0580_w23_qp

Question 1

Topic – E1.4

Write 24.07839

(a) correct to 2 decimal places

(b) correct to the nearest 10.

▶️ Answer/Explanation

Solution

(a) 24.08

Look at the third decimal (8) which rounds up the second decimal (7) to 8.

(b) 20

The digit in the tens place is 2, and the next digit (4) is less than 5, so we round down to 20.

Question 2

Topic – E1.6

Write down the number that is 9 greater than −23.

▶️ Answer/Explanation

Solution

-14

Simply add 9 to -23: -23 + 9 = -14.

Question 3

Topic – E2.5

$v = u + at$

Find the value of $v$ when $u = 30$, $a = -2$ and $t = 7$.

▶️ Answer/Explanation

Solution

Substitute the values into the formula: $v = 30 + (-2) \times 7$

Calculate: $30 – 14 = 16$

Question 4

Topic – E5.1

Change 62 000 millimetres into kilometres.

▶️ Answer/Explanation

Solution

0.062 km

First convert mm to m: 62000 mm ÷ 1000 = 62 m

Then convert m to km: 62 m ÷ 1000 = 0.062 km

Question 5

Topic – E4.6

The diagram shows two intersecting straight lines crossing two parallel lines.

Find the value of $x$.

▶️ Answer/Explanation

Solution

Using the property of parallel lines, the angle adjacent to x is equal to 50°.

Then, x + 50° = 114° (straight line), so x = 114° – 50° = 64°.

Question 6

Topic – E1.1

(a) Explain why 111 is not a prime number.

(b) Find a prime number between 110 and 120.

▶️ Answer/Explanation

Solution

(a) Ans: 111 is divisible by 3 (1+1+1=3) and 37 (3×37=111)

(b) Ans: 113

Checking numbers between 110-120: 111 (divisible by 3), 113 (prime), 117 (divisible by 3), 119 (7×17).

Question 7

Topic – E4.6

Find the bearing of Q from P.

▶️ Answer/Explanation

Solution

Ans: 231°

Bearings are measured clockwise from North.

180° (South) + (90° – 39°) = 180° + 51° = 231°

Question 8

Topic – E1.4

Without using a calculator, work out $3\frac{1}{8} – 1\frac{3}{4}$.

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

▶️ Answer/Explanation

Solution

Ans: $1\frac{3}{8}$

Convert to improper fractions: $\frac{25}{8} – \frac{7}{4}$

Common denominator (8): $\frac{25}{8} – \frac{14}{8} = \frac{11}{8}$

Convert to mixed number: $1\frac{3}{8}$

Question 9

Topic – E1.1

Write 90 as a product of its prime factors.

▶️ Answer/Explanation

Solution

Ans: $2 \times 3^2 \times 5$

Break down 90: 90 = 2 × 45

45 = 3 × 15

15 = 3 × 5

Final form: 2 × 3 × 3 × 5 or $2 \times 3^2 \times 5$

Question 10

Topic – E2.2

Expand and simplify: $2(t + w) + 3(w – t)$

▶️ Answer/Explanation

Solution

Ans: $5w – t$

First expand: $2t + 2w + 3w – 3t$

Combine like terms: $(2t – 3t) + (2w + 3w)$

Simplify: $-t + 5w$ or $5w – t$

Question 11

Topic – E4.4

The two shapes are mathematically similar.

(a) Find the value of h.

(b) The area of the smaller shape is 16 cm². Calculate the area of the larger shape.

▶️ Answer/Explanation

Solution

(a) Ans: 3.5

Using similarity ratio: $\frac{9}{5} = \frac{6.3}{h}$

$h = \frac{6.3 \times 5}{9} = 3.5$ cm

(b) Ans: 51.84

Area ratio is square of length ratio: $(\frac{9}{5})^2 = \frac{81}{25}$

Larger area = $16 \times \frac{81}{25} = 51.84$ cm²

Question 12

Topic – E2.9

The diagram shows a speed-time graph for 16 seconds of a car journey.

(a) Find the deceleration of the car in the final 4 seconds.

(b) Find the total distance travelled during the 16 seconds.

▶️ Answer/Explanation

Solution

(a) Ans: 2.5 m/s²

Deceleration = $\frac{10}{4} = 2.5$ m/s²

(b) Ans: 140 m

Distance = area under graph = rectangle + triangle

$(10 \times 12) + (\frac{1}{2} \times 4 \times 10) = 120 + 20 = 140$ m

Question 13

Topic – E1.7

(a) $3^{3p} \times 3^{2p} = 729$

Find the value of p.

(b) Simplify $(32x^{10})^{\frac{1}{5}}$

▶️ Answer/Explanation

Solution

(a) Ans: 1.2

$3^{5p} = 3^6$ (since $729 = 3^6$)

$5p = 6 \Rightarrow p = \frac{6}{5} = 1.2$

(b) Ans: $2x^2$

$(32)^{\frac{1}{5}} = 2$ and $(x^{10})^{\frac{1}{5}} = x^2$

Final answer: $2x^2$

Question 14

Topic – E2.2

$y = 2w^2 – x$

Rearrange the formula to make w the subject.

▶️ Answer/Explanation

Solution

Ans: $w = \pm\sqrt{\frac{y + x}{2}}$

Start: $y = 2w^2 – x$

Add x to both sides: $y + x = 2w^2$

Divide by 2: $\frac{y + x}{2} = w^2$

Square root both sides: $w = \pm\sqrt{\frac{y + x}{2}}$

Question 15

Topic – E1.2

(a) On the Venn diagram, shade the region P ∪ Q′.

(b) n(ξ) = 20, n(A ∪ B)′ = 1, n(A) = 12, n(B) = 10

Complete the Venn diagram.

▶️ Answer/Explanation

Solution

(a) Ans:

P ∪ Q′ means all elements in P or not in Q.

(b) Ans:

n(A ∩ B) = n(A) + n(B) – n(A ∪ B) = 12 + 10 – 19 = 3

A only = 12 – 3 = 9, B only = 10 – 3 = 7, outside = 1

Question 16

Topic – E1.3

Find the lowest common multiple (LCM) of $ 12x^8 $ and $ 8x^{12} $.

▶️ Answer/Explanation

Solution

Ans: $ 24x^{12} $

1. Find LCM of coefficients: LCM of 12 and 8 is 24

2. For variables, take highest power: $ x^{12} $

3. Combine them: $ 24x^{12} $

Question 17

Topic – E4.7

(a)

A, B and C are points on a circle, centre O.
Angle $ OBA = 28^\circ $.

Find angle $ ACB $.

(b)

P, Q and R are points on a circle.
TU is a tangent to the circle at P.
Angle $ TPR = 47^\circ $ and angle $ PRQ = 52^\circ $.

Find angle $ RPQ $.

▶️ Answer/Explanation

Solution

Ans: (a) 62° (b) 81°

(a) Triangle OAB is isosceles (OA=OB=radius). Angle AOB = 180° – 2×28° = 124°. Angle ACB is half of angle AOB = 62°.

(b) Angle QPU = angle PRQ = 52° (alternate segment theorem). Angle RPQ = 180° – 47° – 52° = 81°.

Question 18

Topic – E5.4

A solid cylinder has radius 5 cm and height 8 cm.

Calculate the total surface area of the cylinder.

▶️ Answer/Explanation

Solution

Ans: 408 cm² (or 408.4 to 408.5)

1. Curved surface area: $ 2πrh = 2×π×5×8 = 80π $

2. Two circular ends: $ 2×πr² = 2×π×25 = 50π $

3. Total surface area: $ 80π + 50π = 130π ≈ 408.4 $ cm²

Question 19

Topic – E2.7

Find the nth term of each sequence.

(a) 11, 8, 5, 2, -1, …

(b) 1, 5, 25, 125, 625, …

▶️ Answer/Explanation

Solution

Ans: (a) $ 14 – 3n $ (b) $ 5^{n-1} $

(a) This is an arithmetic sequence. First term (a)=11, common difference (d)=-3. nth term = a + (n-1)d = 11 + (n-1)(-3) = 14 – 3n.

(b) This is a geometric sequence where each term is multiplied by 5. First term=1, common ratio=5. nth term = $ 5^{n-1} $.

Question 20

Topic – E1.10

The area of a rectangle is 55.2 cm², correct to 1 decimal place.
The length of the rectangle is 9 cm, correct to the nearest cm.

Calculate the upper bound of the width of the rectangle.

▶️ Answer/Explanation

Solution

Ans: 6.5 cm

1. Upper bound of area: 55.25 cm²

2. Lower bound of length: 8.5 cm

3. Upper bound of width = upper bound area / lower bound length

4. Calculation: 55.25 ÷ 8.5 = 6.5 cm

Question 21

Topic – E2.10

The line $y = x + 1$ intersects the curve $y = x^2 + x – 3$ at two points.

Find the coordinates of the two points.

▶️ Answer/Explanation

Solution

Ans: (2, 3) and (-2, -1)

Set equations equal: $x + 1 = x^2 + x – 3$

Simplify to $x^2 – 4 = 0$

Solve: $x = 2$ or $x = -2$

Find y-values: When $x=2$, $y=3$; when $x=-2$, $y=-1$

Question 22

Topic – E1.11

$x$ is inversely proportional to the square root of $w$. When $w = 16$, $x = 3$.

Find $x$ in terms of $w$.

▶️ Answer/Explanation

Solution

Ans: $x = \frac{12}{\sqrt{w}}$

General form: $x = \frac{k}{\sqrt{w}}$

Substitute given values: $3 = \frac{k}{\sqrt{16}}$ → $3 = \frac{k}{4}$

Solve for k: $k = 12$

Final equation: $x = \frac{12}{\sqrt{w}}$

Question 23

Topic – E9.4

Some students record their reaction times.
The table shows the results.

Reaction time (t seconds)	$0 < t \leq 6$	$6 < t \leq 10$
Frequency	18	16

On a histogram, the height of the block for the $0 < t \leq 6$ interval is 7.5 cm.

Calculate the height of the block for the $6 < t \leq 10$ interval.

▶️ Answer/Explanation

Solution

Ans: 10 cm

Find frequency density for first interval: $18 \div 6 = 3$ per second

Height corresponds to frequency density: $7.5 \text{cm} = 3$ per second

Find frequency density for second interval: $16 \div 4 = 4$ per second

Calculate height: $(4/3) \times 7.5 = 10$ cm

Question 24

Topic – E2.2

Simplify.

$\frac{ax – 2a – x + 2}{a^2 – 1}$

▶️ Answer/Explanation

Solution

Ans: $\frac{x – 2}{a + 1}$

Factor numerator: $a(x – 2) – (x – 2) = (x – 2)(a – 1)$

Factor denominator: $a^2 – 1 = (a – 1)(a + 1)$

Cancel common factor $(a – 1)$

Final simplified form: $\frac{x – 2}{a + 1}$

Question 25

Topic – E2.4

The derivative of $2ax^7 + 3x^k$ is $42x^6 + 15x^{k-1}$.

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

▶️ Answer/Explanation

Solution

Ans: $a = 3$, $k = 5$

Find derivative: $\frac{d}{dx}(2ax^7 + 3x^k) = 14ax^6 + 3kx^{k-1}$

Set equal to given derivative: $14a = 42$ and $3k = 15$

Solve for a: $a = 42/14 = 3$

Solve for k: $k = 15/3 = 5$

Question 26

Topic – E7.1

The diagram shows a parallelogram $OPQT$.
The position vector of $P$ is $\mathbf{a}$ and the position vector of $T$ is $\mathbf{b}$.

$K$ is on $PQ$ so that $PK : KQ = 3 : 1$.
The lines $OK$ and $TQ$ are extended to meet at $X$.

Find the position vector of $X$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
Give your answer in its simplest form.

▶️ Answer/Explanation

Solution

Ans: $\mathbf{b} + \frac{4}{3}\mathbf{a}$

Find vector $\overrightarrow{OK} = \mathbf{a} + \frac{3}{4}\mathbf{b}$ (since PK:KQ=3:1)

Find vector equation of line OX: $\lambda(\mathbf{a} + \frac{3}{4}\mathbf{b})$

Find vector equation of line TX: $\mathbf{b} + \mu(\mathbf{a})$

Solve for intersection point to get $\lambda = \frac{4}{3}$

Final position vector: $\mathbf{b} + \frac{4}{3}\mathbf{a}$

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