathbf{k}+t\left(3\mathbf{i}\right)
[3 marks]Vectors, lines and planes
The plane π\pi is 2x−z=32x-z=3 and A is the point with position vector 12i+5j+6k12\mathbf{i}+5\mathbf{j}+6\mathbf{k}. B is the foot of the perpendicular from A to π\pi, so that the line AB is r=12i+5j+6k+t(2i−k)\mathbf{r}=12\mathbf{i}+5\mathbf{j}+6\mathbf{k}+t(2\mathbf{i}-\mathbf{k}). Find the position vector of B.

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Section a, Question 6

[2 marks]Complex numbers and polynomial roots
The complex number a=p+iqa=p+iq, with pp and qq real and conjugate aˉ\bar{a}, satisfies 4aaˉ+12i=8a+164a\bar{a}+12i=8a+16. Find the value of qq.

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[3 marks]Complex numbers and polynomial roots
The complex number a=p+iqa=p+iq, with pp and qq real and conjugate aˉ\bar{a}, satisfies 4aaˉ+12i=8a+164a\bar{a}+12i=8a+16, and comparing imaginary parts gives q=32q=\dfrac{3}{2}. Which equation must pp satisfy?
  1. A4p2−8p−7=04p^{2}-8p-7=0
  2. B4p2+8p+7=04p^{2}+8p+7=0
  3. C4p2−8p+25=04p^{2}-8p+25=0
  4. D4p2−8p−16=04p^{2}-8p-16=0
[2 marks]Complex numbers and polynomial roots
The polynomial p(x)=2x4+x3+17x2+9x−9p(x)=2x^{4}+x^{3}+17x^{2}+9x-9. Evaluate p(3i)p(3i).

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[1 marks]Complex numbers and polynomial roots
The polynomial p(x)=2x4+x3+17x2+9x−9p(x)=2x^{4}+x^{3}+17x^{2}+9x-9 has real coefficients and 3i3i is a root of p(x)=0p(x)=0. State the other complex root.

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[2 marks]Complex numbers and polynomial roots
The polynomial p(x)=2x4+x3+17x2+9x−9p(x)=2x^{4}+x^{3}+17x^{2}+9x-9 has 3i3i and −3i-3i as roots, so x2+9x^{2}+9 is a factor. What is the other quadratic factor?
  1. Ax2+2x−9x^{2}+2x-9
  2. B2x2−x−12x^{2}-x-1
  3. C2x2+x−12x^{2}+x-1
  4. D2x2+x+12x^{2}+x+1
[3 marks]Complex numbers and polynomial roots
The polynomial p(x)=2x4+x3+17x2+9x−9p(x)=2x^{4}+x^{3}+17x^{2}+9x-9 factorises as (x2+9)(2x2+x−1)(x^{2}+9)(2x^{2}+x-1). Find the two real roots of p(x)=0p(x)=0.

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Section a, Question 7

[3 marks]Matrices, transformations and simultaneous equations
The transformation with matrix (−2132)\begin{pmatrix}-2&1\\3&2\end{pmatrix} maps a line onto the line y′=6x′+8y'=6x'+8. Find the equation of the original line, giving your answer in the form ax+by=cax+by=c where aa, bb and cc are integers.

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[3 marks]Matrices, transformations and simultaneous equations
Given M=(321−220100)M=\begin{pmatrix}3&2&1\\-2&2&0\\1&0&0\end{pmatrix} and N=(01314−2102)N=\begin{pmatrix}0&1&3\\1&4&-2\\1&0&2\end{pmatrix}, find MNMN.
  1. A(023−280100)\begin{pmatrix}0&2&3\\-2&8&0\\1&0&0\end{pmatrix}
  2. B(32011617−103)\begin{pmatrix}3&2&0\\11&6&1\\7&-10&3\end{pmatrix}
  3. C(311726−10013)\begin{pmatrix}3&11&7\\2&6&-10\\0&1&3\end{pmatrix}
  4. D(3117−2610013)\begin{pmatrix}3&11&7\\-2&6&10\\0&1&3\end{pmatrix}
[2 marks]Matrices, transformations and simultaneous equations
Given M=(321−220100)M=\begin{pmatrix}3&2&1\\-2&2&0\\1&0&0\end{pmatrix} and N=(01314−2102)N=\begin{pmatrix}0&1&3\\1&4&-2\\1&0&2\end{pmatrix}, find the determinant of MNMN.

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[3 marks]Matrices, transformations and simultaneous equations
Given that MN=(311726−10013)MN=\begin{pmatrix}3&11&7\\2&6&-10\\0&1&3\end{pmatrix} and det⁡(MN)=32\det(MN)=32, find (MN)−1(MN)^{-1}.
  1. A32(28−26−152−69442−3−4)32\begin{pmatrix}28&-26&-152\\-6&9&44\\2&-3&-4\end{pmatrix}
  2. B132(28−26−152−69442−3−4)\dfrac{1}{32}\begin{pmatrix}28&-26&-152\\-6&9&44\\2&-3&-4\end{pmatrix}
  3. C132(28−62−269−3−15244−4)\dfrac{1}{32}\begin{pmatrix}28&-6&2\\-26&9&-3\\-152&44&-4\end{pmatrix}
  4. D132(311726−10013)\dfrac{1}{32}\begin{pmatrix}3&11&7\\2&6&-10\\0&1&3\end{pmatrix}
[3 marks]Matrices, transformations and simultaneous equations
Given that (MN)−1=132(28−26−152−69442−3−4)(MN)^{-1}=\dfrac{1}{32}\begin{pmatrix}28&-26&-152\\-6&9&44\\2&-3&-4\end{pmatrix}, find the values of xx, yy and zz for which MN(xyz)=(3−63)MN\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\-6\\3\end{pmatrix}.

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Section b

Section b, Question 8

[2 marks]Kinematics, velocity-time graphs
A car passes a fixed point A at 10 ms−110\ \mathrm{ms^{-1}} and holds that speed for t1t_1 seconds. It then accelerates uniformly over the next t2t_2 seconds to 15 ms−115\ \mathrm{ms^{-1}}, and then decelerates uniformly to rest in a further t3t_3 seconds. Given that the acceleration and the deceleration are equal in magnitude, express t3t_3 in terms of t2t_2.

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[2 marks]Kinematics, velocity-time graphs
A car passes a fixed point A at 10 ms−110\ \mathrm{ms^{-1}} and holds that speed for t1t_1 seconds, then accelerates uniformly over t2t_2 seconds to 15 ms−115\ \mathrm{ms^{-1}}, then decelerates uniformly to rest in a further t3t_3 seconds. What does its velocity-time graph look like?
  1. AA horizontal segment at v=10v=10, then a segment rising to v=15v=15, then a segment falling to v=0v=0.
  2. BA single straight line rising from v=10v=10 to v=15v=15 and then falling to zero.
  3. CA curve rising from v=10v=10 to v=15v=15, then a horizontal segment, then a curve to zero.
  4. DA horizontal segment at v=10v=10, then a segment falling to zero, then a segment rising to v=15v=15.

Section b, Question 9

[2 marks]Projectile motion
A ball is projected from the top of a building 40 m high with an initial speed of 20 ms−120\ \mathrm{ms^{-1}} at an angle of 30∘30^\circ to the horizontal, and lands on the ground at P. Taking g=9,81 ms−2g=9,81\ \mathrm{ms^{-2}}, find the time of flight of the ball, correct to 3 significant figures.

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[3 marks]Projectile motion
A ball is projected from the top of a building 40 m high with an initial speed of 20 ms−120\ \mathrm{ms^{-1}} at an angle of 30∘30^\circ to the horizontal, and lands on the ground at P after 4,05 s. Taking g=9,81 ms−2g=9,81\ \mathrm{ms^{-2}}, find the angle, to the nearest degree, that the ball's direction of motion at P makes with the horizontal.

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Section b, Question 10

[3 marks]Newton's laws, connected particles
Two particles of mass m1m_1 kg and m2m_2 kg, where m1>m2m_1>m_2, are connected by a light inelastic string passing over a smooth fixed pulley. Find, in terms of gg, m1m_1 and m2m_2, the acceleration of the system.

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[2 marks]Newton's laws, connected particles
Two particles of mass m1m_1 kg and m2m_2 kg, where m1>m2m_1>m_2, hang either side of a smooth fixed pulley on a light inelastic string. Why is the tension taken to be the same throughout the string?
  1. ABecause the two masses are equal, so the pull on one side balances the pull on the other side exactly.
  2. BBecause the string is inelastic, and an inelastic string cannot transmit any force along its own length.
  3. CBecause the string is light and the pulley smooth, so neither weight nor friction can change the tension.
  4. DBecause the acceleration has the same magnitude on both sides, which forces the two tensions to be equal.
[2 marks]Newton's laws, connected particles
Two particles of mass m1m_1 kg and m2m_2 kg, where m1>m2m_1>m_2, are connected by a light inelastic string over a smooth fixed pulley, and the system accelerates at a=(m1−m2)gm1+m2a=\dfrac{(m_1-m_2)g}{m_1+m_2}. What is the tension in the string?
  1. A(m1−m2)gm1+m2\dfrac{(m_1-m_2)g}{m_1+m_2}
  2. Bm1g(2m2m1+m2)m_1g\left(\dfrac{2m_2}{m_1+m_2}\right)
  3. C(m1+m2)g2\dfrac{(m_1+m_2)g}{2}
  4. Dm2g(2m1m1−m2)m_2g\left(\dfrac{2m_1}{m_1-m_2}\right)

Section b, Question 11

[2 marks]Friction and equilibrium of forces
A particle of weight 20 N rests on a rough horizontal surface. A force PP N at an angle θ\theta to the horizontal is applied until the particle is on the point of moving. At that instant the normal force is R=16jR=16\mathbf{j} and the frictional force is F=−9iF=-9\mathbf{i}. Find the coefficient of friction.

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[2 marks]Friction and equilibrium of forces
A particle rests on a rough horizontal surface with normal force R=16jR=16\mathbf{j} and frictional force F=−9iF=-9\mathbf{i} acting on it. Find the magnitude of the contact force between the particle and the surface, correct to 3 significant figures.

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[2 marks]Friction and equilibrium of forces
A particle of weight 20 N rests on a rough horizontal surface. A force PP N at an angle θ\theta above the horizontal is applied until the particle is on the point of moving, at which instant the normal force is 16 N and the frictional force is 9 N. Find θ\theta, to the nearest degree.

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[2 marks]Friction and equilibrium of forces
A particle of weight 20 N rests on a rough horizontal surface. A force PP N at an angle θ\theta above the horizontal is applied until the particle is on the point of moving, at which instant the normal force is 16 N and the frictional force is 9 N. Find PP, correct to 3 significant figures.

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Section c

Section c, Question 12

[2 marks]Geometric distribution
The random variable WW has a geometric distribution, W∼Geo(p)W\sim\mathrm{Geo}(p), with Var(W)=30\mathrm{Var}(W)=30. Find the value of pp.

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[2 marks]Geometric distribution
The random variable WW has a geometric distribution, W∼Geo(p)W\sim\mathrm{Geo}(p), with Var(W)=30\mathrm{Var}(W)=30. Find E(W)\mathrm{E}(W).

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Section c, Question 13

[3 marks]Conditional probability, tree diagrams
A court passes only two verdicts, convicted or discharged, and 80% of its verdicts are convictions. When the verdict is convicted the probability that the accused is innocent is 0,07, and when the verdict is discharged the probability that the accused is innocent is 0,4. Find the probability that a person tried by this court is innocent.

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[3 marks]Conditional probability, tree diagrams
A court passes only two verdicts, convicted or discharged, and 80% of its verdicts are convictions. When the verdict is convicted the probability that the accused is innocent is 0,07, and when the verdict is discharged the probability that the accused is innocent is 0,4, so that the probability a person tried is innocent is 0,136. Find the conditional probability that an innocent person tried by this court is convicted, correct to 3 significant figures.

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Section c, Question 14

[2 marks]Continuous random variables
The continuous random variable XX has probability density function f(x)=2−2x\mathrm{f}(x)=2-2x for 0≤x≤10\le x\le1 and f(x)=0\mathrm{f}(x)=0 otherwise. What is the cumulative distribution function F(x)\mathrm{F}(x) on 0≤x≤10\le x\le1?
  1. AF(x)=2x−2x2\mathrm{F}(x)=2x-2x^{2}
  2. BF(x)=2−2x\mathrm{F}(x)=2-2x
  3. CF(x)=x2−2x\mathrm{F}(x)=x^{2}-2x
  4. DF(x)=2x−x2\mathrm{F}(x)=2x-x^{2}
[2 marks]Continuous random variables
The continuous random variable XX has probability density function f(x)=2−2x\mathrm{f}(x)=2-2x for 0≤x≤10\le x\le1 and f(x)=0\mathrm{f}(x)=0 otherwise. Find P(X>13)P\left(X>\dfrac{1}{3}\right), as an exact fraction.

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[2 marks]Continuous random variables
The continuous random variable XX has probability density function f(x)=2−2x\mathrm{f}(x)=2-2x for 0≤x≤10\le x\le1 and f(x)=0\mathrm{f}(x)=0 otherwise, so its cumulative distribution function is F(x)=2x−x2\mathrm{F}(x)=2x-x^{2} on that interval. Find the value of pp for which P(X<p)=15P(X<p)=\dfrac{1}{5}, correct to 3 decimal places.

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Section c, Question 15

[2 marks]Data representation, stem and leaf and box plots
Twenty four students scored the following marks in Biology: 36, 45, 40, 60, 71, 66, 53, 42, 35, 54, 35, 43, 72, 37, 39, 34, 49, 43, 75, 58, 67, 59, 36, 67. Find the median mark.

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[2 marks]Data representation, stem and leaf and box plots
Twenty four students scored the following marks in Biology: 36, 45, 40, 60, 71, 66, 53, 42, 35, 54, 35, 43, 72, 37, 39, 34, 49, 43, 75, 58, 67, 59, 36, 67. Find the interquartile range.

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[2 marks]Data representation, stem and leaf and box plots
Twenty four students scored the following marks in Integrated Science: 88, 89, 30, 34, 48, 49, 59, 65, 67, 78, 41, 70, 54, 66, 39, 49, 37, 59, 45, 63, 52, 75, 38, 38. Find the median mark.

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[2 marks]Data representation, stem and leaf and box plots
Twenty four students sat Biology and Integrated Science. For Biology the minimum is 34, the quartiles are 38, 47 and 63, and the maximum is 75. For Integrated Science the minimum is 30, the quartiles are 40, 53 and 66,5, and the maximum is 89. Which comment on the two sets of marks is correct?
  1. AThe two subjects have the same median, but Biology marks are more spread out.
  2. BIntegrated Science marks are higher on average and more spread out than Biology marks.
  3. CIntegrated Science marks are higher on average but less spread out than Biology marks.
  4. DBiology marks are higher on average and more spread out than Integrated Science marks.

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