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Question 1

Topic – E1.15

A concert starts at 19 50 and finishes 2 hours 42 minutes later.

Work out the time the concert finishes.

▶️ Answer/Explanation

Solution

Ans: 22 32 or 10:32 pm

19:50 + 2 hours = 21:50

21:50 + 42 minutes = 22:32

Question 2

Topic – E1.4

Find the reciprocal of $ 1\frac{1}{4}$

▶️ Answer/Explanation

Solution

Ans: 4/5 or 0.8

Convert 1 1/4 to improper fraction: 5/4

Reciprocal is 4/5

Question 3

Topic – E1.5

Use one of the symbols <, >, or = to make each statement true:

$ \frac{2}{7}$ ………….. 0.2861

$ \frac{99}{900}$ ………….. 11%

1³ ………….. 4⁰

▶️ Answer/Explanation

Solution

Ans: <, =, =

2/7 ≈ 0.2857 < 0.2861

99/900 = 0.11 = 11%

1³ = 1
4⁰ = 1

1³ = 4⁰

Question 4

Topic – E1.11

Safia has a piece of fabric of length 5.6 m.
She cuts the fabric into two parts, with lengths in the ratio 3:4.

Calculate the length of the longer part.

▶️ Answer/Explanation

Solution

Ans: 3.2 m

Total parts = 3 + 4 = 7

Value per part = 5.6 ÷ 7 = 0.8 m

Longer part = 4 × 0.8 = 3.2 m

Question 5

Topic – E2.2

Work out:

(a) $3\begin{pmatrix}
6 \\-4
\end{pmatrix}$

(b) $\begin{pmatrix}
4 \\ -1
\end{pmatrix} + \begin{pmatrix}
-7 \\ 5
\end{pmatrix}$

▶️ Answer/Explanation

Solution

Ans:

(a) (18, -12)

(b) (-3, 4)

For (a): Multiply each component by 3

For (b): Add corresponding components (4 + -7 = -3, -1 + 5 = 4)

Question 6

Topic – E4.6

The diagram shows a right-angled triangle ABC and a quadrilateral AEDC.

Find the values of x, y, and z.

▶️ Answer/Explanation

Solution

Ans: x = 58, y = 39, z = 251

x = 90° – 32° = 58° (right angle in triangle ABC).

y = 180° – 122° – 19° = 39° (angles on straight line).

z = 360° – 34° – 17° – 58° = 251° (angles in quadrilateral).

Question 7

Topic – E2.2

Factorise: 28x – 35

▶️ Answer/Explanation

Solution

Ans: 7(4x – 5)

Find the greatest common factor of 28 and 35, which is 7.

Divide both terms by 7: 28x ÷ 7 = 4x, 35 ÷ 7 = 5.

Write as product: 7(4x – 5).

Question 8

Topic – E1.16

Edith invests $3000 in a savings account.
The account pays simple interest at a rate of 2.6% per year.

Calculate the total interest earned at the end of 3 years.

▶️ Answer/Explanation

Solution

Ans: $234

Calculate annual interest: 3000 × 0.026 = $78.

Multiply by 3 years: 78 × 3 = $234 total interest.

Question 9

Topic – E4.4

The diagram shows part of a regular polygon.
The interior angle is 132° larger than the exterior angle.

Calculate the number of sides of this polygon.

▶️ Answer/Explanation

Solution

Ans: 15 sides

Let exterior angle = x, then interior = x + 132°.

They sum to 180°: x + (x + 132) = 180 → x = 24°.

Number of sides = 360° ÷ 24° = 15.

Question 10

Topic – E8.1

Jacinda plays a game with her friend.
She can win, lose or draw the game.
The probability that she wins the game is 0.28 .

(a) Jacinda is twice as likely to draw the game as to lose the game.
Work out the probability that she loses the game.

(b) Jacinda plays the game 150 times.
Find the expected number of times that she wins.

▶️ Answer/Explanation

Solution

Ans: (a) 0.24 (b) 42

(a) Let P(lose) = x, then P(draw) = 2x.

0.28 + x + 2x = 1 → x = 0.24.

(b) Expected wins = 150 × 0.28 = 42.

Question 11

Topic – E1.4

Without using a calculator, work out $ 5\frac{1}{3} – 3\frac{4}{7} $.

You must show all your working and give your answer as a mixed number in its simplest form.

▶️ Answer/Explanation

Solution

Ans: $ 1\frac{16}{21} $

Convert to improper fractions: $ \frac{16}{3} – \frac{25}{7} $

Find common denominator (21): $ \frac{112}{21} – \frac{75}{21} $

Subtract: $ \frac{37}{21} $

Convert to mixed number: $ 1\frac{16}{21} $

Question 12

Topic – E2.5

Solve the simultaneous equations.
You must show all your working.

$ 5x + 6y = 9 $
$ 3x – 2y = -17 $

▶️ Answer/Explanation

Solution

Ans: x = -3, y = 4

Multiply second equation by 3: $ 9x – 6y = -51 $

Add to first equation: $ 14x = -42 $ → $ x = -3 $

Substitute x into first equation: $ -15 + 6y = 9 $ → $ y = 4 $

Question 13

Topic – E2.7

(a) A sequence has nth term $ 3n^2 – 1 $. Find the second term in this sequence.

(b) The table shows the first five terms of sequences A and B.

Complete the table to show the nth term of each sequence.

▶️ Answer/Explanation

Solution

Ans: (a) 11 (b) A: 4n-10, B: 2n³+1

(a) Substitute n=2: $ 3(4)-1 = 11 $

(b) For A: linear pattern → 4n-10

For B: cubic pattern → 2n³+1

Question 14

Topic – E4.4

Two solid steel statues are mathematically similar.
The smaller statue has height 12 cm and the larger statue has height 15 cm.
The larger statue has a mass 2.5 kg.
The density of steel is 8 g/cm³.

Calculate the volume of the smaller statue.
[Density = mass ÷ volume.]

▶️ Answer/Explanation

Solution

Ans: 160 cm³

Volume ratio = (12/15)³ = 0.512

Large volume = 2500g ÷ 8g/cm³ = 312.5 cm³

Small volume = 312.5 × 0.512 = 160 cm³

Question 15

Topic – E9.3

Students in class P take a test.
These statistics show information about their marks.

lower quartile = 38
median = 53
interquartile range = 28
range = 81
highest mark = 96

(a) Draw a box-and-whisker plot to represent this information.

(b) Students in class Q take the same test. For class Q, the median is 49 and the interquartile range is 35.
Make two comments comparing the distribution of marks for class P with that of class Q.

▶️ Answer/Explanation

Solution

Ans: (a) Box plot with L=15, LQ=38, M=53, UQ=66, H=96

(b) 1. Class P performed better (higher median)

2. Class Q has more variation (larger IQR)

(a) Calculate UQ: 38+28=66, lowest: 96-81=15

(b) Compare medians and IQRs

Question 16

Topic – E5.4

The diagram shows a sphere of radius 6 cm and a cylinder of height 18 cm and radius R cm.
The volume of the sphere is equal to the volume of the cylinder.

Calculate the curved surface area of the cylinder.
Give your answer in terms of r.
[The volume, V, of a sphere with radius r is V = $\frac{4}{3}\Pi r^{3}$]

▶️ Answer/Explanation

Solution

Ans: 144π cm²

1. Calculate sphere volume: (4/3)π(6)³ = 288π cm³

2. Set equal to cylinder volume: πR²×18 = 288π

3. Solve for R²: R² = 16 ⇒ R = 4 cm

4. Calculate curved surface area: 2π×4×18 = 144π cm²

Question 17

Topic – E2.5

Solve
3x² – 7x – 16 = 0.

You must show all your working and give your answers correct to 2 decimal places.

▶️ Answer/Explanation

Solution

Ans: x = 3.75 or x = -1.42

1. Use quadratic formula: x = [7±√(49+192)]/6

2. Calculate discriminant: √241 ≈ 15.52

3. First solution: (7+15.52)/6 ≈ 3.75

4. Second solution: (7-15.52)/6 ≈ -1.42

Question 18

Topic – E2.13

g(x) = 4^x+3

(a) Find x when g(x) = 1

(b) Find g^-1$\frac{1}{16}$

▶️ Answer/Explanation

Solution

Ans: (a) -3 (b) -5

(a) 1 = 4^x+3 ⇒ x+3 = 0 ⇒ x = -3

(b) Set y = 4^x+3, swap x and y: x = 4^y+3

1/16 = 4^y+3 ⇒ 4^-2 = 4^y+3

Thus y+3 = -2 ⇒ y = -5

Question 19

Topic – E1.2

ℰ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
P = {odd numbers}
Q = {multiples of 3}
R = {square numbers}

(a) Find P ∩ Q ∩ R

(b) (i) Find Q ∪ R

(ii) Find n(P ∩ (Q ∪ R)’)

▶️ Answer/Explanation

Solution

Ans: (a) {9} (b)(i) {1,3,4,6,9} (ii) 2

(a) P∩Q∩R: odd multiples of 3 that are squares → only 9

(b)(i) Q∪R: {3,6,9} ∪ {1,4,9} = {1,3,4,6,9}

(ii) First find (Q∪R)’: all elements not in Q∪R → {2,5,7,8,10}

Then P ∩ (Q∪R)’: odd numbers in this set → {5,7}

Number of elements n(P ∩ (Q∪R)’) = 2

Question 20

Topic – E6.2

The diagram shows two right-angled triangles PQS and RQT.
PQR and QTS are straight lines.

Calculate angle QTR.

▶️ Answer/Explanation

Solution

Ans: 63.7°

1. Find QS in triangle PQS (right-angled at Q):

$ QS = PQ \times \sin(28°) = 18 \times \sin(28°) \approx 8.45 \text{ cm} $

2. Calculate QT (since QTS is a straight line and TS = 4 cm):

$ QT = QS – TS = 8.45 – 4 = 4.45 \text{ cm} $

3. In triangle RQT (right-angled at R), find angle QTR:

$ \tan(\text{angle QTR}) = \frac{QR}{QT} = \frac{9}{4.45} \approx 2.022 $

4. Compute angle QTR using inverse tangent:

$ \text{angle QTR} = \tan^{-1}(2.022) \approx 63.7° $

Question 21

Topic – E6.4

Solve the equation $3\tan x + 5 = 1$ for $0° \leq x \leq 360°$.

▶️ Answer/Explanation

Solution

Ans: 126.9° and 306.9°

1. Rearrange: $3\tan x = -4$ → $\tan x = -\frac{4}{3}$

2. Find reference angle: $\tan^{-1}(\frac{4}{3}) ≈ 53.1°$

3. Solutions in 2nd (180°-53.1°=126.9°) and 4th (360°-53.1°=306.9°) quadrants

Question 22

Topic –E2.10

The graph of $y = (x+2)(x-1)^2$ is shown on the grid.

(a) Show that $y = (x+2)(x-1)^2$ can be written as $y = x^3 – 3x + 2$.

(b) By drawing a suitable straight line, solve the equation $2x^3 – 5x = 0$.

▶️ Answer/Explanation

Solution

Ans:

(a) Expand $(x-1)^2$ to get $x^2-2x+1$, then multiply by $(x+2)$:

$(x+2)(x^2-2x+1) = x^3 – 2x^2 + x + 2x^2 – 4x + 2 = x^3 – 3x + 2$

(b) To solve $2x^3 – 5x = 0$ using the graph:

1. Rearrange to match the given graph form: $2x^3 = 5x$ → $x^3 = 2.5x$

2. Since the graph shows $y = x^3 – 3x + 2$, we rewrite as:

$x^3 = 2.5x$ → $x^3 – 3x + 2 = 2.5x – 3x + 2$ → $y = -0.5x + 2$

3. Draw the straight line $y = -0.5x + 2$ on the graph

4. The solutions are at the intersection points:

x ≈ -1.58 (accept -1.5 to -1.6)

x = 0

x ≈ 1.58 (accept 1.5 to 1.6)

Question 23

Topic – E2.2

Find the value of $p$ and the value of $k$.

$(x-5)^2 + k = x^2 – px – 21$

▶️ Answer/Explanation

Solution

Ans: p = 10, k = -46

1. Expand left side: $x^2 – 10x + 25 + k$

2. Compare coefficients with right side

3. $-10x = -px$ → $p = 10$

4. $25 + k = -21$ → $k = -46$

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