0580_m24_qp

Question 1:

Q1:
(a) E1.16
(b)
(i) E1.13
(ii) E1.11
(iii) E1.16
(iv) E1.13
(c) E1.10

1 A grocer sells potatoes, mushrooms and carrots.
(a) A customer buys 3 kg of mushrooms at $1.04$ per kg and 4kg of carrots at $1.28 per kg$.
Calculate the total cost.

$ …………………………………………

(b) In one week, the ratio of the masses of vegetables sold by the grocer is
potatoes : mushrooms : carrots = 11 : 8 : 6.
(i) Work out the mass of mushrooms sold as a percentage of the total mass.

……………………………………… %

(ii) The total mass of potatoes, mushrooms and carrots sold is 1500kg.
Find the mass of carrots the grocer sells this week.

…………………………………….. kg

(iii) The profit the grocer makes selling 1kg of carrots is $0.75 .
Find the total profit the grocer makes selling carrots this week.

$ ………………………………………… $
(iv) On the last day of the week, the grocer reduces the price of 1kg of potatoes by 8% to $1.15$ .
Calculate the original price of 1kg of potatoes.

$ ………………………………………… $

(c) The grocer buys 620kg of onions, correct to the nearest 20kg.
He packs them into bags each containing 5kg of onions, correct to the nearest 1kg.
Calculate the upper bound for the number of bags of onions that he packs.

▶️Answer/Explanation

Ans :

1(a) 8.24 cao

1(b)(i) 32

1(b)(ii) 360

1(b)(iii) 270

1(b)(iv) 1.25 cao

1(c) 140 nfww
Detailed Solution

(a)
3 kg of mushrooms at $\$1.04$ per kg
$
3 \times 1.04 = 3.12
$
4 kg of carrots at $\$1.28$ per kg
$
4 \times 1.28 = 5.12
$

Total cost
$
3.12 + 5.12 = 8.24
$

(b)
$
\text{Potatoes : Mushrooms : Carrots} = 11:8:6
$
The total parts in the ratio
$
11 + 8 + 6 = 25
$
(i)
The fraction of mushrooms sold
$
\frac{8}{25} \times 100 = 32\%
$

(ii)
Total mass of vegetables = 1500 kg
Mass of carrots
$
\frac{6}{25} \times 1500 = 360 \text{ kg}
$

(iii)
Profit per kg of carrots = $\$0.75$
Total profit
$
360 \times 0.75 = 270
$

(iv)
The price after an $8\%$ reduction is $\$1.15$.
Let the original price be $P$.

Since the new price is $92\%$ of the original price:
$
P \times 0.92 = 1.15
$
$
P = \frac{1.15}{0.92} = 1.25
$

Each bag contains: $ 5 $ kg, correct to nearest 1 kg
Lower bound per bag: $ 5 – 0.5 = 4.5 $ kg

Upper bound for the number of bags:
$
\frac{630}{4.5} = 140
$

Question 2:

Q2:
(a) E4.7
(b) E4.4
(c)
(i) E4.4
(ii) E4.4

A, B, C and D are points on a circle.
ADX and BCX are straight lines.
Angle BAD = x° and angle DCX = y°.
(a) Explain why x = y.
Give a geometrical reason for each statement you make.

(b) Show that triangle ABX is similar to triangle CDX.
(c) AD = 15cm, DX = 9 cm and CX = 12cm.
(i) Find BC.

BC = ……………………………………. cm

(ii) Complete the statement.

The ratio area of triangle ABX : area of triangle CDX = ………….. : 1.

▶️Answer/Explanation

Ans :

2(a) y+BCD = angle 180 oe

AND
angles on a straight line
AND
x+ angle BCD = 180 oe

AND
opposite angles of a cyclic quadrilateral
are supplementary
OR
angles in opposite segments are supplementary
leading to x = y with no errors

2(b) Allow any two statements from:
CXD is common angle
or angle AXB = angle CXD
x = y or angle BAX = angle DCX
angle ABX= angle CDX

States all three equal pairs of angles
OR
2/all angles equal so triangles similar

2(c)(i) 6 nfww

2(c)(ii) 4
Detailed Solution

(a)

The alternate segment theorem states that the angle between a tangent and a chord (such as $ \angle DCX $) is equal to the angle in the opposite segment (such as $ \angle BAD $).
$ x^{\circ} = y^{\circ}$

(b)

Since $ ADX $ and $ BCX $ are straight lines, they form vertically opposite angles.
Angle $ BAD = x $ and Angle $ DCX = y $

Two angles in $ \triangle ABX $ are equal to two angles in $ \triangle CDX $, the third angle must also be equal by the angle sum property.

Thus, by the AA (Angle-Angle) similarity criterion,
$
\triangle ABX \sim \triangle CDX
$

(c)(i)
$
\frac{AX}{CX} = \frac{BX}{DX}
$
$
\frac{24}{12} = \frac{BC + 12}{9}
$
$
2 =\frac{BC + 12}{9}
$
$
18 = BC + 12
$
$
BC = 6 \text{ cm}
$

(c)(ii)

From the similarity of △ABX and △CDX, the corresponding sides AX and CX are in the ratio
$
\frac{AX}{CX} = \frac{24}{12} = 2
$
Since the ratio of areas of similar triangles is the square of the ratio of their corresponding sides
$
\text{Ratio of areas} = \left(\frac{AX}{CX}\right)^2 = 2^2 = 4:1
$

Question 3:

Q3:
(a)
(i) E9.3
(ii) E9.3
(iii) E9.3
(iv) E9.3
(b) E9.3
(c) E9.7

3 (a) The table shows information about the marks gained by each of 10 students in a test.

(i) Calculate the range.

(ii) Calculate the mean.

(iii) Find the median.

(iv) Write down the mode.

(b) Paulo’s mean mark for 7 homework tasks is 17.
After completing the 8th task, his mean mark is 17.5 .
Calculate Paulo’s mark for the 8th task.

On the grid, draw a histogram to show this information.

▶️Answer/Explanation

Ans :

3(a)(i) 5
3(a)(ii) 16.8

3(a)(iii) 16.5
3(a)(iv) 15

3(b) 21

3(c) 5 correct blocks, with correct widths,

Detailed Solution

(a) (i)
$
\text{Range} = \text{Highest Mark} – \text{Lowest Mark}
$
From the table, the highest mark is 20, and the lowest mark is 15.
$
\text{Range} = 20 – 15 = 5
$

(a)(ii)
$
\text{Mean} = \frac{\sum (\text{Mark} \times \text{Frequency})}{\sum \text{Frequency}}
$
$
(15 \times 4) + (16 \times 1) + (17 \times 2) + (18 \times 1) + (19 \times 0) + (20 \times 2)
$
$
= 60 + 16 + 34 + 18 + 0 + 40 = 168
$

Total number of students
$
4 + 1 + 2 + 1 + 0 + 2 = 10
$
$
\text{Mean} = \frac{168}{10} = 16.8
$

(a)(iii)
The median is the middle value when arranged in ascending order.
The total frequency is 10, so the median is the average of the 5th and 6th values.
So, the 5th value is 16 and the 6th value is 17.

$
\text{Median} = \frac{16 + 17}{2} = 16.5
$

(a)(iv)
The mode is the mark that appears most frequently.
From the table, 15 appears 4 times, which is the highest frequency.

(b)
Paulo’s mean for 7 tasks = 17.
$
\text{Total marks for 7 tasks} = 7 \times 17 = 119
$
After the 8th task, the mean becomes 17.5.
$
\text{Total marks for 8 tasks} = 8 \times 17.5 = 140
$
$
\text{Mark for 8th task} = 140 – 119 = 21
$

(c)

$
\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}
$

Question 4:

Q4:
(a)
(i) E5.4
(ii) E5.4
(b) E5.4

4 (a)

The diagram shows a pyramid with a square base BCDE.
The diagonals CE and BD intersect at M, and the vertex F is directly above M.
BE = 12 cm and FM = 9 cm.
(i) Calculate the volume of the pyramid.
[The volume, V, of a pyramid with base area A and height h is V=$\frac{1}{3}Ah$.]

………………………………….. cm$^{3}$

(ii) Calculate the total surface area of the pyramid.

………………………………….. cm$^{2}$
(b)

The diagram shows a toy made from a cone and a hemisphere.
The base radius of the cone and the radius of the hemisphere are both r cm.
The slant height of the cone is 3r cm.
The total surface area of the toy is 304 cm$^{2}$

Calculate the value of r.
[The curved surface area, A, of a cone with radius r and slant height l is A r = r l.]
[The curved surface area, A, of a sphere with radius r is A 4πr$^{2}$ .]

r = …………………………………………

▶️Answer/Explanation

Ans :

4(a)(i) 432

4(a)(ii) 404 or 403.5 to 403.7

4(b) 4.4[0] or 4.398 to 4.399… nfww

Detailed Solution

(a)(i)
The formula for the volume of a pyramid
$
V = \frac{1}{3} \times A \times h
$

The base of the pyramid is a square with side length $ BE = 12 $ cm.

$
\text{Area of square} = 12 \times 12 = 144 \text{ cm}^2
$

The height of the pyramid is given as $ FM = 9 $ cm.
$
V = \frac{1}{3} \times 144 \times 9
$
$
V = \frac{1}{3} \times 1296
$
$
V = 432 \text{ cm}^3
$

(a)(ii)
The total surface area consists of:
1. The base area (square).
2. The 4 triangular faces.

Each triangular face has a base of 12 cm and a height of $ s = 10.8 $ cm.
Base = 12 cm
Height (slant height) = 10.8 cm
$
\text{Area} = \frac{1}{2} \times 12 \times 10.8
$
$
= \frac{12 \times 10.8}{2} = \frac{129.6}{2} = 64.8 \text{ cm}^2
$
$
\text{Total area of triangles} = 4 \times 64.8 = 259.2 \text{ cm}^2
$
$
\text{Base Area} = 12 \times 12 = 144 \text{ cm}^2
$

$
\text{Total Surface Area} = \text{Base Area} + \text{Total area of triangles}
$
$
= 144 + 259.2
$
$
= 403.2 \text{ cm}^2
$

$ r $ is the base radius,
$ l $ is the slant height given as $ 3r $.
$
\pi r (3r) = 3\pi r^2
$

The curved surface area of a hemisphere
$
\text{Curved Surface Area} = 2\pi r^2
$

$
3\pi r^2 + 2\pi r^2 = 304
$
$
5\pi r^2 = 304
$
$
r^2 = \frac{304}{5\pi}
$
$
r^2 = \frac{304}{15.708}
$
$
r^2 \approx 19.36
$
$
r \approx 4.4
$

Question 5:

Q5:
(a)
(i) E2.2
(ii) E2.3
(b) E2.2
(c) E2.3
(d) E2.2
(e) E2.5

5 (a) (i) Factorise.
$x^{2}-x-12$

(ii) Simplify.

$\frac{x^{2}-16}{x^{2}-x-12}$

(b) Simplify.

(2x-3) $^{2}$-(x+1) $^{2}$

$\frac{2x+4}{x+1}-\frac{x}{x-3}$

(d) Expand and simplify.
(x-3) (x -5) (2x+1)

(e) Solve the simultaneous equations.
You must show all your working.

$x-3y=13$

$2x^{2}-9y=116$

▶️Answer/Explanation

Ans :

5(a)(i) ( x-4)(x+3 ) final answer

5(a)(ii) $\frac{x+4}{x+3}$ final answer

5(b) 3x $^{2}$-14x+8 or (x-4) (3x-2) final answer

5(c) $\frac{x^{2}-3x-12}{(x+1) (x-3)}$ or $\frac{x^{2}-3x-12}{(x^{2}-2x-3)}$ final answer

5(d) $2x^{3}-15x^{2}+22x+15$

5(e) $2x^{2}-3x-77[=0] \mathbf{oe}$
$(6x^{2}9x-231[=0])$
$\mathbf{or}$
$18y^{2}+147y+222[=0] \mathbf{oe}$
$(6y^{2}+49y+74[=0])$
$(2x+11)(x-7)[=0]$
$\mathbf{oe}$
$\mathbf{or}\frac{[–]3+\sqrt{([-]3)^{2}-4\times 2\times -77}}{2\times 2}\mathbf{oe}$
$\mathbf{or}(6y+37)(3y+6)[=0]$
$\mathbf{or} \frac{-147\sqrt{147^{2}-4\times 18\times 222}}{2\times 18}\mathbf{oe}$
$x=7\mathbf{ and } y=-2$
$x=-5\frac{1}{2} \mathbf{oe}\; \mathbf{and} y=-6\frac{1}{6}\mathbf{oe}$
Detailed Solution

(a) (i)
$
x^2 – x – 12
$
Two numbers that multiply to -12 and add to -1
numbers are -4 and 3.
$
x^2 – x – 12 = (x – 4)(x + 3)
$

(a) (ii)
$
\frac{x^2 – 16}{x^2 – x – 12}
$
$
x^2 – 16 = (x – 4)(x + 4)
$
$
x^2 – x – 12 = (x – 4)(x + 3)
$
$
\frac{(x – 4)(x + 4)}{(x – 4)(x + 3)}
$
$
\frac{x + 4}{x + 3}
$

(b)
$
(2x – 3)^2 – (x + 1)^2
$
$
(2x – 3)(2x – 3) = 4x^2 – 12x + 9
$
$
(x + 1)(x + 1) = x^2 + 2x + 1
$
$
(4x^2 – 12x + 9) – (x^2 + 2x + 1)
$
$
4x^2 – 12x + 9 – x^2 – 2x – 1
$
$
(4x^2 – x^2) + (-12x – 2x) + (9 – 1)
$
$
3x^2 – 14x + 8
$

(c)
$
\frac{2x+4}{x+1} – \frac{x}{x-3}
$
$
\frac{(2x+4)(x-3)}{(x+1)(x-3)} – \frac{x(x+1)}{(x+1)(x-3)}
$
$
\frac{(2x^2 – 2x – 12) – (x^2 + x)}{(x+1)(x-3)}
$
$
\frac{(2x^2 – x^2) + (-2x – x) + (-12)}{(x+1)(x-3)}
$
$
\frac{x^2 – 3x – 12}{(x+1)(x-3)}
$

(d)
$
(x-3)(x-5)(2x+1)
$
$
(x – 3)(x – 5) = x^2 – 5x – 3x + 15 = x^2 – 8x + 15
$
$
(x^2 – 8x + 15)(2x + 1)
$
$
x^2(2x) + x^2(1) + (-8x)(2x) + (-8x)(1) + 15(2x) + 15(1)
$
$
2x^3 + x^2 – 16x^2 – 8x + 30x + 15
$
$
2x^3 – 15x^2 + 22x + 15
$

(e)

$
x – 3y = 13
$
$
2x^2 – 9y = 116
$
$
x = 3y + 13
$
$
2(3y + 13)^2 – 9y = 116
$
$
(3y + 13)(3y + 13) = 9y^2 + 78y + 169
$
$
18y^2 + 156y + 338 – 9y = 116
$
$
18y^2 + 147y + 338 = 116
$
$
6y^2 + 49y + 74 = 0
$
$
y = \frac{-49 \pm \sqrt{49^2 – 4(6)(74)}}{2(6)}
$
$
y = \frac{-49 \pm \sqrt{625}}{12}
$
$
y = \frac{-49 \pm 25}{12}
$

$ y = \frac{-49 + 25}{12} = \frac{-24}{12} = -2 $
$ y = \frac{-49 – 25}{12} = \frac{-74}{12} = -\frac{37}{6} $

For $ y = -2 $:
$
x = 3(-2) + 13 = -6 + 13 = 7
$
For $ y = -\frac{37}{6} $:
$
x = 3\left(-\frac{37}{6}\right) + 13 = -\frac{111}{6} + \frac{78}{6} = -\frac{33}{6} = -\frac{11}{2}
$

$
(x, y) = (7, -2) \quad \text{or} \quad \left(-\frac{11}{2}, -\frac{37}{6}\right)
$

Question 6:

Q6:
(a) E6.5
(b) E6.5
(c) E6.6

The diagram shows triangle ABC with AB = 17.2cm.
Angle ABC = 54° and angle ACB = 68°.
(a) Calculate AC.

AC = ……………………………………. cm

(b) M lies on BC and MC = 12.8 cm.
Calculate AM.

AM = ……………………………………. cm

………………………………………cm

▶️Answer/Explanation

Ans:

6(a) 15[.0] or 15.00 to 15.01

6(b) 15.7 or 15.65 to 15.66

6(c) 13.9 or 13.90 to 13.92
Detailed Solution

(a)
Sine Rule
$
\frac{AC}{\sin 54^\circ} = \frac{AB}{\sin 68^\circ}
$
$
AC = \frac{AB \times \sin 54^\circ}{\sin 68^\circ}
$
$
AC = \frac{17.2 \times \sin 54^\circ}{\sin 68^\circ}
$
$
AC = \frac{17.2 \times 0.809}{0.927}
$
$
AC \approx \frac{13.915}{0.927}
$
$
AC \approx 15.01 \text{ cm}
$

(b)
Using the Cosine Rule:
$
AM^2 = AC^2 + MC^2 – 2(AC)(MC) \cos \angle AMC
$
$
AM^2 = (15.01)^2 + (12.8)^2 – 2(15.01)(12.8) \cos 68^\circ
$
$
AM^2 = 225.3 + 163.84 – 2(15.01)(12.8)(0.374)
$
$
AM^2 = 225.3 + 163.84 – 143.7
$
$
AM^2 =245.42
$
$
AM \approx 15.6 \text{ cm}
$

(c)

The shortest distance is the perpendicular height $ h $ from $ A $ to $ BC $
$
h = AB \times \sin \angle ABC
$
$
h = 17.2 \times \sin 54^\circ
$
$
h = 17.2 \times 0.809
$
$
h \approx 13.91 \text{ cm}
$

Question 7:

Q7:
(a)
(i) E7.2
(ii) E7.2
(iii) E7.2
(b) E7.4

7 (a) $p=\binom{8}{-5}q=\binom{-4}{5}$
(i) Find 3q.

(ii) (a) Find p- q .

(ii)(b) Find |p – q |.

(b)

In triangle OMN, O is the origin, $\overrightarrow{OM}=a \overrightarrow{ON}$ .
S is a point on MN such that $MS : SN = 5 :3$ .
Find, in terms of a and/or b, the position vector of S.
Give your answer in its simplest form.

▶️Answer/Explanation

Ans :

7(a)(i) $\binom{-12}{15}$

7(a)(ii)(a) $\binom{12}{-10}$

7(a)(ii)(b) 15.6 or 15.62…

7(b)$ \frac{3}{8}a+\frac{5}{8}b$ final answer

Detailed Solution

(a)(i)
$
p = \begin{pmatrix} 8 \\ -5 \end{pmatrix}, \quad q = \begin{pmatrix} -4 \\ 5 \end{pmatrix}
$
$
3q = 3 \times \begin{pmatrix} -4 \\ 5 \end{pmatrix}
= \begin{pmatrix} -12 \\ 15 \end{pmatrix}
$

(ii) (a)
$
p – q = \begin{pmatrix} 8 \\ -5 \end{pmatrix} – \begin{pmatrix} -4 \\ 5 \end{pmatrix}
$
$
= \begin{pmatrix} 8 – (-4) \\ -5 – 5 \end{pmatrix}
= \begin{pmatrix} 8 + 4 \\ -10 \end{pmatrix}
= \begin{pmatrix} 12 \\ -10 \end{pmatrix}
$

(ii) (b)

The magnitude of a vector $ v = \begin{pmatrix} x \\ y \end{pmatrix} $
$
|v| = \sqrt{x^2 + y^2}
$
$ p – q = \begin{pmatrix} 12 \\ -10 \end{pmatrix} $:
$
|p – q| = \sqrt{12^2 + (-10)^2}
$
$
= \sqrt{144 + 100}
$
$
= \sqrt{244}
$
$
\sqrt{244} \approx 15.62
$

(b)
$
\overrightarrow{OM} = \mathbf{a}, \quad \overrightarrow{ON} = \mathbf{b}
$
Point $ S $ divides $ MN $ in the ratio $ MS:SN = 5:3 $.

$
\overrightarrow{OS} = \frac{n \overrightarrow{OM} + m \overrightarrow{ON}}{m+n}
$
$
\overrightarrow{OS} = \frac{3\mathbf{a} + 5\mathbf{b}}{5+3}
$
$
\overrightarrow{OS} = \frac{3\mathbf{a} + 5\mathbf{b}}{8}
$

Question 8:

Q8:
(a) E2.11
(b) E2.11
(c)
(i) E2.12
(ii) E2.12

8 (a) On the axes, sketch the graph of $y = 4 -3x$ .

(b) On the axes, sketch the graph of $y=-x ^{2}$

(c) (i) Find the coordinates of the turning points of the graph of y=10+9x $^{2}$-2x $^{3}$
You must show all your working.

( ………….. , ………….. ) and ( ………….. , ………….. )

(ii) Determine whether each turning point is a maximum or a minimum.
Show how you decide.

▶️Answer/Explanation

Ans :

8(a) Ruled line with negative gradient and positive y-intercept

8(b) Negative quadratic, with vertex at origin

8(c)(i) 18x – 6x$^{2}$ isw

setting their derivative = 0 or $\frac{dy}{dx}=0$

(0, 10) and (3.37)

8(c)(ii) (0, 10) minimum with correct reason

AND
(3, 37) maximum with correct reason
Detailed Solution
(a)

(b)

(c)(i)
$
y = 10 + 9x^2 – 2x^3
$
$
\frac{dy}{dx} = \frac{d}{dx} (10 + 9x^2 – 2x^3)
$
$
\frac{dy}{dx} = 18x – 6x^2
$

$ \frac{dy}{dx} = 0 $

$
18x – 6x^2 = 0
$
$
6x (3 – x) = 0
$
$
x = 0 \quad \text{or} \quad x = 3
$

For $ x = 0 $
$
y = 10 + 9(0)^2 – 2(0)^3 = 10
$

For $ x = 3 $
$
y = 10 + 9(3)^2 – 2(3)^3
$
$
= 10 + 9(9) – 2(27)
$
$
= 10 + 81 – 54
$
$
= 37
$

The turning points are
$
(0,10) \quad \text{and} \quad (3,37)
$

(c)(ii)

Differentiate $ \frac{dy}{dx} = 18x – 6x^2 $
$
\frac{d^2y}{dx^2} = 18 – 12x
$

For $ x = 0 $:
$
\frac{d^2y}{dx^2} = 18 – 12(0) = 18
$

Since $ 18 > 0 $, this means the graph is concave up at $ x = 0 $, so $ (0,10) $ is a minimum.

For $ x = 3 $:
$
\frac{d^2y}{dx^2} = 18 – 12(3) = 18 – 36 = -18
$
Since $ -18 < 0 $, this means the graph is concave down at $ x = 3 $, so $ (3,37) $ is a maximum.

Question 9:

Q9:
(a)
(i) E1.17
(ii) E1.17
(b) E1.17

9 (a) Janna and Kamal each invest $8000$.
At the end of 12 years, they each have $12800$.
(i) Janna invests in an account that pays simple interest at a rate of r% per year.
Calculate the value of r.

r = …………………………………………

(ii) Kamal invests in an account that pays compound interest at a rate of R% per year.
Calculate the value of R.

R = …………………………………………

(b) The population of a city is growing exponentially at a rate of 1.8% per year.
The population now is 260000.
Find the number of complete years from now when the population will first be more than 300000.

▶️Answer/Explanation

Ans:

9(a)(i) 5

9(a)(ii) 4[.0] or 3.99…

9(b) 9 nfww

Detailed Solution

(a) (i)
The formula for simple interest
$
A = P + P \times r \times t
$
$ A = 12800 $ (final amount)
$ P = 8000 $ (initial investment)
$ r $ is the interest rate (as a decimal)
$ t = 12 $ years

$
12800 = 8000 + (8000 \times r \times 12)
$
$
12800 – 8000 = 8000 \times r \times 12
$
$
4800 = 96000r
$
$
r = \frac{4800}{96000} = 0.05
$
$
r = 5\%
$

(a)(ii)
Compound interest formula
$
A = P (1 + R)^t
$
$ A = 12800 $
$ P = 8000 $
$ R $ is the annual interest rate (as a decimal)
$ t = 12 $

$
12800 = 8000 (1 + R)^{12}
$
$
\frac{12800}{8000} = (1 + R)^{12}
$
$
1.6 = (1 + R)^{12}
$
$
1 + R = 1.6^{\frac{1}{12}}
$
$
1 + R \approx 1.0411
$
$
R \approx 0.0411
$
$
R \approx 4.11\%
$

(b)

The formula for exponential growth
$
P = P_0 (1 + r)^t
$

$ P = 300000 $ (future population)
$ P_0 = 260000 $ (current population)
$ r = 1.8\% = 0.018 $
$ t $ is the number of years
$
300000 = 260000 (1.018)^t
$
$
\frac{300000}{260000} = (1.018)^t
$
$
1.1538 = (1.018)^t
$

Taking natural logarithm on both sides

$
\ln(1.1538) = t \ln(1.018)
$
$
0.1429 = t \times 0.0178
$
$
t = \frac{0.1429}{0.0178}
$
$
t \approx 8.03
$

we need the first complete year when the population exceeds 300,000, the answer is:
$
\mathbf{9 \text{ years}}
$

Question 10:

Q10:
(a) E2.10
(b) E2.10
(c) E2.5
(d) E2.5

The table shows some values for $y =2x^{3}+ 6x^{2}-2.5$

(a) Complete the table.
(b) On the grid, draw the graph of $y =2x^{3}+ 6x^{2}-2.5$= for $-3\leqslant x \leqslant 1$

x = ……………….. or x = ……………….. or x = ………………..

(d) The equation 2x$^{3}$+6x$^{2}$-2.5 =k has exactly two solutions.
Write down the two possible values of k.

k = ………………………… or k = …………………………

▶️Answer/Explanation

Ans :

10(a) −2.5 −1.25 5.5

10(b) Correct graph

10(c) y = 2 drawn

−2.75 to –2.65

–1.1 to −1.05
0.75 to 0.85

10(d) –2.5 5.5

Detailed Solution

(a)
$
y = 2x^3 + 6x^2 – 2.5
$

$ x = -3 $
$
y = 2(-3)^3 + 6(-3)^2 – 2.5
$
$
= 2(-27) + 6(9) – 2.5
$
$
= -54 + 54 – 2.5
$
$
= -2.5
$

$ x = -0.5 $
$
y = 2(-0.5)^3 + 6(-0.5)^2 – 2.5
$
$
= 2(-0.125) + 6(0.25) – 2.5
$
$
= -0.25 + 1.5 – 2.5
$
$
= -0.25 + 1.5 – 2.5 = -1.25
$

$ x = 1 $
$
y = 2(1)^3 + 6(1)^2 – 2.5
$
$
= 2(1) + 6(1) – 2.5
$
$
= 2 + 6 – 2.5
$
$
= 8 – 2.5
$
$
= 5.5
$

(b)

(c)

$2x^{3}+6x^{2}=4.5$

$2x^{3}+6x^{2}-2.5=2$

(d)

at k = 2.5 , 5.5 we can see graph touch at two point means two solution

Question 11:

Q11:
(a)
(i) E2.13
(ii) E2.13
(b) E2.13
(c)
(i) E2.13
(ii) E2.13

$f(x)=\frac{1}{x^{,}}x\neq0$
g(x)=3x-5

h(x)=2$^{x}$

(a) Find.
(i) gf(2)

(ii) g$^{-1}$(x)

g$^{-1}$(x) = – …………………………..

(b) Find in its simplest form g(x-2)

(ii) h(x) – g(7)= 0

▶️Answer/Explanation

Ans :

11(a)(i) −3.5 oe

11(a)(ii) $\frac{x+5}{3}$ oe final answer

11(b) 3 x-11 final answer

11(c)(i) 5

11(c)(ii) 4 nfww
Detailed Solution

(a) (i)
$
f(x) = \frac{1}{x}, \quad g(x) = 3x – 5
$
$
f(2) = \frac{1}{2}
$
$
g\left( \frac{1}{2} \right) = 3 \times \frac{1}{2} – 5
$
$
= \frac{3}{2} – 5 = \frac{3}{2} – \frac{10}{2} = \frac{-7}{2}
$

(a) (ii)
$
g(x) = 3x – 5
$
$ y = g(x) $:
$
y = 3x – 5
$
$
y + 5 = 3x
$
$
x = \frac{y + 5}{3}
$
the inverse function is
$
g^{-1}(x) = \frac{x + 5}{3}
$

(b)
$ g(x) = 3x – 5$
$
g(x – 2) = 3(x – 2) – 5
$
$
= 3x – 6 – 5
$
$
= 3x – 11
$

(c) (i)
$
f(x) = \frac{1}{x}, \quad g(x) = 3x – 5
$
$
f(g(x)) = f(3x – 5) = \frac{1}{3x – 5}
$
$
\frac{1}{3x – 5} = 0.1
$
$
1 = 0.1(3x – 5)
$
$
1 = 0.3x – 0.5
$
$
1.5 = 0.3x
$
$
x = \frac{1.5}{0.3} = 5
$

(c) (ii)
$ h(x) – g(7) = 0 $
$g(x) = 3x – 5$
$
g(7) = 3(7) – 5 = 21 – 5 = 16
$
$
h(x) – 16 = 0
$
$ h(x) = 2^x $,
$
2^x = 16
$
$
x = 4
$

Question 12:

Q12:
(a) E5.3
(b) E5.4

12 (a)

The diagram shows a circle of radius 12cm, with a sector removed.
Calculate the perimeter of the remaining shaded shape.

…………………………………….. cm

(b) The diagram in part(a) shows the top of a cylindrical cake with a slice removed.
The volume of cake that remains is 3510 cm $^{3}$

Calculate the height of the cake.

…………………………………….. cm

▶️Answer/Explanation

Ans :

12(a) 88.9 or 88.92 to 88.93…

12(b) 9.01 or 9.009 to 9.010…

Detailed Solution

(a)
The circumference of a full circle
$
C = 2\pi r
$
$
C = 2\pi \times 12 = 24\pi \text{ cm}
$
$
L = \frac{\theta}{360} \times C
$
$
360^\circ – 50^\circ = 310^\circ
$
$
L_{\text{remaining}} = \frac{310}{360} \times 24\pi
$
$
L_{\text{remaining}} = \frac{310}{360} \times 75.40 = \frac{23370}{360} \approx 64.93 \text{ cm}
$
Two straight edges of 12 cm each.
$
\text{Total Perimeter} = 64.93 + 12 + 12
$
$
= 88.93 \text{ cm}
$

(b)

The formula for the volume of a cylinder is

$
V = \pi r^2 h
$

$ r = 12 $ cm
The remaining part is $ \frac{310}{360} $ of the full volume.

$
V_{\text{full}} = \pi (12)^2 h = 144\pi h
$
$
\frac{310}{360} \times 144\pi h = 3510
$
$
\frac{310}{360} \times 144\pi h = 3510
$
$
\frac{310}{360} \times 144 \times 3.1416 \times h = 3510
$
$
\frac{310}{360} \times 452.39 \times h = 3510
$
$
\frac{310}{360} \times 452.39 = 389.38
$
$
389.38 h = 3510
$
$
h = \frac{3510}{389.38} \approx 9.01
$

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