Mathematics May/June 2009

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Question 1

(a) Given that ( – 5)( + ) = a + b , find a and b.

(b) If 2^1-y X 2^y-1/ 2^y+2= 8 ^2-3y , find y

ANSWER

1(a) (√3 – 5√(2)) (√3 – √(2))= a + b√6
3 + √6 – 5√6 – 5(2) = a + b√6
3 -4√6 – 10 = a + b√6
-7 -4√6 = a + b√6 ; By comparison,
A = -7 ; 4√6 = b√6 ; ∴ a = -7 ; b = -4

(b) 2 ^1-y x 2 ^y-1 / 2 ^y-1 = 8^{2 – 3y}

2^{(1-y) + (y-1) – (y+2)} = 2^3(2-3y) ; 2^{1-y + y-1 – y-2} = 2^6-9y

2^-y-2 = 2^6-9y; -y -2 = – 9y ; -y + 9y = 6 + 2

8y = 8 ; y = ⁸/₈ = 1

Question 2
(a) If 9 cos x – 7 = 1 and 0° ≤ x ≤ 90°, find x. (b) Given that x is an integer, find the three greatest values of x which satisfy the inequality 7x < 2x – 13.

Question 2

(a) If 9 cos x – 7 = 1 and 0° ≤ x ≤ 90°, find x.

(b) Given that x is an integer, find the three greatest values of x which

satisfy the inequality 7x < 2x – 13.

ANSWER

(a) 9 cos x^o – 7 = 1 ; 9 cos x^o = 1 + 7 = 8

Cos x^o = 8/9 = 0.8889 ; x^o cos^-1 (0.8889)

x^o = 27.3^o

5x < -13 ; x < -13/5 ; x <-2.6

Thus, the three great values of x are -3, -4 ad -5.

Question 3

The table shows the number of children per family in a community.

No. of children	0	1	2	3	4	5
No. of families	3	5	7	4	3	2

(a) Find the:

(i) Mode;

(ii) third quartile;

(iii) probability that a family has at least 2 children.

(b) If a pie chart were to be drawn for the data, what would be the sectoral angle representing families with one child

ANSWER

	No. of Children (Data)	No. of families (frequency)	Cumulative (frequency)

18^th Item falls here	0 1 2 3 4 5	3 5 7 4 3 2	3 8 15 19 22 24
		Σf = 24

(a) (i) Mode = 2(Data with highest frequency)

(ii) 3rd quartile = Q₃ = ?

Position of 3rd quartile = ³/₄ x Ef

= ³/₄ x 24 = 18

\ 18th item in the data = Q₃ ; => Q₃ = 3

(iii) Prob. (At least 2 children)

= = =

(b) Let the sectorial angle = θ₁ ; θ₁ = x 360^o

Where f1 = No. of families with one child

: f₁ = 5 ; Þ Q₁ = ⁵/₂₄ x 360^o = 75^o

Question 4

(a)Out of 30 candidates applying for a post, 17 have degrees, 15 diplomas and 4 neither degree nor diploma. How many of them have both?

(b) In triangle PQR, M and N are points on the sides PQ and PR respectively such that MN is parallel to QR. If <PRQ=75o, PN = QN and

(i) <NQR;

(ii)

ANSWER

Let D – Degree, P – Diplomas

Total no. of candidates = n (μ) = 30 ; n(D) = 17

n(P) = 15 ; n(D È P) = 4 ; n(D Ç P) = ?

Using a venn’s diagram with n(D Ç P) = x

n(μ) = 30

Now,

(17 – x) + x + (15 – x) + 4 = n(μ)

17 – x + x + 15 – x + 4 = 30

36 – x = 30 ; -x = 30 – 36

-x = -6 ; x = 6

\ 6 students have both degrees and diplomas.

(b) (i)

ÐNQR = ? ; ÐMNQ = ÐNQR (Alternate angles)

ÐPNM ÐPRQ = 75^o (Corresponding angles)

From the diagram ÐPNQ = ÐPNM + ÐMNQ

125^o = 75^o + ÐMNQ ; ÐMNQ = 125^o – 75^o = 50^o

\ ÐNQR = 50^o

Alternatively,

ÐPNQ = ÐNQR + ÐNRQ (Sum of two interior angles of a triangle = opposite exterior angle)

(ii) ÐNPM = ? ; ÐNPM = ÐNPQ

In ΔNPQ, let ÐNPQ = x

ÐNPQ = ÐNQP = x

ÐNPQ = ÐNQP = x (Base angles of an isosceles triangle)

But, ÐNPQ + ÐNQP + ÐPNQ = 180^o

(Sum of the interior angles of a triangle)

\ x + x + 125^o = 180^o ; 2x = 180o – 125^o

2x = 55^o ; x = x = 27.5

\ Þ ÐNPM = 27.5^o

From the diagram,

Question 5

In the diagram, ABCDEF is a triangular prism.

(a) /AC/;

(b)The total surface area of the prism.

ANSWER

(a) In Δ ABC,

Using Pythagoras theorem

(AC) ̅^2 = (AB) ̅^2 + (BC) ̅^2

(AC) ̅^2 = 〖24〗^2 + 7^2

= 576 + 49 = 625

= = 25cm

(b) T.S.A. = (Area of ΔABC x 2)

(Area of rectangle ABEF)

(Area of rectangle BCDE)

(Area of rectangle ACDF)

\ T.S.A. = (½ x x x 2) + ( x ) +

( x ) + ( x ) = ( ½ x 7 x 24 x 2) + 24 x 40)

+ (7 x 40) + (25 x 40) = (7 x 24) + (24 x 40) + (7 x 40)

+ (25 x 40) = 168 + 960 + 280 + 1000 = 2408cm²

Question 6

(a)If log5 = 0.6990, log7 = 0.8451 and log8 = 0.9031, evaluate:

log().

(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the

(i) value of x;

(ii) ratio of the number of children to the number of adults;

(iii) average amount paid per person;

(iv) percentage profit if the organizers spent D720 on the show.

ANSWER

Candidates’ performance in part (a) of the question was said to be satisfactory.

In the part (b), candidates’ performance was reportedly fair.

(a) log () = log( )

4 x 0.8451 = 3.3804

(b) No. of Children = x ; No. of Adult = x + 60
Amount paid per adult = D5
Total amount paid by adults = D5 x (x + 60)
Amount paid per Child = D2
Total amount paid by Children = D2 x x
Now,
(D2 x x) + [D5 x (x + 60] = D 1280
2x + 5x + 300 = 1280
7x = 1280 – 300 = 980 ; x = 980/7 ; D = 140

(ii) Ratio required = (No.of Children)/(No.of Adult) = x/(x+60) = 140/(140+60)
= 140/200 = 7/10 ; Thus, required ratio = 7:10

(iii) Average amount paid per person
= (Total amount paid or collected)/(Total no.of people) = D1280/(x+(x+60))

= D1280/(2x+60) = D1280/(2(140)+ 60) = D1280/340 = D3.76

Percentage profit = Profit/Cost x 100%

where profit = D1280 – D720 = D560

(iv)Percentage profit = D560/D720 x 100%
56/72 x 100% = 77.78%

Question 7
(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44º. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman; (ii)height of the flag pole. (b) In the diagram, O is the centre of the circle, (i) <QPR; (ii) <TQO.

Question 7

(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44º. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the

(i) angle of depression of the foot of the pole from the woman;

(ii)height of the flag pole.

(b)

In the diagram, O is the centre of the circle,

(i) <QPR;

(ii) <TQO.

ANSWER

(a)

(i) Let θ = Angle of depression of the foot of the pole from the woman

From the diagram,

Tan θ = 30m/25m = 6/5

θ = Tan^-1 (⁶/₅) = Tan^-1 (1.2)

θ = 50.2 = 50^o (nearest whole number)

(ii) Let h = height of flagpole,

From the diagram,

Tan 44^o = = 30 – h = 25 tan 44^o

30 – 25 tan 44o = h ; h = 30 – 25 (0.9657)

h =7 5.857m = 6m (nearest whole number)

(b) Construction: Join as shown below

(i) <QPR = ?

<QOR = 2 x <QPR

(angle at centre = two x angles at circum.)

then QPR <QOR/2 …..(1)

In ^ QOR, <ORQ = <ORQ = 32^o

(Base angels of an Isosceles triangle)

But <OQR + <ORQ + <QOR = 180^o (Sum of the angles in a Δ)

32^o + 32^o + <QOR = 180^o ; <QOR = 180^o – 64 = 116^o

Put <QOR = 116^o into (1)

<QPR = = 58^o

(iii) <TQR = ?

From the diagram, <TQR = <TQO + <OQR

then <TQO = <TQR = 32^o ………….*

But, <TPR + <TQR = 180^o (Opposite interior angles of a cyclic quadrilateral)

Where <TPR = <TPQ + <QPR = 15 + 58 = 73^o

=73^o + <TQR = 180o

<TQR = 180^o – 73 = 107^o

Substituting into *

<TQO = 107^o – 32^o = 75^o

Question 8
The marks scored by 50 students in a Geography examination are as follows: 60 50 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42 (a) Using class intervals 21 – 30, 31 – 40…, prepare a frequency distribution table. (b) Calculate the mean mark of the distribution. (c) What percentage of the students scored more than 60%?

Question 8

The marks scored by 50 students in a Geography examination are as
follows:

60      50      40      67      53      73      37      55      62      43
44      69      39      32      45      58      48      67      39      51
46      59      40      52      61      48      23      60      59      47
65      58      74      47      40      59      68      51      50      50
71      51      26      36      38      70      46      40      51      42

(a) Using class intervals 21 – 30, 31 – 40…, prepare a frequency
distribution table.

(b) Calculate the mean mark of the distribution.

ANSWER

Class I interval	Class mark (x)	Tally	Frequency (f)	Fx
21-30	25.5	11	2	51
31-40	35.5	~~1111~~ ~~1111~~	10	355
41-50	45.5	~~1111~~ ~~1111~~ 11	12	546
51-60	55.5	~~1111~~ ~~1111~~ ~~1111~~	15	832.5
61-70	65.5	~~1111~~ 111	8	524
71-80	75.5	111	3	226.5
			50	2535

Mean = = =50.7. The percentage number of students who scored more than 60 marks = 11/50 x 100 = 22%.

Question 9
Simplify x + 2 – x + 3 x – 2 x – 1 (a)The graph of the equation y = A x2 + B x + C passes through the points (0, 0), (1, 4) and (2, 10). Find the: (i) value of C; (ii) values of A and B; (iii) co-ordinates of the other point where the graph cuts the x- axis.

Question 9

Simplify x + 2 – x + 3
x – 2 x – 1

(a)The graph of the equation y = A x2 + B x + C passes through the points (0, 0), (1, 4) and (2, 10). Find the:
(i)     value of C;
(ii)    values of A and B;
(iii)    co-ordinates of the other point where the graph cuts the x-
axis.

ANSWER

(a) (x+2) / (x-2 ) – (x+3 ) / (x-1) = 0

= (x+2)(x-1)- (x+3)(x-2) / (x+2)(x-1)

= (x ² + x-2)-(x ²+ x-6) / (x-2)(x-1) = 4/(x-2)(x-1)

(b) y = Ax² + Bx + C

points (0, 0), (1,4), and (2,10)

(i) At (0, 0), x = 0, y = 0 ;

\ 0 = A(0²) + B(0) + C

0 = 0 + 0 + C ; C = 0

(ii) At (1,4), x = 1, y – 4;

\ 4 = A(1²) + B(1) + C

4 = A(1) + B + C

= A + B + C = 4 ; since C = 0

A + B + 0 = 4 ; A + B = 4 ———– (1)

(2, 10), x = 2, y = 10;

\10 = A(2²) + B(2) + C

10 = 4A + 2B + 0

= 4A + 2B = 10 ; 2(2A + B) = 10

2A + B = ¹⁰/₂ ; 2A + B = 5 ——– (2)

Subtracting (1) from 2

2A + B = 5 ———– (2)

————- (1) ; Þ A = 1

Put A = 1 into (1)

Re: A + B = 4 ——— (1) 1 + B = 4

B = 4 – 1 ; B = 3 ; A = 1 ; B = 3

(iii) Re: y = Ax² + Bx + C

Þ y = x² + 3x + 0 ; y = x² + 3x ; y = x(x+3)

when the graph cuts the x-axis

y = 0 ; 0 = x(x + 3)

x = 0 or x + 3 = 0 ; x = 0 or x = -3

Hence, the required coordinates are:

x = -3 ; y = 0 i.e. (-3, 0)

Question 10
(a)Using ruler and a pair of compasses only, construct: (i) quadrilateral PQRS such that /PQ/ = 10cm, /QR/ = 8cm, /PS/ = 6cm, ÐPQR = 60º and ÐQPS = 75º; (ii) the locus 11 of points equidistant from QR and RS; (iii) locus 12 of points equidistant from R and S. (b) Measure /RS/

ANSWER

i, ii and iii

**diagram**

(b) /RS/ = 4

Question 11
(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take p = ] (b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the (i) length; (ii) diagonal of the rectangle.

Question 11

(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take p = ]

(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the

(i) length;

(ii) diagonal

of the rectangle.

ANSWER

(a) Diameter of circle = length of side of square d = s = 2r

Now, P square + P circle = 100

(Given) i.e. 4s + 2pr = 100

4(2r) + 2 x ²²/₇ x r = 100

8r + ^44r/₇ = 100

= 100 ; 100

r = = 7cm

Hence, the radius of the circle is 7cm

(b) Perimeter of rectangle =

Length of rope

P = 60cm

(i) Length L = ? = 2(l + b) = 60 ; L + b = ⁶⁰/₂

L + b = 30; since L = 4b ; Þ 4b + b = 30

5b = 30 ; b = ³⁰/₅cm ; b = 6cm ; L = 4b

L = 4 x 6cm = 24cm ; length = 24cm

(ii) From the right-angled triangle formed

Using Pythagoras theorem

d² = L² + b² ; d² = (4b)² + b²

d² = 16b² + b² = 17b²

d = + = x b

d = 4.123 x 6cm

d = 24.738cm ; d ~ 24.74cm

y = Sin x + 2Cos x

Question 12

(a)Copy and complete the table of values for y = sin x + 2 cos x, correct to one decimal place.

X	0°	30º	60°	90°	120°	150°	180°	210°	240
Y		2.2				-1.2	-2.0		-1.9

(b) Using a scale of 2 cm to 30° on the x – axis and 2 cm to 0.5 units

on the y-axis, draw the graph of y = sin x + 2cos x for 0° ≤ x ≤ 240°
(c)     Use your graph to solve the equation:
          (i)     sin x + 2cos x = 0;
         (ii)     sin x = 2.1 – 2cos x.
(d)     From the graph, find y when x = 171°.

ANSWER

x	0^o	30^o	60^o	90^o	120^o	150^o	180^o	210^o	240^o
Sin x	0	0.5	0.866	1	0.866	0.5	0	-0.5	-0.866
Cos x	1	0.866	0.5	0	-0.5	-0.866	-1	-0.866	-0.5
2Cos x	2	1.732	1.0	0	-1.0	-1.732	-2	-1.732	-1.0
y	2	2.232	1.866	1.0	-0.134	-1.232	-2	-2.232	-1.866

x	0^o	30^o	60^o	90^o	120^o	150^o	180^o	210^o	240^o
y	2	2.2	1.9	1	-0.1	-1.2	-2	-2.2	-1.9

(b) TITLE: The graph of y = sinx + 2cosx

SCALE: 2cm on x-axis = 30^o

2cm on y-axis = 0.5 unit

***Graph diagram **

(ii) sin x = 2.1 -2cosx ; sinx + 2 cosx = 2.1 ; y = 2.1

From the graph, at y = 2.1 ; x = 12^o or 42^o

(d) When x = 171^o, y = -1.7

Question 13
(a)How many numbers between 75 and 500 are divisible by 7? (b) The 8th term of an Arithmetic progression (A.P.) is 5 times the third term while the 7th term is 9 greater than the 4th term. Write the first five terms of the A.P.

Question 13

(a)How many numbers between 75 and 500 are divisible by 7?

(b) The 8th term of an Arithmetic progression (A.P.) is 5 times the third term while the 7th term is 9 greater than the 4th term. Write the first five terms of the A.P.

ANSWER

This means an A.P of 1st term = 77 i.e. a = 77

Common difference, d = 7 ; Last term, L = 497

Number of terms, n = ? Using n = L = a + (n – 1) d

497 = 77 + (n – 1)7 ; 497 – 77 = 7n – 7

420 + 7 = 7n ; 427 = 7n ; n = ; n = 61

(b) T₈ = 5T₃

i.e. a + (8 -1)d = 5[a + (3-1)d]

a + 7d = 5 (a + 2d) ; a + 7d = 5a + 10d

a + 7d = 5a + 10d ; 0 = 5a + 10d – a – 7d

0 = 4a + 3d ; 4a = -3d ; a = -3d/4 ………. (1)

Also, T_n = T₄ + 9

a + (7 – 1)d = a + (3 – 1)d + 9

a + 6d = a + 3d + 9 ; a + 6d – a – 3d = 9

3d = 9 ; d = ⁹/₃ = 3 ; Put d = 3 into (1)

a = -3(3)/4 ; a =-9/4

Thus T₁ = a = –⁹/₄

T₂ = a + d = + 3 + -9+12/4 = 3/4

T₃ = a + 2d = -9/4 + 2(3) =-9+24/4 = 15/4

T₄ = a + 3d = -9/4 + 3(3) = -9+36/4 = 27/4

T₅ = a + 4d = -9/4 + 4(3) = -9+48/4= 39/4

The 1^st five terms of the A.P are

^-9/₄, ¾, ¹⁵/₄, ²⁷/₄ and ³⁹/₄ respectively.

Mathematics May/June 2009

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