Mathematics lesson note for SS1 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.
Mathematics lesson note for SS1 First Term has been provided in detail here on schoolgist.ng
For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.
To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.
Mathematics Lesson note for SS1 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.
The SS1 Mathematics lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.
The sudden increase in the search for SS1 Mathematics lesson note for First Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.
This post is quite a lengthy one as it provides in full detail, the Mathematics-approved lesson note for all topics and sub-topics in Mathematics as a subject offered in SS1.
Please note that Mathematics lesson note for SS1 provided here for First Term is approved by the Ministry of Education based on the scheme of work.
I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.
SS1 Mathematics Lesson Note (First Term) 2024
SS1 GENERAL MATHEMATICS SCHEME OF WORK FOR FIRST TERM
WEEK[S] TOPIC
1Revision of Jss3 work and basic operations of integers; addition, subtraction multiplication and division.
2Number Bases; [a] Conversion from one base number to base ten. [b] Conversion of decimal fraction [bi-decimals] in one base to base ten. [c] Conversion of numbers from one base to another.
3[a] Addition, subtraction, multiplication and division of number bases. [b] Application of number bases to computer programming.
4Concept of Modular Arithmetic; Addition, subtraction, multiplication and division of operations of modular arithmetic.
5Standard form and Approximation
6Indices; [a]Application of laws of indices. [b] Negative, zero and fractional indices.
7Review of first half term and periodic test
8Logarithms of numbers greater than 1 [whole number]→ use of logarithm table for multiplication and division of numbers.
9Logarithms CTD; [a] calculations involving powers and roots. [b] Relationship between indices and logarithms.
10[a] simple equation and variation. [b] Change of subject of formulae. [c] Type of variation; direct, inverse, joint and partial variation. [d] Application of variation to practical problems.
11Revision.
12Examination.
13Examination
REFERENCE BOOKS
vMAN MATHEMATICS FOR SENIOR SECONDARY SCHOOLS 1
vESSENTIAL MATHEMATICS FOR SENIOR SECONDARY SCHOOLS 1 BY A.J.S OLUWASANMI
vNEW GENERAL MATHEMATICS FOR SENIOR SECONDARY SCHOOL 1 BY M.F.MACRAE ETAL
Week 1.
REVISION AND BASIC OPERATIONS OF INTEGER
RULES OF DIVISIBILITY TEST
A number is divisible by:
Ø2: if the last digit of the number is even or zero.
Ø3: if the sum of the digits is divisible by 3.
Ø4: if the number formed by the last 2 digits is divisible by 4.
Ø5: if the numbers end in 0 or 5.
Ø6: if the number is divisible by both 2and 3.
Ø7: no rule to it yet.
Ø8: if the number formed by the last three digits of the numbers is divisible by 8.
Ø9: if the sum of the digits is divisible by 9 and 3.
Ø10: if the last digit is zero.
Ø11: if the sum of the digits in the odd positions is equal to the sum of digits in the even positions or difference is a multiple of 11.
DIVISIBILITY test is a rule for determining whether one whole number is divisible by another.
EXAMPLE
Determine whether7168 is divisible by2, 3, 4,5,6,8,9,10 and 11
SOLUTION
7168 Is not divisible by 3, 5, 6, 9, 10 and 11. But 2 because the last digit is even, 4 since the last twodigits is divisible by 4. 8 since the last three digit is divisible by 8.
EXAMPLE
Which number is divisible by 35120?
SOLUTION
35120 is divisible by 2,4,5,8and 10.
2 can divide it since it end with zero, divisible by 4, since the last two digit 20 is divisible by 4, 5 since it end with zero, 8 since the last digit 120 is divisible by 8. 10 since the last digit is zero.
EXAMPLE
Determine whether 24739 is divisible by 11?
SOLUTION
24739→sum of digit in odd position 2+7+9=18, sum of digit in even position 4+3=7. Thedifference18-7=11. Then 11 is divisible by 11 and a multiple of 11.
CLASS WORK
Identify the following numbers that can be divisible by 2,4 ,5 , 6,8, 9.
a.3591 b. 2408 c. 7700 d. 18054 e.2032 f. 1827 g.23624 h. 468
INTEGERS: integers are whole numberse.g. 1,3, 6,7,9. etc. not 1.5,,5.
SIMPLIFY: Simplify is to make easier to do or understand.
EVALUATE:To Evaluate is to form a value or quality after thinking , resolving or working a problem.
EXAMPLE
Find the sum of the following number :
a.961,86 and 422. B. 4312,504,614 and 24
Solution
a.961 b. 4 3 1 2
(a) 8 6 ++ 5 0 4
4 2 2 6 1 4
14 6 9 5 4 3 0
Subtract 287 from 306.
Solution
3 0 6
- 2 8 7
0 1 9
Find the product of 452 and 219
SOLUTION
452
X 219 4 0 6 8
- 452
904
98988
Find the value of the following:
a.8(+7). b. 5 (9). (C )
Solution
a.8 (+7 )= 87 =+1
b.-5(-9)=-5+9 =+4
- = =9.6
ASSESSMENT: The students are to work the following questions:
1.Find the product of the following numbers:
a.2184×11 b. 5412×99 c. 217×405
2.Subtract the following numbers:
a.23 from 36 b. 94 from 104
3.Find the values of :
a.336÷4 b. 867÷17 c. 1848÷12 d. (-18)÷(-3) e. -25÷4
4.find the values of the following:
- b.
Week 2.
NUMBER BASES /BASE NUMBER.
Base number is the basis of which each place value column in a number system or the classification of numbers to which one or more other numbers are appended or added.
TYPES OF BASE NUMBERS.
OCTAL BASE; Octal base are numbers express in base eight. E.g. 25
DENARY/DECIMAL BASE: These are numbers express in base ten. E.g. 18
BINARY: These are numbers express in base two. E.g. 1100
BICIMAL: This is the fractional binary number or fraction in base two. E.g.() =()= 0.10101… in base two.
DUODECIMAL BASE: This is the number system that is express in base 12.
HEXADECIMAL: Is system of numbers which is express in base 16. I.e base 2,3,4,5,6,7,8,9,A,B,C,D,E,F.
HINT: No number must be equal or greater than the base number in operation. If you are working in base two, the highest digit will be 1 and the lowest number is 0
EXPRESSION NUMBERS IN BASE TEN.
450 = 4 × + 5× + 0× in base ten.
CONVERTION OF NUMBERS TO BASE TEN
EXAMPLE;
Convert the following numbers to denary base:
- c. .
Solution
a.101111 =1x +0x +1x +1x +1x +1x +1x
=1×64+0x32+1×16+1×8+1×4+1×2+1×1
=64+0+16+8+4+2+1
=9
- 43 = 4x+ 3x +2x
= 4×25+3×5+2×1
= 100+15+2
=11.
C.. 43 = 4x + 3x +1x
= 4+3X+1
CONVERTION OF BASE NUMBERS FROM BASE TEN TO ANOTHER BASE.
Express the following base ten numbers to each base giving:
- 1007 to i. octal base ii.Binary base.
- 761 to ( i).Base 12 (ii). Base16
SOLUTION
a.100 =8 1007 2 1007
8 125 r 7 2 503 r 1
8 15 r 5 2 251 r 1 2 125 r 1
8 0 r 1 2 62 r 1
100= 175 2 31 r 0
2 15 r 1
2 7 r 1
2 3 r 1
2 1 r 1
2 0 r 1
100= 111110111
b.76= 12 761 16 761
12 63 r 5 16 47 r 9
12 5 r 3 16 2 r F
12 0 r 5 16 0 r 2
76 = 53 76= 2F
CONVERTION FROM ONE BASE TO ANOTHER
HINT: First express the number to base ten and then convert from base ten to the required base.
EXAMPLE
Express 31 to octal base
Solution
31 = 3 X + 1 X + 3 X
= 3 X 36 + 1 X 6 + 3 X 1
=108 +6 +3 = 11
117 base ten to Octal base8 117
8 14 r 5
8 1 r 6
31= 16
FRACTIONAL BASE NUMBER
EXAMPLE: Convert 1011.0 to denary base.
SOLUTION
1011.0 = 1 X + 0 X + 1x + 1 X + 0 X + 1 X
= 1X8 + 0X4 +1X2 + 1X1 +1X +1X
= 8 + 0 + 2 + 1 + +
=11 .
EXAMPLE:
Express as bicimal number.
SOLUTION
(=( .= 0.10101010…
ASSESSMENT: Students should work the following questions
1.Express the following base numbers to base ten.
a.312.2 b. 1051.1 c .2341
- Convert the following base ten numbers to bicimals:
(a). (b). (c).(d). (e).
- Convert the following to base; I. Base 5 ii. Base12. iii. Base 15
- 5 b. 12 c. 1000 d. 12110.
WEEK 3
RULES OF BASE NUMBER
1.Numbers must not be equal to or greater than the base number under consideration.
2.Base numbers of the same base can be added,subtracted, multiplied and divided otherwise it must first be converted to base ten or equal base before the required operation is done.
3.When subtracting base numbers , the number carried from nearby to support the other becomes the base in operation added to the original number in that position.
BASIC OPERATIONS OF BASE NUMBER.
EXAMPLE
(A)Find the sum of the octal numbers 174 and 233. (B). Simplify 23121. (c). find the product of 214 and 23 both in base five.(D). if 10= 68, find the value of x?
SOLUTION
A 1 7 4 b. 2 3 1 1 C. 2 1 4
- 2 3 – 2 1 3 x 2
4 2 2 0 3 1 2 0 2 . + 4 3 3
1 1 0 3
c.10 = 68
1 X + 0 X + 4 X = 68
- 0 + 4 = 68
= 68 – 4 : = 64
X = ±
APPLICATION OF BASE NUMBER TO COMPUTER PROGRAMMING
In computer programming the punched cards uses the binary numbers instead of the letters.
A = 1. B = 2. C = 3. D = 4. E = 5. F = 6. P = 16. U =21. Z = 26. The binary equivalent of the number code of letters in binary, such as:
A = 00001, B = 00010, C = 00011, p = 10000, Z = 11010.
Yes = 1 and No = 0
ASSESSMENT: The students are to do the following questions:
1.If 410 = 211 + . Findx?
2.Simplify the following number bases:
I.1101x 10 ii. 61 50 iii. If 12 = 83, find y?
- Represent I LOVE MATHEMATICS in binary code.
ASSIGNMENT: MAN Mathematics for senior secondary schools 1. Page 8, Exercise C4.Numbers,1,2,3 and 7. And miscellaneous Exercises number 3, 6, 10, 14 and 15
MORAL OBJECTIVE: PSALM 90:12. Teach us to number our days so that we may grow in wisdom.
WEEK 4
MODULAR ARITHMETIC
Modular arithmetic is a branch of Mathematics use to predict the outcomes of cyclic events such as days of the week, marketdays, months of the year, time etc.
RULES OF MODULAR ARITHMETIC
ØThe modulo value must be greater than the number worked upon.
ØWhen using cyclic pattern in adding numbers, you must count clock wise direction.
ØIn subtraction of numbers , you must count anti -clock wise direction
EXAMPLE:
The shorter hand of a clock points to 5 on a clock face. What number does it point to after 30 hours?
Solution
30 hours after = 11 o” clock.
(B)Find the following numbers in their simplest form in modulo 4
a.15 b. 102
Solution
15(mod4) = 15÷4
3 remainder 3, therefore 3 the remainder is taken as 3 mod 4
102mod4 = 102 ÷4 = 25 remainder 2
102 (mod 4) = 2 mod 4
ADDITION OF MODULO ARITHMETIC
EXAMPLE
Find the following modulo addition
:a.42 28 (mod 8) b. 54 25 (mod 5)
Solution
a.42 + 28 = 70 mod 8
70 mod 8 = 6 mod 8
b.54 + 25 = 79 mod 5
79 mod 5 = 4 mod 5
SUBTRACTION OF MODULO NUMBERS
Find the simplest form of the following in their giving moduli.
a.-5 mod 6 b. -17 mod 10 c. -75 mod 7
SOLUTION
a.-5 mod 6 = -6×1+1 = 1 mod 6 the value added to the negative number to give the require result becomes your result.
b.-17 mod 10 = -10 x 2 + 3 = 3mod 10.
c.-75 mod 7 = -7 x 11+2 =2 mod 7.
MULTIPLICATION OF MODULO NUMBERS
Evaluate the following in their moduli.
a.16 7 mod 5 b. 21 65 mod 4
SOLUTION
A 16 X 7 = 117 mod 5, which is 2 mod 5
b.21 x 65 = 1365 mod 4 = 1mod 4
EQUATION OF MODULO
Solve the following equations in their giving moduli
a.3x =5mod 7 b. 2x + 3 =1 mod 6 c.
SOLUTION
a.3x =5+7 ,. 3x =12
X = 4 mod 7
b.2x +3 =1+6 , 2x +3 =7
2x 7 -5, then 2x = 4
X =2 mod 6
ASSESSMENT: Solve the following questions;
1.Use the cycle number in modulo 6 to simplify the following.
(i)2 + 10 (ii). 5 + 5 (iii) 15 + 37 (iv) 2 – 9 (v) 0 – 22
2.A toy car starts at a point 0 and runs around a circular track of 2 meters. How far is the car from its starting point along the track when it has gone :
(a)6m (b) 15m (c) 21m (d) 87m
3.Find the following numbers in their simplest form in modulo 4:
(i). 62(ii). 102 (iii) -56 (iv) -78 (v) -202
4.Solve the following equations in the set of positive integers of each modular arithmetic:
i.3X + 4 = 7 mod 8 ii. 4X – 3 = 6 mod 7 iii. – 2X + 2 = 0 mod 5 IV. = 2 in (a). Mod 5 (b). Mod 6 (c). mod 9
MORAL OBJECTIVE: EXODUS 15:25and he cried unto the Lord and there he proved them.
WEEK 5
STANDARD FORM AND APPROXIMATION
STANDARD FORM: Is a convenient way of writing very large or small numbers. It is the product of the numbers in powers of 10 to determine the position of the decimal pointand it´s writing between 1 and 9 i.e. a X . Where ais the number between 1and 9 and n is the position of the decimal point.
EXAMPLE:
Express the following numbers in standard form;
(a)9 (b) 54.6 (c) 570200
SOLUTION.
(a)9 = 9 x (b) 54.6 = 5.46 x (c) 570200 = 5.702 x
NEGATIVE POWERS (NUMBERS LESS THAN 1 )
Express the following numbers in standard form;
(a)0.02 (b) 0.000175
SOLUTION
(a)0.02 = 2 x (b) o.000175 =1.75 x .
ADDITION AND SUBTRACTION OF NUMBERS IN STANDARD FORM
Simplify the following;
(a)5.14 x + 2.842 x (b) 5.24 x 4.33 x
SOLUTION
(a)5.14 x = 5.14 x 10x10x10x10x10x10x10 = 51400000
2.842 x = 2.842 x 10x10x10x10x10 = 284200 +
51684200
51684200 =5.16842 x
(b)5.24 x = 5.24 x= = 0.0000524
4.33 x = 4.33 x = = 0.00000433
0.0000524 0.00000433 =0.00004803
0.00004807 =4.807 x
MULTIPLICATION AND DIVISION OF STANDARD FORM
Simplify the following:
(a)8.222 x x 6.32 x (b) 2.8 x 1.5 x
SOLUTION
(a)(8.222 x 6.32) x () = 51.96304 x
5.196304 x =5.196304 x
(b) (2.8 1.5) x ( ) = 1.86667 x
1.86667 x
APPROXIMATION (SIGNIFICANT FIGURE, DECIMAL PLACES AND ROUNDING OFF)
PPROXIMATION: Is calculating values or numbers only to a certain degree of accuracy.
SIGNIFICANT FIGURE: Is a process in which each of the digits in a number that are needed to a given accuracy is presented.
DECIMAL: is the writing of numbers to a required fractional part.
EXAMPLE:
Approximate the following to the nearest (i) ten (ii) hundred (iii) thousand
(a)7562 (b)907235 (c)8991
SOLUTION
(a)7562 = 7560 (nearest ten)
= 7600 (nearest hundred)
=8000 (nearest thousands)
(b)907235 = 907240 (nearest ten)
= 907200 (nearest hundred)
=907000 (nearest thousand)
© 8991 = 8990 (nearest ten)
= 9000 (nearest hundred)
= 9000(nearest thousands)
SIGNIFICANT FIGURE: Express the following numbers to (i) 1s.f (ii) 2s.f (iii) 3s.f
(a)78602 (b) 702.976 (c) 0.000057849
SOLUTION
(a)78602 = 80000 (1s.f), 79000 (2s.f), 78600(3s.f)
(b) 702.976 = 700 (1s.f), 700(2s.f) 703 (3s.f)
(c) 0.000057849 = 0.00006(1s.f), 0.000058(2s.f), 0.0000578(3s.f)
DECIMAL PLACES: Express the following numbers to ; (i) 2d.p (ii)3d.p
(a)67.3994 (b) 0.00749
SOLUTION
(a)67.3994 = (i) 67.40(2d.p) , (ii) 67.399(3d.p)
(b) 0.00749 = (i) 0.01(2d.p), (ii) 0.007(3d.p)
ASSESSMENT: Solve the following questions;
1.Write the following in standard form:
(i)7560 (ii)7560000000 (iii) 0,00756 (iv) 0.000000756
2.Find each of the following ,leaving your answers in standard form:
(i)(4 x ) x (2.6 x ) (ii) (6.4 x (8 x
3.Evaluate (9.05 x ) + (6.5 x ) – (6.5 x ) giving your answer in standard form.
4.Express each of the following numbers to : (i) 1s.f, (ii) 3s.f, (iii) 2d.p (iv) 3d.p
(a)799.8094 (b) 6.3006 (c) 0.0014698
.
WEEK 6
INDICES
INDICES: are numbers expressed in powers on ten i.e. . The analysis and simplification of indices depends on the basic interpretation and rules of indices as enumerated below.
LAWS OF INDICES
1.
2.
- = 1
- = (
5.( =
- =
EXAMPLES:
Write down the values of the following in index form:
(i)x x 4x 2 (iii) 16 (iv) ()
SOLUTION
(I)
(II)x 4x 2 = (5 x 4 x 2) = 40 =
(III) = (16 = 4
(IV) =
(V) = = = 1÷ = 1 x = or 2.25 or 2
Simplify the following:
(a) x ( (b) 3÷ 6
SOLUTION
(a) x = x =
(b)(3÷6) = () = .
ASSESSMENT
Simplify the following questions:
(1(2)(3). x ÷ (4) -10÷ (-5) (5) x x (6)
ASSIGNMENT: MAN Mathematics for senior secondary school 1
1.Page 11 Exercise B1 numbers 8, 10, 17,20 and 30.
2.Page 12, exercise B3 e,f,I,k,r,t ,v and z
3.Page 13 exercise B4 a, b, c, d, e, g, h and i.
WEEK 7 Review of first half and periodic test
Week 8
LOGARITHMS OF WHOLE NUMBERS
The logarithms of any number N to any base M is the index or power to which the base must be raised, to equal the number N.
The logarithms of any given number consist of two parts called the characteristics and the mantissa.The characteristics is a whole number which can either be positive, zero or negative integers, While the Mantissa is the decimal (fractional) part of the integers always from the table values.
EXAMOLE
399 = 2.6010. 2 Is the characteristics of the number and 6010 from table is the Mantissa or 3.99 x
Find the Logarithms of the following numbers:
(a)8615 (b) 690460 (c) 1.607
SOLUTION
(a)8615 = 8.615 x : in mathematics table, check logarithm of 86 under 1 difference 5 = 9350+3
(b)690460 = 6.90460 x
(c)1.607 = 1.607 x
ANTILOGARITHM: Is the opposite of logarithm.
Find the original number of the following logarithms numbers:
(a) (b) (c) 6.3892
SOLUTION
(a) = 1.862, from antilogarithm table check 27 under zero since there is no third value and the zero before the point (characteristics) determines where the point occupies in the number. Add onto every positive characteristics to determine your value
(b) = 3698.0 or 3698
(c)2450000.0
MULTIPLICATION OF NUMBERS
When multiplying numbers in logarithms, their table values are been added before checking antilogarithms for its solutions.
EXAMPLE
Evaluate the following using table:
(a)143.8 x 23.46 (b) 8234 x 70000
SOLUTION
(a)143.8 x 23.46 = NUMBER LOGARITHM
143.8
23.46 +
Antilogarithm of 5280 = 3374
143.8 x 23.46 = 3374.0
(b) 8234 x70000 = NO LOG
8234
70000 +
=
Antilog of 7609 = 5766 characteristics is 8+1 =9 numbers before point
8234 x70000 = 576600000
DIVISION OF NUMBERS IN LOGARITHMS: When dividing numbers in logarithms we subtract their values
EXAMPLE
Evaluate the following numbers using table:
(a)912.4 ÷ 30.42 (b) 36.75 x 284.7 ÷ 26.45
SOLUTION
(a)912.4 ÷ 30.42 = NO LOG
912.4
30.42 –
=
Antilog of 4770 = 2999
912.4 ÷ 30.42 = 29.99.
(b) 36.75 x284.7 ÷26.45 = NO LOG
36.75
284.7 +
26.45 –
Antilog of 5973 =3957
36.75 x 284.7 ÷ 26.45 = 395.7
ASSESSMENT: Using table evaluate the following numbers:
1(a)497.2 x 8.789 (b) 89 x34.56 x2.094 (c) 8050 ÷ 20.15 (d) 45.08 ÷ 5.462
2(a) 98.45 x 56 ÷ 30.8 (b) (c)
- Find the antilogarithms of the following numbers:
(a) (b) (c) 0.5971 (d) 7.8903 (e) 2.0079
WEEK 9
SQUARES OR POWERS OF NUMBERS IN LOGARITHMS
When squaring numbers in logarithms you multiply the power by the logarithm values
EXAMPLE
Evaluate the following numbers using logarithm table:
(a)(18.42 (b) (
SOLUTION
(a)(18.42 = NO LOG
(18.42 1.2653 x 4 = 5.0612
Antilog of 0612 =1151
(18.42 = 115100
(b) ( = NO LOG
67.9 1.8319
5.23 0.7185
1.1134 X 3 = 3.3402
Antilog of 3402 = 2189
( = 2189
SQUARE ROOTS OF NUMBERS IN LOGARITHMS
When finding the square root of any number with logarithm table , you divide the table value by the value of the square root.
EXAMPLE
Evaluate the following number using logarithm table:
(a) (b)
(a) = NO LOG
0.7551÷ 5 = 0.15102
Antilog of 1510 = 1.416
(b) = NO LOG
=
= +
=
161.5 –
=
Antilog of 1094 = 12.86
RELATIONSHIP BETWEEN INDICES AND LOGARITHMS
Considering numbers in ten (10)
NUMBERS INDICES LOGARITHM
10 = 1
100 2
1000 3
-2
0.00001 -5
ASSESSMENT: Using tables, evaluate the following numbers;
(1)(a) (56.89 (b) (3.9562 (c) (d)
(2)(a) x 45.1 (b) x 92.6
(3)(a) (b) (c)
MORAL OBJECTIVES: JOHN 3:33 the man who has accepted it has certified that God is truthful
WEEK 10
SIMPLE EQUATION AND VARIATION
SIMPLE EQUATION: is any algebraic equation with one unknown.
EXAMPLE
Solve for p in the equation p – 7 = 24
SOLUTION
.IF P – 7 = 24, then add 7 to both sides of the equation
P -7 +7 = 24 + 7
P = 31
Solve the equation 5(c +2) – 3(3c -5) = 1
Solution
5c+10-9c+15 =1. First open the bracket, collect like terms and simplify.
5c-9c+10+15 = 1
-4c+25 = 1, subtract 25 from both sides of the equation
-4c = 1-25,
-4c = -24, divide -4 by both sides
C = 6.
CHANGE OF SUBJECT OF FORMULAE
A formula is an equation consisting of letters which represent quantities.
EXAMPLE
Make each of the following letters giving the subject of formula:
(a)A= ax + b, x (b) T = a + (n-1)d, a (c) T = 2, g
SOLUTION
(a)A=ax + b. make x the subject of formula
Subtract b from both sides
A – b = ax divide both sides by a
= x
(b)T = a + (n-1) d, a. subtract (n-1)d from both sides
T – (n-1) d = a or a =T – nd + d
©T = 2, g. Divide both sides by 2
= , cross multiply
Tg =2l, divide both sides by T
g =
VARIATION: is a change or difference in condition or amount or level etc. within certain limits.
TYPES OF VARIATION
Direct variation, indirect or inverse variation, joint variation and partial variation
Direct variation is the proportional increase in x with a corresponding increase in y or a decrease in x with a corresponding decrease in y when considering two quantities X and Y. that is X Y, where is sigh of proportionality and the equation becomes X = kY where k is constant.
EXAMPLE
The number of bottles of wine drinks is directly proportional to the cost of the bottles of wine drinks. If 10 bottles of wine drink cost ₦400
(a)What is the cost of 18 bottles?
(b)How many bottles can₦200 buy?
SOLUTION
Let N = numbers of wine bottles and C = cost of wine drinks
N C. then N = Ck, N = 10 ,C = ₦400
10 = 400x k . k = =
Therefore the equation connecting N and C is N =
(a)N = = 18 =
C = 18 x 40 = 720. The cost of 18 bottles of wine drinks is ₦720
(b)N = = 5. The numbers of bottles ₦200 can buy is 5 bottles.
INDIRECT OR INVERSE VARIATION
Given two quantities X and Y such that Y increases with a corresponding decrease in X or a
Decrease inY with a corresponding increase in x then Y varies inversely as X. Y then,
the equation becomes Y
EXAMPLE
Y is inversely proportional to x. if y = 9 when x = 4, find the equation connecting x and y
SOLUTION
Y then y =
9 = , then k = 9×4 = 36
The equation connecting x and y is y = .
JOINT VARIATION
This involves three or more variables or quantities in a relationship which occur in many forms. It involves the combination of two direct variations or the combination of one direct and one inverse.
EXAMPLE
Z and z y that is two direct variables. Which is z xy. Equation is z = kxy
V varies directly as T and inversely as P can be written as V
EXAMPLE
Y varies jointly as x and y. W x= 2 and z = 3, y = 30. Find the equation connecting the relationship xyz
SOLUTION
Y xz y = kxz
30 = k x 2 x3 30 = 5k
K = = 6
Equation of the relationship is y = 6xz
PARTIAL VARIATION
Partial or part variation consists of two or more parts of quantities added together. One part
may be constant while the others can vary either directly, indirectly or jointly.
S is partly constant and partly varies directly as T
This statement can be written as :
S k + T. Then the equation is S = k + aT where k and a are constants.
EXAMPLE
X is partly constant and partly varies as y. When y = 5, x = 7 and when y = 7, x = 8. Find
(a)The law of the variation. (b) x when y = 11.
SOLUTION
x k + ay x = k + ay where a and k are constants.
When y = 5, x = 7 : 7 = k + 5a …………(1)
When y = 7, x = 8: 8 = k + 7a …………..(2)
Solving the equation simultaneously, subtract (1) from (2)
2k = 1, then a = .
Substitute for a in (2), 8 =k + 7 x
16 = 2k + 7, 16 – 7 = 2k
K = or 4.5
The law of variation is x = or 2x =9 + y.
(b)When y = 11, 2x = 9 + 11
2x = 20, x = 10.
ASSESSMENT
Evaluate the following questions,
1.The speed s km/h of a car is partly constant and partly varies as the time t the brake is applied. When t = 0, s = 40 and when t = 8, s =30, find s when t = 10 and t when s = 24.
2.A quantity Q is the sum of two quantities, one of which is constant while the other varies inversely as the square of R. when R =1, Q =-1 and when R =2, Q = 2. Find the positive value of R when Q = 2.
3.Y , y = 27 when x =9 and z = 2. Find
(a)The relation between x, y and z.
(b)Find y when x =14 and z = 12.
4.The price of a material in the market varies indirectly with the number of people demanding the material. When there are 80 people, the price of the material is ₦3.50. what is the price when there are 56 people?
5.Make the given letters the subject of the formula of the following equations:
(a) = , Q (b) A = ) , h,d (c) A = r, r
6.Solve the following equations:
(a)8y -19 = 5 +3y (b) 12 – 3t – 9 = 3 – 5t (c) 2 = 5(5w – 2) – 9 (3w – 2) (d) + = 6 (e) – =
MORAL OBJECTIVES: JAMES 1:17 Every good gift and every perfect gift is from above, and cometh down from the Father of lights, with whom is no variableness, neither shadow of turning.
Week 11 Revision
Hope you got what you visited this page for? The above is the lesson note for Mathematics for SS1 class. However, you can download the free PDF file for record purposes.
If you have any questions as regards Mathematics lesson note For SS1 class, kindly send them to us via the comment section below and we shall respond accordingly as usual.
