Question 1
Here are the first four terms of an arithmetic sequence:
1, 4, 7, 10
(a) Find an expression, in terms of \( n \), for the \( n \)th term of this sequence.
The \( n \)th term of a different arithmetic sequence is \( 5n + 17 \).
(b) Find the 12th term of this sequence.
▶️ Answer/Explanation
Solution :-
(a) Working not required, so the correct answer scores full marks (unless from obvious incorrect working): \( 3n – 2 \).
(b) 77
Question 2
450 students were asked how they travelled to school on Monday.
Each student walked, travelled by bus, travelled by car, or travelled by bicycle.
Each student used just one method of travel.
One of these students is chosen at random.
The table shows information about the probability of each method of travel.
Work out how many of the 450 students travelled by car.
▶️ Answer/Explanation
Solution :-
\( 1 – (0.20 + 0.26) \) or \( 0.54 \) or
\( x + 2x + 0.26 + 0.20 = 1 \) or
\( x + 2x = 0.54 \) or
\( \frac{0.54}{3} (= 0.18) \) or
\( \frac{2}{3} \times 0.54 (= 0.36) \) or
\( 0.54 \times 450 (= 243) \)
\( (2x) \, 0.18 \times 450 \) or \( 81 \) or
\( 0.36 \times 450 \)
Final Answer: 162 (Working not required, so correct answer scores full marks unless from obvious incorrect working).
\( (0.2 \times 450) + (0.26 \times 450) (= 207) \) or
\( 90 + 117 (= 207) \) or \( 0.46 \times 450 (= 207) \)
\( 450 – 207 (= 243) \)
\( \frac{1}{3} \times 243 \) or \( 81 \) or \( \frac{2}{3} \times 243 \)
Final Answer: 162 (Working not required, so correct answer scores full marks unless from obvious incorrect working).
Question 3
Find the highest common factor (HCF) of 72 and 108.
Show your working clearly.
▶️ Answer/Explanation
Solution :-
Factors of 72: \( 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 \)
Factors of 108: \( 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108 \)
Prime factorization of 72: \( 2^3 \times 3^2 \)
Prime factorization of 108: \( 2^2 \times 3^3 \)
HCF is found by taking the lowest powers of common factors:
\( 2^2 \times 3^2 = 4 \times 9 = 36 \)
or using a factor tree:
Final Answer: HCF = 36
Question 4
Ava records the number of kilometres she drives each month.
In April, Ava drove 943 kilometres.
This is 15% more than the number of kilometres she drove in March.
Work out the number of kilometres Ava drove in March.
▶️ Answer/Explanation
Solution :-
Since April’s distance is 15% more than March, we set up the equation:
\( x + 0.15x = 943 \) or \( 1.15x = 943 \)
Rearranging to solve for \( x \):
\( x = \frac{943}{1.15} \)
\( x = 820 \)
Alternatively, using percentage calculations:
\( 100\% + 15\% = 115\% \) or \( 115\% = 1.15 \)
\( \frac{943}{115} \times 100 = 820 \)
Final Answer: Ava drove 820 kilometres in March.
Question 5
In the diagram, \(ABCDE\) is a regular pentagon.
Angle \(AEF = 96^\circ\)
Work out the size of the obtuse angle \(FED\).
Show your working clearly.
▶️ Answer/Explanation
Solution :-
The sum of interior angles of a regular pentagon is calculated as:
\( (5-2) \times 180 = 540^\circ \)
The size of each interior angle in a regular pentagon:
\( \frac{540}{5} = 108^\circ \)
The exterior angle of a regular pentagon:
\( 180^\circ – 108^\circ = 72^\circ \)
Using the given information, we calculate angle \(FED\):
\( 180^\circ – 96^\circ = 84^\circ \)
\( FED = 72^\circ + 84^\circ = 156^\circ \)
Final Answer: \( \mathbf{156^\circ} \)
Question 6
(a) Expand and simplify \( (m+5)(m-8) \).
(b) Solve \( 3n – 4 = \frac{5n+6}{3} \).
Show clear algebraic working.
▶️ Answer/Explanation
Solution :-
(a) Expanding the expression:
\[ (m+5)(m-8) = m^2 – 8m + 5m – 40 \]
\[ = m^2 – 3m – 40 \]
Final Answer: \( m^2 – 3m – 40 \)
(b) Solving for \( n \):
Given equation:
\[ 3n – 4 = \frac{5n+6}{3} \]
Multiply everything by 3 to eliminate the fraction:
\[ 9n – 12 = 5n + 6 \]
Rearrange terms:
\[ 9n – 5n = 12 + 6 \]
\[ 4n = 18 \]
Divide by 4:
\[ n = \frac{18}{4} = \frac{9}{2} \]
Final Answer: \( \mathbf{\frac{9}{2}} \)
Question 7
$\mathcal{E} = \{23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34\}$
$A = \{\text{even numbers}\}$
$B = \{23, 29, 31\}$
$C = \{\text{multiples of 3}\}$
(a) List the members of the set
(i) $B \cup C$
(ii) $A’ \cap C$
(b) Is it true that $B \cap C = \emptyset$?
Tick ($\checkmark$) one of the boxes below.
Yes $\square$ No $\square$
Give a reason for your answer.
The set $D$ has 4 members and is such that $D \cap (A \cup C) = \emptyset$
(c) List the members of set $D$
▶️Answer/Explanation
Solution :-
(a)(i) 23, 24, 27, 29, 30, 31, 33
(a)(ii) 27, 33
(b) eg Yes, there are no multiples of 3 in set $B$
1. Yes, no members/numbers/values in common
2. Yes, nothing in common
3. Yes, no common members/numbers/values
4. Yes, they share no common members/numbers/values
5. Yes, there is not the same members/numbers/values in both sets
6. Yes, there is no intersection or there is nothing in B and C
7. Yes, as there are no members/numbers/values the same (in B and C)
8. Yes, no members/numbers/values in B are in C or vice versa
9. Yes, there are no members/numbers in B that are multiples of 3
10. Yes, there are no members/numbers/values in that empty set
11. Yes, 23, 29, 31 not in C
12. Yes, 24, 27, 30, 33 are not in B
Allow sector for set
This is not an exhaustive list
Allow element(s) for members/numbers/values
(c) 23, 25, 29, 31
Question 8
A cylinder is placed on a table.
The volume of the cylinder is $1575 \text{ cm}^3$
The force exerted by the cylinder on the table is $84 \text{ newtons.}$
Work out the pressure on the table due to the cylinder.
▶️Answer/Explanation
Solution :-
$1575 = (area) \times 21$ oe
or
$(area = ) 75$
or
$1575 = \pi \times r^2 \times 21$ oe
or
$r^2 = \frac{1575}{21\pi} (= 23.8(732…))$ oe
or
$r = \sqrt{\frac{1575}{21\pi}} (= 4.88(602…))$ oe
$\frac{84}{“75”}$ oe or $\frac{84}{\pi “4.88”^2}$ oe or $\frac{84}{\pi “23.8”}$ oe
Working not required, so correct answer scores full marks (unless from obvious incorrect working) & 1.12
Question 9
The table gives the amount of rice produced by each of two countries in 2020
(a) Write $3.5 \times 10^7$ as an ordinary number.
In 2020, Japan produced 6780000 more tonnes of rice than Argentina.
(b) Work out the amount of rice Japan produced in 2020
Give your answer in standard form.
▶️Answer/Explanation
Solution :-
(a) 35 000 000
(b) $8.2 \times 10^5 + 6 780 000$ oe or $820 000 + 6 780 000$ oe
or
$7 600 000$ or $76 \times 10^5$ oe
or
$7.6 \times 10^n$ where $n \ne 6$
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $7.6 \times 10^6$
Question 10
(a) Simplify $(2p)^0$ where $p > 0$
$y^9 \times y^{-3} = y^n$
(b) Find the value of $n$
(c) Simplify fully $(5a^4c^2)^3$
▶️Answer/Explanation
Solution :-
(a) 1
(b) 6
(c) $125a^2c^6$
Question 11
The diagram shows a roof support.
The roof support is made from four lengths of wood, $AB$, $AC$, $BC$ and $MC$
$AC = BC = 9$ m $AB = 12$ m
angle $AMC = 90^\circ$
Lewis is going to buy lengths of wood to make the roof support.
The wood costs 21.50 euros per metre.
Each length of wood he buys has to be a whole number of metres.
Work out the total cost of the wood Lewis needs to buy.
Show your working clearly.
▶️Answer/Explanation
Solution :-
$(CM)^2 + (12 \div 2)^2 = 9^2$ oe or
$9^2 – (12 \div 2)^2 (= 81 – 36 = 45)$
$\sqrt{9^2 – (12 \div 2)^2}$ oe
$(=\sqrt{81-36} = \sqrt{45} = 3\sqrt{5} = 6.7(08…))$
$(“7” + 9 + 9 + 12) \times 21.5(0)$ or
$37 \times 21.5(0)$
Working required & 795.5(0)
Question 12
(a) Factorise fully $6y^2 – 5y – 4$
(b) Express $\frac{2x+1}{4x} + \frac{7-5x}{3x}$ as a single fraction in its simplest form.
▶️Answer/Explanation
Solution :-
(a) $(2y \pm 1)(3y \pm 4)$
or
$(2y \pm 4)(3y \pm 1)$
or
$2y(3y-4) + 1(3y-4)$
or
$3y(2y+1) – 4(2y+1)$
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $(2y+1)(3y-4)$
(b) $\frac{3(2x+1)}{12x} + \frac{4(7-5x)}{12x}$ or $\frac{3x(2x+1)}{12x^2} + \frac{4x(7-5x)}{12x^2}$ or
$\frac{3(2x+1)+4(7-5x)}{12x}$ oe or $\frac{3x(2x+1)+4x(7-5x)}{12x^2}$ oe
$\frac{6x+3+28-20x}{12x}$ oe or $\frac{6x^2+3x+28x-20x^2}{12x^2}$ oe or
$\frac{31x-14x^2}{12x^2}$ oe
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $\frac{31-14x}{12x}$
Question 13
Harman has two bags of beads.
In bag A, there are 3 white beads and 7 black beads.
In bag B, there are 5 white beads and 4 black beads.
Harman takes at random a bead from bag A and a bead from bag B.
(a) Complete the probability tree diagram.
(b) Work out the probability that Harman takes two beads of the same colour.
▶️Answer/Explanation
Solution :-
(a) $\frac{3}{10}, \frac{7}{10}$
$\frac{5}{9}, \frac{4}{9}$
$\frac{5}{9}, \frac{4}{9}$
(b) “$\frac{3}{10}” \times “\frac{5}{9}”$ oe or “$\frac{7}{10}” \times “\frac{4}{9}”$ oe or
“$\frac{3}{10}” \times “\frac{4}{9}”$ oe or “$\frac{7}{10}” \times “\frac{5}{9}”$ oe or
“$\frac{3}{10}” \times “\frac{5}{9}” + “\frac{7}{10}” \times “\frac{4}{9}”$ oe or
$1 – (“\frac{3}{10}” \times “\frac{4}{9}” + “\frac{7}{10}” \times “\frac{5}{9}”)$ oe
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $\frac{43}{90}$
Question 14
The combined savings of Abel and Bahira are 15435 dinars.
The savings of Bahira are 45% more than the savings of Abel.
The savings of Bahira are $\frac{3}{2}$ times the savings of Chanda.
Work out the savings of Chanda.
▶️Answer/Explanation
Solution :-
1 + 1.45 (= 2.45) or $1 + \frac{29}{20} (= \frac{49}{20})$ or $B = 1.45A$ oe or $B = \frac{29}{20}A$ oe or
$A + 1.45A$ or $A + \frac{29}{20}A$ or $2.45A$ or
$(A:B=) 100:145$ oe or $100 + 145 (= 245)$ oe or
$(B:C=) 3:2$ oe or $B = 1.5C$ oe
$A + 1.45A = 15435$ or & M2 for
$15435 \div “2.45”$ or $15435 \div \frac{49}{20}$ or & $15435 \div (\frac{1}{1.45} + 1) (= 9135)$ oe or
$15435 \div “245” \times 100$ or $63 \times 100 (= 6300)$ & $15435 \div (\frac{49}{29}) (= 9135)$ oe
$15435 – “6300”$ or &
$1.45 \times “6300”$ or &
$145 \times “63” (= 9135)$ &
$”9135″ \div \frac{3}{2}$ oe or $”9135″ \times \frac{2}{3}$ oe &
Working not required, so correct answer scores full marks (unless from obvious incorrect working) & 6090
Question 15
The function f is defined as
$f:x \mapsto \frac{3x+1}{x-2}$
(a) State the value of $x$ that cannot be included in any domain of the function f
(b) Express the inverse function $f^{-1}$ in the form $f^{-1}(x) = …$
▶️Answer/Explanation
Solution :-
(a) $(x=) 2$
(b) $y(x-2) = 3x+1$ oe or $x(y-2) = 3y+1$ oe or
$yx – 2y = 3x+1$ oe $yx – 2x = 3y+1$ oe
$x(y-3) = 1+2y$ oe $y(x-3) = 1+2x$ oe
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $\frac{1+2x}{x-3}$
Question 16
There are 20 sweets in a box.
15 of the sweets are red
5 of the sweets are yellow
Fred takes at random 3 sweets from the box.
Work out the probability that Fred takes at least one sweet of each colour from the box.
▶️Answer/Explanation
Solution :-
$\frac{15}{20}\times\frac{14}{19}\times\frac{5}{18}(=\frac{35}{228})$ oe or $\frac{5}{20}\times\frac{4}{19}\times\frac{15}{18}(=\frac{5}{114})$ oe
or
$\frac{15}{20}\times\frac{14}{19}\times\frac{13}{18}(=\frac{91}{228})$ oe or $\frac{5}{20}\times\frac{4}{19}\times\frac{3}{18}(=\frac{1}{114})$ oe
$3\times\frac{15}{20}\times\frac{14}{19}\times\frac{5}{18}$ oe or $3\times\frac{5}{20}\times\frac{4}{19}\times\frac{15}{18}$ oe
or
$\frac{15}{20}\times\frac{14}{19}\times\frac{13}{18}$ oe and $\frac{5}{20}\times\frac{4}{19}\times\frac{3}{18}$ oe
$3\times\frac{15}{20}\times\frac{14}{19}\times\frac{5}{18}+3\times\frac{5}{20}\times\frac{4}{19}\times\frac{15}{18}$ oe
or
$(\frac{15}{20}\times\frac{14}{19}\times\frac{5}{18})+(\frac{5}{20}\times\frac{4}{19}\times\frac{15}{18})+(\frac{15}{20}\times\frac{5}{19})+(\frac{5}{20}\times\frac{15}{19})$ oe
or
$1-(\frac{15}{20}\times\frac{14}{19}\times\frac{13}{18}+\frac{5}{20}\times\frac{4}{19}\times\frac{3}{18})$ oe
Working not required, so correct answer scores full marks (unless from obvious incorrect working) & $\frac{45}{76}$
Question 17
Show that $\frac{1+\sqrt{5}}{3-\sqrt{5}}$ can be written in the form $a + \sqrt{b}$ where $a$ and $b$ are integers.
Show each stage of your working clearly.
▶️Answer/Explanation
Solution :-
$\frac{1+\sqrt{5}}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}$ oe or
$\frac{1+\sqrt{5}}{3-\sqrt{5}}\times\frac{-3-\sqrt{5}}{-3-\sqrt{5}}$ oe
$\frac{3+\sqrt{5}+3\sqrt{5}+\sqrt{5}\sqrt{5}}{9+3\sqrt{5}-3\sqrt{5}-\sqrt{5}\sqrt{5}}$ oe or
$\frac{3+\sqrt{5}+3\sqrt{5}+\sqrt{5}\sqrt{5}}{9-\sqrt{5}\sqrt{5}}$ oe or
$\frac{3+\sqrt{5}+3\sqrt{5}+5}{9+3\sqrt{5}-3\sqrt{5}-5}$ oe or
$\frac{8+\sqrt{5}+3\sqrt{5}}{9-5}$ oe or
$\frac{3+4\sqrt{5}+5}{9-5}$ oe or
$\frac{8+4\sqrt{5}}{4}$
Working required & $2+\sqrt{5}$
Question 18
A curve $C$ has equation $y = x^3 – 40x + 1$
Find the coordinates of both the points on $C$ at which the gradient is 8
▶️Answer/Explanation
Solution :-
$3x^2$ or $-40$
$3x^2 – 40$
“$3x^2 – 40” = 8$
$(y=)”4″^3 – 40 \times “4” + 1 (=-95)$ or
$y = (“-4”)^3 – 40 \times “-4” + 1 (=97)$
Working not required, so correct answer scores full marks (unless from obvious incorrect working) & (4, -95), (-4, 97)
Question 19
Here is quadrilateral \( ABCD \).
Work out the value of \( x \).
Give your answer correct to 3 significant figures.
▶️ Answer/Explanation
Solution :-
Step 1: Use the sine rule to find \( BD \)
Applying the sine rule:
\[ \frac{BD}{\sin 62^\circ} = \frac{12.8}{\sin 40^\circ} \]
Rearranging for \( BD \):
\[ BD = \frac{12.8}{\sin 40^\circ} \times \sin 62^\circ \]
\[ BD = 17.582 \]
Step 2: Use the cosine rule to find \( x \)
Using the cosine rule:
\[ BD^2 = 13.4^2 + 15.2^2 – 2 \times 13.4 \times 15.2 \times \cos x \]
Substituting values:
\[ 17.582^2 = 13.4^2 + 15.2^2 – 2 \times 13.4 \times 15.2 \times \cos x \]
\[ 309.139 = 179.56 + 231.04 – 407.36 \cos x \]
Step 3: Solve for \( \cos x \)
\[ \cos x = \frac{179.56 + 231.04 – 309.139}{407.36} \]
\[ \cos x = \frac{410.6 – 309.139}{407.36} \]
\[ \cos x = 0.247 \]
Step 4: Find \( x \)
\[ x = \cos^{-1}(0.247) \]
\[ x = 75.6^\circ \]
Final Answer: \( 75.6^\circ \) (to 3 significant figures).
Question 20
The diagram shows a sector \( OABC \) of a circle with center \( O \).
\(\angle AOC = 60^\circ\)
The area of the shaded segment \( ABC \) is \( 38 \text{ cm}^2 \).
Work out the perimeter of the shaded segment \( ABC \).
Give your answer correct to one decimal place.
▶️ Answer/Explanation
Solution :-
Step 1: Express the area of the segment
The area of the segment is given by:
\[ \text{Area of sector} – \text{Area of triangle} = 38 \]
Using formulas:
\[ \frac{\pi r^2}{6} – \frac{1}{2} r^2 \sin 60^\circ = 38 \]
Rewriting:
\[ \frac{\pi r^2}{6} – \frac{\sqrt{3}}{4} r^2 = 38 \]
Step 2: Solve for \( r \)
\[ r^2 = \frac{38}{\left(\frac{\pi}{6} – \frac{\sqrt{3}}{4} \right)} \]
\[ r^2 = \frac{38}{0.09058} = 419.49 \]
\[ r = \sqrt{419.49} = 20.48 \]
Step 3: Find the arc length \( BC \)
The formula for arc length is:
\[ \frac{\pi}{6} \times 2r \]
Substituting \( r = 20.48 \):
\[ \frac{\pi}{6} \times (2 \times 20.48) = 21.45 \]
Step 4: Find the perimeter
Perimeter of segment \( ABC \) = Arc length \( BC \) + 2 × chord length \( AB \)
Final answer: \( 41.9 \) cm (rounded to 1 decimal place).
Question 21
A curve has the equation \( y = f(x) \).
There is one minimum point on this curve.
The coordinates of this minimum point are \( (5, -4) \).
Write down the coordinates of the minimum point on the curve with equation:
(i) \( y = f(x+7) \)
(ii) \( y = f(x) – 6 \)
▶️ Answer/Explanation
Solution :-
Understanding Transformations:
(i) \( y = f(x+7) \)
– This transformation represents a **horizontal shift**.
– \( x+7 \) means shifting **left** by 7 units.
– The new coordinates: \( (5 – 7, -4) = (-2, -4) \).
(ii) \( y = f(x) – 6 \)
– This transformation represents a **vertical shift**.
– Subtracting 6 means shifting **down** by 6 units.
– The new coordinates: \( (5, -4 – 6) = (5, -10) \).
Final Answers:
(i) \( (-2, -4) \)
(ii) \( (5, -10) \)
Question 22
The incomplete histogram shows some information about the distances, in kilometres, that 100 adults ran last week.
The given conditions are:
- All of the adults ran at least 5 kilometres.
- None of the adults ran more than 55 kilometres.
- 14 adults ran between 15 kilometres and 20 kilometres.
Complete the histogram.
▶️ Answer/Explanation
Solution :-
Step 1: Understanding Frequency Density (FD)
The frequency density is calculated as:
\[ \text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} \]
For the class 15–20 km:
\[ \frac{14}{5} = 2.8 \]
Thus, the frequency density for this range is **2.8**.
Step 2: Identifying the Histogram Scale
- 10 small squares = 1 adult
- 1 large square = 2.5 adults
Step 3: Identifying Missing Frequencies
The missing frequencies are calculated using:
\[ 14 + (15 \times 3.4) + (20 \times 0.4) = 73 \]
or
\[ 100 – [14 + (15 \times 3.4) + (20 \times 0.4)] = 27 \]
which simplifies to:
\[ 14 + 51 + 8 = 73 \]
or
\[ 100 – [14 + 51 + 8] = 27 \]
Step 4: Assigning Correct Bar Heights
Using scaling calculations:
\[ (140 + 510 + 80) \times 0.1 = 73 \]
\[ [1000 – (140 + 510 + 80)] \times 0.1 = 27 \]
or
\[ (5.6 + 20.4 + 3.2) \times 2.5 = 73 \]
\[ [40 – (5.6 + 20.4 + 3.2)] \times 2.5 = 27 \]
Final Answer: The histogram is completed with a **bar height of 2.7 and the correct width**.
Question 23
A solid shape is made by removing a hemisphere (shaded) from a cone, as shown in the diagram.
The given dimensions are:
- Radius of the hemisphere: \( 2x \) cm
- Radius of the base of the cone: \( 5x \) cm
- Vertical height of the cone: \( 6x \) cm
The volume of the remaining solid is given as \( 6948\pi \) cm³.
Work out the total surface area of the solid hemisphere that has been removed from the cone.
Give your answer correct to the nearest integer.
▶️ Answer/Explanation
Solution :-
Step 1: Calculate the volume of the cone
The formula for the volume of a cone is:
\[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \]
Substituting \( r = 5x \) and \( h = 6x \):
\[ V_{\text{cone}} = \frac{1}{3} \pi (5x)^2 (6x) = \frac{1}{3} \pi (25x^2) (6x) \]
\[ V_{\text{cone}} = 50\pi x^3 \]
Step 2: Calculate the volume of the removed hemisphere
The volume of a sphere is:
\[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \]
For a hemisphere:
\[ V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi (2x)^3 \]
\[ V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi (8x^3) \]
\[ V_{\text{hemisphere}} = \frac{16}{3} \pi x^3 \]
Step 3: Solve for \( x \)
Using the given equation:
\[ 50\pi x^3 – \frac{16}{3} \pi x^3 = 6948\pi \]
\[ \frac{134}{3} \pi x^3 = 6948\pi \]
Cancel \( \pi \) and solve for \( x^3 \):
\[ x^3 = \frac{6948 \times 3}{134} \]
\[ x^3 = \frac{10422}{67} \approx 155.55 \]
\[ x \approx \sqrt[3]{155.55} \approx 5.38 \]
Step 4: Compute the surface area of the hemisphere
The total surface area of a hemisphere includes:
- The curved surface area: \( 2\pi r^2 \)
- The circular base area: \( \pi r^2 \)
\[ A_{\text{hemisphere}} = 3\pi (2x)^2 \]
\[ A_{\text{hemisphere}} = 3\pi (4x^2) \]
\[ A_{\text{hemisphere}} = 12\pi x^2 \]
Substituting \( x \approx 5.38 \):
\[ A_{\text{hemisphere}} = 12\pi (5.38)^2 \]
\[ A_{\text{hemisphere}} \approx 12\pi (28.96) \]
\[ A_{\text{hemisphere}} \approx 1090 \]
Final Answer: \( \mathbf{1090} \) cm² (nearest integer)
Question 24
A polygon has \( n \) sides, where \( n > 5 \).
The interior angles of the polygon form an arithmetic sequence.
The smallest angle of the polygon is \( 84^\circ \).
The common difference of the sequence is \( 4^\circ \).
Work out the sum of the interior angles of the polygon.
Show clear algebraic working.
▶️ Answer/Explanation
Solution :-
The sum of the interior angles of an \( n \)-sided polygon is given by:
\[ S = (n-2) \times 180 \]
The sum of an arithmetic sequence is given by:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
Substituting \( a = 84 \) and \( d = 4 \):
\[ \frac{n}{2} [2(84) + (n-1)(4)] \]
\[ \frac{n}{2} [168 + 4n – 4] = \frac{n}{2} [164 + 4n] \]
\[ 82n + 2n^2 = (n-2) \times 180 \]
Expanding:
\[ 82n + 2n^2 = 180n – 360 \]
Rearrange to form a quadratic equation:
\[ 2n^2 – 98n + 360 = 0 \]
Divide by 2:
\[ n^2 – 49n + 180 = 0 \]
Solving for \( n \) using factorization:
\[ (n – 45)(n – 4) = 0 \]
Since \( n > 5 \), we take \( n = 45 \).
Sum of interior angles:
\[ S = (45 – 2) \times 180 \]
\[ S = 43 \times 180 \]
\[ S = 7740 \]
Final Answer: \( \mathbf{7740^\circ} \)
Question 25
Given function:
\[ f(x) = 17 – 3x^2 + 12x \]
Write \( f(x) \) in the form \( a – b(x-c)^2 \), where \( a \), \( b \), and \( c \) are constants.
▶️ Answer/Explanation
Solution :-
Rearrange the quadratic expression:
\[ f(x) = -3x^2 + 12x + 17 \]
Factor out \(-3\) from the quadratic terms:
\[ f(x) = -3(x^2 – 4x) + 17 \]
Complete the square inside the brackets:
\[ x^2 – 4x = (x – 2)^2 – 4 \]
Substituting back:
\[ f(x) = -3[(x – 2)^2 – 4] + 17 \]
Distribute \(-3\):
\[ f(x) = -3(x – 2)^2 + 12 + 17 \]
\[ f(x) = 29 – 3(x – 2)^2 \]
Final Answer: \( \mathbf{29 – 3(x – 2)^2} \)