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Quiz 10 of 16

2017 Mathematics WAEC Theory Past Questions

WAEC: MATHEMATICS 2017 Mathematics WAEC Theory Past Questions

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Question 1a

If (y−1)log104 = ylog1016, without using Mathematics tables or calculator, find the value of y.

Solution

Step 1:

(y−1)log104 = ylog1016, re-write as

(y−1)log104 = ylog104²

(y−1)log104 = 2ylog104

Step 2: divide through by log104

\( \frac{(y \: -\:1) \log_{10} 4}{\log_{10} 4} = \frac{2y\log_{10} 4}{\log_{10} 4} \)

y – 1 = 2y

y – 2y = 1

-y = 1

y = -1

Question 1b

(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.

Solution

Step 1: Speed = \( \frac{distance}{time}\)

Step 2: let the distance covered be x km

At 4km/h, the time taken to get to the office from the house = \( \frac{x}{4}\scriptsize hr\)

At 5km/h, the time taken to get to the office from the house = \( \frac{x}{5}\scriptsize hr\)

\( \frac{x}{4} \: – \: \frac{x}{5}\scriptsize = 30mins\)

Step 3: Changing minutes to hour

\( \frac{x}{4} \: – \: \frac{x}{5} = \frac{30}{60}\) \( \frac{x}{4} \: – \: \frac{x}{5} = \frac{1}{2}\)

Step 4: multiply through by 20

\(\normalsize \frac{x}{4} \scriptsize \: \times \: 20 \: – \: \normalsize  \frac{x}{5}\scriptsize \: \times \: 20  = \normalsize\frac{1}{2}\scriptsize \: \times \: 20  \)

5x – 4x = 10

x = 10

Hence the distance is 10km.

Question 2a

Solve the equation: \( \normalsize \frac{2}{3} \scriptsize (3x \: -\:5) \: – \: \normalsize  \frac{3}{5} \scriptsize (2x \: -\: 3) = 3 \)

Solution

Step 1: Multiply the equation through by 15 (the LCM of the denominators)

⇒ \( \normalsize \frac{2}{3} \scriptsize (3x \: -\:5) \: \times \: 15\: – \: \normalsize  \frac{3}{5} \scriptsize (2x \: -\: 3)\: \times \: 15 = 3\: \times \: 15 \)

⇒ 10(3x – 5) – 9(2x -3) = 45

Step 2: Open the brackets and simplify

  30x – 50 – 18x + 27 = 45

12x – 23 = 45

12x = 45 + 23

12x = 68

Step 3: Make x the subject

x = \( \frac{68}{12}  =  \frac{17}{3} \)

x = \( 5 \frac{2}{3}  \)

or

x = 5.66

Question 2a

(b) In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).

Solution

Step 1: From the diagram \( \scriptsize n \: + \: \hat{UQT} \: + \: 88^o = 180^o \)

 [sum of angles on a straight line]

\( \scriptsize \hat{UQT}  =180^o\: – \: 88^o  \: – \: n \) 

\( \scriptsize \hat{UQT}  = 92^o  \: – \: n \) 

Step 2: But m = \( \scriptsize  80^o \: + \: \hat{UQT}  \) 

{exterior angle of a triangle is equal to the sum of the opposite interior angles}

m = 800 + 920 – n

m + n = 800 + 920

m + n = 1720

Question 3a

The angle of depression of a point P on the ground from the top T of a building is 23.6º. If the distance, from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height of the building.

Solution

Step 1: Sketch the diagram as shown below

Step 2: From the triangle

Tan 23.60 = \( \frac{h}{50} \)

h = 50 x Tan 23.60

h = 21.844

h = 22m (to the nearest meter)

Question 3b

(b)In the diagram, PT||SU, QS||TR, |SR| = 6cm and |RU| = 10cm. If the area of  ΔTRU = 45cm2, Calculate the area of trapezium QTUS.

Solution

Step 1: |SR| = |QT| = 6cm [Opposite sides of parallelogram SQTR]

Step 2: Determine the height of the triangle TRU, Using

Area = \( \frac{1}{2} \scriptsize\: \times \: base \: \times \: height \)

45 = \( \frac{1}{2} \scriptsize\: \times \: 10 \: \times \: h \)

h = \( \frac{45}{5} \)

h = 9cm

Step 3: This height of 9cm is also the height of the trapezium QTUS, Area of the trapezium is given as

\( \frac{1}{2} \scriptsize (a \: + \: b)h\)

Step 4: Substitute the given parameters into the formula

Area = \( \frac{1}{2} \scriptsize (16 \: + \: 6) \: \times \: 9\)

Area = \( \frac{1}{2} \scriptsize \: \times \: 22 \: \times \: 9\)

= 11 x 9

99cm2

Question 4

If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the

a. First term

Solution

Step 1: The nth term and the sum of the first n-terms of an A.P is given by

Tn = a + (n – 1)d

Sn = \( \frac{n}{2} \scriptsize [2a \: + \: (n \: – \: 1)d] \)

Step 2: T6 = 37

  37 = a + (6 – 1) d

37 = a + 5d ———- (1)

Step 3: S6 = 147

147 = \( \frac{6}{2} \scriptsize [2a \: + \: (6 \: – \: 1)d] \)

147 = \(  \scriptsize 3 [2a \: + \: 5d] \)

49 = 2a + 5d —————- (2)

Step 4: solving equations (1) and (2) simultaneously we have

  37 = a + 5d

  49 = 2a + 5d

-12 = -a

a = 12

 

b. Sum of the first fifteen terms

 Step 5: the sum of the first fifteen terms is obtained from

Sn = \( \frac{n}{2} \scriptsize [2a \: + \: (n \: – \: 1)d] \)

Substitute a = 12 into equation (1)

37 = 12 + 5d

5d = 37 – 12

5d = 25

d = \( \frac{25}{5} \)

d = 5

Step 6: substitute a = 12, d = 5 and n = 15 into the formula

S15 = \( \frac{15}{2} \scriptsize [2(12) \: + \: (15 \: – \: 1)5] \)

S15 = \( \frac{15}{2} \scriptsize [24 \: + \: (14 \: \times \: 5)] \)

S15 = \( \frac{15}{2} \scriptsize [24 \: + \: 70] \)

S15 = \( \frac{15}{2}\: \times \: \frac{94}{1}\)

S15 = 705

Question 5

Out of 120 customers in a shop 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:

a. Illustrate this information in a diagram

b. Find the number of customers who bought shoes:

c. Calculate the probability that a customer selected at random bought bags.

Solution

Step 1: Let the following notations represent those customers

e.g. B – those who bought bags

      S – those who bought shoes

Step 2: and let

n (B/S) = x

n (S/B) = (11 + x)

n (B n S) = 45

 

a.

 

b. Step 3:  from the diagram

  120 = x + 45 + x + 11

120 = 2x + 56

120 – 56 = 2x

64 = 2x

x = \( \frac{64}{2}\)

x = 32

Step 4: therefore the number of customers that bought shoes is

= 45 + 11 + 32

= 88

 

c.

Step 5: the number of customers that bought bags

= 32 + 45

= 77

Step 6: the required probability

= \( \frac{77}{120} \)

Question 6a

A manufacturing company requires 3 hours of direct labour to process every N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget for direct labour at N18.25 per hour?

Solution

Step 1: No of hours to process N87.00 worth of raw materials = 3 hours

No of hours required to process raw materials that worth N30,450.00

⇒ \( \frac{30,450.00}{47}  \scriptsize \: \times \: 3 \)

= 1,050 hours

Step 2: the amount the company should budget for direct labour at N18.25 per hour

= 1,050 x 18.25

= N19,162.50

Question 6b

An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the

(i). Values of x and y

Solution

Step 1: From the equation

x + y = 8,000,000 .00 ——– (1)

Step 2: The interest on Nx in bank M at the rate of 6% is

I1 = \( \frac{x \: \times \: 6 \: \times \: 4}{100} \)

I1 = 0.24x

Step 3: The interest on Ny in bank N at the rate of 8% is

I2 = \( \frac{y \: \times \: 8 \: \times \: 4}{100} \)

I2 =0.32y

Step 4: But Interest = I1 + I2

  0.24x + 0.32y = N2,320,000

Step 5: Solving equations (1) and (2) simultaneously we have

From equation (1)

y = 8,000, 000 – x —— (3)

Step 6: substitute for y in equation (2)

  0.24x + 0.32 (8,000,000 – x) = 2,320,000

  0.24x + 2560,000 – 0.32x = 2,320,000

-0.08x = 2,320,000 – 2560,000

-0.08x = -240,000

x = \( \frac{-240,000}{-0.08} \)

x = N3,000,000

Step 7: Substitute the value of x in equation (3)

  y = 8,000,000 – 3,000,000

y = N5,000,000

 

(ii) Interest paid by the second bank

Solution

Step 8: the interest paid by the second bank is obtained from 0.32y

  0.32 x N5,000,000

= N1,600,000

Question 7

a. Copy and complete the table of values for the equation

y = 2x2 – 7x – 9 for -3 ≤  x ≤ 6

x

-3

-2

-1

0

1

2

3

4

5

6

y

  

13

  

-9

-14

-12

  

6

    

 

Solution

Step 1: Copy and complete the table as shown below

x

-3

-2

-1

0

1

2

3

4

5

6

2x2

18

8

2

0

2

8

18

32

50

72

-7x

21

14

7

0

-7

-14

-21

-28

-35

-42

-9

-9

-9

-9

-9

-9

-9

-9

-9

-9

-9

y

30

13

0

-9

-14

-15

-12

-5

-6

21

 

 

(b) Using scales of 2cm to 1 unit on the x-axis and 2cm to 4 units on the y – axis, draw the graphs of

y = 2x2 – 7x – 9   for -3 ≤  x ≤ 6

 

(c) Use the graph to estimate the

(i) roots  of the equation 2x2 – 7x = 26

(ii) coordinates of the minimum point of y;

(iii) range of values for which 2x2 – 7x < 9.

Solution

Step 1: Given that 2x2 – 7x  = 26. Subtract “+9” from both sides

2x2 – 7x – 9 = 26 – 9

2x2 – 7x – 9 = 17 ———– (*)

Step 2: Compare equation (*) with the graph

y = 2x2 – 7x – 9

y = 17

Step 3: the solution to the equation 2×2 – 7x = 26 are the values of x where the curve intersect the line y = 17 i.e. from the graph the points are labelled points A and B

i.e. x = -2.2 or 5.7

(ii) Step 4: The coordinates of the minimum point of y are (2, -15) from the graph.

(iii) Step 5: the range of values for which 2×2 – 7x < 9 is the region shaded on the graph.

i.e. -1 < x < 4.5

Question 8

Marks 1 2 3 4 5
Number of students m+2 m-1 2m-3 m+5 3m-4

The table shows the distribution of marks scored by some students in a test.

(a) If the mean mark is \( \scriptsize 3\frac{6}{23} \), find the value of m

Solution

Step 1: mean is defined mathematically as

\(\scriptsize  \bar{x} = \normalsize \frac{\sum fx}{\sum f} \)

\(\scriptsize \sum fx \) = (m + 2) + 2 (m – 1)  + 3 (2m – 3) + 4(m + 5) + 5 (3m – 4)

\(\scriptsize \sum fx \) = m + 2 + 2m – 2 + 6m – 9 + 4m + 20 + 15m – 20

\(\scriptsize \sum fx \) = 28m – 9

Step 2: Determine \(\scriptsize \sum f \)

\(\scriptsize \sum f \) = (m + 2) + (m – 1) + (2m – 3) + (m + 5) + (3m – 4)

\(\scriptsize \sum f \) = m + 2 + m – 1 + 2m – 3 + m + 5 + 3m – 4

\(\scriptsize \sum f \) = 8m – 1

Step 3: But mean \(\scriptsize  \bar{x} = \normalsize \frac{\sum fx}{\sum f} \)

\(\scriptsize  \bar{x} = \normalsize \frac{28m \: – \: 9}{8m \: – \: 1} = \scriptsize 3 \frac{6}{23} \)

\(\scriptsize  \bar{x} = \normalsize \frac{28m \: – \: 9}{8m \: – \: 1} =  \frac{75}{23} \)

Step 4: Solving for m

  75 (8m – 1) = 23 (28m – 9)

600m – 75 = 644m – 207

-75 + 207 = 644m – 600m

132 = 44m

m = \( \frac{132}{44} \)

m = 3

 

(b) Find the

i. Interquartile range

Solution

Step 1: Interquartile range

Q = Q3 – Q1

Step 2: Q3 = 75th percentile

Q3 

= \( \left( \frac{75}{100} \scriptsize \: \times \: 23 \right)^{th} \)

= 17.25th

Observation

Q1 = 25th percentile

= \( \left( \frac{25}{100} \scriptsize \: \times \: 23 \right)^{th} \)

= 5.75th

Observation

Step 3: from the frequency table, with m = 3, i.e.

Marks x

1

2

3

4

5

Frequency

5

2

3

8

5

Q3 = 4 and Q1 = 2

Step 4: Interquartile range

= 4 – 2

= 2

ii. Probability of selecting a student who scored at least 4 marks in the test.

Solution

Step 1: the probability of selecting students that scored at least 4 marks in the test is

= \( \frac{No \: of \:those\: who \: score \: at \: least \: one}{23} \)

Step 2: Probability = \( \frac{8 \: + \: 5}{23} \)

= \( \frac{13}{23} \)

= 0.5652

Question 9a

(a) PQ is a tangent to a circle RST at the point S. PRT is a Straight line, <TPS = 34º and <TSQ = 65º

(i) Illustrate the information in a diagram

Solution

 Step 1: the information can be illustrated as shown below

(ii) Find the value of (a) <RTS (b) <SRP

Step 2: From the diagram \( \scriptsize \hat{SRT} =\hat{TSQ} = 65^o \) 

[angle in alternate segment]

Step 3: \( \scriptsize \hat{PRS}\) = 180º – 65º

\( \scriptsize \hat{PRS}\) = 115º

[sum of angles on a straight line]

Step 4: \( \scriptsize \hat{RSQ}\) = 34º + 115º

\( \scriptsize \hat{RSQ}\) = 149º

But \( \scriptsize \hat{RST} =\hat{RSQ} \: – \: \hat{TSQ}  \)

\( \scriptsize \hat{RST} =149^o\: – \: 65^o  \)

\( \scriptsize \hat{RST} =84^o  \)

Step 5: Hence \( \scriptsize \hat{RTS} = 180^o\: – \: (65^o \: + \: 84^o)  \)

\( \scriptsize \hat{RTS} = 180^o\: – \: 149^o  \)

[sum of angles in Δ]

\( \scriptsize \hat{RTS} = 31^o  \)

 

(b) \( \scriptsize \hat{SRP} = 115^o  \)

(Calculated above)

Question 9b

(b) In the diagram, |VZ| = |YZ|, <YXZ = 200 and <ZVY = 520, calculate the size of <WYZ.

Solution

Step 1: From the diagram

\( \scriptsize \hat{YWZ} = \hat{ZVY} = 52^o \)

(Angles in the same segment)

⇒ \( \scriptsize \hat{YWZ}  = 52^o \)

Step 2: \( \scriptsize \hat{VYZ} = \hat{YVZ} = 52^o \)

(base angles of an Isosceles triangle)

\( \scriptsize \hat{VYZ}  = 52^o \)

Step 3: But \( \scriptsize \hat{YZW} = 180^o  \: – \: (20^o \: + \: 52^o )\)

\( \scriptsize \hat{YZW} = 180^o  \: – \: 72^o\)

\( \scriptsize \hat{YZW} = 108^o\)

Step 4: Therefore

\( \scriptsize \hat{WYZ} = 180^o  \: – \: (52^o \: + \: 108^o )\)

\( \scriptsize \hat{WYZ} = 180^o  \: – \: 160^o\)

\( \scriptsize \hat{WYZ} = 20^o\)

Question 10a

Given that Sin x = \( \frac{5}{13}\scriptsize, O^o < x < 90^o\)

Find \( \frac{cosx \: – \: 2 sinx}{2 tan x} \)

Solution

Step 1: Using right angled triangle, we have

Step 2: determine the third side using Pythagoras theorem.

  y2 = 13 – 5

y2 = 169 – 25 = 144

y = √144

y = 12

Step 3: from the triangle

Cos x = \( \frac{12}{13} \)

Tan x = \( \frac{5}{12} \)

Step 4: Substituting the trigonometry ratios into

\( \frac{cos x \: – \: 2sin x}{2 tan x} = \frac{\frac{12}{13} \: -\: 2 \left(\frac{5}{13} \right)}{2 \left(\frac{5}{12} \right)} \)

Step 5: Simplifying further we have

 = \( \left( \frac{12}{13} \: -\: \frac{10}{13} \right ) \: \div \: \left( \frac{10}{12}\right) \)

 = \( \frac{2}{13}  \: \times\:\frac{12}{10}\)

 = \( \frac{12}{65} \)

Question 10b

A ladder, LA, leans against a vertical pole at a point L which is 9.6meters above the ground. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the;

(i) Length of ladder LA 

Solution

Step 1: Sketch the diagram as shown below

Step 2: Let LA makes angle  θ  with the horizontal

Step 3: From  ΔLBD

BD2 = 122 – 9.62

BD2 = 144 – 92.16

BD2 = 51.84

BD = \( \scriptsize \sqrt{51.84}\)

BD = 7.2m

Step 4: from the diagram

|AD| = |AB| – |BD|

|AD| = 10 – 7.2

|AD| = 2.8m

Step 5: From  ΔADL, by Pythagoras

LA² = 9.62 + 2.82

LA² = 92.16 + 7.84

LA² = 100

LA = √100

LA = 10m

 

(ii) angle which LA makes with the ground

Solution

(ii) Step 1: from ΔADL

Sinθ = \( \frac{9.6}{10} \)

Sinθ = 0.96

θ = Sin -1(0.96)

θ = 73.7º

θ = 74º (Correct to 2 significant figures)

Question 11a

It takes 8 students two-thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one-third of an hour at the same rate?

Solution

Step 1: Two-third of an hour = \( \frac{2}{3} \scriptsize \: \times \: 60 \\ \scriptsize = 40 \: minutes \)

Step 2: Using proportion method

8 students (use 40mins) to fill 12 tanks

∴  4 students (will use 80mins) to fill 12 tanks

Step 3:

80mins is required by the 4 students to fill 12 tanks

1min is required by the 4 students to fill \( \frac{12}{80} \) tanks

Step 4: Therefore,

20mins will be required by the 4 students to fill

\( \left( \frac{12}{80} \scriptsize \: \times \: 20\right) \)

⇒ \(\frac{12}{80} \scriptsize \: \times \: 20\)

= \(\frac{12}{4} \scriptsize = 3\)

= 3 tanks

Question 11b

A chord, 20cm long is 12cm from the centre of circle. Calculate, correct to one decimal place, the;

(i) Angle subtended by the chord at the centre of the circle;

Solution

Step 1: Sketch the diagram as shown below:

Step 2: Let θ be the angle \( \scriptsize \hat{BOC} \), from ΔOBC

Tanθ = \( \frac{opp}{hyp} \)

Tanθ = \( \frac{10}{12} \)

Tanθ   = 0.8333

θ = \( \scriptsize tan^{-1} (0.8333) \)

θ   = 39.80º

Step 3: the angle subtends by the chord at the centre of the circle is \( \scriptsize \hat{AOC} \)

\( \scriptsize \hat{AOC} = 2 \: \times \: \hat{BOC} \)

   = 2 x 39.800

        = 79.60

      = 79.60

 

(ii) Perimeter of the minor segment cut off by the chord. [Take π = 3.142]

Solution

Step 1: the perimeter of the minor segment = Length of the arc + the length of the chord |AC|.

Step 2: Length of the arc

L =  \( \frac{\theta}{360}  \scriptsize \: \times \: 2 \pi r \)

L =  \( \frac{79.6}{360} \: \times \: \frac{3.142}{1} \scriptsize \: \times \: 2r \)…(1)

Step 3: the radius (r) is obtained by Pythagoras theorem

i.e. OC2 = OB2 + BC2

r2 = 122 + 102

r2 = 144 + 100

r2 = 244

r = √244

r = 15.62cm

Step 4: substitute r = 15.62 in equation (1)

L =  \( \frac{79.6}{360} \: \times \: \frac{3.142}{1} \scriptsize \: \times \: 15.62 \: \times \: 2 \)

L = 21.703cm

Length of chord = 10 + 10 = 20cm

Step 5: therefore the perimeter of the minor segment is

21.703 + 20

= 41.703

= 41.7cm (to one decimal place)

Question 12a

Using completing the square method, solve, correct to 2 decimal places, the equation

3y2 – 5y + 2 = 0

Solution

Step 1: move the constant to the RHS

3y2 – 5y = -2

Step 2: divide through by the coefficient of y2 (i.e. 3)

\( \frac{3y^2}{3} \: – \: \frac{5y}{3} = \frac{-2}{3} \) \( \scriptsize y^2 \normalsize \: – \: \frac{5y}{3} = \frac{-2}{3} \)

Step 3: add to both sides the square of half of the coefficient of the middle term

\( \scriptsize y^2 \normalsize \: – \: \frac{5y}{3} \: + \: \left( \frac{5}{6} \right)^2 = \frac{-2}{3}\: + \: \left( \frac{5}{6} \right)^2 \)

Step 4: the LHS is now a complete square, factorise and solve for the unknown:

\( \left(  \scriptsize y \: – \: \normalsize \frac{5}{6}\right)^2 = \frac{-2}{3}\: + \: \frac{25}{36} \) \( \left(  \scriptsize y \: – \: \normalsize \frac{5}{6}\right)^2 = \frac{-24 \: + \: 25}{36} \) \( \left(  \scriptsize y \: – \: \normalsize \frac{5}{6}\right)^2 = \frac{1}{36} \)

Step 5: taking the square root of both sides

\( \sqrt{\left(  \scriptsize y \: – \: \normalsize \frac{5}{6}\right)^2} = \scriptsize \pm \normalsize \sqrt{\frac{1}{36}} \) \(   \scriptsize y \: – \: \normalsize \frac{5}{6} = \scriptsize \pm \normalsize \frac{1}{6} \) \(   \scriptsize y  =  \normalsize \frac{5}{6} \scriptsize \pm \normalsize \frac{1}{6} \) \(   \scriptsize y  =  \normalsize \frac{5}{6} \scriptsize \: + \: \normalsize \frac{1}{6} \)

or

\(   \scriptsize y  =  \normalsize \frac{5}{6} \scriptsize \: – \: \normalsize \frac{1}{6} \) \(   \scriptsize y  =  \normalsize \frac{6}{6} \scriptsize \: or \: \normalsize \frac{4}{6} \) \(   \scriptsize y  =  1 \: or \: \normalsize \frac{2}{3} \)

Question 12b

Given that

M = \( \scriptsize \begin{pmatrix}1& 2\\4& 3\end{pmatrix} \)

N = \( \scriptsize \begin{pmatrix}m& x\\n& y\end{pmatrix} \)

MN = \( \scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix} \)

Find the matrix N.

Solution

Given that

MN = \( \scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix} \)

∴

M x N = MN

\( \scriptsize \begin{pmatrix}m+2n & x + 2y\\4m + 3n & 4x + 3y\end{pmatrix} \scriptsize  = \scriptsize \begin{pmatrix}2& 1\\3& 4\end{pmatrix}\)

Step 2: Simplifying the above we have

Step 3: using equality of matrix, then we have

m + 2n = 2 ————– (1)

x + 2y = 1 ————— (2)

4m + 3n = 3 ————- (3)

4x + 3y = 4 ————– (4)

Solving equations (1) and (3)

m + 2n = 2

m = 2 – 2n ————- (*)

Substitute m in equation (3)

4(2 – 2n) + 3n = 3

8 – 8n + 3n = 3

8 – 5n = 3

-5n = 3 – 8

-5n = -5

n = \( \frac{-5}{-5} \)

n = 1

Step 4: substitute n = 1 into (*)

m = 2 – 2 (1)

m = 2 – 2

m = 0

  n = 1 and m = 0

Step 5: Solving equations (2) and (4) from (2)

x = 1 – 2y ———– (**)

Step 6: substitute x = 1 – 2y in equation (4)

  4 (1 -2y) + 3y = 4

4 – 8y + 3y = 4

4 – 5y = 4

-5y = 4 – 4

-5y = 0

y = \( \frac{0}{-5} \)

y = 0

Step 7: substitute y = 0 in equation (**)

x = 1 – 2(0)

x = 1 – 0

x = 1

Step 8: hence the matrix

N = \( \scriptsize \begin{pmatrix}m& x\\n& y\end{pmatrix} \)

= \( \scriptsize \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix} \)

Question 13a

(a) The operation (*) is defined on the set of real number, R, by

x (*) y = \( \frac{x \: +\: y}{2}\scriptsize ,\: x,\: y \:  \epsilon \:  R \)

(i) Evaluate \( \scriptsize 3 (\ast )\normalsize \frac{2}{5} \)

Solution

Step 1: given that x (*) y = \( \frac{x \: +\: y}{2}\scriptsize ,\: x,\: y \:  \epsilon \:  R \)

⇒ \( \scriptsize 3 (\ast )\normalsize \frac{2}{5} = \frac{3 \: + \frac{2}{5}}{2} \)

\( \frac{17}{5} \: \times \: \frac{1}{2} \\ = \frac{17}{10} \)

 

(ii) If \( \scriptsize 8(\ast )y =  8 \frac{1}{4}\)

find the value of y

Solution

⇒ \( \frac{8+y}{2} = \frac{33}{4} \)

4(8 + y) = 66

Step 1: solve for y, by removing the brackets

32 + 4y = 66

4y = 66 – 32

4y = 34

y = \( \frac{34}{4} \)

y = \( \frac{17}{2} \)

y = \(\scriptsize 8 \frac{1}{4}\)

Question 13b

In ΔABC,

\( \scriptsize \bar{AB} = \normalsize \binom{-4}{\:\:6} \) \( \scriptsize \bar {AC} = \normalsize \binom{\:\:3}{-8} \)

If P is the midpoint of  \( \scriptsize \bar{AB}\), express \( \scriptsize \bar{CP}\)  as a column vector.

Solution

Step 1:  \( \scriptsize \bar{AP} = \normalsize \frac{1}{2}\scriptsize \bar{AB}\)

 \( \scriptsize \bar{AP} = \normalsize \frac{1}{2} \binom{-4}{\:\:6}\)

 \( \scriptsize \bar{AP} = \normalsize \binom{-2}{\:\:3}\)

Step 2: But \( \scriptsize \bar{AC} = \bar{AP} \: + \: \bar{CP} \)

\( \scriptsize \bar{CP} = \bar{AP} \: – \: \bar{AC} \)

\( \scriptsize \bar{CP} = \normalsize \binom{-2}{\:\:3} \: – \: \binom{\:\:3}{-8} \)

 \( \scriptsize \bar{CP} = \normalsize \binom{-5}{\:\:11}\)

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