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Quiz 16 of 16

2014 Mathematics WAEC Theory Past Questions

WAEC: MATHEMATICS 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

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.

Solution: To find the value of <GHB

Step1:– produce \(\scriptsize \bar{AB} \:  to \: W \: on \: \bar{GH} \)

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)

Add equation 1 and 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)

On the bad surface: 

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 sector – Area 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 sector – Area of the triangle XOY.

504 = \( \frac{90}{360} \: \times \: \frac{22}{7} \: \times \: r^2 \: – \: \frac{r^2}{2} \)

504 = \( \frac{11}{14}r^2 \: – \: \frac{r^2}{2} \)

:- \( \scriptsize 504 = 0.7857r^2 \: – \: 0.5r^2 \)

:- \( \scriptsize 504 = 0.2857r^2 \)

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(760 – 38)

 =  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 7b

 

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.

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