# JSS 3 FIRST TERM LESSON NOTE MATHEMATICS

SCHEME OF WORK FOR MATHEMATICS JSS

SESSION

WEEK TOPIC

1 Revision of JSS 2 work

2 The Binary number system

3 Binary number system continued

4 Algebraic Processes

5 Word problems

6 Change of subject of formulae

7 Revision of first half terms work and periodic test

8 Statistics

9 Statistics Continued

10 Simple equations involving fraction and simultaneous equations

11 Revision of 2nd half term’s lesson and periodic test

12-13 First term examination

REFERENCE MATERIALS

ESSENTIAL MATHEMATICS for junior secondary school, book 3 by A. J. S Oluwasanmi

EFFECTIVE MATHEMATICS for junior secondary school book 3 by M.K.Akinsola, M.C.Ejike and A.Tella

JSS 1 SECOND MATHEMATICS LESSON NOTE

WEEK 1

REVISION OF JS S 2 WORK

WEEK TWO

BINARY NUMBERS

Numbers in base two are called binary numbers at is made up two digit is 0 and 1

Converting base 10 numbers to base two number

We do this by dividing the base ten number repeatedly by 2, writing down the remainder until we get to zero and reading the remainder upwards.

Example: (a) Write 810 to a number in base two

b) Express 85 in a binary number

c) Convert 10710 to a number in the base two

d) Convert 152ten to a number in base two

e) Convert 3/8ten to a binary fraction (bicimal)

f) Express 15.12510 in binary notation

SOLUTION

(a) 2 8

2 4 R 0

2 2 R 0

0 R 1

810 = 10002

(b) 2 85

2 42 R 1

2 21 R 0

2 10 R 1

2 5 R 0

2 2 R 1

1 R 0

0 R 1

8510 = 1010101two

(c) 2 107

2 53 R 1

2 26 R 1

2 13 R 0

2 6 R 1

2 3 R 0

2 1 R 1

0 R 1

10710 = 110100112

(d) 2 152

2 76 R 0

2 38 R 0

2 19 R 0

2 9 R 1

2 4 R 1

2 2 R 0

2 1 R 0

0 R 1

152ten = 100110002

(e) 2 3

2 1 R 1

0 R 1

310 = 112

2 8

2 4 R 0

2 2 R 0

2 1 R 0

2 0 R 1

810 = 1000two

First express 3 and 8 in binary, 3810 = 112/10002= 0.0112

(f) 15.125 = 151251000 = 1518= 121810

2 121

2 60 R 1

2 30 R 0

2 15 R 0

2 7 R 1

2 3 R 1

1 R 1

0 R 1

12110 = 11110012

2 8 R

2 4 0

2 2 0

2 1 0

2 0 1

810 = 10002

(1218)10 = 1111001/10002 = 1111.0012

Exercise: Convert the following binary numbers.

(a) 72 (b) 34 (c) 0.875 (d) 32

Converting Base Two Numbers to Base 10 Numbers

We express the given binary numbers as a sum of multiples of powers of two 20, 21, 22, 23 etc.

Example: Convert (i) 101two (ii) 10.10012 (iii) 1112

SOLUTION

1012 = 1×22 + 0x21 + 1×20

= 4 + 0 + 1

= 510

1112 = 1×22 + 1×21 + 1×20

= 4 + 2 + 1

= 910

10.10012 = 1×21 + 0x20 + 1 x 2 -1 + 0 x 2-2 + 0 x 2-3 + 1 x 2-4

= 2 + 0 + 12 + 0 + 0 + 116

= 4116

= 291610

101012 = 1 x 24 + 0 x 23 + 1 x 22 + 0 x 21 + 1 x 20

= 16 + 0 + 4 + 0 + 1

= 2710

Exercise: If 1102 = P10. Find the value of P

Assignment

Write 1-10 in binary numbers

Convert to base 10 (a) 111012 (b) 11.01012 (c) 10110012

Convert to binary number (a) 43ten (b) 1280ten (c) 17610

ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION IN BINARY NUMBERS

Examples

Add 1011112 and 100112

Subtract 100112 from 1011112

11012 x 1112

100100012 1012

Solution

1. 1011112 2. 1011112 3. 11012

100112 +100112 x1112

111002 10000102 1101

1101

1101

1011011

4.Convert to base 10 to have 14510 510 = 2910 = 111012 or

111

1012 10010012

101

1000

101

101

101

0

Exercise: (1) add 1001102, 1010102 and 1110112

(2) Multiply 100112 x 112

(3) 10110102 – 1001112

(4) 10101112 1112

WEEK THREE

APLLICATION OF BINARY NUMBERS

Bunch Cards: it is used in business farms, examining from boards and other organization as a solution for sorting information.

Example: suppose that a failure in a subject is represented by the digit zero. Then for any 5 subjects English, Mathematics, French, Agric and C.R.K. The number 10110 represent a pass in English, a failure in mathematics, a pass in French, a pass in agric and a failure in C.R.K.

Bunch Tape

Letters Decimal No Binary Code

A 1 1

B 2 10

C 3 11

D 4 100

E 5 101

F 6 110

G 7 111

H 8 1000

I 9 1001

J 10 1010

K 11 1011

L 12 1100

M 13 1101

N 14 1110

O 15 1111

P 16 1000

Q 17 10001

R 18 10010

S 19 10011

T 20 10100

U 21 10101

V 22 10110

W 23 10111

X 24 11000

Y 25 11001

Z 26 11010

French tapes are used to enter information and instruction into some computers.

Exercise:

Using a letter per line code and with the words represented by the following

(a) 1 (b) 10000 (c) 100

100 101 101

100 10100 1100

10010 101 1001

101 10010 10110

10011 101

10011 10010

Write the following in a letter per line code.

(a) S (b) D (c) A (d) S (e) E

U I N C L

N V G H I

D I E O A

A N L O S

Y E L

Code your surname in binary

WEEK 4

ALGEBRAIC PROCESSES

Objective: Use letter to generalize statements

Interpret mathematical symbols

Solve simple and problems

Generating statement

The following symbols are generally used in mathematics

“=” means is equal to”

Example: 13 – 6 = 7 means thirteen minus six is equal to seven

“>” means is greater than”

Example: 18 > 10 means eighteen is greater than ten”

“<” means is less than”

Example: -10 < 2 mean negative ten is less than two

“ ≅” means is approximately equal to”

Example: 227 is approximately equal to 3.142 that is 227≅ 3.14

“” means not equal to”

Example: 2 3 means two is not equal to three.

Exercise

Write the following using symbols

M is equal to 8 (ii) y lies between 8 and 12

(iii) Twice y equals fourteen, therefore y equals seven.

OPERATION IN ALGEBRA

EXAMPLE

Write down the meaning of the following and find the values if x = 8, y = 3, z = 9

(a) 2x (b) 2 + x (c) x2 (d) x – 5 (e) x2 (f) x + yz

Solution

(a) Multiply x by 2 (b) Add two to x (c) Divide x by 2

I.e. 2 x 8 = 16 i.e. 8 + 2 = 10 x2 = 82 = 4

(d) Subtract five from x (e) Square x (f) multiply y by z

x– 5 = 8 – 5 = 3 x 2 = 8 x 8 = 64 x + yz = 8 + 3(9) = 8 + 27 = 35

Exercise: If a = 5, b = -3 and c = 2. Find the value of (a) 2a + b – c (b) 5a – 8 (c) 8ac – 2b (d) a+b+c4

LIKE AND UNLIKE TERMS (WORD PROBLEMS)

Examples 1: A fence is made up of 3 different bundles of length 3m each, 5 different hurdles of lengths 5m each and 2 different hurdles of length 2m each. What is the total length?

SOLUTION

Let the 3 hurdles bex, y and z

Then total length force = 3x + 5y – 2z

2) A tourist walked xkm an hour for 3 hours on the first day and xkm an hour for y hours on the second day. How far did he walk in the two days?

Ans. (3x + xy) km

Exercise

A family eats a loaves of bread a day. How many loaves are need for (a) M-days (b) Kweeks (c) How long will p loaves last? (a) 4m loaves (b) 28k Loaves (c) pd days

Use of brackets

Examples 1: The angles of a triangle measured in decrees’ are 2 x, 3 x+ 5, 4 (x+ 10). What are the values of the angles?

SOLUTION

2 x + (3 x + 5) + 4 (x + 10) = 180

2 x + 3 x + 5 + 4 x+ 40 = 180

9x + 45 + 180 – 45

9×9 = 1359

x= 150

The angles are 300, 500 and 1000

2. The results of adding three consecutive numbers is one hundred and twenty. Find the number.

SOLUTION

Let the numbers bex, x + 1, x+ 2

x+ x+ 1 + x+ 2 = 120

3 x + 3 = 120

3 (x+1) = 120

x + 1 + 40

x + 40 -1

x= 39o

Exercise: The greatest of 5 consecutive even numbers is 2a. What is the sum of all the five numbers?

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WEEK 5

WORD PROBLEMS

Sum and Difference

When 30 is added to a number the result is -18. Find the number.

SOLUTION

The number be x

x + 30 = -18

x = – 48

The difference between a number and 5 is 20 what are the two positive numbers

SOLUTION

Let the number be x

x – (-5) = 20

x + 5 = 20

x – 15

If xis less than -5

-5 – x = 20

x = -5 – 20

x = -25

The sum of three consecutive integers is 138. Find the numbers.

SOLUTION

Let the numbers be n, n + r, and n + 2

n + (n + 1) + (n + 2) = 138

3n + 3 + 138

3n = 138

n = 45

n + 1 = 46

n + 2 = 47

The numbers are 45, 46, and 47

Exercise: Ex 2.1 No 6, 7, 8

Product

Examples: (1) Find the product of 134, -0.8 and – 212 (2) what number must be multiplied by 25 to make 34 ? (3) The product of 3 numbers is 3600. If two of the numbers are equal and the third number are 25. Find the two equal numbers.

SOLUTION

1. 74x (-810) x -52 (2) Let the number be x (3) 25 x x x x = 3600

= 74 25 ×x = 34 25×2 = 3600

= 312 100x = 3 x2 = 360025

x = 3100 x2 = 144

x = 0.03 x = 12

Solving combine products with sum and difference

Examples: (1) Add the sum of the 20 and 30.5 to the position difference between 25 and 45. (2) From the sum of 8 ad 7 subtract the negative difference between 15 and 24 (3) find the product of 334 and 56

SOLUTION

1. (20 + 30.5) + (45 – 25) (2) -8 + 7 – 115 – 25 (3) Sum = 154 + 56

= 50.5 + 20 = -8 + 7 + 9 45+1012

= 70.5 = 8 = 5512

Product = 25 x 5512

= 116 = 156

Exercise: Ex. 2.3 No 1, 2, 3, page 17.

Word problems with fractions

Examples:

Find the three-fifth of the sum of 45 and -60. 35 (45 – 60) = 35 (-15) = -9

Divide the difference between 25 and 10 by the product of 6 and 5. 25-106 x 5 = 1530 = 0.5

Find one-quarter of the sum of the product of 2 18 and 3 15 and the product of 112 and 49

Solution: 14178 x 165+ 32 x 49 = 2810 = 11315

Exercise: Ex 2.4 No 1, 2, 3, page 18

Problems leading to equations

Examples

When 35 of a number is added to 30. The result is 20 added to the number find the number.

When the sum of 28 and a certain number is divided by 5. The result is equal to treble the original number. What is the number?

SOLUTION

(1) Let the number be x (2) Let the number be x

Hence 35x + 30 = x + 30 28+x5 = 3x

Multiply each term by 5 15x = 28 + x

3x + 150 = 5x + 100 14×14 = 2814

5x – 3x = 150 -100 x = 2

2x = 50

x = 25

ASSIGNMENT

EXERCISE 2.6; NO 1, 2, AND 3 PAGE 20.

WEEK 6

CHANGE OF SUBJECT OF A FORMULA

Subject of formula: A formula is always written in terms of the subject e.g. V = 13πr2h is a formula for volume of a cone, r is the radius, height (h) =227.

Examples 1: Make r the subject of the formula in A = πr2 (2) Make T the subject of the formula in PTV = K (3) Make x the subject of the formula in y =km/x

SOLUTION

a) A = πr2 b) PTV = K c) y = kmx

r2 =A PT = KV xy = km

r2 = √ A T = KVp x = kmy

Exercise

Make b the subject of the formula in A = 12 (a + b) x

Make x the subject of the formula in a = b ( 1 – x)

Make w the subject of the formula in L = wha(w+p)

Substitution in formula

Examples: The total surface area of a closed cylinder of base radius rcm and height hcm is given by A = 2 πr(r+h)

Write in terms of A and r

Find the height of the cylinder of base radius 7cm and the total surface area of 396cm2. = 227.

SOLUTION

A = 2 πr(r+h) b) h = 396227 x 7 x 2 – 7

r + h = A2 πr h = 2m

h = A2 πr – r

Exercise: The mass of water in a rectangular tank 1m long. 6m wide and him deep is Mkg where M = 1000 lbh (a) What is the mass of water in a tank 5m long, 4m wide and 5m deep? (b) What is the depth of the water in a tank 5m long and 3m wide of its mass is 24000kg?

Assignment

if T = 2 π√mk (a) Express m in terms of T an K (b) Find m when T = 20 and K = 50, = 227

make h the subject of the formula V = 13πr2h

hence find the value of h when V = 256, = 227, r = 21

WEEK 8

EVERYDAY STATISTICS

Data Presentation

Frequency Table

Example: The raw data below show the number of mobile phone calls made by a group of students in a certain day.

5 6 7 8 4 4 5 7 8 10

7 6 5 8 5 7 8 7 10 7

6 5 6 7 7 5 4 5 7 8

Use a tally mark to prepare a frequency table for this data

What calls occur most often?

What percentage of students made 8 calls?

SOLUTION

No of calls made Tally Frequency

4 3

5 7

6 4

7 9

8 5

9 0

10 2

7 calls

530 x 100

= 1623%

Pictogram

Example: The following table shows the colour of cars in a car park one morning. Draw pictogram to illustrate this data.

Colours of car frequency

Black 20

White 17

Red 8

Yellow 5

Green 10

SOLUTION

Colour of cars

Black

White

Red

Yellow

Green

Key: = 2 = 1

Bar Chart

Bar charts consist of series of bars with equal width.

Example: Draw bar chart to illustrate the data of the example above

Frequency colour of cars of the park

Compound bar chart

It is used to compare two or more different sets of information.

Example:

The following table shows the number of candidates who gained admission into higher institutions at a certain town over a period of years.

Year Boys Girls

1997 65 46

1998 50 55

1999 80 73

2000 70 92

2001 45 64

Illustrate this information on dual bar chart

Illustrate what year did girls leave the highest admission?

Illustrate what year did boys have the least admission?

How many more boys had admission than girls in 1999?

How many more candidates gained admission in 2000 than 1998?

SOLUTION

No of candidates

ASSIGNMENT

EXERCISE 22.1; NO 1, 2, AND 3 PAGE 226.

Pie charts

A pie chart is a circle divided into sectors whose angle are used to display data

Example:

In a certain year, the expenditure of a university is shown in the table below.

Items Expenditure in Million Naira

Equipment 20

Salaries and wages 25

Building projects 70

Maintenance 25

Miscellaneous 10

Draw a pie chart to illustrate the information

What percentage of total expenditure goes on project

SOLUTION

Items Expenditure in Million Naira Angles

Equipment 20 20150 x 360°1 = 48o

Salaries and wages 25 25150 x 360°1 = 60o

Building Project 70 70150 x 360°1 = 168o

Maintenance 25 25150 x 360°1 = 60o

Miscellaneous 10 10150 x 360°1 = 24o

Total 150 360o

b) 70150×100°1 = 46.7o

Example 2

History 40% , Geography 30% , Further Mathematicss10% and Physics 20%

The pie chart shows the percentage of students taking Further Mathematics, Physics, History and Geography

What angle represented subject?

What fraction of students are taking history

If the total number of students is 500, how many students are taking physics?

SOLUTION

Subject Percentage Angles

Further Maths 10% 10100 x 360°1 = 36o

Physics 20% 20100 x 360°1 = 72o

History 40% 40100 x 360°1 = 144o

Geography 30% 30100 x 360°1 = 108o

Total 100% 360o

(b) Fraction for history = 40100 = 25

(c) No of Physics Students = 20100 x 500 = 100 Students

Exercise: Ex. 22.2 No 1 and 5 page 228 and 229

WEEK 9

MEASURE OF CENTRAL TENDENCY (MEAN, MEDIAN, MODE)

Example

In a test 10 pupils obtained the following marks 5, 7, 4, 8, 5, 7, 10, 9, 3. Find (a) The mean mark (b) Median mark (c) Modal mark.

SOLUTION

Mean = sum of mark No of marks = 5+7+4+8+5+7+10+9+6+3 10 = 64 10

Mean = 6.4

Arrange the marks in ascending order of magnitude 3, 4, 5, 5, 6, 7, 8, 9, 10

Median = 6+7 2 = 132 = 6.5

The mode is the value that occurs most the mode are 5 and 7 this is bimodal.

Calculating average from frequency tables

Examples: In a science test. The following score shown in the table below were obtained out of 10 by some students.

Marks No of Students (Frequency)

0 2

1 1

2 2

3 4

5 1

6 7

7 3

8 4

9 1

Find (i) the mode (ii) the median (iii) the mean of the frequency distribution

How many students scored at least 5 marks

SOLUTION

(i) Mode = 8

(ii) 0, 0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9

Median = 5+6 2 = 5.5

(iii) Sum of values = 0 + 0 + 1 + 2 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 + 7 + 7 + 8 + 8 + 8 + 8 + 9 = 99

Mean = sum of values No of values = 9920 = 4.95

ALTERNATIVELY

Scores (x) frequency (f) frequency x score (fx)

0 2 2 x 0 = 0

1 1 1 x 1 = 1

2 2 2 x 2 = 4

3 1 1x 3 = 9

4 3 3 x 4 = 12

5 1 1 x 5= 5

6 2 2 x 6 = 12

7 3 3 x 7 = 21

8 4 4 x 8 = 32

9 1 1 x 9 = 99

f = 20 fx = 99

Mean = fxf = 9920 = 4.95

Exercise: Ex 22.3 No 1 and 2 page 231

Range: It gives a measure of how spread and the values are. Range = Highest value – Lowest value

Examples

Find the range of these numbers 9, 4, 7, 6, 12, 8, 15, 10

Solution: Range = 15 – 4 = 11

A student obtained the following marks each out of 100 in different geography test 42, 44, 50 40, 54, 48, 10 88. Find (a) The mean (b) the range (c) make a comment why the range in this case is not good to measure the spread.

SOLUTION

Mean = sum of values No of values = 42+44+50+40+54+48+10+88 8

Mean = 48

Range = 88 – 10

= 78

The two extreme values i.e. 10 and 88 affects the range, so at it not a good measure of spread in this particular case.

Exercise

Find the range of the following

35cm, 50cm, 45cm, 90cm, 30cm

67km, 50km, 20km, 48km, 55km

5.2, 4.7, 8.2, 9.3, 6.4, 5.5

ASSIGNMENT

EXERCISE 22.4; NO 8, 9 AND 10. PAGE 240.

WEEK 10

SIMPLE EQUATIONS INVOLVING FRACTIONS

Examples: Solve the following equations

2×5 – 8- x10

P = 35-3p2

a-52 = 5 + a3

SOLUTION

The L.C.M of 5 and 10 is 10 multiply both sides by 10

10 x2x5 = 10 (3- x10) OR we cross multiply

2 x 2x = 8 – x 2×5 = 8-x10

5×5 = 85 10 x 2x = 5 x (8 –x)

x = 135 20x = 40 – 5x

25×25 = 4025

x = 135

(b) P = 35-3p2 (c) Multiply both sides by 6

2p = 35 – 3p 6 x (a-5)2 = 6 x 5 + 6 x a3

5p5 = 355 3 (a-5) = 30 + 2a

P = 7 3a – 15 = 30 + 2a

a = 45

Exercise: Solve the following equations

(a) x12 = 5 (b) x9 = 43 (c) y-54 = 8 (d) x-72 + x-215 = 0

Fraction with binomial denominator

Examples

a)25-x – 4 = 0 (b) 42y-3 – 35 = 0 (c) 3y + 2 = 52y-1

SOLUTION

25-x – 4 = 0 (b) 42y-3 – 35 = 0 (c) 3y + 2 = 52y-1

25-x – 4 42y-3 = 35 3 (2y – 1) = 5 (y + 2)

4 (5 – x) = 2 3 (2y – 3) = 20 6y – 3 = 5y + 10

20 – 4x = 2 6y – 9 = 20 Y = 13

4x = 18 6y6 = 296

x = 45 y = 456

EXERCISE

a) 1 x+ 13 + 8 = -3 (b) 112x – 5 = 4 (c) 128 + 3y = 1 (d) 81 + 5x = 37 + x

Simultaneous linear Equations

These are equations such as 4x – y = 8 and 3x – y = 6

Graphical Method

To solve simultaneous equations graphically

Make a table of values for both equations

Draw the graphs of both equations on the same axes

Find the coordinate where both graph interest. This values (x-y) are the solutions of both axes

Examples: Solve the following simultaneous equations graphically (a) x-2y=4 and 2x-y=5 (b) 2x-y=10 and 4x-2y=4r

SOLUTION

Y = -2 + 0.5 x y=2x- 5

x 0 2 4 x -1 0 5 x = 2

y -2 -1 0 y -7 -5 5 x = -1

y=10-2x y=-2+2x

x -1 0 2 x -2 0 2 x = 2

y 12 10 6 y -6 -2 2 x = -1

Exercise

2x-3y=8 (b) x+y=7

3x+ 2y=7 x-y=1

SUBSTITUTION METHOD

Examples (a) x-2y=4 , 2X – Y = 52 (b) 4x+y=8

3x+ 2y=7

SOLUTION

x-2y=4 …………1

2x- y=5 ………….2

Step 1: Rearrange one of the equations so that are variable is made the subject

That is from eqn I x=2y+4………..3

Step 2: Substitute into the second equation. That is substitute and solves the resulting equation.

x=2y+4 Into eqn…………2

2(2y+4) – y = 5 4y+8-y=2

3y3 = -63

Y = -1

Step 3: Substitute your answer into 3 to find the other variable

That is x=2y+4

x=2(-1)+4

x=-2+4

x=2

∴x, y=(2,-1)

4x+y=8 ,3x+ 2y=7

From eqn……..1 y=8-4x ……… 3

Substitute eqn3 into eqn 2 gives

3x-8-4x= 6

3x-8+4x= 6

7×7= 147

x = 2

From Eqn 3

y=8- 4x

=8- 4x

y =8- 4 (2)

y =0

∴x, y=(2, 0)

EXERCISE: 2x+y=10, 4x-2y=4 (b) y=5x+2, x+2y=15

ELIMINATION METHOD

Examples: (a) 5x+2y=10, 3x-2y=6 b 3a+3b=15, 4a+3b=20

When one of the unknown has equal coefficient

SOLUTION

5x+2y=10 …………. 1 3a+3b=15

+3x-2y=6…………… 2 4a+3b=20

8×8 = 168 – a = -5

x =2 a = 5

From eqn 1 , 5x+2y=10 from Eqn 1 ,3 5+ 3b=15

10+2y=10 15+8b=15

2y2 = 08

y=0 b=0

x, y=(2, 0) a, b=(5, 0)

Example 2 (When none of the unknown has equal coefficient)

Example: x-2y=4 (b) 3c+4d= -9

2x-y=5 4c+5d= -1

SOLUTION

To make the coefficients of x equal multiply eqn 1 by 2 and eqn 2 by 1

2 x 1 2x-4y=8……….. 3

1 x 2 2x-y=5…………. 4

-3y-3 = 33

y=-1

From Eqn 1 x=2y+4

x=2 -1+ 4

x= -2+4

x=2

3c+4d= -9 ——-1

4c+5d= -11 ——-2

4 x 1 12c+16d= -36 ——–3

3 x 2 12c+15d= -33 d= -3 …………….4

From Eqn (1) 3c+4d= -9

3c+4-3= -9

3c-12= -9

3c= -9+12

3c 3 = 33

c=1

c, d=(1, -3)

EXERCISE: 3x+2y=13, 2x+3y=12; 4x-4y=44 ,5x+2y=34 ; 6x+9y=31 , 4x+3y=6

WORD PROBLEMS

EXAMPLES

The sum of two numbers is 30 and their difference is 15. Find the two numbers

3 boxes and 2 packages weigh 1240g while 5 boxes and 7 packages weigh 2800g. What is the weight of a box and a package?

SOLUTION

x+y=30 (2) 3x+2y=1240 —–(1)

-x-y=15 5x+7y=2800 – – – -(2)

2y=15 5 x 1 15x+10y=6200

y=7.5 3 x 2 15x+21y=8400

x=30-7.5 11y 11 = 220011

x-22.5 y=200

From (1) 3x=1240-2200

3x=1240-400

3x=840

x=260

Exercise:

The sum of two numbers is 18 and their difference is 12. Find the two numbers

This shape is an equilateral triangle with dimension show finds its perimeter.

2x+5y-2 4x – y + 1

x+y+5

Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob. How much does each one actually have?

In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.

If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions.

ASSIGNMENT:

EXERCISE 15.5; NO 2 – 5.PAGE 127

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WEEK 11

REVISION and EXAMINATION