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Question 1

Topic – E1.1

Write down a prime number between 30 and 40.

▶️ Answer/Explanation

Solution

Ans: 31 or 37

Prime numbers between 30 and 40 are numbers divisible only by 1 and themselves. Checking each number: 31 (prime), 32 (not), 33 (not), 34 (not), 35 (not), 36 (not), 37 (prime), 38 (not), 39 (not).

Question 2

Topic – E1.7

Calculate $ 4^5 – 5^4 $.

▶️ Answer/Explanation

Solution

Ans: 399

Calculate each term: $4^5 = 1024$ and $5^4 = 625$. Then subtract: $1024 – 625 = 399$.

Question 3

Topic – E1.15

Jason starts a run at 10.05 am and finishes at 1.02 pm.

Work out the time Jason takes to complete the run.

▶️ Answer/Explanation

Solution

Ans: 2h 57min

From 10:05 am to 1:02 pm is 2 hours and 57 minutes (10:05 to 1:05 would be 3 hours, subtract 3 minutes).

Question 4

Topic – E1.4

Calculate $ \frac{1 – 0.7}{0.45 – 0.38} $, giving your answer correct to 4 significant figures.

▶️ Answer/Explanation

Solution

Ans: 4.286

Numerator: $1 – 0.7 = 0.3$. Denominator: $0.45 – 0.38 = 0.07$. Divide: $0.3 ÷ 0.07 ≈ 4.2857$. Round to 4 significant figures: 4.286.

Question 5

Topic – E1.16

Kirsty changes $380.80 into pounds (£) when £1 = $1.19.

Calculate the amount Kirsty receives.

▶️ Answer/Explanation

Solution

Ans: 320

Divide dollars by exchange rate: $380.80 ÷ 1.19 = 320$. So Kirsty receives £320.

Question 6

Topic – E1.1

Write 180 as a product of its prime factors.

▶️ Answer/Explanation

Solution

Answer: $2 \times 2 \times 3 \times 3 \times 5$ or $2^2 \times 3^2 \times 5$

1. Start dividing by the smallest prime number, 2: 180 ÷ 2 = 90

2. Continue dividing by 2: 90 ÷ 2 = 45

3. Now divide by 3: 45 ÷ 3 = 15

4. Again divide by 3: 15 ÷ 3 = 5

5. Finally divide by 5: 5 ÷ 5 = 1

The prime factors are all the divisors used: 2, 2, 3, 3, 5

Question 7

Topic – E1.4

Without using a calculator, work out $\frac{3}{7} – \frac{2}{21}$.

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

▶️ Answer/Explanation

Solution

Answer: $\frac{1}{3}$

1. Find common denominator: 7 and 21 → LCD is 21

2. Convert $\frac{3}{7}$ to $\frac{9}{21}$ (multiply numerator and denominator by 3)

3. Now subtract: $\frac{9}{21} – \frac{2}{21} = \frac{7}{21}$

4. Simplify $\frac{7}{21}$ by dividing numerator and denominator by 7 → $\frac{1}{3}$

Question 8

Topic – E2.5

$s = \frac{1}{2} at^2$

(a) Work out the value of $s$ when $a = 0.9$ and $t = 4$.

(b) Rearrange the formula to find $t$ in terms of $s$ and $a$.

▶️ Answer/Explanation

Solution

8(a) Answer: 7.2

Substitute values: $s = \frac{1}{2} \times 0.9 \times 4^2 = 0.45 \times 16 = 7.2$

8(b) Answer: $t = \pm \sqrt{\frac{2s}{a}}$

1. Start with $s = \frac{1}{2} at^2$

2. Multiply both sides by 2: $2s = at^2$

3. Divide both sides by $a$: $\frac{2s}{a} = t^2$

4. Take square root of both sides: $t = \pm \sqrt{\frac{2s}{a}}$

(Note: The ± is included because time can be considered positive in physical contexts, but mathematically both roots exist)

Question 9

Topic – E2.2

Factorise completely: $14xy – 7y^2$

▶️ Answer/Explanation

Solution

Answer: $7y(2x – y)$

1. Identify common factors in both terms: 7 and y

2. Factor out 7y from both terms: $7y(2x) – 7y(y)$

3. This becomes $7y(2x – y)$

Check by expanding back: $7y \times 2x = 14xy$ and $7y \times (-y) = -7y^2$

Question 10

Topic – E2.7

22, 17, 12, 7, 2, …

(a) Find the next term of the sequence.

(b) Find the nth term of the sequence.

▶️ Answer/Explanation

Solution

10(a) Answer: -3

The sequence decreases by 5 each time: 22, 17 (-5), 12 (-5), 7 (-5), 2 (-5), so next term is 2 – 5 = -3

10(b) Answer: $27 – 5n$

1. The common difference is -5

2. The 0th term would be 22 + 5 = 27

3. So nth term formula is: first term + (n-1)×difference

Or: 0th term + n×difference → 27 + n×(-5) = 27 – 5n

Question 11

Topic – E4.4

Triangle ABC is mathematically similar to triangle PQR.

(a) Calculate QR.

(b) The two triangles are the cross-sections of two mathematically similar prisms. The volume of the larger prism is 320 cm³.

Calculate the volume of the smaller prism.

▶️ Answer/Explanation

Solution

Ans:

(a) 4.5 cm (Using similarity ratio: 8/6 = 6/QR → QR = 6×6/8)

(b) 135 cm³ (Volume ratio is cube of linear ratio: (6/8)³ × 320)

Question 12

Topic – E4.6

The interior angles of a pentagon are in the ratio 4 : 5 : 5 : 7 : 9.

Find the size of the largest angle.

▶️ Answer/Explanation

Solution

Ans: 162°

Sum of interior angles = (5-2)×180° = 540°

Total ratio parts = 4+5+5+7+9 = 30

Largest angle = (9/30)×540° = 162°

Question 13

Topic – E1.8

Work out $2 \times 10^{100} – 2 \times 10^{98}$, giving your answer in standard form.

▶️ Answer/Explanation

Solution

Ans: $1.98 \times 10^{100}$

Factor out $2 \times 10^{98}$ to get $2 \times 10^{98}(100 – 1)$

This equals $2 \times 10^{98} \times 99 = 198 \times 10^{98}$

Convert to standard form: $1.98 \times 10^{100}$

Question 14

Topic – E1.12

A train passes through a station at a speed of 108 km/h.
The length of the station is 120 m.
The train takes 7 seconds to completely pass through the station.

Work out the length of the train.

▶️ Answer/Explanation

Solution

Ans: 90 m

Convert speed to m/s: 108 km/h = 108000 m/3600 s = 30 m/s

Total distance covered = speed × time = 30 × 7 = 210 m

Train length = total distance – station length = 210 – 120 = 90 m

Question 15

Topic – E2.4

Find the value of x.

$4^{x} = \frac{1}{64}$

▶️ Answer/Explanation

Solution

Ans: -3

Express both sides with base 4: $4^x = 4^{-3}$ (since $64 = 4^3$)

Therefore, $x = -3$

Question 16

Topic – E7.1

Describe fully the single transformation that maps triangle T onto triangle P.

▶️ Answer/Explanation

Solution

Ans: Enlargement with scale factor ½, center (4,4)

1. Observe triangle P is half the size of triangle T

2. Check lines from center point (4,4) to corresponding vertices

3. Confirm all distances are halved from center (4,4)

Question 17

Topic – E5.4

Find the radius of a hemisphere of volume 80 cm³.

[The volume, V, of a sphere with radius r is $V = \frac{4}{3}πr^3$]

▶️ Answer/Explanation

Solution

Ans: 3.37 cm

1. Hemisphere volume = ½ sphere volume: $\frac{2}{3}πr^3 = 80$

2. Rearrange: $r^3 = \frac{80×3}{2π} = \frac{120}{π}$

3. Calculate: $r = \sqrt[3]{\frac{120}{π}} ≈ 3.37$ cm

Question 18

Topic – E4.7

A, B, C and D are points on a circle.
TU is a tangent to the circle at D.
DA is parallel to CB.

Find the value of x and the value of y.

▶️ Answer/Explanation

Solution

Ans: x = 38, y = 22

1. x = angle between tangent and chord = angle in alternate segment (38°)

2. Since DA ∥ CB, angle CBA = 180° – 38° = 142°

3. In triangle ABC: y = 180° – 38° – 120° = 22°

Question 19

Topic – E4.4

In the diagram, AB is parallel to PQ.
AQ and PB intersect at X with .

Complete the following statements.

In triangles ABX and QPX,
AX XQ = is given information.

Angle BAX Angle = ….. because …………….
Angle AXB Angle = ….. because ……………..
Triangle ABX is congruent to triangle QPX because of the congruency criterion ………….
PX = ……. because the triangles are congruent.

▶️ Answer/Explanation

Solution

Ans:

Angle BAX = Angle PQX (alternate angles)

Angle AXB = Angle PXQ (vertically opposite angles)

Triangle ABX ≡ triangle QPX (ASA congruency)

PX = XB (corresponding sides of congruent triangles)

Question 20

Topic – E2.9

The diagram shows the speed-time graph for 24 seconds of a car journey.

(a) Calculate the deceleration of the car in the final 8 seconds.

(b) Calculate the total distance travelled during the 24 seconds.

▶️ Answer/Explanation

Solution

Ans:

(a) 1.5 m/s² (Deceleration = (12-0)/8 = 1.5 m/s²)

(b) 240 m (Area under graph: (16×12) + ½(8×12) = 192 + 48 = 240 m)

Question 21

Topic – E2.2

Factorise completely: $1 – q – a + aq$

▶️ Answer/Explanation

Solution

Ans: $(1-q)(1-a)$

1. Group terms: $(1 – q) + (-a + aq)$

2. Factorise: $(1 – q) – a(1 – q)$

3. Common factor: $(1 – q)(1 – a)$

Question 22

Topic – E2.4

Simplify fully $(216y^{216})^{\frac{2}{3}}$.

▶️ Answer/Explanation

Solution

Ans: $36y^{144}$

First simplify $216^{\frac{2}{3}}$ = $(6^3)^{\frac{2}{3}} = 6^2 = 36$.

Then $(y^{216})^{\frac{2}{3}} = y^{216 \times \frac{2}{3}} = y^{144}$.

Combine both parts to get $36y^{144}$.

Question 23

Topic – E2.2

(a) $x^2 + 8x + 10 = (x + p)^2 + q$

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

(b) Solve $x^2 + 8x + 10 = 30$.

▶️ Answer/Explanation

Solution

Ans (a): p = 4, q = -6

Complete the square: $(x^2 + 8x) + 10 = (x + 4)^2 – 16 + 10 = (x + 4)^2 – 6$.

Ans (b): x = -10 or x = 2

Using part (a): $(x + 4)^2 – 6 = 30$ → $(x + 4)^2 = 36$ → $x + 4 = ±6$.

So $x = -4 ±6$ giving solutions -10 and 2.

Question 24

Topic – E6.1

A cuboid measures 24 cm by 12 cm by 8 cm.

Calculate the length of a diagonal of the cuboid.

▶️ Answer/Explanation

Solution

Ans: 28 cm

Use 3D Pythagoras: $\sqrt{24^2 + 12^2 + 8^2}$.

Calculate: $\sqrt{576 + 144 + 64} = \sqrt{784} = 28$ cm.

Question 25

Topic – E1.11

$w$ is proportional to the square root of $y$.
$y$ is inversely proportional to $x$.
When $x = 4$, $y = 16$ and $w = 8$.

Find $w$ in terms of $x$.

▶️ Answer/Explanation

Solution

Ans: $\frac{16}{\sqrt{x}}$

First find $y = \frac{k}{x}$ → $16 = \frac{k}{4}$ → $k = 64$, so $y = \frac{64}{x}$.

Then $w = m\sqrt{y}$ → $8 = m\sqrt{16}$ → $m = 2$.

Substitute: $w = 2\sqrt{\frac{64}{x}} = \frac{16}{\sqrt{x}}$.

Question 26

Topic – E7.1

The diagram shows a triangle $OAB$ and a parallelogram $OALK$.
The position vector of $A$ is a and the position vector of $B$ is b.
$K$ is a point on $AB$ so that $AK : KB = 1 : 2$.

Find the position vector of $L$, in terms of a and b.
Give your answer in its simplest form.

▶️ Answer/Explanation

Solution

Ans: $\frac{5}{3}a + \frac{1}{3}b$

Find $\overrightarrow{AK} = \frac{1}{3}\overrightarrow{AB} = \frac{1}{3}(b – a)$.

Since $OALK$ is parallelogram, $\overrightarrow{AL} = \overrightarrow{OK} = \overrightarrow{OA} + \overrightarrow{AK} = a + \frac{1}{3}(b – a)$.

Thus $\overrightarrow{OL} = \overrightarrow{OA} + \overrightarrow{AL} = a + \frac{1}{3}(b – a) + a = \frac{5}{3}a + \frac{1}{3}b$.

Question 27

Topic – E2.10

The line $y = x + 1$ intersects the graph of $y = x^2 – 3x – 11$ at the points $A$ and $B$.
Find the coordinates of $A$ and the coordinates of $B$.

▶️ Answer/Explanation

Solution

Ans: A(-2, -1) and B(6, 7)

Set equations equal: $x^2 – 3x – 11 = x + 1$ → $x^2 – 4x – 12 = 0$.

Factorize: $(x + 2)(x – 6) = 0$ → $x = -2$ or $6$.

Find y-coordinates: when $x = -2$, $y = -1$; when $x = 6$, $y = 7$.

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