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Non-Calculator Paper: Track 2: Year 11

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These are the solutions to the Mathematics past questions for Non-Calculator Paper: Track 2: Year 11.
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(1.) Estimate the value of: $96 \div \sqrt{25.4}$


$ 25.4 \approx 25 \\[3ex] \sqrt{25} = 5 \\[3ex] 96 \div 5 \\[3ex] = 19 + \dfrac{1}{5} \\[5ex] = 19 + 0.2 \\[3ex] = 19.2 \\[3ex] \implies \\[3ex] 96 \div \sqrt{25.4} \approx 19.2 $
(2.) The area of France is about 552 thousand square kilometres.
Write this number in standard form.


$ 552000 \;km^2 \\[4ex] = 552 * 1000 \\[3ex] = 5.52 * 100 * 1000 \\[3ex] = 5.52 * 10^5\;km^2 $
(3.) Circle the value of the digit 6 in the number 12.368

$ \dfrac{6}{1000} \hspace{5em} 6 \hspace{5em} \dfrac{6}{100} \hspace{5em} \dfrac{6}{10} \\[5ex] $

$ 12.368 \\[3ex] 6 \rightarrow 6 \;hundredth \\[3ex] = \dfrac{6}{100} $
(4.) Solve: $\dfrac{x}{7} - 6 = 3$


$ \dfrac{x}{7} - 6 = 3 \\[5ex] \dfrac{x}{7} = 3 + 6 \\[5ex] \dfrac{x}{7} = 9 \\[5ex] x = 7(9) \\[3ex] x = 63 \\[3ex] $ Check
$x = 63$
LHS RHS
$ \dfrac{x}{7} - 6 \\[5ex] \dfrac{63}{7} - 6 \\[5ex] 9 - 6 \\[3ex] 3 $ $3$
(5.) Number 5

Use the graph above to tick (✓) the incorrect statement:

The bear population has increased over the past five years.

The lynx population rose sharply, and the lynx is frequently sighted in the Rocky Mountains.

The moose population remained relatively stable during the past five years.


From the graph, we notice that the population of lynx decreased exponentially (not rose sharply) from 2019 to 2024
Hence, the incorrect statement is the second one. The second box should be ticked.
(6.) Calculate the simple interest earned on €5000 invested for 4 years at a rate of 3.5% per annum.


$ P = €5000 \\[3ex] r = 3.5\% = \dfrac{3.5}{100} = \dfrac{35}{10 * 100} = \dfrac{35}{1000} \\[5ex] t = 4\;years \\[5ex] SI = P * r * t \\[3ex] = 5000 * \dfrac{35}{1000} * 4 \\[5ex] = 5 * 35 * 4 \\[3ex] = 5 * 4 * 35 \\[3ex] = 20 * 35 \\[3ex] = 2 * 10 * 35 \\[3ex] = 2 * 35 * 10 \\[3ex] = 70 * 10 \\[3ex] = €700 $
(7.) Express 750 as a product of its prime factors.


$ \begin{array}{c} & & 750 \\ & \swarrow & & \searrow \\ \large{\bigcirc} \hspace{-5.5mm} \small{2} & & & & 375 \\ & & & \swarrow & & \searrow \\ & & \large{\bigcirc} \hspace{-5.5mm} \small{3} & & & & 125 \\ & & & & & \swarrow & & \searrow \\ & & & & \large{\bigcirc} \hspace{-5.5mm} \small{5} & & & & 25 \\ & & & & & & & \swarrow & & \searrow \\ & & & & & & \large{\bigcirc} \hspace{-5.5mm} \small{5} & & & & \large{\bigcirc} \hspace{-5.5mm} \small{5} \\ \end{array} \\ 750 = 2 * 3 * 5 * 5 * 5 $
(8.) The function f is defined as $f(x) = x^2 - 6$.
Work out the value of f(7)


$ f(x) = x^2 - 6 \\[3ex] f(7) = 7^2 - 6 \\[4ex] = 49 - 6 \\[3ex] = 43 $
(9.) Work out the value of 𝑥.

Number 9


$ 15^2 = 9^2 + x^2 ...Pythagorean\;\;Theorem \\[4ex] 225 = 81 + x^2 \\[4ex] x^2 + 81 = 225 \\[4ex] x^2 = 225 - 81 \\[4ex] x^2 = 144 \\[4ex] x = \sqrt{144} \\[3ex] x = 12\;cm $
(10.) The volume of a cube is 125 000 cm³.
Work out the length of one side of the cube.


$ \text{Side Length of the cube} = L^3 \\[4ex] Volume = L^3 = 125000\;cm^3 \\[4ex] Volume = 125000\;cm^3 \\[4ex] \implies \\[3ex] L^3 = 125000 \\[4ex] L = \sqrt[3]{125000} \\[3ex] L = 50\;cm $
(11.) This is a circle of radius 14 cm.
Work out the area of the shaded part.
(Take π as $\dfrac{22}{7}$)

Number 11


$ radius,\;r = 14\;cm \\[3ex] Area,\;A = \pi r^2 \\[4ex] = \dfrac{22}{7} * 14 * 14 \\[5ex] = 22 * 2 * 14 \\[5ex] \text{Area of shaded part} = \dfrac{3}{4} * A \\[5ex] = \dfrac{3}{4} * 22 * 2 * 14 \\[5ex] = 3 * 22 * 7 \\[3ex] = 66 * 7 \\[3ex] = 462\;cm^2 $
(12.) Work out, giving your answer in standard form:

$ \left(5 * 10^4\right) \div \left(2 * 10^{-2}\right) $


$ \left(5 * 10^4\right) \div \left(2 * 10^{-2}\right) \\[4ex] \dfrac{5 * 10^4}{2 * 10^{-2}} \\[6ex] \dfrac{5}{2} * \dfrac{10^4}{10^{-2}} \\[6ex] 2.5 * 10^{4 - (-2)} \\[4ex] 2.5 * 10^{4 + 2} \\[4ex] 2.5 * 10^6 $
(13.) Expand: $x(2x^2 + 5)$


$ x(2x^2 + 5) \\[4ex] 2x^3 + 5x $
(14.) Work out: $\sqrt{1\dfrac{11}{25}}$


$ \sqrt{1\dfrac{11}{25}} \\[6ex] \sqrt{\dfrac{36}{25}} \\[6ex] \dfrac{\sqrt{36}}{\sqrt{25}} \\[6ex] \dfrac{6}{5} \\[5ex] 1\dfrac{1}{5} $
(15.) Number 15

Use the grid above to solve the simultaneous equations: $y = 2x + 4$ and $y = 6$


$ y = 2x + 4...eqn.(1) \\[3ex] y = 6 ...eqn.(2) \\[3ex] y = y \implies eqn.(1) = eqn.(2) \\[3ex] 2x + 4 = 6 \\[3ex] 2x = 6 - 4 \\[3ex] 2x = 2 \\[3ex] x = \dfrac{2}{2} \\[5ex] x = 1 \\[3ex] \therefore x = 1;\;\;y = 6 \\[5ex] \underline{Check} \\[3ex] 2x + 4 \\[3ex] 2(1) + 4 \\[3ex] 2 + 4 \\[3ex] 6 $
(16.) One euro (€) is equivalent to 1.63 Australian Dollars ($)
How many Australian Dollars ($) are obtained for €3000?


$ €1 = 1.63\;A\$ \\[3ex] €3000 = ? \\[5ex] 1 * 3000 = 3000 \\[3ex] Similarly: \\[3ex] 1.63 * 3000 \\[3ex] 1.63 * 3 * 1000 \\[3ex] 1.63 * 1000 * 3 \\[3ex] 1630 * 3 \\[3ex] 4890\;A\$ $
(17.) A flight from Malta to Serbia takes 1 hour 55 minutes.
An aeroplane leaves Malta at 8:35 am.
At what time does it land in Serbia?


$ 1\;hr \;\;\;55\;min = 2\;hrs - 5\;min \\[3ex] \implies \\[3ex] \begin{array}{c@{}c@{}c@{}c} & 8 & : & 35 \, \text{AM} \\ + & 2 & : &\;\; 00 \, \text{hours} \\ \hline & 10 & : & 35 \, \text{AM} \\ - & & : & \,\;\;\;\;\;\; 05 \, \text{minutes} \\ \hline & 10 & : & 30 \, \text{AM} \\ \end{array} $
(18.) The nth term of a sequence is 7 − 3n
Find the value of the 25th term.


Let the nth term = T(n)

$ T(n) = 7 - 3n \\[3ex] T(25) = 7 - 3(25) \\[3ex] = 7 - 75 \\[3ex] = -68 $
(19.) Simplify: $\dfrac{21x^5}{7x^2}$


$ \dfrac{21x^5}{7x^2} \\[6ex] = \dfrac{21}{7} * \dfrac{x^5}{x^2} \\[6ex] 3 * x^{5 - 2} \\[4ex] = 3x^3 $
(20.) Work out the value of: 7.5 × (8.8 + 1.2) ÷ 15


$ 7.5 * (8.8 + 1.2) \div 15 \\[3ex] 7.5 * 10 \div 15 \\[3ex] 75 \div 15 \\[3ex] 5 $




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