 (a) Mr Sarfo borrowed $25,000.00 from AFLAK Financial Services at: 21% simple interest per annum for 3 years. If he was able to pay back the loan in 2 years at equal yearly installments, how much did he pay each year?
(b) Two consecutive numbers are such that the sum of thrice the smaller and twice the larger is 17. Find, correct to three, significant figures, the smaller number as a percentage of the sum of the two numbers.
ANSWER
 (a) P = $25,000, R = 21%, T = 3 years
Using A = P(I + RT%) = 25000
= 25000(I + 0.63)
= 25000×1.63 =$40750
Amount after 3 years is $40,750
Yearly installment = 40750 = $20,375
2
 A man left town M at 10:00am and travelled by car to town N at an average speed of 72km/h. He spent 2 hours for a meeting and returned to town M by bust at an average speed of 40km/h. If the distance covered by the bus was 2km longer than that of the car and he arrived At town M at 1:55pm. Calculate the distance from M to N.
ANSWER
 Let distance from M to N be x km
Time taken from M to N = (x/72)h
Let distance for the return journey be (x + 2)
Time taken = x+2
Total time 3hr 55minutes – 2 hours
For meeting = 1hr 55mins
Time spent from M to N and back to M is
+ = 1 ; + =
Multiply through by 360, the hem
5x + 9(x + 2) = 690 ; 5x + 9x + 18 = 690
14x = 672 ; x = 672 = 48km
14
Distance from M to N is 48km
 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 270^{o}. calculate, correct to:
 Two significant figures, {YZ}:
The nearest degree, the bearing of Y from Z
ANSWER
 a) In ZXY, ∠ZXY = 90^{o} Using Pythagoras Theorem, YZ^{2} = ZX^{2} + XY^{2} = 202 + 152 = 400 + 225
YZ = = 25km(2 s.f)
 In the diagram, AD is a diameter of a circle with centre O. if ABD is a triangle in a semicircle and <OAB = 34^{o}, find
 <OBD
 <OCB
ANSWER
 a) In ABD, ∠ABD = 90^{o} (Angle in a semicircle) ABD is an isosceles AO = OB (radii) ∠BAO = ∠ABO = 34o (base angles) ∠OBD – ∠ABD
∠OBD = 90^{o} – 34^{o} = 56^{o}
(b) ∠BOD = 34^{o} + 34^{o} = 68^{o} (Ext. angle is sum of int. opp. Angles) BC is the tangent at B
∠OBC = 90^{o} (radius 1 BC) In AOBC, ∠OCB + 158^{o }= 180^{o}. ∠OCB = 180^{o} – 158^{o} = 22^{o}
 A man shared his property among his children as follows:
Child’s name  Ann  Afia  Kolo  Nuno  Akosua 
Percentage share  5  15  10  45  25 
 Represent the information on a pie chart.
 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.
ANSWER
 (a) Total percentage = 5 + 15 + 10 + 45 + 25 = 100 Angle representing the various percentage is as shown below.
Child’s name  Share  Sectorial angle 
Ann  5%  0.05 x 360 = 18^{o} 
Afia  15%  0.15 x 360 = 54^{o} 
Kojo  10%  0.1 x 360 = 36^{o} 
Nuno  45%  0.45 x 360 = 162^{o} 
Akosua  25%  0.25 x 360 = 90^{o} 
PIE CHART diagram
Showing a man’s property was shared
 Total number of beads = 5+3+4=12
Number of red beads = 5
P(1^{st} bead picked red) = 5/12
P(2^{nd} bead picked red) = 4/12
P(both beads picked red) = x =
 (a) In a class of 80 students, ¾ study Biology and 3/5 study Physics. If each student studies at least one of the subjects:
 draw a venn diagram to represent this information:
 how many students study both subjects:
 find the fraction of the class that study Biology but not Physics.
 Johnson and Jocatol Ltd, owned a business office with floor measuring 15m by 8m which was to be carpeted. The cost of carpeting was GH¢ 890.00 per square metre. If a total of GH¢ 216,120.00 was spent on painting and carpeting, how much was the cost of painting?
ANSWER
 (a) number of students = 80
3/4 of 80 = 60 study Biology
3/5 of 80 = 48 study Physics
 SET DRAWING
 Let those that study both subjects be y with the usual notation y+60y+48y=80 ; y = 80 – 108 y = 28 ; 28 students study both subjects
 Students who study Biology only is represented by 60 – y (where y = 28)
60 – 28 = 32 students
The fraction = 32/80 = 2/5
(b) Area of floor = 15m x 18m = 120m^{2 }
1m^{2} costs GH¢ 890.00
120m^{2} will 890 x 120 = GH¢ 106800.00
Total cost on painting and carpeting = GH¢21620.00
Cost of painting = GH¢ 109320.00
 (a) Copy and complete the table of values for the relation y = 2x^{2} – x – 2 for 4 < x <
x  4  3  2  1  0  1  2  3  4 
y 
 19 

 2 


 26 
(b) using a scale of 2cm to 1 unit on the xaxis and 2cm to 5 units on the yaxis, draw the graph of y= 2×2x2 for 4 < x < 4.
 On the same axes, draw the graph of y = 2x + 3.
 Use the graph to find the:
 Roots of the equation 2x^{2}3x5=0
(ii) Range of values of x for which 2x^{2}x2<0.
ANSWER
 (a) 7=2x^{2} – x – 2
x  4  3  2  1  0  1  2  3  4 
Y  24  19  8  1  2  1  4  13  26 
(b) the graph of y = 2×2 – x – 2
 (a) In PQR, <PQR = 90^{o}. If its area is 216cm^{2} and PQ: QR is 3:4, find PR.
 The present ages of a man and his sons are 47years and 17 years respectively. In, how many years would the man’s age be twice that of the son?
ANSWER
 (a) Area of PQR = 216cm2 i.e, 1/2base x height = 216
1/2 x (4x)(3x) = 216
6x^{2} = 216; x^{2} = 36
x = = 6
using pythagoras theorem
PR^{2} = 18^{2} + 24^{2} = 324 + 576
PR^{2} = 900 ; PR = = 30cm^{2}
(b) Man’s age = 47 years
Son’s age = 17 years
Let number of years be y
In y years time,
Man’s age = 47+y
Son’s age = 17+y
i.e 47+7 = 2(17+y) ; 47+7 = 34 + 2y
y2y = 3447 ; y = 13
y = 13
The man’s age would be twice his son’s age in 13 years time
 In the diagram PQRS is trapezium with QR//PS. U and T are points on PS such that PU = 5cm, QU = 12cm and <PUQ = <STR = 90^{o}. If the PQR = 20cm2, calculate, correct to
nearest whole number, the (a) perimeter (b) area of the trapezium
ANSWER
 In PQR, QP2 = 52 + 122
(pythagoras theorem)
QP2 = 25 + 144
QP = = 13cm
Area of PQR = ½ Area of QUTR
20 = 1/2 x b x 12; 6b = 20 ; b = 3.3cm
In TRS, 12/RS = sin50o
RS = = ; RS = 15.67cm
Also, 12/TS = tan50^{o }
TS = = ; TS = 10.07cm
PS = 5cm + 3.33cm + 10.07cm = 18.40cm
 Perimeter of trapezium PQRS = 18.40cm + 15.67cm + 3.33cm + 13cm = 50.40cm ~ 50cm
 Area of trapezium PQRS = 1/2(QR + PS) x QU =1/2(3.33 + 18.40) x 12
= 21.73 x 6 = 130.38cm^{2} ~ 130cm^{2 }
 a) A cottage is on a bearing of 200^{o} and 110^{o} from Dogbe’s and Manu’s forms respectively. If Dogbe walked 5km and Manu 3km from the cottage to their farms, find, correct to:
 two significant figures, the distance between the two farms;
 the nearest degree, the bearing of Manu’s farm from Dogbe’s
 a ladder 10m long leaned against a vertical wall x m 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.
ANSWER
 (a)
 Distances between D and M Using Pythagoras theorem.
DM^{2} = 5^{2} + 3^{2} = 25 + 9 = 34
DM = ~ 5.6km
 Let ∠PMD =
The hearing of Mami’s farm from Dogbe’s in this case will be + 20^{o}
Find , using sine formula = ; sin =
θ = sin^{1} (0.5329) = 32.2^{o}
The bearing of Mamu’s farm from Dogbe’s is
021.20^{o} + 020^{o} = 052.2^{o}
(b) Using pythagoras theorem.
In ΔABC x^{2} + (x+2)^{2} = 10^{2}
x^{2} + x^{2} + 4x = 100
2x^{2} + 4x – 96 = 0
X^{2} + 2x – 48 = 0
(x + 8)(x – 6) = 0
Either x + 8 = 0 or x – 6 = 0
X = 8 or 6 ∴ x = 6m
 The table shows the distribution of the number of hours per day spent in studying by 50 students.
Number of hours per day  4  5  6  7  8  9  10  11 
Number of students  5  7  5  9  12  4  3  5 
Calculate, correct to two decimal places, the;
 Mean: (b) Standard Deviation
ANSWER
x  f  f  d = x –  (x – )^{2}  fd^{2} 
4 5 6 7 8 9 10 11  5 7 5 9 12 4 3 5  20 35 30 63 96 36 30 55  3.3 2.3 1.3 0.3 0.7 1.7 2.7 3.7  10.89 5.29 1.69 0.09 0.49 2.89 7.29 13.69  54.45 37.03 8.45 0.81 5.68 11.56 21.87 68.45 
 Σf=50  Σfx=365 

 Σfd^{2} = 28.3 
 Mean x =
 D = = 4.17 (2 d.p)
 (a) In the diagram PQRS is a circle. PQ = QS, <SPR = 26^{o} of PQS are in the ration 2:3:3, Calculate:
 <PQR: (b) <RPQ: (c) <PRQ
 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 PQ = units, find the value of x
ANSWER
 (a) 2x + 3x + 3x = 180
(Angles in ΔPOS)
8x = 180; x = 22.5^{o}
2x = 45^{o}
3x = 67.5^{o}
∠SQR = ∠SPR = 26^{o}
(Angles in the same segment)
 ∠PQR = 45^{o} + 26^{o} = 71^{o}
 ∠RPQ = 180^{o} – (67.5^{o} + 71^{o}) (Angles in ΔPQR)
∠RPQ = 41.5^{o}
 ∠PQR = PSQ = 3x = 67.5^{o} (Angles in the same segment)
(b) x, y = (7, 3) and x_{2}, y_{2} = (5, x)
Using PQ^{2} = (x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}
= (5 – 7)^{2} + (x – 3)^{2}
PQ = = 29
4 + x^{2} – 6x + 9 = 29; x^{2} – 6x + 13 = 29
X^{2} – 6x – 16 = 0; (x + 2)(x – 8) = 0
Either x + 2 = o or x – 8 = 0; x = 2 or 8
Since 2 is unsuitable, 8 = 8
 (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.
(b) In the diagram, ABCD are points on the circle centre O. if AB = BC and <ADC = 50^{o}, find <BAD.
ANSWER
 (a) Amount at A years is
A = P ^{n} (compound interest formula)
Amount at 4 years is
A = 100 ^{4} 1000(1 + 0.04)^{2}
= 1000 (1.04)4 = 1000 x 1.16985 = $1169.85
Amount in 4 years is $1169.85
(b) ∠CAD + 50^{o} + 90^{o} = 180^{o}
(<s in ΔCAD)
∠CAD = 180^{o} – 140^{o} = 40^{o}
∠ABC + ∠ADC = 180
(opp <s of a cyclic quad. ABCD)
∠ABC + 50^{o} = 180^{o}; ∠ABC – 180^{o} – 50^{o} = 130^{o}
∠BAC = BCA (base angles of ΔABC)
∠BAC =
∠BAD = ∠BAC + ∠CAD = 25^{o} + 40^{o} = 65^{o}