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Question
The points X, Y and Z are located such that Y is 15km south of X, Z is 20km from X on a bearing of 270º. Calculate, correct to:
(a) two significant figures, Â |YZ|
(b) the nearest degree, the bearing of Y from Z
(a) to two significant figures, Â |YZ|
Solution
Using Pythagoras Theorem
⇒ \(\scriptsize |YZ|^2 = 15^2 \: + \: 20^2 \\ \scriptsize = 225 \: + \: 400 \\ \scriptsize = |YZ|^2 = 625 \\\scriptsize = |YZ| = \sqrt{625} \\ \scriptsize|YZ| = 25 \: km \)
Answer ⇒ |YZ| = 25km
(b) to the nearest degree, the bearing of Y from Z
Solution
tan θ = \( \frac{15}{20} \)
θ = \( \scriptsize tan^{-1} \left( \frac{15}{20}\right) \)
θ = 26.8699º
Bearing of Y from Z
= 90° + 36.8699
= 126.8699
Answer⇒ \( \scriptsize \simeq 127^o\)
Question
NOT DRAWN TO SCALE
In the diagram, \( \scriptsize \bar{AD} \) is a diameter of a circle with centre O. If ABD is a triangle in a semi-circle and ∠OAB = 34º,
find: (a) ∠OBD (b) ∠OCB
(a) ∠OBD
Solution
⇒ \( \scriptsize \hat{OAB} = \hat{OBA} = 34^o \)
(base angles of an isosceles triangle)
⇒ \( \scriptsize \hat{ABD} = 90^o \)
(angles in a semi-circle)
∴ \( \scriptsize \hat{OBD} = \hat{ABD} \: – \: \hat{OBA} \)
= 90º – 34º
= 56°
Answer:  ∠OBD = 56°
(b) ∠OCB
Solution
⇒ \( \scriptsize \hat{BOC} \: + \: \hat{OBC} \: + \: \hat{OCB} = 180^o \)
(sum of angles in a triangle)
⇒ \( \scriptsize \hat{BOD} = \hat{OAB} \: + \: \hat{OBA} \)
(sum of two opp. interior = exterior)
⇒ \( \scriptsize \hat{BOD} = 34 + 34 \)
⇒ \( \scriptsize \hat{BOD} = 68^o \)
Consider ΔOCB
⇒ \( \scriptsize \hat{OBC} = 90^o \)
(angles in a semi-circle)
∴ \( \scriptsize 68^o \: + \: 90^o \: + \: \hat{OCB} = 180^o \)
⇒ \( \scriptsize \hat{OCB} = 180^o \: -\: 158^o \)
⇒ \( \scriptsize \hat{OCB} = 22^o \)
Answer ⇒  \( \scriptsize \hat{OCB} = 22^o \)
Question
(a) A man shared his property among his children as follows:
Represent the information on a pie chart
Solution
| Child’s Name | Percentage | Angle (Deg) |
| Ann | 5 | \( \frac{5}{100} \scriptsize \: \times \: 360^o = 18^o\) |
| Afia | 15 | \( \frac{15}{100} \scriptsize \: \times \: 360^o = 54^o\) |
| Kojo | 10 | \( \frac{10}{100} \scriptsize \: \times \: 360^o = 36^o\) |
| Nuno | 45 | \( \frac{45}{100} \scriptsize \: \times \: 360^o = 162^o\) |
| Akosua | 25 | \( \frac{25}{100} \scriptsize \: \times \: 360^o = 90^o\) |
| Total | 100 | \( \scriptsize 360^o \) |
Answer
(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability that a girl takes away two red beads, one after the other, from the box.
Solution
5 Red, 3 Green, 4 Blue
Total beads = 5 + 3 + 4 = 12
Pro. (taking the 1st red) = \( \frac{5}{12} \)
Pro. (taking the 2nd red) = \( \frac{4}{11} \)
Pro. (two red beads) = \(\frac{5}{12} \: \times \: \frac{4}{11} \)
Answer = \(\frac{5}{33} \)
Question
(a) In a class of 80 students, \( \frac{3}{4} \) study Biology and \( \frac{3}{5} \) study Physics. if each student studies at least one of the subjects.
(i) draw a venn diagram to represent this information
Solution
U = 80
n(B) = \(\frac{3}{4}\)
n(P) = \(\frac{3}{5}\)
Answer
Biology = \( \frac{3}{4} \scriptsize \: \times \: 80 = 60 \)
Physics = \( \frac{3}{5} \scriptsize \: \times \: 80 = 48 \)
(ii) how many students study both subjects;
Solution
80 = 60 – x + x + 48 – x
80 = 108 – x
x = 108 – 80
x = 28
Answer = 28
(iii) find the fraction of the class that study Biology but not Physics.
Solution
fraction = \( \frac{60 \:-\:28}{80}\)
= \( \frac{32}{80}\)
Answer = \( \frac{2}{5}\)
Question
(a) Copy and complete the table of values for the relation:
⇒ \( \scriptsize y = 2x^2 \: -\: x \: -\: 2 \\ \scriptsize\: for \: -4 \leq x \leq 4 \)
Solution
(b) Using a scale of 2cm to 1 unit on the x-axis, and 2cm to 5 units on the y – axis, draw the graph of
⇒ \( \scriptsize y = 2x^2 \: -\: x \: -\: 2 \\ \scriptsize\: for \: -4 \leq x \leq 4 \)
Solution
(c) On the same axes, draw the graph of  y = 2x + 3
Solution
(d) Use the graph to find the:
(i) roots of the equation \( \scriptsize 2x^2 \: – \: 3x \: – \: 5 = 0 \)
Solution
Answer⇒ x = -1 and x = 2.5
(ii) range of values for which \( \scriptsize 2x^2 \: – \: x \: – \: 2 < 0 \)
range of values of x
Answer⇒ \( \scriptsize -0.8 < x < 1.2 \)
Question
In ΔPQR, ∠PQR = 90º. If its area is 216cm² and |PQ|:|QR| is 3:4
Find |PR|
Solution
A = \( \frac{1}{2} \scriptsize |PQ||QR| sin \theta \)
216 = \( \frac{1}{2} \scriptsize |PQ||QR| sin 90 \)
⇒ \( \frac{|PQ|}{|QR|} = \frac{3}{4}\)
⇒ \( \scriptsize |PQ| = \normalsize \frac{3}{4} \scriptsize |QR| \)
216 = \( \frac{1}{2} \: \times \: \frac{3}{4} \scriptsize |QR| |QR| \: \times \: 1 \)
216 = \( \frac{3}{8}Â \scriptsize |QR|^2Â \)
⇒ \( \scriptsize |QR|^2 = \normalsize \frac{216 \: \times \: 8}{3} \)
⇒ \( \scriptsize |QR| = \sqrt{576} \)
⇒ \( \scriptsize |QR| = 24 \: cm\)
⇒ \( \scriptsize |PQ| = \normalsize \frac{3}{4} \: \scriptsize |QR| \)
⇒ \( \scriptsize |PQ| = \normalsize \frac{3}{4} \scriptsize \: \times \: 24 = 18cm \)
⇒ \( \scriptsize |PR|^2 = 18^2 \: + \: 24^2 \)
⇒ \( \scriptsize |PR| = \sqrt{18^2 \: + \: 24^2}\)
⇒ \( \scriptsize |PR| = \sqrt{900}\)
⇒ \( \scriptsize |PR| = 30 \: cm\)
Answer⇒ \( \scriptsize |PR| = 30 \: cm\)
Question
NOT DRAWN TO SCALE
In the diagram, PQRS is a trapezium with \( \scriptsize \bar{QR} || \bar{PS} \)
U and T are points on \( \scriptsize \bar{PS} \) such that \( \scriptsize \bar{PU}Â = 5cm, \:\bar{QU} = 12cm\)
and ∠PUQ = ∠STR =90°. If the area of ΔUQR = 20 cm2, calculate to the nearest whole number, the:
(a) Perimeter of the trapezium.
Solution
Area of ΔUQR = \( \frac{1}{2} \scriptsize \: \times \: b \: \times \: h \)
20 = \( \frac{1}{2} \scriptsize \: \times \: b \: \times \: 12 \)
40 = 12b
b = \( \frac{40}{12} \scriptsize = 3.33\:cm \)
b = |QR| = |UT| = 3.33cm
⇒ \( \scriptsize |PQ|^2 = 12^2 \: + \: 5^2 \)
⇒ \( \scriptsize |PQ|^2 = 144 \: + \: 25 \)
⇒ \( \scriptsize |PQ|^2 = 169 \)
⇒ \( \scriptsize |PQ| = \sqrt{169} \)
⇒ \( \scriptsize |PQ| = 13\:cm \)
⇒ \( \scriptsize |QU| = |RT| = 12\:cm \)
ΔRTS
sin 50º = \( \frac{12}{RS} \)
|RS| = \( \frac{12}{sin\:50^o} \)
|RS| = 15.6649cm
tan 50º = \( \frac{12}{TS} \)
|TS| = \( \frac{12}{tan\:50^o} \)
|TS| = 10.069cm
Perimeter = addition of all boundaries
p = |PQ| + |QR| + |RS| + |TS| + |UT| + |PU
P = 13 + 3.33 + 15.6649 + 10.069 + 3.33 + 5
p = 50.3939
p ≅ 50 cm ( nearest whole number)
Answer⇒ Perimeter ≅ 50 cm
(b) Area of the trapezium.
Solution
Area of trapezium = \( \frac{1}{2} \scriptsize h(a \: + \: b)\)
A = \( \frac{1}{2} \scriptsize \: \times \:Â 12(QR \: + \: PS)\)
|PS| = 5 + 3.33 + 10.069 = 18.399
A = \( \frac{1}{2} \scriptsize \: \times \:Â 12(3.33 \: + \: 18.399)\)
A = 6(21.729)
= 130.374cm2
A ≅ 130cm2 (nearest whole number)
Answer⇒ Area ≅ 130cm2
Question
(a) A cottage is on a bearing of 200° and 110° from Dogbe’s and Manu’s farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find:
(i) correct to two significant figures, the distance between the two farms;
Solution
Let Dogbe = D, Cottage = C and Manu = M
Note: The angles in purple and red are alternate angles
|DM| = the distance between the two farms
From the sketch, ∠DCM = 90º
That means we can use Pythagoras’ theorem to find |DM|
|DM| = \( \scriptsize \sqrt{3^2 + 5^2} \)
i.e. \( \scriptsize |DM|^2 = |CD|^2 \: + \: |CM|^2 \)
|DM| = \( \scriptsize \sqrt{3^2 \: + \: 5^2} \\ = \scriptsize \sqrt{9 \: + \: 25} \\ = \scriptsize \sqrt{34}\\ = \scriptsize 5.831 \: km \\ \scriptsize \approx 5.8 \: km\)
Answer⇒  Distance between two farms ≅ 5.8km
(ii) correct to the nearest degree, the bearing of Manu’s farm from Dogbe’s.
tan θ = \( \frac{opp}{adj} \\ = \frac{3}{5} \\ = \scriptsize 0.6\)
θ = \( \scriptsize tan^{-1} 0.6 \)
θ = 30.9638º
Bearing = 30.9638 + 200
= 230.9638
Answer ⇒ the bearing of Manu’s farm from Dogbe’s ≅ 231º (nearest degree)
(b) A ladder 10m long leaned against a vertical wall high. The distance between the wall and the foot of the ladder is 2m longer than the height of the wall. Calculate the value of x
Solution
⇒ \( \scriptsize 10^2 = (x \: + \: 2)^2 \: + \: x^2 \)
⇒ \( \scriptsize 100 = (x \: + \: 2)(x \:+\:2) \: + \: x^2 \)
⇒ \( \scriptsize 100 = x^2 \: + \: 2x \: + \: 2x \: + \: 4\: + \: x^2 \)
⇒ \( \scriptsize 100 = 2x^2 \: + \: 4x \: + \: 4 \)
⇒ \( \scriptsize 2x^2 \: + \: 4x \: + \: 4 \: -\: 100 = 0\)
⇒ \( \scriptsize 2x^2 \: + \: 4x \: – \: 96= 0\)
divide through by 2
⇒ \( \scriptsize x^2 \: + \: 2x \: – \: 48 = 0\)
factorise
⇒ \( \scriptsize x^2 \: + \: 8x \: – \: 6x \: – \: 48 = 0\)
⇒ \( \scriptsize x(x \: + \: 8) \: – \: 6(x \: + \: 8) = 0\)
⇒ \( \scriptsize (x \: – \: 6)(x \: + \: 8) \)
x – 6 = 0 or x + 8 = 0
x = 6 or -8
Length can’t be negative
∴ x = 6m
Answer⇒ x = 6m
Question
The table shows the distribution of the number of hours per day spent in studying by 50 students 
Calculate, correct to two decimal places, the:
(a) Mean;
Solution
| X | F | fx | Fx2 |
| 4 | 5 | 20 | 80 |
| 5 | 7 | 35 | 175 |
| 6 | 5 | 30 | 180 |
| 7 | 9 | 63 | 441 |
| 8 | 12 | 96 | 768 |
| 9 | 4 | 36 | 324 |
| 10 | 3 | 30 | 300 |
| 11 | 5 | 55 | 605 |
|  | ΣF=50 | ΣFx=365 | ΣFx2=2873 |
⇒ \(\scriptsize mean \: \bar{X} = \normalsize \frac{\sum Fx}{\sum F} \)
=\( \frac{365}{50} \)
⇒ \(\scriptsize mean \: \bar{X} = 7.30\: (2d.p) \)
Answer ⇒ mean – 7.30 (2d.p) Â
(b) Standard deviation
Solution
S.D = \( \sqrt{\frac{\sum Fx^2}{\sum F} \: -\: \left( \frac{\sum Fx}{\sum F}\right)^2} \)
S.D = \( \sqrt{\frac{2873}{50} \: -\: \left( \scriptsize 7.3 \right )^2} \)
= \( \scriptsize \sqrt{57.46 \: -\: 53.29} \)
= \( \scriptsize \sqrt{4.17} \)
= 2.0421
≅ 2.04 (2 d.p)
Answer ⇒ ≅ 2.04 (2 d.p)
Question
(a)
In the diagram, PQRS is a circle. |PQ| = |QS|, ∠SPR = 26º and the interior of angles of ΔPQS are in the ratio 2:3:3.
Calculate:
(i) ∠PQR
(ii) ∠RPQ
(iii) ∠PRQ
Solution
(a. i) ∠PQR
∠PSQ = ∠QPS
(base angles of isosceles)
Total ratio = 2 + 3 + 3
= 8
⇒ \( \scriptsize \angle PQS = \normalsize \frac{2}{8} \scriptsize \: \times \: 180^o \\ \scriptsize = 45^o \)
⇒ \( \scriptsize \angle SPQ = \angle PSQ = \normalsize \frac{3}{8}\scriptsize \: \times \: 180^o \\ \scriptsize = 67.5^o \)
⇒ \( \scriptsize \angle SQR = \angle SPQ = 26^o \)
⇒ \( \scriptsize \angle PQR = \angle PQS\: +\:\angle SRQ \)
= 45º + 26º
Answer ⇒  = 71º
(a.ii ) ∠RPQ
∠RPQ = ∠SPQ – ∠SPR
= 67.5º – 26º
Answer ⇒  = 41.5º
(a.iii) ∠PRQ
∠PRQ = 180 – 41.5 – 71
Answer ⇒ = 67.5º
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where x is a real number.
If \( \scriptsize |PQ| = \sqrt{29} \: units, \) find the value of x
Solution
⇒ \( \scriptsize PQ^2 = (x_2 \: -\: x_1)^2 \: + \: (y_2 \: -\: y_1)^2 \)
⇒ \( \scriptsize PQ = (x_2 \: -\: x_1)^2 \: + \: (y_2 \: -\: y_1)^2 \)
⇒ \( \scriptsize x_2 = 5, \: x_1 = 7 \)
⇒ \( \scriptsize y_2 = x, \: y_1 = 3 \)
⇒ \( \scriptsize \sqrt{29} = \sqrt{(5\:-\:7)^2 \: +\: (x \: -\: 3)^2} \)
⇒ \( \scriptsize 29 = (-2)^2 \: + \: (x \: – \: 3)(x\:-\:3) \)
⇒ \( \scriptsize 29 = 4 \: + \: x^2 \: -\: 3x \: -\: 3x \: +\: 9 \)
⇒ \( \scriptsize 29 = 4 \: + \: x^2 \: -\: 6x \: +\: 9 \)
⇒ \( \scriptsize  x^2 \: -\: 6x \: +\: 9\: + \: 4 \: -\: 29 = 0 \)
⇒ \( \scriptsize  x^2 \: -\: 6x \: – \: 16= 0 \)
⇒ \( \scriptsize  x^2 \: -\: 8x \: + \: 2x \: – \: 16= 0 \)
⇒ \( \scriptsize  x(x \: -\: 8) \: +\: 2(x \: -\: 8) = 0 \)
⇒ \( \scriptsize  (x \: + \: 2)(x\:-\:8) = 0\)
x + 2 = 0 or x – 8 = 0
Answer ⇒ x = -2 or x = 8
Question
(a) On Sam’s first birthday celebration, his grandfather deposited an amount of $1,000.00 in a bank compounded at 4% interest annually. Find how much is in the account if Sam is 4 years old.
Solution
Amount at end of 2nd year
= \( \frac{104}{100}\scriptsize \: \times \: 1000 \)
= $1040
Amount at end of 3rd year
= \( \frac{104}{100}\scriptsize \: \times \: 1040 \)
= $1081
Amount at end of 4th year
= \( \frac{104}{100}\scriptsize \: \times \: 1081 \)
Answer ⇒ = $1124.86
(b)
NOT DRAW TO SCALE
In the diagram, ABCD are points on the circle O.
If |AB| = |BC| and ∠ADC = 50º, find ∠BAD
Solution
⇒ \( \scriptsize \angle DCA =90^0 \)
(angles in a semi-circle)
⇒ \( \scriptsize \angle DCA \: + \: \angle ADC \: + \: \angle CAD =180^0 \)
(angles in a triangle)
⇒ \( \scriptsize 90^o\: + \: 50^o \: + \: \angle CAD =180^0 \)
⇒ \( \scriptsize \angle CAD =180^0 \: -\: 140^o \)
= 40º
⇒ \( \scriptsize \angle ADC \: + \: \angle ABC = 180^0 \)
(cyclic quadrilateral)
⇒ \( \scriptsize 50 \: + \: \angle ABC = 180^0 \)
⇒ \( \scriptsize \angle ABC = 180^0 \: – \: 50^o \)
⇒ \( \scriptsize \angle ABC = 130^o \)
⇒ \( \scriptsize \angle BAC = \angle BCA \)
(base angles of isosceles)
⇒ \( \scriptsize \angle BAC \: + \: \angle BCA \: + \: \angle ABC = 180^o \)
(sum of angles in a triangle)
⇒ \( \scriptsize 2 \angle BAC \: + \: 130^o = 180^o \)
⇒ \( \scriptsize 2 \angle BAC = 180^o \: – \: 130^o \)
⇒ \( \scriptsize 2 \angle BAC = 50^o \)
⇒ \( \scriptsize \angle BAC = \normalsize \frac{50}{2} \)
⇒ \( \scriptsize \angle BAC = 25^o \)
⇒ \( \scriptsize \angle BAD = \angle BAC \: + \: \angle CAD\)
⇒ \( \scriptsize \angle BAD = 25 \: + \: 40\)
⇒ \( \scriptsize \angle BAD = 65^o\)
Answer ⇒ \( \scriptsize \angle BAD = 65^o\)
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