Mathematics May June 2021

1. (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

1. (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

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1. 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

1. 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

1. 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

1. a) In ZXY, ∠ZXY = 90^o Using Pythagoras Theorem, YZ² = ZX² + XY² = 202 + 152 = 400 + 225

|YZ| =   = 25km(2 s.f)

1. In the diagram, AD is a diameter of a circle with centre O. if ABD is a triangle in a semi-circle and <OAB = 34^o, find

• <OBD
• <OCB

ANSWER

1. a) In  ABD, ∠ABD =  90^o (Angle in a semi-circle)   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

1. A man shared his property among his children as follows:

• 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

1. (a) Total percentage = 5 + 15 + 10 + 45 + 25 = 100 Angle representing the various percentage is as shown below.

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  =

1. (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

1. (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+60-y+48-y=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 = 120m²

1m² costs GH¢ 890.00

120m² will 890 x 120 = GH¢ 106800.00

Total cost on painting and carpeting = GH¢21620.00

Cost of painting = GH¢ 109320.00

1. (a) Copy and complete the table of values for the relation y = 2x² – x – 2 for -4 < x <

(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 y= 2×2-x-2 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 2x²-3x-5=0

(ii) Range of values of x for which 2x²-x-2<0.

ANSWER

1. (a) 7=2x² – x – 2

(b) the graph of y = 2×2 – x – 2

1. (a) In PQR, <PQR = 90^o. If its area is 216cm² 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

1. (a) Area of PQR = 216cm2 i.e, 1/2base x height = 216

1/2 x (4x)(3x) = 216

6x² = 216; x² = 36

x =    = 6

using pythagoras theorem

PR² = 18² + 24² = 324 + 576

PR² = 900 ; PR =   = 30cm²

(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

y-2y = 34-47 ; -y = -13

y = 13

The man’s age would be twice his son’s age in 13 years time

1. 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

1. 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 = 130.38cm² ~ 130cm²

1. 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

1. (a)

• Distances between D and M Using Pythagoras theorem.

DM² = 5² + 3² = 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 x² + (x+2)² = 10²

      x² + x² + 4x = 100

      2x² + 4x – 96 = 0

      X² + 2x – 48 = 0

      (x + 8)(x – 6) = 0

      Either x + 8 = 0 or x – 6 = 0

      X = -8 or 6  ∴  x = 6m

1. 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;

• Mean: (b) Standard Deviation

ANSWER

• Mean x =

• D = = 4.17 (2 d.p)

1. (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

1. (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 x₂, y₂ = (5, x)

            Using PQ² = (x₂ – x₁)² + (y₂ – y₁)²

            = (5 – 7)² + (x – 3)²

            PQ =    = 29

            4 + x² – 6x + 9 = 29;  x² – 6x + 13 = 29

            X² – 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

1. (a) On Sam’s first birthday celebration, his grand-father 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

1. (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