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Quiz 13 of 13

# 2014 Mathematics WAEC Theory Past Questions

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## Question 1a

(a) Without using tables or calculator, simplify :

$$\frac{0.6 \: \times \: 32 \: \times \: 0.004}{1.2 \: \times \: 0.008 \: \times \: 0.16}$$

leaving the answer in standard form (scientific notation).

Solution: Step 1: Rewrite each term of the fraction in standard form

:- $$\frac{6 \: \times \: 10^{-1} \: \times \: 3.2 \: \times \: 10^{1} \: \times \: 4 \: \times \: 10^{-3}}{1.2 \: \times \: 8 \: \times \: 10^{-3} \times 1.6 \: \times \: 10^{-1} }$$

Step 2:- Divide through by common factors

= $$\frac{3 \: \times \: 2 \: \times \: 10^1}{1.2}$$

= $$\frac{6 \times \: 10^1}{1.2}$$

= $$\frac{6 \times \: 10^1}{1.2}$$

=  $$\frac{10^1}{0.2}$$

Step 3:- Write in standard form

=  $$\frac{10^1}{2 \: \times \: 10^1}$$

=  $$\scriptsize 10 \: \times \: \normalsize \frac{1}{2}\scriptsize \: \times \: 10^1$$

=  $$\scriptsize 5 \: \times \: 10^1$$

## Question 1b

(b) In the diagram, $$\scriptsize\overline{EF}$$ is parallel to $$\scriptsize\overline{GH}$$.

If < AEF = 3x°,< ABC = 120° and < CHG = 7x°, find the value of < GHB.

Step 2:- Determine the size of $$\scriptsize \hat{WBC}$$

:- $$\scriptsize \hat{WBC} = 180^o \: - \: 120^o = 60^o$$

(sum of angle on a straight line)

Step 3:-  $$\scriptsize \hat{BWH}$$ = 3xº (corresponding angle, since $$\scriptsize \bar{EF} || \bar{GH}$$

Step 4:- But $$\scriptsize \hat{WBC} \: + \: \hat{BWH} = 7x^o$$

(sum of the opposite angles of an exterior angle of a triangle)

60º + 3xº = 7xº

60º = 7xº - 3xº

60º = 4xº

xº = $$\frac{60}{4}$$

xº = 15º

Step 5:-

:- $$\scriptsize \hat{GHB} = 180 \: -\: 7x$$

(sum of angle on a straight line)

:- $$\scriptsize \hat{GHB} = (180 \: -\: 7 \: \times \: 15)$$

:- $$\scriptsize \hat{GHB} = (180 \: -\: 105)$$

:- $$\scriptsize \hat{GHB} = 75^o$$

## Question 2a

(a) Simplify : 3√75 - √12 + √108, leaving the answer in surd form (radicals).

Step 1:- First write each term in the base form.

:- $$\scriptsize 3\sqrt{25 \: \times \: 3} \: - \: \sqrt{4 \: \times \: 3} \: + \: \sqrt{36 \: \times \: 3}$$

:- $$\scriptsize 3 \: \times \: 5\sqrt{3} \: - \: 2 \sqrt{3} \: + \: 6\sqrt{3}$$

=  $$\scriptsize 19 \sqrt{3}$$

## Question 2b

If 124n = 232five, find n.

Solution: Step 1:- Express each number in base ten.

Step 2:- Simplify this we have

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 = 50\: + \: 15\: + \: 2$$

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 = 67$$

Step 3:- Move 67 to the LHS

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 - 67 = 0$$

:- $$\scriptsize n^2 \: + \: 2n \: - \: 63 = 0$$

Step 4:- Factorize the expression and solve for the unknown

:- $$\scriptsize n^2 \: + \: 9n \: - \: 2n \: - \: 63 = 0$$

:- $$\scriptsize n(n \: + \: 9) \: - \: 7(n \: + \: 9) = 0$$

:- $$\scriptsize n = -9 \: or \: 7$$

Note: the correct value of n is 7, n cannot be expressed as a negative number

n = 7

## Question 3a

(a) Solve the simultaneous equation:

$$\frac{1}{x} \: + \: \frac{1}{y} \scriptsize = 5$$

$$\frac{1}{y} \: - \: \frac{1}{x} \scriptsize = 1$$

Solution: Step 1:- First transform the equation by using:

p = $$\frac{1}{x}$$

and

q = $$\frac{1}{y}$$

Step 2:- Now transform the equation:

:- $$\scriptsize p \: + \: q = 5$$

:- $$\scriptsize q \: + \: p = 1$$

Step 3:- Re-arrange the equation

:- $$\scriptsize p \: + \: q = 5$$.............(1)

:- $$\scriptsize -p \: + \: q = 1$$.............(2)

2q  = 6

q = $$\frac{6}{2}$$

q = 3

Step 4:- Substitute q in equation (1)

:- $$\scriptsize p \: + \: q = 5$$.............(1)

:- $$\scriptsize p \: + \: 3 = 5$$

:- $$\scriptsize p = 5 \: - \: 3$$

:- $$\scriptsize p = 2$$

∴ p = 2, q = 3

Step 5:- Now find the value of x and y from the transformation

p = $$\frac{1}{x}$$

and

q = $$\frac{1}{y}$$

For p = $$\frac{1}{x}$$

2 = $$\frac{1}{x}$$

∴x = $$\frac{1}{2}$$

Similarly for q = $$\frac{1}{y}$$

3 = $$\frac{1}{y}$$

y = $$\frac{1}{3}$$

## Question 3b

A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.

Step 1:-  Find the value of x and y from the transformation change the word problem into a mathematical expression.

Total distance travelled = 48 km

Total time taken = 45 minutes

= $$\frac{45}{60} \scriptsize = 0.75hour$$

Let the man travel a distance of x km on the good surface. Then he travels a distance of (48 - x) km on the bad surface.

Similarly, Let the time taken to travel on the good surface be t hours, then the time taken to travel on the bad surface = (0.75 - t) hrs.

Step 1:- Using Speed = $$\frac{distance}{time}$$

On the good surface:

72 = $$\frac{x}{t}$$

x = 72t ..............(1)

48 = $$\frac{48 \: -\: x}{0.75 \: - \: t}$$

48(0.75 − t) = 48 − x

36 − 48t = 48 − x--------------- (2)

Step 3:- Substitute equation (1) into equation (2)

36 − 48t = 48 − 72t

72t − 48t = 48 - 36

24t = 12

t = $$\frac{12}{24}$$

t = 0.5hours

Step 4:- Substitute for t in equation (1)

x = 72 × 0.5

=36km

Therefore, the man travels a distance of 36 km on the good road surface.

## Question 4a

(a) In the diagram, O is the centre of the circle radius r cm and < XOY = 90°. If the area of the shaded part is 504cm2, calculate the value of r. [Take π=22/7].

Solution:

Step 1: state the formula of the area of a sector

Area of a sector = $$\frac{\theta}{360} \scriptsize \: \times \: \pi r^2$$

Step2:- Area of the shaded region = Area of the sectorArea of the triangle XOY.

Area of  ΔXOY = $$\frac{1}{2} \scriptsize \: \times \: b \: \times \: h$$

= $$\frac{1}{2} \scriptsize \: \times \:r \: \times \: r$$

= $$\frac{r^2}{2}$$

Step 3: Simplify the expression below:

Area of the shaded region = Area of the sectorArea of the triangle XOY.

504 = $$\frac{90}{360} \: \times \: \frac{22}{7} \: \times \: r^2 \: - \: \frac{r^2}{2}$$

504 = $$\frac{11}{14} \: - \: \frac{r^2}{2}$$

:- $$\scriptsize 504 = 0.7857r^2 \: - \: 0.5r^2$$

:- $$\scriptsize 504 = 0.2857r^2$$

r² = $$\frac{504}{0.2857}$$

r² = $$\scriptsize 1764$$

r² = $$\scriptsize \sqrt{1764}$$

r = 42cm

## Question 4b

(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If <PQR = 66° and <PSQ = 109°, calculate the value of <RQS.

Step 1:- Sketch the diagram as shown above

Step 2:- $$\scriptsize \hat{RQP} = 66^0$$

(base angle of an isosceles triangle)

Similarly,

:- $$\scriptsize \hat{SQP} = \normalsize\frac{180 \: - \: 109}{2}$$

(base angle of an isosceles triangle)

:-$$\scriptsize \hat{SQP} = \normalsize \frac{71}{2} \scriptsize = 35.5$$

Step 3:- $$\scriptsize \hat{RQS} =\hat{RQP} \: + \: \hat{SQP}$$

:- $$\scriptsize \hat{RQS} = 66^o \: + \: 35.5^o$$

:- $$\scriptsize \hat{RQS} = 101.5^o$$

## Question 5

A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :

(a) both contracts ;

(b) exactly one of the contracts ;

(c) neither of the contracts?

Solution:

Prob (X) = 0.5

Prob (Y) = 0.3

Step 1:- Determine the prob(X1) and prob(Y)

Prob(X1) = 1 - 0.5 = 0.5

Prob(Y1) = 1 - 0.3 = 0.7

Step 2:- Determine the probability that he will win both contracts

:- $$\scriptsize Prob( X \cap Y) = Prob(X^1) \: \times \: Prob(Y^1)$$

(Since the two events are independent events)

:- $$\scriptsize Prob( X \cap Y) = 0.5 \: \times \: 0.7$$

:- $$\scriptsize Prob( X \cap Y) = 0.35$$

Step 3:- Determine the probability that the man wins exactly one of the contracts.

$$\scriptsize Prob(X \cap Y^1) = Prob(X^1 \cap Y)$$

= (0.5 × 0.3) + (0.5 × 0.7)

= 0.15 + 0.35

= 0.5

## Question 6a

(a) If $$\frac{3}{2p\: - \: \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p \: + \: 1}$$
find p

Solution:

Step 1:- Multiply through by the LCM of the denominator, i.e

:- $$\left( \scriptsize 2P \: -\: \normalsize \frac{1}{2} \right) \: - \: \left( \frac{1}{4} \scriptsize P \: + \: 1 \right)$$

= $$\scriptsize 3 \left( \normalsize \frac{1}{4} \scriptsize \: + \: 1 \right) = \normalsize \frac{1}{3} \scriptsize \left (2P \: - \: \normalsize \frac{1}{2} \right)$$

Step 2:- Open the bracket

:- $$\normalsize \frac{3}{4} \scriptsize P \: + \: 3 = \normalsize \frac{2}{3} \scriptsize P \: - \: \normalsize \frac{1}{6}$$

Step 3:- Multiply through by 12 (LCM of the denominators)

:- $$\left ( \normalsize\frac{3}{4} \scriptsize P \: \times\: 12 \right) \scriptsize \: + \: (3 \: \times\: 12) = \left( \normalsize \frac{2}{3} \scriptsize P \: \times\: 12 \right) \: - \: \left(\normalsize \frac{1}{6} \scriptsize \: \times\: 12 \right)$$

:- $$\scriptsize 9P \: + \: 36 = 8P \: - \: 2$$

Step 4:- collecting like terms

:- $$\scriptsize 9P \: - \: 8P = \: - \: 2 \: - \: 36$$

:- $$\scriptsize P = -38$$

## Question 6b

A television set was marked for sale at GH ȼ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.

Step 1:- Determine the cost price of the television set:

:- $$\scriptsize 120 \% \equiv$$ Ghȼ 760.00

:- $$\scriptsize 110 \% \equiv \normalsize \frac{760}{12} \scriptsize \: \times \: 100$$

Cost price  =  GHȼ633

Step 2:- Calculate the amount the television is eventually sold

Discount  = 5%  of 760

:- $$\frac{5}{100}\scriptsize \: \times \: 760$$

= GHȼ38.00

The amount the television was sold:

= GHȼ(76038)

=  GHȼ 722

Step3:- The actual percentage profit

:- $$\frac{Profit}{Actual \: Cost \: Price}\scriptsize \: \times \: 100$$

:- $$\frac{722 \: - \: 633.33}{633.33}\scriptsize \: \times \: 100$$

= $$\frac{88.67}{633.33}\scriptsize \: \times \: 100$$

= 14.0006

≅ 14% (2 sig. figs.)

## Question 7a

Solution:   y = 2sinx + 1

 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 2sin 0 1.0 1.7 2.0 1.7 1.0 0.0 -1.0 -1.7 -2.0 +1 +1 +1 +1 +1 +1 +1 +1 1.0 +1 +1 Y 1.0 2.0 2.7 3.0 2.7 2.0 1.0 0.0 -0.7 -1.0

Step 1:- Fill the cells in the second row by substituting the value of x  into 2sinx  i.e

when x = 0º

2sinx = 2sin0º =  2 x 0 = 0

when x = 30º

2sinx = 2sin30º =  2 x 0.5 = 1.0

when x = 60º

2sinx = 2sin60º =  2 x 0.866 = 1.7

when x = 90º

2sinx = 2sin90º =  2 x 1 = 2

when x = 60º

2sinx = 2sin120º =  2 x 0.5 = 1

when x = 180º

2sinx = 2sin180º =  2 x 0 = 0

when x = 210º

2sinx = 2sin210º =  2 x -0.5 = -1

when x = 240º

2sinx = 2sin240º =  2 x -0.866 = -1.7

when x = 270º

2sinx = 2sin270º =  2 x -1 = -2

Step 2:- Fill the cells in the third row with +1(because it is a constant)

Step 3:- Fill the last row by summing up items in the 2nd and 3rd cells.

## Question 7c

Use the graph to find the values of x for which $$\scriptsize Sinx = \normalsize \frac{1}{4}$$

Solution:

Sinx = \normalsize \frac{1}{4} [/latex]

Step1:-  Multiply the above equation through by 2

:- $$\scriptsize 2 Sinx \: + \: 1 = \normalsize \frac{1}{2} \scriptsize \: + \: 1$$

:- $$\scriptsize 2 Sinx \: + \: 1 = \scriptsize 1.5$$.............(1)

Step 3:- compare equation 1 with y = 2Sinx + 1

y = 1.5

Step 4:- The solution to

$$\scriptsize Sinx = \normalsize \frac{1}{4}$$ are the values of  x where the line y  cuts the graph. The points are labelled A and B on the graph.

Step 5:- Read the values of A and B

x = 12º(Point A) and

x = 168º (Point B)

x = 12º or 168º

## Question 8a

(a) Copy and complete the following table for multiplication modulo 11.

 ⊗ 1 5 9 10 1 1 5 9 10 5 5 9 9 10 10

Solution:

 ⊗ 1 5 9 10 1 1 5 9 10 5 5 3 1 6 9 9 1 4 2 10 10 6 2 1

To complete the above table in modulo 11.

Step 1:- Fill the empty cells in the 3rd row as follows

:- $$\scriptsize 5 \bigotimes 5 \\ = \scriptsize 5 \: \times \: 5 \\ = \scriptsize 25\: \div \: 11 \\ = \scriptsize 2 \: remainder \: 3$$

Write the remainder, i.e 3

:- $$\scriptsize 5 \bigotimes 9 \\ = \scriptsize 5 \: \times \: 9 \\ = \scriptsize 45\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 1$$

Write the remainder, i.e 1

:- $$\scriptsize 5 \bigotimes 10 \\ = \scriptsize 5 \: \times \: 10 \\ = \scriptsize 50\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 6$$

Write the remainder, i.e 6

Step 2:- Move to the 4th row repeat the process i. e

:- $$\scriptsize 9 \bigotimes 5 \\ = \scriptsize 9 \: \times \: 5 \\ = \scriptsize 45\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 1$$

Write the remainder, i.e 1

:- $$\scriptsize 9 \bigotimes 9 \\ = \scriptsize 9 \: \times \: 9 \\ = \scriptsize 81\: \div \: 11 \\ = \scriptsize 7 \: remainder \: 4$$

Write the remainder, i.e 4

:- $$\scriptsize 9 \bigotimes 10 \\ = \scriptsize 9 \: \times \: 10 \\ = \scriptsize 90\: \div \: 11 \\ = \scriptsize 8 \: remainder \: 2$$

Write the remainder, i.e 2

Step 3:- Move to the 5th row repeat the process i. e

:- $$\scriptsize 10 \bigotimes 5 \\ = \scriptsize 10 \: \times \: 5 \\ = \scriptsize 50\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 6$$

Write the remainder, i.e 6

:- $$\scriptsize 10 \bigotimes 9 \\ = \scriptsize 10 \: \times \: 9 \\ = \scriptsize 10\: \div \: 11 \\ = \scriptsize 8 \: remainder \: 2$$

Write the remainder, i.e 2

:- $$\scriptsize 10 \bigotimes 10 \\ = \scriptsize 10 \: \times \: 10 \\ = \scriptsize 100\: \div \: 11 \\ = \scriptsize 9 \: remainder \: 1$$

Write the remainder, i.e 1

Use the table to:

i. Evaluate (9 5) (10 10);

Step 1:- From the table above

$$\scriptsize (9 \bigotimes 5) = 1$$ .............(1)

$$\scriptsize (10 \bigotimes 10) = 1$$ .............(2)

Step 2:- Substitute the corrrect values from the previous question into equation (1) and (2)

(9 5) (10 10)

= 1 ⊗ 1

= 1

ii. Find the truth set of

(a) 10 m = 2

Step1:- study the above and trace the value of m

From the above table, 10 9 = 2

By comparison, therefore

m   = 9

(b) n n = 4.

Step1:- From the above, it is only

9 9 = 4

By comparison, therefore

n = 9

## Question 8b

When a fraction is reduced to its lowest term, it is equal to $$\frac{3}{4}$$ . The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.

Solution:

Step 1: First transform the world problem into mathematical sentences

Let the numerator of the fraction be x

Let the denominator by y, then from the first sentence we have

⇒ $$\frac{x}{y} = \frac{3}{4}$$

Step 2:- Clearing the fraction we have

4x =  3y  ------------------(1)

Step 3:- the second sentence of the equation can be interpreted as

2x = y + 34 ----------------(2)

Step 4:- Solve the equation (1) and (2) simultaneously. Substitute equation (2) into equation (1)

$$\scriptsize 4 \left(\normalsize \frac{y \: + \: 34}{2}\right ) \scriptsize = 3y$$

= $$\scriptsize 2 (y \: + \: 34) \scriptsize = 3y$$

:-  $$\scriptsize 2 (y \: + \: 68) \scriptsize = 3y$$

:-  $$\scriptsize 68 = 3y \: - \: 2y$$

:-  $$\scriptsize y = 68$$

Step 5:- Now substitute for y in equation (1)

:-  $$\scriptsize 4x = 3 \: \times \: 68$$

x = $$\frac{3 \: \times \: 68}{4}$$

x = 3 x 17

x = 51

Hence the fraction is $$\frac{51}{68}$$

## Question 9a

In the Venn diagram, P, Q and R are subsets of the universal set U. If n(U) = 125, find:

i. the value of x

Solution: Given that  n(U) = 125

(i). To calculate x

Step 1:- Sum up all what we have in the cells and the rectangle

125 = (16 - 2x) + 5x + 4x + 8x + (6 + x) + 7x + (19 - 3x) + 4

Step 2:- Simplifying and solving for x we have

125 = 45 + 20x

20x = 125 - 45

20x = $$\frac{80}{20}$$

x = 4

ii.  $$\scriptsize n( P \cup Q \cap R' )$$

Step 1: From the diagram,

$$\scriptsize n( P \cup Q \cap R' )$$ = (16 - 2x) + 5x + (6 + x)

Step 2:- Substitute x = 4  into the expression in step 1

(16 - 8) + 20 + 10

= 8 + 20 + 10

∴$$\scriptsize n( P \cup Q \cap R' ) = 38$$

## Question 9b

In the diagram, O is the centre of the circle. If WX is parallel to YZ and <WXY = 500, find the value of :

i. <WYZ

Solution: To find the value of $$\scriptsize \hat{WYZ}$$

Step 1:- We first find the angle $$\scriptsize \hat{XWY}$$

Since $$\scriptsize \hat{XYW} = 90^o$$

(angle in a semi-circle)   then,

:- $$\scriptsize \hat{XYW} \: + \: \hat{XWY} \: + \: \hat{WXY} = 180^o$$

(Sum of angles in a triangle)

:- $$\scriptsize 90^o \: + \: \hat{XWY} \: + \: 50^o = 180^o$$

:- $$\scriptsize \hat{XWY} \: + \: 50^o = 180^o - 140^o$$

:- $$\scriptsize 40^o$$

Step 2:- Determine $$\scriptsize \hat{WYZ}$$

$$\scriptsize \hat{WYZ} = \hat{XWY} = 40^o$$

(alternate angles since YZ || XW)

$$\scriptsize \hat{WYZ} = 40^o$$

ii. <YEZ

Step 1:- Determine $$\scriptsize \hat{ZOW}$$

:- $$\scriptsize \hat{ZOW} = 2 \: \times \: \hat{ZYW}$$

(angle subtends at the center of a circle by an Arc)

:- $$\scriptsize \hat{ZOW} = 2 \: \times \: 40^o = 80^0$$

Step 2:- Determine the size of the angle $$\scriptsize \hat{WEO}$$

From ΔWEO

:- $$\scriptsize 80^o \: + \: 40^o \: + \: \hat{WEO} = 180^o$$

(Sum of angles in a triangle)

:- $$\scriptsize \hat{WEO} = 180^o \: - \: 120$$

:- $$\scriptsize \hat{WEO} = 60^o$$

Step 3:- Now determine $$\scriptsize \hat{YEZ}$$

:- $$\scriptsize \hat{YEZ} = \hat{WEO} = 60^o$$

Vertically opposite angles)

Hence $$\scriptsize \hat{YEZ} = 60^o$$

## Question 10a

Solve $$\scriptsize (x \: - \: 2)(x \: - \: 3) = 12$$

Solution:

Step 1:- First expand the expression on the LHS of the equation

:- $$\scriptsize (x \: - \: 2)(x \: - \: 3) = 12$$

:- $$\scriptsize x^2 \: - \: 3x \: - 2x \: + \: 6 = 12$$

:- $$\scriptsize x^2 \: - \: 5x \: + \: 6 = 12$$

Step 2:- Move the item of the RHS to the left

:- $$\scriptsize x^2 \: - \: 5x \: + \: 6 \: - \: 12 = 0$$

:- $$\scriptsize x^2 \: - \: 5x \: - \: 6 = 0$$

Step3:- Factorize the expression on the left hand and solve for the unknown

:- $$\scriptsize x^2 \: - \: 6x \: + \: x \: - \: 6 = 0$$

:- $$\scriptsize x(x \: - \: 6) + 1(x \: - \: 6) = 0$$

:- $$\scriptsize (x \: + \: 1) (x \: - \: 6) = 0$$

x = 6 or - 1

## Question 10b

In the diagram, M and N are the centres of two circles of equal radii 7cm. The circles intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion.

$$\left [ \scriptsize Take \: \pi \: = \normalsize \frac{22}{7} \right ]$$

Step 1:- Join P to Q

Step 2:- Find the area of the minor segment in the circle

i.e.

Step 3:- Area of minor segment =   Area of the sector -  Area of triangle PNQ

Area of the sector = $$\frac{60}{360} \: \times \: \frac{22}{7} \scriptsize \: \times \: 7 \: \times \: 7$$

= $$\frac{77}{3} \scriptsize cm^2 = 25.66cm^2$$

Area of triangle PNQ = $$\frac{1}{2} \scriptsize \: \times \: 7 \: \times \: 7 \: \times \: sin60$$

= $$\frac{49 \: \times \: 0.8660}{2} \scriptsize cm^2 = 21.27cm^2$$

Area of the minor segment = (25.66 - 21.27)cm²

= 4.44cm²

Step 4:- the shaded region in the figure, is

2 x 4.44

= 8.88cm²

## Question 11

 Scores 1 2 3 4 5 6 Frequency 2 5 13 11 9 10

The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the

i. Mean deviation of the distribution

Solution:

 Scores (x) Frequency (f) f Deviation $$\scriptsize (d) = x \: - \: \bar{x}$$ |d| f|d| 1 2 2 -3 3 6 2 5 10 -2 2 10 3 13 39 -1 1 13 4 11 44 0 0 0 5 9 45 1 1 9 6 10 60 2 2 20 ∑50 ∑200 ∑69

Step 1:- Set out the table, as shown above, find the third column by taking the product of “f" and "x "

e.g.

1 x 2 = 2

2 x 5 = 10

e.t.c.........

Step 2:- Determine the cumulative frequency by summing the 2nd column, i.e $$\scriptsize \sum f = 50$$

Step 2:-

Find the mean using the formula $$\scriptsize \bar{x} = \frac{\sum fx }{f}$$

:- $$\scriptsize \sum fx = 200$$

:- $$\scriptsize \bar{x} = \normalsize\frac{200 }{5} \scriptsize = 4$$

Step 4:- Determine the value in the 4th column, using $$\scriptsize d = x \: - \: \bar{x}$$

e.g

1 - 4 = -3

2 - 4 = -2

3 - 4 = -1

e.t.c .......

Step 5:- Find the value in the 5th column by finding the absolute value of d, i.e |d|,  By simply removing the negative signs.

Step 6:- Determine the value in the 6th column by taking the product of |d|

e.g

2 x 3 = 6

5 x 2 = 10

13 x 1 = 13

e.t.c .......

Step 7:- Take the sum of the items in the 6th column i. e

:- $$\scriptsize \sum f |d| = 6+10+13+0+9+20 = 69$$

Step 8:- Mean deviation is obtained from:

:- $$\frac{\sum f |d|}{\sum f}$$

= $$\frac{69}{50}$$

= 1.38

ii. Probability that a score selected at random is at least a 4.

Step 1:- let the event a score selected at random is at least 4 be T

Step 2:- Determine the value of T.

T = 9 + 11 + 10 = 30

Step 3:- Determine the probability of T

Prob(T) = $$\frac{30}{50} = \frac{3}{5}$$

= 0.6

## Question 12a

(a) Given that $$\scriptsize 5cos(x \: + \: 8.5)^o \: - \: 1 = 0,\: 0^o \leq x \leq 90^o$$

calculate, correct to the nearest degree, the value of x.

Solution:

:- $$\scriptsize 5cos(x \: + \: 8.5)^o \: - \: 1 = 0$$

Step 1:- Move "-1" to the RHS

:- $$\scriptsize 5cos(x \: + \: 8.5)^o = 1$$

Step 2:- Divide through by 5

:- $$\scriptsize cos(x \: + \: 8.5)^o = \normalsize \frac{1}{5}$$

:- $$\scriptsize cos(x \: + \: 8.5)^o = 0.2$$

Step 3:- take the cosine inverse of both sides

:- $$\scriptsize cos^{-1}cos(x \: + \: 8.5)^o = cos^{-1}(0.2)$$

:- $$\scriptsize (x \: + \: 8.5)^o =78.46^0$$

Step 4:- Make x  the subject of the formula

:- $$\scriptsize x = 78.46^0 \: - \: 8.5^o$$

:- $$\scriptsize x = 69.96^o$$

:- $$\scriptsize x \approx 69.96^o$$

(to the nearest degree)

## Question 12b

(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P

(i) represent this information in a diagram;

Solution:

Step 1:- The diagram should be sketched as shown above

ii. To calculate the distance between Q and R

Step 1:- Use cosine rule

$$\scriptsize p^2 = q^2 \: + \: r ^2 \: - \: 2qrcosp$$

Step2:- Substitute for q, r and p in the formula

:- $$\scriptsize p^2 = 32^2 \: + \: 24 ^2 \: - \: 2\: \times \: 32 \: \times \: 24cos 45^o$$

:- $$\scriptsize p^2 = 1024 \: + \: 576 \: - \: 1536cos 45^o$$

:- $$\scriptsize p^2 = 1600 \: - \: (1536 \: \times \: 0.7071)$$

:- $$\scriptsize p^2 = 1600 \: - \: 1086.11$$

:- $$\scriptsize p^2 = 513.89$$

:- $$\scriptsize p = \sqrt{513.89}$$

:- $$\scriptsize p = 22.669$$

:- $$\scriptsize p = 22.67 \: km$$

to 2 decimal places

(iii) To calculate the bearing of R from Q

Step1:- First find the angle Q using sine rule

:- $$\frac{q}{SinQ} = \frac{p}{SinP}$$

Step 2:- Substitute the value of P and Q into the formula

We have $$\frac{32}{SinQ} = \frac{22.67}{Sin45^o}$$

Step 3:- Make sin Q  the subject

SinQ = $$\frac{32 \: \times \: sin45^o}{22.67}$$

SinQ = $$\frac{22.6724}{22.67}$$

SinQ = 0.9981

Q = $$\scriptsize sin^{-1}(0.9981)$$

Q = 86.48

:- $$\scriptsize Q \approx 86^o$$

Step 4:- From the diagram, the bearing of R from Q is

= 270º - (86 - 60)º

= 270º - 26º

= 244º

(To the nearest degree)

## Question 13a

Two functions, f and g, are defined by

$$\scriptsize f : x \rightarrow 2x^2 \: - \: 1 \: and \: g: x \rightarrow 3x \: + \: 2$$

where x is a real number.

(i) If f(x - 1) – 7 = 0, find the values of x

Solution:

$$\scriptsize f(x) = 2x^2 \: - \: 1$$ ...............(1)

$$\scriptsize g(x) = 3x^2 \: + \: 2$$ ...............(2)

To find x in f(x - 1) - 7 = 0

Step 1:- First from an expression for f(x - 1), by substitution (x -1) for x in equation (1)

i.e $$\scriptsize f(x \: - \: 1) = 2(x \: - \: 1)^2 \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2(x^2\: - \: 2x \: + \: 1) \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2x^2\: - \: 4x \: + \: 2 \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2x^2\: - \: 4x \: + \: 1$$ ............(3)

Step 2:- Substitute equation (3) into f(x - 1) - 7 = 0

:- $$\scriptsize 2x^2\: - \: 4x \: + \: 1 - 7 = 0$$

:- $$\scriptsize 2x^2\: - \: 4x \: - \: 6 = 0$$

Step 3:- Solve the quadratic equation, first reduce the equation by dividing through by 2

:- $$\scriptsize x^2\: - \: 2x \: - \: 3 = 0$$

:- $$\scriptsize x^2\: - \: 3x \: + \: x \: - \: 3 = 0$$

:- $$\scriptsize x(x \: - \: 3) \: + \: 1(x \: - \: 3) = 0$$

:- $$\scriptsize (x \: - \: 3) (x \: - \: 1) = 0$$

x = 3 or - 1

(ii) Evaluate $$\frac{f \left( -\frac{1}{2} \right).g(3)}{f(4) \: - \: g(5)}$$

Step 1:- Find the value of f(-½), g(3), f(4), g(5) and substitute the values into the expression

f(-½) = $$\scriptsize 2 \left (\frac{1}{2} \right )^2 \: - \: 1 \\ = \scriptsize 2 \: \times \: \normalsize \frac{1}{4} \scriptsize \: - \: 1 \\ = \frac{1}{2} \scriptsize \: - \: 1 \\ = \: - \frac{1}{2}$$

f(-½) = $$- \frac{1}{2}$$

g(3) = 3(3) + 2

g(3) = 9 + 2

g(3) = 11

f(4) = 2(4)² - 1

f(4) = 2(4)² - 1

f(4) = 32 - 1

f(4) = 31

g(5) = 3(5) + 2 = 15 + 2

g(5) = 17

Step 2:- Substituting the values into the expression we have

:- $$\frac{f \left( -\frac{1}{2} \right).g(3)}{f(4) \: - \: g(5)} \\ = \frac{-\frac{1}{2} \: \times \: 11}{31 \: - \: 17}$$

= $$\frac{-\frac{11}{2}}{14}$$

= $$-\frac{11}{2} \: \times \: \frac{1}{14}$$

= $$-\frac{11}{28}$$

## Question 13b

An operation, $$\scriptsize (\ast)$$ is defined on the set R, of real numbers, by $$\scriptsize m(\ast)n$$=  $$\frac{-n}{m^2 \: + \: 1}$$

where  $$\scriptsize m, n \epsilon R$$

If $$\scriptsize -3, -10 \epsilon R$$ Show whether or not $$\scriptsize (\ast)$$ is commutative

Solution:

$$\scriptsize m(\ast )n = \normalsize \frac{-n}{m^2 \: + \: 1}, \:\scriptsize m, \: n \: \epsilon R$$

To show whether or not is commutative given that $$\scriptsize -3, -10 \epsilon R$$

then,

$$\scriptsize -3(\ast ) \: - \: 10 = \: -10(\ast) \: - \: 3$$

Step 1:- Find the value of $$\scriptsize -3(\ast ) \: - \: 10$$

$$\scriptsize -3(\ast ) \: - \: 10 = \normalsize \frac{-(-10)}{(-3)^2 \: + \: 1} \\ = \normalsize \frac{10}{9 \: + \: 1}$$

$$\scriptsize -3(\ast ) \: - \: 10 = \normalsize \frac{10} {10} \scriptsize = 1$$ ........(1)

Step 2:- Also find the value of  $$\scriptsize -10(\ast ) \: - \: 3$$

$$\scriptsize -10(\ast ) \: - \: 3 = \normalsize \frac{-(-3)}{(-10)^2 \: + \: 1} \\ = \normalsize \frac{3}{100 \: + \: 1}$$

$$\scriptsize -10(\ast ) \: - \: 3 = \normalsize \frac{3} {101}$$ ........(2)

Step 3:- Compare equations (1) and (2). Since equation (1) is not equal to equation (2)

$$\scriptsize 1 \neq \normalsize \frac{3}{101}$$

Hence

$$\scriptsize -3(\ast ) \: - \: 10 \neq -10(\ast ) \: - \: 3$$

Therefore $$\scriptsize (\ast)$$ is not commutative.

## Question 1a

(a) Without using tables or calculator, simplify :

$$\frac{0.6 \: \times \: 32 \: \times \: 0.004}{1.2 \: \times \: 0.008 \: \times \: 0.16}$$

leaving the answer in standard form (scientific notation).

Solution: Step 1: Rewrite each term of the fraction in standard form

:- $$\frac{6 \: \times \: 10^{-1} \: \times \: 3.2 \: \times \: 10^{1} \: \times \: 4 \: \times \: 10^{-3}}{1.2 \: \times \: 8 \: \times \: 10^{-3} \times 1.6 \: \times \: 10^{-1} }$$

Step 2:- Divide through by common factors

= $$\frac{3 \: \times \: 2 \: \times \: 10^1}{1.2}$$

= $$\frac{6 \times \: 10^1}{1.2}$$

= $$\frac{6 \times \: 10^1}{1.2}$$

=  $$\frac{10^1}{0.2}$$

Step 3:- Write in standard form

=  $$\frac{10^1}{2 \: \times \: 10^1}$$

=  $$\scriptsize 10 \: \times \: \normalsize \frac{1}{2}\scriptsize \: \times \: 10^1$$

=  $$\scriptsize 5 \: \times \: 10^1$$

## Question 1b

(b) In the diagram, $$\scriptsize\overline{EF}$$ is parallel to $$\scriptsize\overline{GH}$$.

If < AEF = 3x°,< ABC = 120° and < CHG = 7x°, find the value of < GHB.

Step 2:- Determine the size of $$\scriptsize \hat{WBC}$$

:- $$\scriptsize \hat{WBC} = 180^o \: - \: 120^o = 60^o$$

(sum of angle on a straight line)

Step 3:-  $$\scriptsize \hat{BWH}$$ = 3xº (corresponding angle, since $$\scriptsize \bar{EF} || \bar{GH}$$

Step 4:- But $$\scriptsize \hat{WBC} \: + \: \hat{BWH} = 7x^o$$

(sum of the opposite angles of an exterior angle of a triangle)

60º + 3xº = 7xº

60º = 7xº - 3xº

60º = 4xº

xº = $$\frac{60}{4}$$

xº = 15º

Step 5:-

:- $$\scriptsize \hat{GHB} = 180 \: -\: 7x$$

(sum of angle on a straight line)

:- $$\scriptsize \hat{GHB} = (180 \: -\: 7 \: \times \: 15)$$

:- $$\scriptsize \hat{GHB} = (180 \: -\: 105)$$

:- $$\scriptsize \hat{GHB} = 75^o$$

## Question 2a

(a) Simplify : 3√75 - √12 + √108, leaving the answer in surd form (radicals).

Step 1:- First write each term in the base form.

:- $$\scriptsize 3\sqrt{25 \: \times \: 3} \: - \: \sqrt{4 \: \times \: 3} \: + \: \sqrt{36 \: \times \: 3}$$

:- $$\scriptsize 3 \: \times \: 5\sqrt{3} \: - \: 2 \sqrt{3} \: + \: 6\sqrt{3}$$

=  $$\scriptsize 19 \sqrt{3}$$

## Question 2b

If 124n = 232five, find n.

Solution: Step 1:- Express each number in base ten.

Step 2:- Simplify this we have

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 = 50\: + \: 15\: + \: 2$$

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 = 67$$

Step 3:- Move 67 to the LHS

:- $$\scriptsize n^2 \: + \: 2n \: + \: 4 - 67 = 0$$

:- $$\scriptsize n^2 \: + \: 2n \: - \: 63 = 0$$

Step 4:- Factorize the expression and solve for the unknown

:- $$\scriptsize n^2 \: + \: 9n \: - \: 2n \: - \: 63 = 0$$

:- $$\scriptsize n(n \: + \: 9) \: - \: 7(n \: + \: 9) = 0$$

:- $$\scriptsize n = -9 \: or \: 7$$

Note: the correct value of n is 7, n cannot be expressed as a negative number

n = 7

## Question 3a

(a) Solve the simultaneous equation:

$$\frac{1}{x} \: + \: \frac{1}{y} \scriptsize = 5$$

$$\frac{1}{y} \: - \: \frac{1}{x} \scriptsize = 1$$

Solution: Step 1:- First transform the equation by using:

p = $$\frac{1}{x}$$

and

q = $$\frac{1}{y}$$

Step 2:- Now transform the equation:

:- $$\scriptsize p \: + \: q = 5$$

:- $$\scriptsize q \: + \: p = 1$$

Step 3:- Re-arrange the equation

:- $$\scriptsize p \: + \: q = 5$$.............(1)

:- $$\scriptsize -p \: + \: q = 1$$.............(2)

2q  = 6

q = $$\frac{6}{2}$$

q = 3

Step 4:- Substitute q in equation (1)

:- $$\scriptsize p \: + \: q = 5$$.............(1)

:- $$\scriptsize p \: + \: 3 = 5$$

:- $$\scriptsize p = 5 \: - \: 3$$

:- $$\scriptsize p = 2$$

∴ p = 2, q = 3

Step 5:- Now find the value of x and y from the transformation

p = $$\frac{1}{x}$$

and

q = $$\frac{1}{y}$$

For p = $$\frac{1}{x}$$

2 = $$\frac{1}{x}$$

∴x = $$\frac{1}{2}$$

Similarly for q = $$\frac{1}{y}$$

3 = $$\frac{1}{y}$$

y = $$\frac{1}{3}$$

## Question 3b

A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.

Step 1:-  Find the value of x and y from the transformation change the word problem into a mathematical expression.

Total distance travelled = 48 km

Total time taken = 45 minutes

= $$\frac{45}{60} \scriptsize = 0.75hour$$

Let the man travel a distance of x km on the good surface. Then he travels a distance of (48 - x) km on the bad surface.

Similarly, Let the time taken to travel on the good surface be t hours, then the time taken to travel on the bad surface = (0.75 - t) hrs.

Step 1:- Using Speed = $$\frac{distance}{time}$$

On the good surface:

72 = $$\frac{x}{t}$$

x = 72t ..............(1)

48 = $$\frac{48 \: -\: x}{0.75 \: - \: t}$$

48(0.75 − t) = 48 − x

36 − 48t = 48 − x--------------- (2)

Step 3:- Substitute equation (1) into equation (2)

36 − 48t = 48 − 72t

72t − 48t = 48 - 36

24t = 12

t = $$\frac{12}{24}$$

t = 0.5hours

Step 4:- Substitute for t in equation (1)

x = 72 × 0.5

=36km

Therefore, the man travels a distance of 36 km on the good road surface.

## Question 4a

(a) In the diagram, O is the centre of the circle radius r cm and < XOY = 90°. If the area of the shaded part is 504cm2, calculate the value of r. [Take π=22/7].

Solution:

Step 1: state the formula of the area of a sector

Area of a sector = $$\frac{\theta}{360} \scriptsize \: \times \: \pi r^2$$

Step2:- Area of the shaded region = Area of the sectorArea of the triangle XOY.

Area of  ΔXOY = $$\frac{1}{2} \scriptsize \: \times \: b \: \times \: h$$

= $$\frac{1}{2} \scriptsize \: \times \:r \: \times \: r$$

= $$\frac{r^2}{2}$$

Step 3: Simplify the expression below:

Area of the shaded region = Area of the sectorArea of the triangle XOY.

504 = $$\frac{90}{360} \: \times \: \frac{22}{7} \: \times \: r^2 \: - \: \frac{r^2}{2}$$

504 = $$\frac{11}{14} \: - \: \frac{r^2}{2}$$

:- $$\scriptsize 504 = 0.7857r^2 \: - \: 0.5r^2$$

:- $$\scriptsize 504 = 0.2857r^2$$

r² = $$\frac{504}{0.2857}$$

r² = $$\scriptsize 1764$$

r² = $$\scriptsize \sqrt{1764}$$

r = 42cm

## Question 4b

(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If <PQR = 66° and <PSQ = 109°, calculate the value of <RQS.

Step 1:- Sketch the diagram as shown above

Step 2:- $$\scriptsize \hat{RQP} = 66^0$$

(base angle of an isosceles triangle)

Similarly,

:- $$\scriptsize \hat{SQP} = \normalsize\frac{180 \: - \: 109}{2}$$

(base angle of an isosceles triangle)

:-$$\scriptsize \hat{SQP} = \normalsize \frac{71}{2} \scriptsize = 35.5$$

Step 3:- $$\scriptsize \hat{RQS} =\hat{RQP} \: + \: \hat{SQP}$$

:- $$\scriptsize \hat{RQS} = 66^o \: + \: 35.5^o$$

:- $$\scriptsize \hat{RQS} = 101.5^o$$

## Question 5

A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :

(a) both contracts ;

(b) exactly one of the contracts ;

(c) neither of the contracts?

Solution:

Prob (X) = 0.5

Prob (Y) = 0.3

Step 1:- Determine the prob(X1) and prob(Y)

Prob(X1) = 1 - 0.5 = 0.5

Prob(Y1) = 1 - 0.3 = 0.7

Step 2:- Determine the probability that he will win both contracts

:- $$\scriptsize Prob( X \cap Y) = Prob(X^1) \: \times \: Prob(Y^1)$$

(Since the two events are independent events)

:- $$\scriptsize Prob( X \cap Y) = 0.5 \: \times \: 0.7$$

:- $$\scriptsize Prob( X \cap Y) = 0.35$$

Step 3:- Determine the probability that the man wins exactly one of the contracts.

$$\scriptsize Prob(X \cap Y^1) = Prob(X^1 \cap Y)$$

= (0.5 × 0.3) + (0.5 × 0.7)

= 0.15 + 0.35

= 0.5

## Question 6a

(a) If $$\frac{3}{2p\: - \: \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p \: + \: 1}$$
find p

Solution:

Step 1:- Multiply through by the LCM of the denominator, i.e

:- $$\left( \scriptsize 2P \: -\: \normalsize \frac{1}{2} \right) \: - \: \left( \frac{1}{4} \scriptsize P \: + \: 1 \right)$$

= $$\scriptsize 3 \left( \normalsize \frac{1}{4} \scriptsize \: + \: 1 \right) = \normalsize \frac{1}{3} \scriptsize \left (2P \: - \: \normalsize \frac{1}{2} \right)$$

Step 2:- Open the bracket

:- $$\normalsize \frac{3}{4} \scriptsize P \: + \: 3 = \normalsize \frac{2}{3} \scriptsize P \: - \: \normalsize \frac{1}{6}$$

Step 3:- Multiply through by 12 (LCM of the denominators)

:- $$\left ( \normalsize\frac{3}{4} \scriptsize P \: \times\: 12 \right) \scriptsize \: + \: (3 \: \times\: 12) = \left( \normalsize \frac{2}{3} \scriptsize P \: \times\: 12 \right) \: - \: \left(\normalsize \frac{1}{6} \scriptsize \: \times\: 12 \right)$$

:- $$\scriptsize 9P \: + \: 36 = 8P \: - \: 2$$

Step 4:- collecting like terms

:- $$\scriptsize 9P \: - \: 8P = \: - \: 2 \: - \: 36$$

:- $$\scriptsize P = -38$$

## Question 6b

A television set was marked for sale at GH ȼ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.

Step 1:- Determine the cost price of the television set:

:- $$\scriptsize 120 \% \equiv$$ Ghȼ 760.00

:- $$\scriptsize 110 \% \equiv \normalsize \frac{760}{12} \scriptsize \: \times \: 100$$

Cost price  =  GHȼ633

Step 2:- Calculate the amount the television is eventually sold

Discount  = 5%  of 760

:- $$\frac{5}{100}\scriptsize \: \times \: 760$$

= GHȼ38.00

The amount the television was sold:

= GHȼ(76038)

=  GHȼ 722

Step3:- The actual percentage profit

:- $$\frac{Profit}{Actual \: Cost \: Price}\scriptsize \: \times \: 100$$

:- $$\frac{722 \: - \: 633.33}{633.33}\scriptsize \: \times \: 100$$

= $$\frac{88.67}{633.33}\scriptsize \: \times \: 100$$

= 14.0006

≅ 14% (2 sig. figs.)

## Question 7a

Solution:   y = 2sinx + 1

 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 2sin 0 1.0 1.7 2.0 1.7 1.0 0.0 -1.0 -1.7 -2.0 +1 +1 +1 +1 +1 +1 +1 +1 1.0 +1 +1 Y 1.0 2.0 2.7 3.0 2.7 2.0 1.0 0.0 -0.7 -1.0

Step 1:- Fill the cells in the second row by substituting the value of x  into 2sinx  i.e

when x = 0º

2sinx = 2sin0º =  2 x 0 = 0

when x = 30º

2sinx = 2sin30º =  2 x 0.5 = 1.0

when x = 60º

2sinx = 2sin60º =  2 x 0.866 = 1.7

when x = 90º

2sinx = 2sin90º =  2 x 1 = 2

when x = 60º

2sinx = 2sin120º =  2 x 0.5 = 1

when x = 180º

2sinx = 2sin180º =  2 x 0 = 0

when x = 210º

2sinx = 2sin210º =  2 x -0.5 = -1

when x = 240º

2sinx = 2sin240º =  2 x -0.866 = -1.7

when x = 270º

2sinx = 2sin270º =  2 x -1 = -2

Step 2:- Fill the cells in the third row with +1(because it is a constant)

Step 3:- Fill the last row by summing up items in the 2nd and 3rd cells.

## Question 7c

Use the graph to find the values of x for which $$\scriptsize Sinx = \normalsize \frac{1}{4}$$

Solution:

Sinx = \normalsize \frac{1}{4} [/latex]

Step1:-  Multiply the above equation through by 2

:- $$\scriptsize 2 Sinx \: + \: 1 = \normalsize \frac{1}{2} \scriptsize \: + \: 1$$

:- $$\scriptsize 2 Sinx \: + \: 1 = \scriptsize 1.5$$.............(1)

Step 3:- compare equation 1 with y = 2Sinx + 1

y = 1.5

Step 4:- The solution to

$$\scriptsize Sinx = \normalsize \frac{1}{4}$$ are the values of  x where the line y  cuts the graph. The points are labelled A and B on the graph.

Step 5:- Read the values of A and B

x = 12º(Point A) and

x = 168º (Point B)

x = 12º or 168º

## Question 8a

(a) Copy and complete the following table for multiplication modulo 11.

 ⊗ 1 5 9 10 1 1 5 9 10 5 5 9 9 10 10

Solution:

 ⊗ 1 5 9 10 1 1 5 9 10 5 5 3 1 6 9 9 1 4 2 10 10 6 2 1

To complete the above table in modulo 11.

Step 1:- Fill the empty cells in the 3rd row as follows

:- $$\scriptsize 5 \bigotimes 5 \\ = \scriptsize 5 \: \times \: 5 \\ = \scriptsize 25\: \div \: 11 \\ = \scriptsize 2 \: remainder \: 3$$

Write the remainder, i.e 3

:- $$\scriptsize 5 \bigotimes 9 \\ = \scriptsize 5 \: \times \: 9 \\ = \scriptsize 45\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 1$$

Write the remainder, i.e 1

:- $$\scriptsize 5 \bigotimes 10 \\ = \scriptsize 5 \: \times \: 10 \\ = \scriptsize 50\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 6$$

Write the remainder, i.e 6

Step 2:- Move to the 4th row repeat the process i. e

:- $$\scriptsize 9 \bigotimes 5 \\ = \scriptsize 9 \: \times \: 5 \\ = \scriptsize 45\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 1$$

Write the remainder, i.e 1

:- $$\scriptsize 9 \bigotimes 9 \\ = \scriptsize 9 \: \times \: 9 \\ = \scriptsize 81\: \div \: 11 \\ = \scriptsize 7 \: remainder \: 4$$

Write the remainder, i.e 4

:- $$\scriptsize 9 \bigotimes 10 \\ = \scriptsize 9 \: \times \: 10 \\ = \scriptsize 90\: \div \: 11 \\ = \scriptsize 8 \: remainder \: 2$$

Write the remainder, i.e 2

Step 3:- Move to the 5th row repeat the process i. e

:- $$\scriptsize 10 \bigotimes 5 \\ = \scriptsize 10 \: \times \: 5 \\ = \scriptsize 50\: \div \: 11 \\ = \scriptsize 4 \: remainder \: 6$$

Write the remainder, i.e 6

:- $$\scriptsize 10 \bigotimes 9 \\ = \scriptsize 10 \: \times \: 9 \\ = \scriptsize 10\: \div \: 11 \\ = \scriptsize 8 \: remainder \: 2$$

Write the remainder, i.e 2

:- $$\scriptsize 10 \bigotimes 10 \\ = \scriptsize 10 \: \times \: 10 \\ = \scriptsize 100\: \div \: 11 \\ = \scriptsize 9 \: remainder \: 1$$

Write the remainder, i.e 1

Use the table to:

i. Evaluate (9 5) (10 10);

Step 1:- From the table above

$$\scriptsize (9 \bigotimes 5) = 1$$ .............(1)

$$\scriptsize (10 \bigotimes 10) = 1$$ .............(2)

Step 2:- Substitute the corrrect values from the previous question into equation (1) and (2)

(9 5) (10 10)

= 1 ⊗ 1

= 1

ii. Find the truth set of

(a) 10 m = 2

Step1:- study the above and trace the value of m

From the above table, 10 9 = 2

By comparison, therefore

m   = 9

(b) n n = 4.

Step1:- From the above, it is only

9 9 = 4

By comparison, therefore

n = 9

## Question 8b

When a fraction is reduced to its lowest term, it is equal to $$\frac{3}{4}$$ . The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.

Solution:

Step 1: First transform the world problem into mathematical sentences

Let the numerator of the fraction be x

Let the denominator by y, then from the first sentence we have

⇒ $$\frac{x}{y} = \frac{3}{4}$$

Step 2:- Clearing the fraction we have

4x =  3y  ------------------(1)

Step 3:- the second sentence of the equation can be interpreted as

2x = y + 34 ----------------(2)

Step 4:- Solve the equation (1) and (2) simultaneously. Substitute equation (2) into equation (1)

$$\scriptsize 4 \left(\normalsize \frac{y \: + \: 34}{2}\right ) \scriptsize = 3y$$

= $$\scriptsize 2 (y \: + \: 34) \scriptsize = 3y$$

:-  $$\scriptsize 2 (y \: + \: 68) \scriptsize = 3y$$

:-  $$\scriptsize 68 = 3y \: - \: 2y$$

:-  $$\scriptsize y = 68$$

Step 5:- Now substitute for y in equation (1)

:-  $$\scriptsize 4x = 3 \: \times \: 68$$

x = $$\frac{3 \: \times \: 68}{4}$$

x = 3 x 17

x = 51

Hence the fraction is $$\frac{51}{68}$$

## Question 9a

In the Venn diagram, P, Q and R are subsets of the universal set U. If n(U) = 125, find:

i. the value of x

Solution: Given that  n(U) = 125

(i). To calculate x

Step 1:- Sum up all what we have in the cells and the rectangle

125 = (16 - 2x) + 5x + 4x + 8x + (6 + x) + 7x + (19 - 3x) + 4

Step 2:- Simplifying and solving for x we have

125 = 45 + 20x

20x = 125 - 45

20x = $$\frac{80}{20}$$

x = 4

ii.  $$\scriptsize n( P \cup Q \cap R' )$$

Step 1: From the diagram,

$$\scriptsize n( P \cup Q \cap R' )$$ = (16 - 2x) + 5x + (6 + x)

Step 2:- Substitute x = 4  into the expression in step 1

(16 - 8) + 20 + 10

= 8 + 20 + 10

∴$$\scriptsize n( P \cup Q \cap R' ) = 38$$

## Question 9b

In the diagram, O is the centre of the circle. If WX is parallel to YZ and <WXY = 500, find the value of :

i. <WYZ

Solution: To find the value of $$\scriptsize \hat{WYZ}$$

Step 1:- We first find the angle $$\scriptsize \hat{XWY}$$

Since $$\scriptsize \hat{XYW} = 90^o$$

(angle in a semi-circle)   then,

:- $$\scriptsize \hat{XYW} \: + \: \hat{XWY} \: + \: \hat{WXY} = 180^o$$

(Sum of angles in a triangle)

:- $$\scriptsize 90^o \: + \: \hat{XWY} \: + \: 50^o = 180^o$$

:- $$\scriptsize \hat{XWY} \: + \: 50^o = 180^o - 140^o$$

:- $$\scriptsize 40^o$$

Step 2:- Determine $$\scriptsize \hat{WYZ}$$

$$\scriptsize \hat{WYZ} = \hat{XWY} = 40^o$$

(alternate angles since YZ || XW)

$$\scriptsize \hat{WYZ} = 40^o$$

ii. <YEZ

Step 1:- Determine $$\scriptsize \hat{ZOW}$$

:- $$\scriptsize \hat{ZOW} = 2 \: \times \: \hat{ZYW}$$

(angle subtends at the center of a circle by an Arc)

:- $$\scriptsize \hat{ZOW} = 2 \: \times \: 40^o = 80^0$$

Step 2:- Determine the size of the angle $$\scriptsize \hat{WEO}$$

From ΔWEO

:- $$\scriptsize 80^o \: + \: 40^o \: + \: \hat{WEO} = 180^o$$

(Sum of angles in a triangle)

:- $$\scriptsize \hat{WEO} = 180^o \: - \: 120$$

:- $$\scriptsize \hat{WEO} = 60^o$$

Step 3:- Now determine $$\scriptsize \hat{YEZ}$$

:- $$\scriptsize \hat{YEZ} = \hat{WEO} = 60^o$$

Vertically opposite angles)

Hence $$\scriptsize \hat{YEZ} = 60^o$$

## Question 10a

Solve $$\scriptsize (x \: - \: 2)(x \: - \: 3) = 12$$

Solution:

Step 1:- First expand the expression on the LHS of the equation

:- $$\scriptsize (x \: - \: 2)(x \: - \: 3) = 12$$

:- $$\scriptsize x^2 \: - \: 3x \: - 2x \: + \: 6 = 12$$

:- $$\scriptsize x^2 \: - \: 5x \: + \: 6 = 12$$

Step 2:- Move the item of the RHS to the left

:- $$\scriptsize x^2 \: - \: 5x \: + \: 6 \: - \: 12 = 0$$

:- $$\scriptsize x^2 \: - \: 5x \: - \: 6 = 0$$

Step3:- Factorize the expression on the left hand and solve for the unknown

:- $$\scriptsize x^2 \: - \: 6x \: + \: x \: - \: 6 = 0$$

:- $$\scriptsize x(x \: - \: 6) + 1(x \: - \: 6) = 0$$

:- $$\scriptsize (x \: + \: 1) (x \: - \: 6) = 0$$

x = 6 or - 1

## Question 10b

In the diagram, M and N are the centres of two circles of equal radii 7cm. The circles intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion.

$$\left [ \scriptsize Take \: \pi \: = \normalsize \frac{22}{7} \right ]$$

Step 1:- Join P to Q

Step 2:- Find the area of the minor segment in the circle

i.e.

Step 3:- Area of minor segment =   Area of the sector -  Area of triangle PNQ

Area of the sector = $$\frac{60}{360} \: \times \: \frac{22}{7} \scriptsize \: \times \: 7 \: \times \: 7$$

= $$\frac{77}{3} \scriptsize cm^2 = 25.66cm^2$$

Area of triangle PNQ = $$\frac{1}{2} \scriptsize \: \times \: 7 \: \times \: 7 \: \times \: sin60$$

= $$\frac{49 \: \times \: 0.8660}{2} \scriptsize cm^2 = 21.27cm^2$$

Area of the minor segment = (25.66 - 21.27)cm²

= 4.44cm²

Step 4:- the shaded region in the figure, is

2 x 4.44

= 8.88cm²

## Question 11

 Scores 1 2 3 4 5 6 Frequency 2 5 13 11 9 10

The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the

i. Mean deviation of the distribution

Solution:

 Scores (x) Frequency (f) f Deviation $$\scriptsize (d) = x \: - \: \bar{x}$$ |d| f|d| 1 2 2 -3 3 6 2 5 10 -2 2 10 3 13 39 -1 1 13 4 11 44 0 0 0 5 9 45 1 1 9 6 10 60 2 2 20 ∑50 ∑200 ∑69

Step 1:- Set out the table, as shown above, find the third column by taking the product of “f" and "x "

e.g.

1 x 2 = 2

2 x 5 = 10

e.t.c.........

Step 2:- Determine the cumulative frequency by summing the 2nd column, i.e $$\scriptsize \sum f = 50$$

Step 2:-

Find the mean using the formula $$\scriptsize \bar{x} = \frac{\sum fx }{f}$$

:- $$\scriptsize \sum fx = 200$$

:- $$\scriptsize \bar{x} = \normalsize\frac{200 }{5} \scriptsize = 4$$

Step 4:- Determine the value in the 4th column, using $$\scriptsize d = x \: - \: \bar{x}$$

e.g

1 - 4 = -3

2 - 4 = -2

3 - 4 = -1

e.t.c .......

Step 5:- Find the value in the 5th column by finding the absolute value of d, i.e |d|,  By simply removing the negative signs.

Step 6:- Determine the value in the 6th column by taking the product of |d|

e.g

2 x 3 = 6

5 x 2 = 10

13 x 1 = 13

e.t.c .......

Step 7:- Take the sum of the items in the 6th column i. e

:- $$\scriptsize \sum f |d| = 6+10+13+0+9+20 = 69$$

Step 8:- Mean deviation is obtained from:

:- $$\frac{\sum f |d|}{\sum f}$$

= $$\frac{69}{50}$$

= 1.38

ii. Probability that a score selected at random is at least a 4.

Step 1:- let the event a score selected at random is at least 4 be T

Step 2:- Determine the value of T.

T = 9 + 11 + 10 = 30

Step 3:- Determine the probability of T

Prob(T) = $$\frac{30}{50} = \frac{3}{5}$$

= 0.6

## Question 12a

(a) Given that $$\scriptsize 5cos(x \: + \: 8.5)^o \: - \: 1 = 0,\: 0^o \leq x \leq 90^o$$

calculate, correct to the nearest degree, the value of x.

Solution:

:- $$\scriptsize 5cos(x \: + \: 8.5)^o \: - \: 1 = 0$$

Step 1:- Move "-1" to the RHS

:- $$\scriptsize 5cos(x \: + \: 8.5)^o = 1$$

Step 2:- Divide through by 5

:- $$\scriptsize cos(x \: + \: 8.5)^o = \normalsize \frac{1}{5}$$

:- $$\scriptsize cos(x \: + \: 8.5)^o = 0.2$$

Step 3:- take the cosine inverse of both sides

:- $$\scriptsize cos^{-1}cos(x \: + \: 8.5)^o = cos^{-1}(0.2)$$

:- $$\scriptsize (x \: + \: 8.5)^o =78.46^0$$

Step 4:- Make x  the subject of the formula

:- $$\scriptsize x = 78.46^0 \: - \: 8.5^o$$

:- $$\scriptsize x = 69.96^o$$

:- $$\scriptsize x \approx 69.96^o$$

(to the nearest degree)

## Question 12b

(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P

(i) represent this information in a diagram;

Solution:

Step 1:- The diagram should be sketched as shown above

ii. To calculate the distance between Q and R

Step 1:- Use cosine rule

$$\scriptsize p^2 = q^2 \: + \: r ^2 \: - \: 2qrcosp$$

Step2:- Substitute for q, r and p in the formula

:- $$\scriptsize p^2 = 32^2 \: + \: 24 ^2 \: - \: 2\: \times \: 32 \: \times \: 24cos 45^o$$

:- $$\scriptsize p^2 = 1024 \: + \: 576 \: - \: 1536cos 45^o$$

:- $$\scriptsize p^2 = 1600 \: - \: (1536 \: \times \: 0.7071)$$

:- $$\scriptsize p^2 = 1600 \: - \: 1086.11$$

:- $$\scriptsize p^2 = 513.89$$

:- $$\scriptsize p = \sqrt{513.89}$$

:- $$\scriptsize p = 22.669$$

:- $$\scriptsize p = 22.67 \: km$$

to 2 decimal places

(iii) To calculate the bearing of R from Q

Step1:- First find the angle Q using sine rule

:- $$\frac{q}{SinQ} = \frac{p}{SinP}$$

Step 2:- Substitute the value of P and Q into the formula

We have $$\frac{32}{SinQ} = \frac{22.67}{Sin45^o}$$

Step 3:- Make sin Q  the subject

SinQ = $$\frac{32 \: \times \: sin45^o}{22.67}$$

SinQ = $$\frac{22.6724}{22.67}$$

SinQ = 0.9981

Q = $$\scriptsize sin^{-1}(0.9981)$$

Q = 86.48

:- $$\scriptsize Q \approx 86^o$$

Step 4:- From the diagram, the bearing of R from Q is

= 270º - (86 - 60)º

= 270º - 26º

= 244º

(To the nearest degree)

## Question 13a

Two functions, f and g, are defined by

$$\scriptsize f : x \rightarrow 2x^2 \: - \: 1 \: and \: g: x \rightarrow 3x \: + \: 2$$

where x is a real number.

(i) If f(x - 1) – 7 = 0, find the values of x

Solution:

$$\scriptsize f(x) = 2x^2 \: - \: 1$$ ...............(1)

$$\scriptsize g(x) = 3x^2 \: + \: 2$$ ...............(2)

To find x in f(x - 1) - 7 = 0

Step 1:- First from an expression for f(x - 1), by substitution (x -1) for x in equation (1)

i.e $$\scriptsize f(x \: - \: 1) = 2(x \: - \: 1)^2 \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2(x^2\: - \: 2x \: + \: 1) \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2x^2\: - \: 4x \: + \: 2 \: - \: 1$$

:- $$\scriptsize f(x \: - \: 1) = 2x^2\: - \: 4x \: + \: 1$$ ............(3)

Step 2:- Substitute equation (3) into f(x - 1) - 7 = 0

:- $$\scriptsize 2x^2\: - \: 4x \: + \: 1 - 7 = 0$$

:- $$\scriptsize 2x^2\: - \: 4x \: - \: 6 = 0$$

Step 3:- Solve the quadratic equation, first reduce the equation by dividing through by 2

:- $$\scriptsize x^2\: - \: 2x \: - \: 3 = 0$$

:- $$\scriptsize x^2\: - \: 3x \: + \: x \: - \: 3 = 0$$

:- $$\scriptsize x(x \: - \: 3) \: + \: 1(x \: - \: 3) = 0$$

:- $$\scriptsize (x \: - \: 3) (x \: - \: 1) = 0$$

x = 3 or - 1

(ii) Evaluate $$\frac{f \left( -\frac{1}{2} \right).g(3)}{f(4) \: - \: g(5)}$$

Step 1:- Find the value of f(-½), g(3), f(4), g(5) and substitute the values into the expression

f(-½) = $$\scriptsize 2 \left (\frac{1}{2} \right )^2 \: - \: 1 \\ = \scriptsize 2 \: \times \: \normalsize \frac{1}{4} \scriptsize \: - \: 1 \\ = \frac{1}{2} \scriptsize \: - \: 1 \\ = \: - \frac{1}{2}$$

f(-½) = $$- \frac{1}{2}$$

g(3) = 3(3) + 2

g(3) = 9 + 2

g(3) = 11

f(4) = 2(4)² - 1

f(4) = 2(4)² - 1

f(4) = 32 - 1

f(4) = 31

g(5) = 3(5) + 2 = 15 + 2

g(5) = 17

Step 2:- Substituting the values into the expression we have

:- $$\frac{f \left( -\frac{1}{2} \right).g(3)}{f(4) \: - \: g(5)} \\ = \frac{-\frac{1}{2} \: \times \: 11}{31 \: - \: 17}$$

= $$\frac{-\frac{11}{2}}{14}$$

= $$-\frac{11}{2} \: \times \: \frac{1}{14}$$

= $$-\frac{11}{28}$$

## Question 13b

An operation, $$\scriptsize (\ast)$$ is defined on the set R, of real numbers, by $$\scriptsize m(\ast)n$$=  $$\frac{-n}{m^2 \: + \: 1}$$

where  $$\scriptsize m, n \epsilon R$$

If $$\scriptsize -3, -10 \epsilon R$$ Show whether or not $$\scriptsize (\ast)$$ is commutative

Solution:

$$\scriptsize m(\ast )n = \normalsize \frac{-n}{m^2 \: + \: 1}, \:\scriptsize m, \: n \: \epsilon R$$

To show whether or not is commutative given that $$\scriptsize -3, -10 \epsilon R$$

then,

$$\scriptsize -3(\ast ) \: - \: 10 = \: -10(\ast) \: - \: 3$$

Step 1:- Find the value of $$\scriptsize -3(\ast ) \: - \: 10$$

$$\scriptsize -3(\ast ) \: - \: 10 = \normalsize \frac{-(-10)}{(-3)^2 \: + \: 1} \\ = \normalsize \frac{10}{9 \: + \: 1}$$

$$\scriptsize -3(\ast ) \: - \: 10 = \normalsize \frac{10} {10} \scriptsize = 1$$ ........(1)

Step 2:- Also find the value of  $$\scriptsize -10(\ast ) \: - \: 3$$

$$\scriptsize -10(\ast ) \: - \: 3 = \normalsize \frac{-(-3)}{(-10)^2 \: + \: 1} \\ = \normalsize \frac{3}{100 \: + \: 1}$$

$$\scriptsize -10(\ast ) \: - \: 3 = \normalsize \frac{3} {101}$$ ........(2)

Step 3:- Compare equations (1) and (2). Since equation (1) is not equal to equation (2)

$$\scriptsize 1 \neq \normalsize \frac{3}{101}$$

Hence

$$\scriptsize -3(\ast ) \: - \: 10 \neq -10(\ast ) \: - \: 3$$

Therefore $$\scriptsize (\ast)$$ is not commutative.