L.C. MATHS

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## Question 1

A bank issues a unique six-digit password to each of its online customers. The password may contain any of the numbers $$0$$ to $$9$$ in any position and numbers may be repeated. For example, the following is a valid password.

\begin{align}\fbox{0}&&\fbox{7}&&\fbox{1}&&\fbox{7}&&\fbox{3}&&\fbox{7}\end{align}

(a) How many different passwords are possible?

$$1{,}000{,}000$$

Solution

\begin{align}10\times10\times10\times10\times10\times10=1{,}000{,}000\end{align}

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(b)

(i) How many different passwords do not contain any zero?

(ii) One password is selected at random from all the possible passwords. What is the
probability that this password contains at least one zero?

(i) $$531{,}441$$

(ii) $$0.0468559$$

Solution

(i)

\begin{align}9\times9\times9\times9\times9\times9=531{,}441\end{align}

(ii)

\begin{align}P&=\frac{1{,}000{,}000-531{,}441}{1{,}000{,}000}\\&=0.0468559\end{align}

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(c) John is issued with one such password from the bank. Each time John wants to access his account online, the bank’s website requires him to input three of his password digits into the boxes provided. For example, he may be asked for the 2nd, 4th and 5th digits.

In how many different ways can the bank select the three required boxes?

$$20$$

Solution

\begin{align}{6\choose3}=20\end{align}

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## Question 2

The line $$p$$ makes equal intercepts on the axes at $$A$$ and at $$B$$, as shown.

The line $$p$$ makes equal intercepts on the axes at $$A$$ and at $$B$$, as shown.

(a)

(i) Write down the slope of $$p$$.

(ii) The point $$(1,5)$$ is on $$p$$. Find the equation of $$p$$.
Write your answer in the form $$ax+by+c=0$$, where $$a$$, $$b$$ and $$c\in\mathbb{Z}$$.

(i) $$m=1$$

(ii) $$x-y+4=0$$

Solution

(i) $$m=1$$

(ii)

\begin{align}y-y_1=m(x-x_1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-5=1(x-1)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y-5=x-1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x-y+4=0\end{align}

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(b) The line $$q$$ is perpendicular to $$p$$ and contains the point $$O(0,0)$$. Find the equation of $$q$$.

$$x+y=0$$

Solution

\begin{align}m_q=-1&&(x_1,y_1)=(0,0)\end{align}

\begin{align}y-0=-1(x-0)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x+y=0\end{align}

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(c) The lines $$p$$ and $$q$$ intersect at the point $$C$$. Explain why the triangles $$OCA$$ and $$OBC$$ are congruent.

The $$RHS$$ rule.

Solution

$$|OA|=|OB|$$ (equal intercepts)

$$|\angle ACO|=|\angle OCB|$$ (right angles)

and they share $$|OC|$$. Therefore, they are congruent according to the $$RHS$$ rule.

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## Question 3

(a) Draw the circle $$c:x^2+y^2=25$$. Show your scale on both axes.

Solution
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(b) Verify, using algebra, that $$A(-4,3)$$ is on $$c$$.

Solution

\begin{align}x^2+y^2&=(-4)^2+3^2\\&=16+9\\&=25\end{align}

as required.

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(c) Find the equation of the circle with centre $$(-4,3)$$ that passes through the point $$(3,4)$$.

$$(x+4)^2+(y-3)^2=50$$

Solution

\begin{align}r&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\&=\sqrt{(4-3)^2+(3-(-4))^2}\\&=\sqrt{1+49}\\&=\sqrt{50}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(x+4)^2+(y-3)^2=50\end{align}

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## Question 4

(a) The diagram shows a parallelogram, with one side produced.
Use the data on the diagram to find the value of $$x$$, of $$y$$ and of $$z$$.

$$x=30^{\circ}$$ (alternate angles)

$$y=40^{\circ}$$ (angles in a triangle add to $$180^{\circ}$$)

$$z=70^{\circ}$$ (sum of $$x$$ and $$y$$)

Solution

$$x=30^{\circ}$$ (alternate angles)

$$y=40^{\circ}$$ (angles in a triangle add to $$180^{\circ}$$)

$$z=70^{\circ}$$ (sum of $$x$$ and $$y$$)

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(b) The area of the parallelogram $$ABCD$$ is $$480 \mbox{ m}^2$$.

(i) Find the area of the triangle $$ABD$$.

(ii) $$E$$ is the midpoint of $$[CD]$$. Find the area of the triangle $$BCE$$.

(i) $$240\mbox{ m}^2$$

(ii) $$120\mbox{ m}^2$$

Solution

(i)

\begin{align}|\Delta ABD|&=\frac{1}{2}(480)\\&=240\mbox{ m}^2\end{align}

(ii)

\begin{align}|\Delta BCE|&=\frac{1}{2}(240)\\&=120\mbox{ m}^2\end{align}

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## Question 5

The diagram shows the triangles $$BCD$$ and $$ABD$$, with some measurements given.

(a)

(i) Find $$|BC|$$, correct to two
decimal places.

(ii) Find the area of the triangle $$BCD$$, correct to two decimal places.

(i) $$11.39\mbox{ m}$$

(ii) $$42.78\mbox{ m}^2$$

Solution

(i)

\begin{align}\frac{|BC|}{\sin42^{\circ}}=\frac{16}{\sin110^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|&=\frac{16\sin42^{\circ}}{\sin110^{\circ}}\\&\approx11.39\mbox{ m}\end{align}

(ii)

\begin{align}A&=ab\sin C\\&=(11.39)(16)\sin(180^{\circ}-110^{\circ}-42^{\circ})\\&\approx42.78\mbox{ m}^2\end{align}

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(b) Find $$|AB|$$, correct to two decimal places.

$$16.53\mbox{ m}$$

Solution

\begin{align}|AB|&=\sqrt{16^2+10^2-2(10)(16)\cos(180^{\circ}-63^{\circ}-42^{\circ})}\\&\approx16.53\mbox{ m}\end{align}

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## Question 6

(a) Mark ten or more points on each of the scatter graphs below to show an example of the type of correlation named under each graph.

(i) Strong positive correlation

(ii) Strong negative correlation

(ii) No correlation

(i)

(ii)

(iii)

Solution

(i)

(ii)

(iii)

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(b) A few days before the Scottish Independence Referendum in September $$2014$$ a YouGov poll estimated the support for the ‘No’ campaign to be $$54\%$$.

(i) If YouGov sampled $$1000$$ people, find the margin of error.

(ii) Create a $$95\%$$ confidence interval for the level of support for the ‘No’ campaign in the population.

(i) $$3.2\%$$

(ii) $$50.8\leq\hat{p}\leq57.2$$

Solution

(i)

\begin{align}\mbox{Margin of error}&=\frac{1}{\sqrt{n}}\\&=\frac{1}{\sqrt{1{,}000}}\\&=0.03162…\\&\approx3.2\%\end{align}

(ii)

\begin{align}54-3.2\leq\hat{p}\leq54+3.2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50.8\leq\hat{p}\leq57.2\end{align}

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## Question 7

The diagram below shows the right-angled triangle $$ABC$$, which is used in the logo for a company called Deane Construction Limited ($$DCL$$). The triangle $$PQR$$ is the image of $$ABC$$ under an enlargement.

(a)

(i) Construct the centre of enlargement and label it $$O$$.

(ii) Measure, in centimetres, $$|OB|$$ and $$|OQ|$$.

(iii) Use your measurements to find the scale factor of the enlargement, correct to one
decimal place.

(i)

(ii) $$|OB|=15\mbox{ cm}$$ and $$|OQ|=22.5\mbox{ cm}$$

(iii) $$1.5$$

Solution

(i)

(ii)

\begin{align}|OB|=15\mbox{ cm}&&|OQ|=22.5\mbox{ cm}\end{align}

(iii)

\begin{align}k&=\frac{|OQ|}{|OB|}\\&=\frac{22.5}{15}\\&=1.5\end{align}

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(b) The area of the triangle $$ABC$$ is $$7.5\mbox{ cm}^2$$.
Use the scale factor to find the area of the image triangle $$PQR$$ under the enlargement.

$$16.875\mbox{ cm}^2$$

Solution

\begin{align}A&=7.5\times(1.5^2)\\&=16.875\mbox{ cm}^2\end{align}

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(c)

(i) Given that $$|AB|=5\mbox{ cm}$$, use the scale factor to find $$|PQ|$$.

(ii) Given that $$|QR|=8.7\mbox{ cm}$$, use the scale factor to find $$|BC|$$.

(iii) Hence, show that $$|\angle ABC|=|\angle PQR|$$.

(i) $$7.5\mbox{ cm}$$

(ii) $$5.8\mbox{ cm}$$

Solution

(i)

\begin{align}|PQ|&=5\times1.5\\&=7.5\mbox{ cm}\end{align}

(ii)

\begin{align}|BC|&=\frac{8.7}{1.5}\\&=5.8\mbox{ cm}\end{align}

(iii)

\begin{align}\cos|\angle ABC|&=\frac{5}{5.8}\\&=0.862…\end{align}

and

\begin{align}\cos|\angle PQR|&=\frac{7.5}{8.7}\\&=0.862…\end{align}

Therefore, both angles must be the same (as they are both also less than $$90^{\circ}$$).

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## Question 8

(a) A company has a spherical storage tank. The diameter of the tank is $$12\mbox{ m}$$.

(i) Write down the radius of the tank.

(ii) Find the volume of the tank, correct to the nearest $$\mbox{m}^3$$.

(i) $$6\mbox{ m}$$

(ii) $$905\mbox{ m}^3$$

Solution

(i)

\begin{align}r&=\frac{d}{2}\\&=6\mbox{ m}\end{align}

(ii)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(6^3)\\&\approx905\mbox{ m}^3\end{align}

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(b) The company paints the outside curved surface of the spherical tank.

(i) Find the curved surface area of the tank, correct to one decimal place.

(ii) The curved surface is painted with a special paint. One litre of paint will cover $$3.5\mbox{ m}^2$$.
Find how many litres of paint are used, correct to the nearest litre.

(iii) The paint is sold in $$25$$ litre tins. Each tin costs €$$180$$. Find the total cost of the paint.

(i) $$452.4\mbox{ m}^2$$

(ii) $$129\mbox{ litres}$$

(iii) $$1{,}080\mbox{ euro}$$

Solution

(i)

\begin{align}CSA&=4\pi r^2\\&=4\pi(6^2)\\&\approx452.4\mbox{ m}^2\end{align}

(ii)

\begin{align}\frac{452.4}{3.5}\approx129\mbox{ litres}\end{align}

(iii)

\begin{align}\frac{129}{25}=5.16\end{align}

\begin{align}\downarrow\end{align}

$$6$$ cans of paint

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Cost}&=6\times180\\&=1{,}080\mbox{ euro}\end{align}

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(c) At another site the company has a differently shaped tank with the same volume.
This tank has hemispherical ends and a cylindrical mid-section of length $$h\mbox{ m}$$, as shown.
The radius of each hemispherical end is $$4.5\mbox{ m}$$.

(i) Find the volume of one hemispherical end, correct to the nearest $$\mbox{m}^3$$.

(ii) Find the length, $$h$$, of the cylindrical section, correct to one decimal place.

(i) $$191\mbox{ m}^3$$

(ii) $$8.2\mbox{ m}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi(4.5^3)\\&\approx191\mbox{ m}^3\end{align}

(ii)

\begin{align}2(191)+\pi(4.5)^2h=905\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\frac{905-2(191)}{\pi(4.5)^2}\\&\approx8.2\mbox{ m}\end{align}

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## Question 9

The heights of a random sample of $$1000$$ students were collected and recorded.

(a) Which of the following indicates how you would categorise the type of data collected? Explain your choice.

• Categorical Nominal
• Categorical Ordinal
• Numerical Discrete
• Numerical Continuous

Numerical continuous as height has a non-discrete numerical value.

Solution

Numerical continuous as height has a non-discrete numerical value.

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(b) The sample of $$1000$$ students was made up of $$500$$ boys and $$500$$ girls. The data from the $$500$$ girls was used to create the information shown in Table $$1$$.

Height (cm) $$145$$ - $$150$$ $$150$$ - $$155$$ $$155$$ - $$160$$ $$160$$ - $$165$$ $$165$$ - $$170$$ $$170$$ - $$175$$ $$175$$ - $$180$$ $$180$$ - $$185$$

Number of girls

$$15$$

$$48$$

$$80$$

$$112$$

$$125$$

$$81$$

$$29$$

$$10$$

(i) Use the information in Table 1 to estimate the mean height of the girls, using mid-interval values.

(ii) What is the largest possible value for the range of the heights of the girls in this sample?

(iii) The median height of the girls in the sample is $$164.5\mbox{ cm}$$. Explain what this means in the context of the heights of the $$500$$ girls.

(i) $$164.43\mbox{ cm}$$

(ii) $$40\mbox{ cm}$$

(iii) The height of $$250$$ of the girls is $$164.5\mbox{ cm}$$ or more.

Solution

(i)

\begin{align}\frac{147.5\times15+152.5\times48+157.5\times80+162.5\times112+167.5\times125+172.5\times81+177.5\times29+182.5\times10}{15+48+80+112+125+81+29+10}=164.43\mbox{ cm}\end{align}

(ii)

\begin{align}185-145=40\mbox{ cm}\end{align}

(iii) The height of $$250$$ of the girls is $$164.5\mbox{ cm}$$ or more.

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(c)

(i) Use the data in Table $$1$$ to complete Table $$2$$ by finding the percentage of girls in each of the height categories.

Height (cm) $$145$$ - $$150$$ $$150$$ - $$155$$ $$155$$ - $$160$$ $$160$$ - $$165$$ $$165$$ - $$170$$ $$170$$ - $$175$$ $$175$$ - $$180$$ $$180$$ - $$185$$

Percentage of girls

$$22.4$$

$$25$$

(ii) Use the data in Table $$2$$ to draw a histogram showing the percentage of girls in each height category.

(iii) A histogram showing the percentage of boys in each height category is given above.
John examines both histograms and comments that “There are roughly twice as many boys as girls in the $$175$$ to $$180\mbox{ cm}$$ category”. Do the histograms support his claim?

(iv) Mary examines both histograms and comments that “I see that there are more tall girls than tall boys”. Do the two histograms support her claim? Explain your answer.

(i)

Height (cm) $$145$$ - $$150$$ $$150$$ - $$155$$ $$155$$ - $$160$$ $$160$$ - $$165$$ $$165$$ - $$170$$ $$170$$ - $$175$$ $$175$$ - $$180$$ $$180$$ - $$185$$

Percentage of girls

$$3$$

$$9.6$$

$$16$$

$$22.4$$

$$25$$

$$16.2$$

$$5.8$$

$$2$$

(ii)

(iii) Yes as the area of the boys’ column is roughly twice the size of the corresponding girls’ column.

(iv) No as the sum of the areas of the columns to the right of $$165\mbox{ cm}$$ is larger for boys than for girls.

Solution

(i)

Height (cm) $$145$$ - $$150$$ $$150$$ - $$155$$ $$155$$ - $$160$$ $$160$$ - $$165$$ $$165$$ - $$170$$ $$170$$ - $$175$$ $$175$$ - $$180$$ $$180$$ - $$185$$

Percentage of girls

$$3$$

$$9.6$$

$$16$$

$$22.4$$

$$25$$

$$16.2$$

$$5.8$$

$$2$$

(ii)

(iii) Yes as the area of the boys’ column is roughly twice the size of the corresponding girls’ column.

(iv) No as the sum of the areas of the columns to the right of $$165\mbox{ cm}$$ is larger for boys than for girls.

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(d)

(i) The mean height of the boys in the sample is $$166.7\mbox{ cm}$$ and the standard deviation of their height is $$8.9\mbox{ cm}$$. Assuming that boys’ heights are normally distributed, use the Empirical Rule to find an interval that will contain the heights of approximately $$95\%$$ of all boys.

(ii) The standard deviation of the heights of the girls in the sample is $$7.7\mbox{ cm}$$ while the standard deviation of the heights of the boys is $$8.9\mbox{ cm}$$. Interpret this difference in the context of the data.

(i) $$148.9\mbox{ cm}<h<184.5\mbox{ cm}$$

(ii) There is a larger spread in the height of the boys compared with the height of the girls.

Solution

(i)

\begin{align}166.7-2(8.9)<h<166.7-2(8.9)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}148.9\mbox{ cm}<h<184.5\mbox{ cm}\end{align}

(ii) There is a larger spread in the height of the boys compared with the height of the girls.

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