SCHEME OF WORK
WEEKS TOPICS
(1) Revision of S.S 1work (ii) Revision of approximations significant figure and place values.
(2) Logarithm of number less than 1 (ii) Reciprocal (iii) Accuracy of result using logarithms tables and straight calculation.
(3) Approximation and percentage error.
(4) Ratio, proportions and rates (b) Percentages
(5) Sequence and Series (a) concept of sequence and series (b) Terms A.P and sum (c) Solving problems on A.P
(6) (a) Terms of G.P and sum (b) Problem solving on G.P. (C) Geometric Mean.
(7) Review of the first half term’s work and periodic test.
(8) Simultaneous equation- one linear one quadratic solution by substitution method (ii) Solving more problems on the topic. Word problems on simultaneous equation
(9) Graphical solutions to simultaneous equations, one linear and one quadratic.
(10) Surds: (i) addition and subtraction (ii) Multiplication and division (iii) Rationalization of surd.
(11) Review of second half term lesson and periodic test.
1ST TERM
WEEK 1
MAIN TOPIC: Revision of S.S 1 works
SPECIFIC TOPIC: Revisions
REFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Revise the topics they have been thought in S.S 1
CONTENT: CALCULATING MODE USING GRAPHICAL SOLUTION
Example 1 : Find the mode of the following graphically
Marks No. Of pupils
0 – 9 5
10 – 19 4
20- 29 8
30 – 39 12
40 – 49 6
14
12
10 ∆1 4
8 6 ∆2
6
4
2
0 9 19 29 39 49
Therefore 29 + 4 = 33
Hence Mode = 33
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the mode of the following by using Graphical method
CLASS FREQUENCY
10 – 12 2
13 – 15 2
16 – 18 4
19 – 21 6
22 – 24 3
25 – 27 2
28 – 30 1
SPECIFIC TOPIC: SIGNIFICANT FIGURE
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on significant figure
CONTENT: SIGNIFICANT FIGURE S
Significant figures are the positioning of non-zero digits in a numeral. It is usually applicable to both decimals and whole numbers. The first significant figure in any numerical is thus the first non-zero digit when counted from the left-hand side. For example in 0.007368, the first significant figure (s.f) is 7, the second is 3, the third is 6, while the fourth is 8.
Any time a figure is taken to any number of significant figure, the convention is to round off digit 5 and above into a whole number and add to the next digit.
Example 1: Express 0.006457 to (a) 3 s.f (b) 2 s.f (c) 1 s.f
SOLUTION
(a) 0.00646 = 3 s.f
(b) 0.0065 = 2 s.f
(c) 0.006 = 1 s.f
Example 2: Express 45682 to (a) 2 s.f (b) 3 s.f (c) 4 s.f
SOLUTION
(a) 46000 = 2 s.f
(b) 45700 = 3 s.f
(c) 45680 = 4 s.f
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Write each of the following correct to the number of significant figure shown in front of each of them.
(a) 0.0973 (2 s.f) (b) 14.0675 (3 s.f) (c) 2.00584 (4 s.f) (d) 0.008406 (2 s.f)
ASSIGNMENT: Correct the following into the number of significant figure shown in front of them
(a) 4384467 (3 s.f) (b) 28.0059 (4 s.f) (c) 0.35481 (2 s.f) (d) 0.002635 (1 s.f)
SPECIFIC TOPIC: Approximations
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximations
CONTENT: APPROXIMATIONS
Approximations are often linked to such terms as decimal places, significant figures, standard forms and so on. But since all these aforementioned above are definite terms with their own unique names.
Hence, approximation is a method of assuming precise values to figures. It is a method or system of counting to the nearest whole. Such wholes here could be tens, hundreds, thousands, distances, weights, and other values.
Example 1: 2 hours 12mins
Solution
2hours 12mins = 2 12/60 hours
= 2.2hours
= 2hours (to the nearest hour)
Example 2: Round off 241.863 to the nearest whole number)
Solution
241.863 = 242 (to the nearest whole number)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(a) Take 173.245 to the nearest hundred.
(b) Take 146.8432 to the nearest hundred
ASSIGNMENT: Take the following to the nearest whole number:
(a) 6.32km
(b) 486.4m
(c) 827.305
(d) 154.81cm
SPECIFIC TOPIC: Decimal Places
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on decimal places
CONTENT: DECIMAL PLACES
In order to determine the number of decimal places contained in a figure, we count the number of digits after the decimal point. Zeros after the decimal are counted if found in between non-zero digits. The last zero in a decimal number is not counted. Decimal places are shortened to d.p
Example 1: Write the following in 4 d.p (i) 0.00630427 (ii) 15.300649
Solution
(i) 0.00630427 = 0.0063 4 d.p
(ii) 15.300649 = 15.3006 4 d.p
Example 2: Add up 3.42, 4.761, 3.04, 6.3, 11.304
Solution
3.42
4.761
3.04
6.3
11.304
28.825
Example 2: Simplify 17.36 x 4.65
Solution
17.36
X 4.65
8680
10416
6944
80.7240
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Add up 5.67, 0.453,14.056,4.780
(ii) Simply 15.326 x 2.15
ASSIGNMENT:
Evaluation 0.071685 ÷ 5.36 without using tables
SPECIFIC TOPIC: Place values
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on place values
CONTENT: PLACE VALUES AND ESTIMATION
Sometime we just need to have an idea of the size or value of an operation involving multiplication, addition, division or subtraction. We can just use the method of estimation to determine a rough value for the operation.
Example 1: Estimate the following (i) 2.17 x 5.21 (ii) 43.26 + 19.06
Solution
(i) The estimate will be 2 x5 = 10
The real value is 11.3 to 3 d.p
(ii) 43.26 + 19.06
The estimate is 43 + 19 = 62
The real value is 62.32
Example 2: Write out the place value of 8 and 5 in the numbers given below
264153.078
Solution
The place value of 5 = tens
The place value of 8 = thousandth
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
WEEKLY TEST
Solve the following questions below
(i) Subtract 0.003649 from0.0316
(ii) Correct the following to 2 significant figures (a) 0.86439 (b) 35864
(iii) Correct the following figures to the number of decimal places indicated (i) 3.0561(2 d.p) (ii) 0.008153 (1 d.p) (iii) 0.09634 (3 d.p) (iv) 0.0006005 (4 d.p)
(iv) Estimate 21.42 x 2.43
ASSIGNMENT;
Solve the following questions
(i) Express 0.0006457 to (i) 3 s.f (ii) 2 s.f (iii)1 s.f (iv) 4 s.f
(ii) Round off the following to the nearest hundred (a) 243.36 (ii) 643.79 (iii) 3428km (iv) 236.95m
SPECIFIC TOPIC: Revisions
REFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Revise the topics they have been thought in S.S 1
CONTENT: CALCULATING MODE USING GRAPHICAL SOLUTION
Example 1 : Find the mode of the following graphically
Marks No. Of pupils
0 – 9 5
10 – 19 4
20- 29 8
30 – 39 12
40 – 49 6
14
12
10 ∆1 4
8 6 ∆2
6
4
2
0 9 19 29 39 49
Therefore 29 + 4 = 33
Hence Mode = 33
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the mode of the following by using Graphical method
CLASS FREQUENCY
10 – 12 2
13 – 15 2
16 – 18 4
19 – 21 6
22 – 24 3
25 – 27 2
28 – 30 1
SPECIFIC TOPIC: SIGNIFICANT FIGURE
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on significant figure
CONTENT: SIGNIFICANT FIGURE S
Significant figures are the positioning of non-zero digits in a numeral. It is usually applicable to both decimals and whole numbers. The first significant figure in any numerical is thus the first non-zero digit when counted from the left-hand side. For example in 0.007368, the first significant figure (s.f) is 7, the second is 3, the third is 6, while the fourth is 8.
Any time a figure is taken to any number of significant figure, the convention is to round off digit 5 and above into a whole number and add to the next digit.
Example 1: Express 0.006457 to (a) 3 s.f (b) 2 s.f (c) 1 s.f
SOLUTION
(a) 0.00646 = 3 s.f
(b) 0.0065 = 2 s.f
(c) 0.006 = 1 s.f
Example 2: Express 45682 to (a) 2 s.f (b) 3 s.f (c) 4 s.f
SOLUTION
(a) 46000 = 2 s.f
(b) 45700 = 3 s.f
(c) 45680 = 4 s.f
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Write each of the following correct to the number of significant figure shown in front of each of them.
(a) 0.0973 (2 s.f) (b) 14.0675 (3 s.f) (c) 2.00584 (4 s.f) (d) 0.008406 (2 s.f)
ASSIGNMENT: Correct the following into the number of significant figure shown in front of them
(a) 4384467 (3 s.f) (b) 28.0059 (4 s.f) (c) 0.35481 (2 s.f) (d) 0.002635 (1 s.f)
SPECIFIC TOPIC: Approximations
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximations
CONTENT: APPROXIMATIONS
Approximations are often linked to such terms as decimal places, significant figures, standard forms and so on. But since all these aforementioned above are definite terms with their own unique names.
Hence, approximation is a method of assuming precise values to figures. It is a method or system of counting to the nearest whole. Such wholes here could be tens, hundreds, thousands, distances, weights, and other values.
Example 1: 2 hours 12mins
Solution
2hours 12mins = 2 12/60 hours
= 2.2hours
= 2hours (to the nearest hour)
Example 2: Round off 241.863 to the nearest whole number)
Solution
241.863 = 242 (to the nearest whole number)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(a) Take 173.245 to the nearest hundred.
(b) Take 146.8432 to the nearest hundred
ASSIGNMENT: Take the following to the nearest whole number:
(a) 6.32km
(b) 486.4m
(c) 827.305
(d) 154.81cm
SPECIFIC TOPIC: Decimal Places
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on decimal places
CONTENT: DECIMAL PLACES
In order to determine the number of decimal places contained in a figure, we count the number of digits after the decimal point. Zeros after the decimal are counted if found in between non-zero digits. The last zero in a decimal number is not counted. Decimal places are shortened to d.p
Example 1: Write the following in 4 d.p (i) 0.00630427 (ii) 15.300649
Solution
(i) 0.00630427 = 0.0063 4 d.p
(ii) 15.300649 = 15.3006 4 d.p
Example 2: Add up 3.42, 4.761, 3.04, 6.3, 11.304
Solution
3.42
4.761
3.04
6.3
11.304
28.825
Example 2: Simplify 17.36 x 4.65
Solution
17.36
X 4.65
8680
10416
6944
80.7240
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Add up 5.67, 0.453,14.056,4.780
(ii) Simply 15.326 x 2.15
ASSIGNMENT:
Evaluation 0.071685 ÷ 5.36 without using tables
SPECIFIC TOPIC: Place values
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on place values
CONTENT: PLACE VALUES AND ESTIMATION
Sometime we just need to have an idea of the size or value of an operation involving multiplication, addition, division or subtraction. We can just use the method of estimation to determine a rough value for the operation.
Example 1: Estimate the following (i) 2.17 x 5.21 (ii) 43.26 + 19.06
Solution
(i) The estimate will be 2 x5 = 10
The real value is 11.3 to 3 d.p
(ii) 43.26 + 19.06
The estimate is 43 + 19 = 62
The real value is 62.32
Example 2: Write out the place value of 8 and 5 in the numbers given below
264153.078
Solution
The place value of 5 = tens
The place value of 8 = thousandth
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
WEEKLY TEST
Solve the following questions below
(i) Subtract 0.003649 from0.0316
(ii) Correct the following to 2 significant figures (a) 0.86439 (b) 35864
(iii) Correct the following figures to the number of decimal places indicated (i) 3.0561(2 d.p) (ii) 0.008153 (1 d.p) (iii) 0.09634 (3 d.p) (iv) 0.0006005 (4 d.p)
(iv) Estimate 21.42 x 2.43
ASSIGNMENT;
Solve the following questions
(i) Express 0.0006457 to (i) 3 s.f (ii) 2 s.f (iii)1 s.f (iv) 4 s.f
(ii) Round off the following to the nearest hundred (a) 243.36 (ii) 643.79 (iii) 3428km (iv) 236.95m
WEEK 2
MAIN TOPIC: LOGARITHM
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
We recall that
Number Standard form Characteristic Mantissa logarithm
0.148 1.48 x 10-1 - or 1 1703 1 .1703
0.065 6.5 x 10-2 -2 or 2 8129 2.8129
Example 1: Simplify 2.3502 + 1.875 + 1.4625
Solution
2.3502 + 1.875 + 1. 4625 = 3.6877
Example 2: Evaluate by using logarithm tables 16.82 x 0.00635 correct to three significant figures.
0.04822
Solution
Number Logarithm
16.82 1.2258
0.00635 3.8028 +
1.0286 = 1.0286
0.04822 2.6830
X 2
3.3660 = 3.3660 -
Anti log = 45.98 1.6626
= 46.0 (3 sig.fig)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Using logarithms table, evaluate 5.876 x 0.0432 correct to three significant figures
0.00034
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Using logarithm table, evaluate 3 1.376 correct to three significant figure.
5 0.007
Solution Number Logarithm
3 1.476 0.1386 ÷ 3 = 0.0462
5 0.007 3.8451 ÷ 5 = 1.56902 -
0.47718 = 0.4772
Anti log = 30.00
= 30.0 (3 sig. Fig)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Using logarithms table, evaluate 3 3.543
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Evaluate 4 0.0763
309 x 0.008465
Solution number log
0.0763 2.8825 = 2.8825
309 2.4900
0.008465 3.9277 +
1.4377 = 1.4377 -
1.4442
Anti log = 2781
= 2780 to 3 sig. Fig.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Evaluate 4 0.3456 x 0.054
4.210 x 0.00065
SPECIFIC TOPIC: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using logarithm and straight calculation
CONTENT: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
The results obtained when calculating with 4-figure tables are accurate only to the first three figures. The 4th digit is not likely to be accurate and used for rounding off a result to 3 significant figures. Three-figure accuracy is sufficient for most practical purposes.
Example 1: Calculate to 3 significant figures, the area of a flat circular washer 6.84cm in diameter with a hole 2.96cm in diameter. Take log ∏ to be 0.4971
Solution
Outer radius = 6.84 cm = 3.42cm
2
Inner radius = 2.96 cm = 1.48cm
2
Area = ∏ x 3.422 - ∏x 1.482 cm2
= ∏(3.422 – 1.482)cm2
= ∏(3.42 + 1.48) (3.42 – 1.48)cm2
= ∏ x4.9 x 1.94cm2
= 29.86cm2
= 29.9cm2 to 3 sig.fig.
Using logarithm
Number logarithm
∏ 0.4971
4.9 0.6902
1.94 0.2878
Anti log = 29.86 1.4751
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area in hectares of a rectangular field 126m long and 97m wide.(1 ha = 10 000m2)
ASSIGNMENT:
(i) 3 0.023 x 0.0041
8.001 x 0.00031
(ii) Calculate the area of a circular washer of diameter 3.42cm if the hole in the centre is 1cm in diameter.
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
We recall that
Number Standard form Characteristic Mantissa logarithm
0.148 1.48 x 10-1 - or 1 1703 1 .1703
0.065 6.5 x 10-2 -2 or 2 8129 2.8129
Example 1: Simplify 2.3502 + 1.875 + 1.4625
Solution
2.3502 + 1.875 + 1. 4625 = 3.6877
Example 2: Evaluate by using logarithm tables 16.82 x 0.00635 correct to three significant figures.
0.04822
Solution
Number Logarithm
16.82 1.2258
0.00635 3.8028 +
1.0286 = 1.0286
0.04822 2.6830
X 2
3.3660 = 3.3660 -
Anti log = 45.98 1.6626
= 46.0 (3 sig.fig)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Using logarithms table, evaluate 5.876 x 0.0432 correct to three significant figures
0.00034
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Using logarithm table, evaluate 3 1.376 correct to three significant figure.
5 0.007
Solution Number Logarithm
3 1.476 0.1386 ÷ 3 = 0.0462
5 0.007 3.8451 ÷ 5 = 1.56902 -
0.47718 = 0.4772
Anti log = 30.00
= 30.0 (3 sig. Fig)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Using logarithms table, evaluate 3 3.543
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Evaluate 4 0.0763
309 x 0.008465
Solution number log
0.0763 2.8825 = 2.8825
309 2.4900
0.008465 3.9277 +
1.4377 = 1.4377 -
1.4442
Anti log = 2781
= 2780 to 3 sig. Fig.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Evaluate 4 0.3456 x 0.054
4.210 x 0.00065
SPECIFIC TOPIC: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using logarithm and straight calculation
CONTENT: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
The results obtained when calculating with 4-figure tables are accurate only to the first three figures. The 4th digit is not likely to be accurate and used for rounding off a result to 3 significant figures. Three-figure accuracy is sufficient for most practical purposes.
Example 1: Calculate to 3 significant figures, the area of a flat circular washer 6.84cm in diameter with a hole 2.96cm in diameter. Take log ∏ to be 0.4971
Solution
Outer radius = 6.84 cm = 3.42cm
2
Inner radius = 2.96 cm = 1.48cm
2
Area = ∏ x 3.422 - ∏x 1.482 cm2
= ∏(3.422 – 1.482)cm2
= ∏(3.42 + 1.48) (3.42 – 1.48)cm2
= ∏ x4.9 x 1.94cm2
= 29.86cm2
= 29.9cm2 to 3 sig.fig.
Using logarithm
Number logarithm
∏ 0.4971
4.9 0.6902
1.94 0.2878
Anti log = 29.86 1.4751
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area in hectares of a rectangular field 126m long and 97m wide.(1 ha = 10 000m2)
ASSIGNMENT:
(i) 3 0.023 x 0.0041
8.001 x 0.00031
(ii) Calculate the area of a circular washer of diameter 3.42cm if the hole in the centre is 1cm in diameter.
WEEK 3
SPECIFIC TOPIC: APPROXIMATION AND PERCENTAGE ERROR
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximation and percentage error
CONTENT:
APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy’s error = 15.6 – 13.4 = 2.2m
Second boy’s error = 16.9 – 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = Absolute error x 100
Actual value
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 – 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 – 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 – 26.00
#374.00
Example 2: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66 x 6500
100 1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: An error of 4% was made in finding the length of a rope that was actually25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximation and percentage error
CONTENT:
APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy’s error = 15.6 – 13.4 = 2.2m
Second boy’s error = 16.9 – 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = Absolute error x 100
Actual value
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 – 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 – 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 – 26.00
#374.00
Example 2: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66 x 6500
100 1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: An error of 4% was made in finding the length of a rope that was actually25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
WEEK 4
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: SERIES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are 3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
SPECIFIC TOPIC: CALCULATION INVOLVING SUM OF AN A.P
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Solve problems on sum of an A.P
(ii) Find the nth term of an A.P
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: SERIES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are 3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
SPECIFIC TOPIC: CALCULATION INVOLVING SUM OF AN A.P
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Solve problems on sum of an A.P
(ii) Find the nth term of an A.P
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
WEEK 5
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: SERIES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are 3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: SERIES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are 3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
WEEK 6
SPECIFIC TOPIC: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on simultaneous equation-one linear, one quadratic
CONTENT: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC
Sometimes, a pair of simultaneous equations may contain one linear and one quadratic.
In order to solve this type of equation we begin with the linear equation. In the linear equation, the substitution method is used, whereby one of the unknown is made the subject of the formula.
The value got is thereafter re-substituted in the other equation which is quadratic, in two unknowns to obtain a simpler quadratic equation in one unknown.
Example 1: solve the equations
2x – y = 10 ..................................(i)
3x + y2 = 22.................................(ii)
Solution
From equation(i), make y the subject of formula
2x – y = 10
Y = 2x – 10
Substitute for y in equation (ii)
3x + y2 = 22
3x + (2x – 10)2 = 22
3x + 4x2 – 40x + 100 = 22
4x2 – 37x + 78 = 0
Using quadratic formula:
X = -b+- b2 - 4ac
2a
= 37+- 372 – 4(4) (78)
2(4)
= 37 +- 1369 – 1248
8
= 37 +- 121
8
= 37 +- 11
8
= 37 + 11 or 37 - 11
8 8
= 48/8 or 26/8
= 6 or 13/4
Substitute for x in equation (i)
2x – y = 10
When x = 6
2(6) – y = 10
12 – y = 10
-y = 10 – 12
-y = -2
Y = 2
When x = 13/4
2(13/4) – y = 10
13/2 – y 10
-y = 10 – 13/2
= 20 - 13
2
Y = - 7/2
Therefore: x = 6, y = 2 or x = 13/4, y = -7/2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equations
3x + 2y = 10 ...............................(i)
X + 2y = 2 ................................(ii)
ASSIGNMMENT;
Solve the equations
3x + 2y = 12 .........................(i)
Xy + 5y = 21 .........................(ii)
Example 1: Solve the equations
3x + 2y = 12 ...................(i)
Xy + 5y = 21 .................(ii)
Solution
From equation (i), make y the subject
3x + 2y = 12
2y = 12 – 3x
Y = 12 – 3x
2
Substitute 12 – 3x for y in equation (ii)
2
X 12 – 3x + 5 12 – 3x = 21
2 2
12x – 3x2 + 60 – 15x = 21
2 2
12x – 3x2 + 60 – 15x = 42
-3x2 – 3x + 18 = 0
Solving the quadratic equation in x
-3x2 – 3x + 18 = 0
-3x2 -9x + 6x + 18 = 0
- (3x2 + 9x) + (6x + 18) = 0
- 3x (x + 3) + 6(x + 3) = 0
(x + 3) (-3x + 6) = 0
X = -3 or 2
Substitute for x in equation (i)
3x + 2y = 12
When x = -3
3(-3) + 2y = 12
-9 + 2y = 12
2y = 12 +9
2y = 21
Y = 21/2
When x = 2
3(2) + 2y = 12
6 + 2y = 12
2y = 12 – 6
2y = 6
Y = 3
Therefore, x =-3, y = 21/2 or x = 2, y = 3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equation
Xy + x = 28........................(i)
X = y+4...........................(ii)
ASSIGNMENT:
Solve the equations
3x + 2y = 14 .....................(i)
2xy + 5 = 21 ....................(ii)
CONTENT: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC
Example 1: Solve the equation 3x + 2y = 10
X2 + 2y = 2
Solution
2y = 10 – 3x
Y = 10 – 3x
2
Substitute y in equation (i)
X2 + 2 10 – 3x = 2
2
X2 + 20 – 6x = 2
2
X2 + 20 – 6x = 4
X2 – 6x + 16 = 0
a= 1, b = -6, c = 16
x = -b ± b2 – 4ac
2a
X = -(-6) ± (-6)2 – 4(1)(16)
2(1)
X = 6 ± 36 - 64
2
X = 6 ± -28
2
X = 6 ± 0
2
X = 6/2 = 3
If x = 3
From equation (i)
3x + 2y = 10
3(3) + 2y = 10
9 + 2y = 10
2y = 10 – 9
2y = 1
Divide both sides by 2
2y/2 = ½
Y = ½
Hence x = 3 and y = 1/2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equation
2x – y = 6
Xy + y = 24
ASSIGNMENT: solve the equation
5x + 4y = 22
3x + 5y = 21
Example 1: solve
Xy = 21
X + 5y = 22
Solution
Y = 21/x
Substitute y in equation (i)
X + 5 21 = 22
x
X + 105 = 22
X
x2 + 105 = 22
x
x2 + 105 = 22x
x2 – 22x + 105 = 0
a= 1, b = -22, c = 105
x = -b ± b2 – 4ac
2a
X = -(-22) ± (-22)2 – 4(1)(105)
2(1)
X = 22 ± 484 + 420
2
X = 22 ± 904
2
X = 22 ± 30.07
2
X = 22 + 30 or 22 - 30
2 2
X = 52/2 or -18/2
X = 26 or – 9
If x = 26
From equation (i)
xy = 21
26(y) = 21
Divide both sides by 26
Y = 21/26
Or
If x = -9
From equation (i)
Xy = 21
-9(y) = 21
-9y = 21
Divide both sides by -9
-9y/-9 = 21/-9
Y = -7/3
Y = - 2 1/3
Hence x = 26 or -9 and y = 21/26 or -7/3
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
Solve the equation
Xy + x = 28
X = y + 4
ASSIGNMENT;
Solve the equation
x/4 – y/6 = -1
x/2 + y/5 = 10
TEST
(1) Solve the equations
3m – 5n = 11.............................(i)
4m – 6n = 16 ...........................(ii)
(2) Solve the equations
3x + 2y = 14......................(i)
2xy + 5= 21......................(ii)
ASSIGNMENT:
Solve the equations
(i) 12m + 12n = -7.................................(i)
4m – 3n = 7 ...................................(ii)
(ii) x/2 –y/5 = 1.........................................(i)
y –x/3 = 8 .........................................(ii)
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on simultaneous equation-one linear, one quadratic
CONTENT: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC
Sometimes, a pair of simultaneous equations may contain one linear and one quadratic.
In order to solve this type of equation we begin with the linear equation. In the linear equation, the substitution method is used, whereby one of the unknown is made the subject of the formula.
The value got is thereafter re-substituted in the other equation which is quadratic, in two unknowns to obtain a simpler quadratic equation in one unknown.
Example 1: solve the equations
2x – y = 10 ..................................(i)
3x + y2 = 22.................................(ii)
Solution
From equation(i), make y the subject of formula
2x – y = 10
Y = 2x – 10
Substitute for y in equation (ii)
3x + y2 = 22
3x + (2x – 10)2 = 22
3x + 4x2 – 40x + 100 = 22
4x2 – 37x + 78 = 0
Using quadratic formula:
X = -b+- b2 - 4ac
2a
= 37+- 372 – 4(4) (78)
2(4)
= 37 +- 1369 – 1248
8
= 37 +- 121
8
= 37 +- 11
8
= 37 + 11 or 37 - 11
8 8
= 48/8 or 26/8
= 6 or 13/4
Substitute for x in equation (i)
2x – y = 10
When x = 6
2(6) – y = 10
12 – y = 10
-y = 10 – 12
-y = -2
Y = 2
When x = 13/4
2(13/4) – y = 10
13/2 – y 10
-y = 10 – 13/2
= 20 - 13
2
Y = - 7/2
Therefore: x = 6, y = 2 or x = 13/4, y = -7/2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equations
3x + 2y = 10 ...............................(i)
X + 2y = 2 ................................(ii)
ASSIGNMMENT;
Solve the equations
3x + 2y = 12 .........................(i)
Xy + 5y = 21 .........................(ii)
Example 1: Solve the equations
3x + 2y = 12 ...................(i)
Xy + 5y = 21 .................(ii)
Solution
From equation (i), make y the subject
3x + 2y = 12
2y = 12 – 3x
Y = 12 – 3x
2
Substitute 12 – 3x for y in equation (ii)
2
X 12 – 3x + 5 12 – 3x = 21
2 2
12x – 3x2 + 60 – 15x = 21
2 2
12x – 3x2 + 60 – 15x = 42
-3x2 – 3x + 18 = 0
Solving the quadratic equation in x
-3x2 – 3x + 18 = 0
-3x2 -9x + 6x + 18 = 0
- (3x2 + 9x) + (6x + 18) = 0
- 3x (x + 3) + 6(x + 3) = 0
(x + 3) (-3x + 6) = 0
X = -3 or 2
Substitute for x in equation (i)
3x + 2y = 12
When x = -3
3(-3) + 2y = 12
-9 + 2y = 12
2y = 12 +9
2y = 21
Y = 21/2
When x = 2
3(2) + 2y = 12
6 + 2y = 12
2y = 12 – 6
2y = 6
Y = 3
Therefore, x =-3, y = 21/2 or x = 2, y = 3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equation
Xy + x = 28........................(i)
X = y+4...........................(ii)
ASSIGNMENT:
Solve the equations
3x + 2y = 14 .....................(i)
2xy + 5 = 21 ....................(ii)
CONTENT: SIMULTANEOUS EQUATIONS-ONE LINEAR, ONE QUADRATIC
Example 1: Solve the equation 3x + 2y = 10
X2 + 2y = 2
Solution
2y = 10 – 3x
Y = 10 – 3x
2
Substitute y in equation (i)
X2 + 2 10 – 3x = 2
2
X2 + 20 – 6x = 2
2
X2 + 20 – 6x = 4
X2 – 6x + 16 = 0
a= 1, b = -6, c = 16
x = -b ± b2 – 4ac
2a
X = -(-6) ± (-6)2 – 4(1)(16)
2(1)
X = 6 ± 36 - 64
2
X = 6 ± -28
2
X = 6 ± 0
2
X = 6/2 = 3
If x = 3
From equation (i)
3x + 2y = 10
3(3) + 2y = 10
9 + 2y = 10
2y = 10 – 9
2y = 1
Divide both sides by 2
2y/2 = ½
Y = ½
Hence x = 3 and y = 1/2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the equation
2x – y = 6
Xy + y = 24
ASSIGNMENT: solve the equation
5x + 4y = 22
3x + 5y = 21
Example 1: solve
Xy = 21
X + 5y = 22
Solution
Y = 21/x
Substitute y in equation (i)
X + 5 21 = 22
x
X + 105 = 22
X
x2 + 105 = 22
x
x2 + 105 = 22x
x2 – 22x + 105 = 0
a= 1, b = -22, c = 105
x = -b ± b2 – 4ac
2a
X = -(-22) ± (-22)2 – 4(1)(105)
2(1)
X = 22 ± 484 + 420
2
X = 22 ± 904
2
X = 22 ± 30.07
2
X = 22 + 30 or 22 - 30
2 2
X = 52/2 or -18/2
X = 26 or – 9
If x = 26
From equation (i)
xy = 21
26(y) = 21
Divide both sides by 26
Y = 21/26
Or
If x = -9
From equation (i)
Xy = 21
-9(y) = 21
-9y = 21
Divide both sides by -9
-9y/-9 = 21/-9
Y = -7/3
Y = - 2 1/3
Hence x = 26 or -9 and y = 21/26 or -7/3
EVALUATION: The lesson is evaluated as the students are asked to solve the problem below.
Solve the equation
Xy + x = 28
X = y + 4
ASSIGNMENT;
Solve the equation
x/4 – y/6 = -1
x/2 + y/5 = 10
TEST
(1) Solve the equations
3m – 5n = 11.............................(i)
4m – 6n = 16 ...........................(ii)
(2) Solve the equations
3x + 2y = 14......................(i)
2xy + 5= 21......................(ii)
ASSIGNMENT:
Solve the equations
(i) 12m + 12n = -7.................................(i)
4m – 3n = 7 ...................................(ii)
(ii) x/2 –y/5 = 1.........................................(i)
y –x/3 = 8 .........................................(ii)
WEEK 7
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on graphical solution of simultaneous, one linear and one quadratic
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
If both equations are linear: To solve simultaneous equations with one linear, one quadratic, the approach is similar to the way we solve simultaneous equation using graphical method.
Equation ax2 + bx + c = 0 and px + q = 0
Can be solved using the graphical method. All we need to do is to plot the graphs of the same axis to locate their points of intersection. The values of x at the points of intersection give the solution of the equations.
Example 1: copy and complete the following table of values for the relation y = 2 + x – x2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
y 0 1.25 2 2 1.25 -4
(i) Draw the graph of the relation, using a scale of 2cm to 1 unit on each axis.
(ii) Using the same axis, draw the graph of y = 1- x
(iii) From your graphs determine the roots of the equation 1 + 2x – x2 = 0
Solution
x -2 -1.5 -1 -.05 0 0.5 1 1.5 2 2.5 3
2 2 2 2 2 2 2 2 2 2 2 2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
X2 -4 -2.25 -1 -0.25 0 -0.25 -1 -2.25 -4 -6.25 -9
y -4 -1.75 0 1.25 2 2.25 2 1.25 0 -1.75 -4
(ii)
x -2 -1.5 -0.5 -1 0 0.5 1 1.5 2 2.5 3
1 1 1 1 1 1 1 1 1 1 1 1
-x 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3
y 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -4
The roots of the equation 1 + 2x – x2 = 0 are found when 2 + x – x2 = 1 – x . This shows the points of intersection of the curve and the straight line at points A and B on the graphs respectively. Hence values of x at these points A and B are -0.4 or 2.4 from the graph
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Copy and complete the table of value of 2cm for the relation y = x2 – 2x -1
x -2 -1 0 1 2 3 4
Y -7 2 7
(i) Draw the graph of the relation using a scale of 2cm to 1unit on both axes
(ii) Using your graph, find (i) root of the equation x2 – 2x – 1 = 0
(iii) Using the same axes, draw the graph of the equation y = 2x -3
(iv) From your graph determine the root of the equation y = 2x -3
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 1
Solve the following simultaneous equations:
Solution:
Note:
The graphical solution can be seen below.
Example 2: copy and complete the following table of values for the relation y = 2x2 – 3x -10.
x -3 -2 -1 0 1 2 3 4 5
Y 17 -10 -1 25
(b) using a scale of 2cm for 5units on the y-axis and 1cm for 1unit on the x-axis, draw the graph of the relation y = 2x2 – 3x – 10 = 0
(c) from your graph, find the (i) roots of the equation 2x2 – 3x – 10 = 0
(ii) value of x for which y is the minimum
(iii)roots of the equation 2x2 – 3x – 6 = 0
(iv)roots of the equation 2x2 -5x – 13 = 0
Solution
x -3 -2 -1 0 1 2 3 4 5
2x2 18 8 2 0 2 8 18 32 50
-3x 9 6 3 0 -3 -6 -9 -12 -15
-10 -10 -10 -10 -10 -10 -10 -10 -10 -10
y 17 4 -5 -10 -11 -8 -1 10 25
(Ci) ROOTS OF THE EQUATION Y = 2X2 -3X -10 are found at the point where the curve intersects with the x axis. These values are -1.6 or 3.2
ii. y is minimum at -11. Therefore the corresponding value of x = 1
iii. Finding the roots of the equation 2x2 -3x -6= 0implies that the original equation 2x2 – 3x – 10 = -4, hence, draw a parallel line to the x-axis to pass through y =-4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Copy and complete the table for the relation y = 2 + 2x – x2 for -1 ≤ x ≤ 3
x -1 -0.5 0 0.5 1 1.5 2 2.5 3
Y -1 0.75 2 -1
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on graphical solution of simultaneous, one linear and one quadratic
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
If both equations are linear: To solve simultaneous equations with one linear, one quadratic, the approach is similar to the way we solve simultaneous equation using graphical method.
Equation ax2 + bx + c = 0 and px + q = 0
Can be solved using the graphical method. All we need to do is to plot the graphs of the same axis to locate their points of intersection. The values of x at the points of intersection give the solution of the equations.
Example 1: copy and complete the following table of values for the relation y = 2 + x – x2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
y 0 1.25 2 2 1.25 -4
(i) Draw the graph of the relation, using a scale of 2cm to 1 unit on each axis.
(ii) Using the same axis, draw the graph of y = 1- x
(iii) From your graphs determine the roots of the equation 1 + 2x – x2 = 0
Solution
x -2 -1.5 -1 -.05 0 0.5 1 1.5 2 2.5 3
2 2 2 2 2 2 2 2 2 2 2 2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
X2 -4 -2.25 -1 -0.25 0 -0.25 -1 -2.25 -4 -6.25 -9
y -4 -1.75 0 1.25 2 2.25 2 1.25 0 -1.75 -4
(ii)
x -2 -1.5 -0.5 -1 0 0.5 1 1.5 2 2.5 3
1 1 1 1 1 1 1 1 1 1 1 1
-x 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3
y 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -4
The roots of the equation 1 + 2x – x2 = 0 are found when 2 + x – x2 = 1 – x . This shows the points of intersection of the curve and the straight line at points A and B on the graphs respectively. Hence values of x at these points A and B are -0.4 or 2.4 from the graph
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Copy and complete the table of value of 2cm for the relation y = x2 – 2x -1
x -2 -1 0 1 2 3 4
Y -7 2 7
(i) Draw the graph of the relation using a scale of 2cm to 1unit on both axes
(ii) Using your graph, find (i) root of the equation x2 – 2x – 1 = 0
(iii) Using the same axes, draw the graph of the equation y = 2x -3
(iv) From your graph determine the root of the equation y = 2x -3
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 1
Solve the following simultaneous equations:
Solution:
Note:
The graphical solution can be seen below.
Example 2: copy and complete the following table of values for the relation y = 2x2 – 3x -10.
x -3 -2 -1 0 1 2 3 4 5
Y 17 -10 -1 25
(b) using a scale of 2cm for 5units on the y-axis and 1cm for 1unit on the x-axis, draw the graph of the relation y = 2x2 – 3x – 10 = 0
(c) from your graph, find the (i) roots of the equation 2x2 – 3x – 10 = 0
(ii) value of x for which y is the minimum
(iii)roots of the equation 2x2 – 3x – 6 = 0
(iv)roots of the equation 2x2 -5x – 13 = 0
Solution
x -3 -2 -1 0 1 2 3 4 5
2x2 18 8 2 0 2 8 18 32 50
-3x 9 6 3 0 -3 -6 -9 -12 -15
-10 -10 -10 -10 -10 -10 -10 -10 -10 -10
y 17 4 -5 -10 -11 -8 -1 10 25
(Ci) ROOTS OF THE EQUATION Y = 2X2 -3X -10 are found at the point where the curve intersects with the x axis. These values are -1.6 or 3.2
ii. y is minimum at -11. Therefore the corresponding value of x = 1
iii. Finding the roots of the equation 2x2 -3x -6= 0implies that the original equation 2x2 – 3x – 10 = -4, hence, draw a parallel line to the x-axis to pass through y =-4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Copy and complete the table for the relation y = 2 + 2x – x2 for -1 ≤ x ≤ 3
x -1 -0.5 0 0.5 1 1.5 2 2.5 3
Y -1 0.75 2 -1
WEEK 8
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 1
Solve the following simultaneous equations:
Solution:
Note:
The graphical solution can be seen below.
Example 2: copy and complete the following table of values for the relation y = 2 + x – x2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
y 0 1.25 2 2 1.25 -4
(i) Draw the graph of the relation, using a scale of 2cm to 1 unit on each axis.
(ii) Using the same axis, draw the graph of y = 1- x
(iii) From your graphs determine the roots of the equation 1 + 2x – x2 = 0
Solution
x -2 -1.5 -1 -.05 0 0.5 1 1.5 2 2.5 3
2 2 2 2 2 2 2 2 2 2 2 2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
X2 -4 -2.25 -1 -0.25 0 -0.25 -1 -2.25 -4 -6.25 -9
y -4 -1.75 0 1.25 2 2.25 2 1.25 0 -1.75 -4
(ii)
x -2 -1.5 -0.5 -1 0 0.5 1 1.5 2 2.5 3
1 1 1 1 1 1 1 1 1 1 1 1
-x 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3
y 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -4
The roots of the equation 1 + 2x – x2 = 0 are found when 2 + x – x2 = 1 – x . This shows the points of intersection of the curve and the straight line at points A and B on the graphs respectively. Hence values of x at these points A and B are -0.4 or 2.4 from the graph
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 2: copy and complete the following table of values for the relation y = 2x2 – 3x -10.
X -3 -2 -1 0 1 2 3 4 5
Y 17 -10 -1 25
(b) using a scale of 2cm for 5units on the y-axis and 1cm for 1unit on the x-axis, draw the graph of the relation y = 2x2 – 3x – 10 = 0
(c) From your graph, find the (i) roots of the equation 2x2 – 3x – 10 = 0
(ii) Value of x for which y is the minimum
(iii) Roots of the equation 2x2 – 3x – 6 = 0
(iv)Roots of the equation 2x2 -5x – 13 = 0
Solution
X -3 -2 -1 0 1 2 3 4 5
2x2 18 8 2 0 2 8 18 32 50
-3x 9 6 3 0 -3 -6 -9 -12 -15
-10 -10 -10 -10 -10 -10 -10 -10 -10 -10
y 17 4 -5 -10 -11 -8 -1 10 25
(Ci) ROOTS OF THE EQUATION Y = 2X2 -3X -10 are found at the point where the curve intersects with the x axis. These values are -1.6 or 3.2
ii. y is minimum at -11. Therefore the corresponding value of x = 1
iii. Finding the roots of the equation 2x2 -3x -6= 0implies that the original equation 2x2 – 3x – 10 = -4, hence, draw a parallel line to the x-axis to pass through y =-4
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
TEST
The incomplete table is for the relation y = 2x2 – 5x + 1
x -3 -2 -1 0 1 2 3 4
y 8 1 -1 26
(i) Copy and complete the table
(ii) Using a scale of 2cm on the axis and 2cm to 10units on the y-axis draw the graph of the relation y = 2x2 – 5x + 1 for -3≤ x ≤ 5
(iii) On the same axes, draw the graph of y = x + 6
(iv) From the graphs, estimate the least value of y and the corresponding value of x
(v) Find also from the graphs the solution of the equation 2x2 – 5x +1 = x + 6
ASSIGNMENT;
Copy and complete the table below for y = 5sin3θ – 3 cos 2θ
θ 0 30 60 90 120 150 180
y -3 1.5
(b) using a scale of 2cm to 300 on θ axis and 2cm to 1 unit on the y-axis, draw the graph of the relation y = 5sin3θ -3cos 2θ for 00≤ θ ≤ 1800
(c) Use your graph to find the (i) solution of y = 5sinθ – 3cos2θ(ii) values of θ for which y is maximum.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 1
Solve the following simultaneous equations:
Solution:
Note:
The graphical solution can be seen below.
Example 2: copy and complete the following table of values for the relation y = 2 + x – x2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
y 0 1.25 2 2 1.25 -4
(i) Draw the graph of the relation, using a scale of 2cm to 1 unit on each axis.
(ii) Using the same axis, draw the graph of y = 1- x
(iii) From your graphs determine the roots of the equation 1 + 2x – x2 = 0
Solution
x -2 -1.5 -1 -.05 0 0.5 1 1.5 2 2.5 3
2 2 2 2 2 2 2 2 2 2 2 2
x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
X2 -4 -2.25 -1 -0.25 0 -0.25 -1 -2.25 -4 -6.25 -9
y -4 -1.75 0 1.25 2 2.25 2 1.25 0 -1.75 -4
(ii)
x -2 -1.5 -0.5 -1 0 0.5 1 1.5 2 2.5 3
1 1 1 1 1 1 1 1 1 1 1 1
-x 2 1.5 1 0.5 0 -0.5 -1 -1.5 -2 -2.5 -3
y 3 2.5 2 1.5 1 0.5 0 -0.5 -1 -1.5 -4
The roots of the equation 1 + 2x – x2 = 0 are found when 2 + x – x2 = 1 – x . This shows the points of intersection of the curve and the straight line at points A and B on the graphs respectively. Hence values of x at these points A and B are -0.4 or 2.4 from the graph
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
Example 2: copy and complete the following table of values for the relation y = 2x2 – 3x -10.
X -3 -2 -1 0 1 2 3 4 5
Y 17 -10 -1 25
(b) using a scale of 2cm for 5units on the y-axis and 1cm for 1unit on the x-axis, draw the graph of the relation y = 2x2 – 3x – 10 = 0
(c) From your graph, find the (i) roots of the equation 2x2 – 3x – 10 = 0
(ii) Value of x for which y is the minimum
(iii) Roots of the equation 2x2 – 3x – 6 = 0
(iv)Roots of the equation 2x2 -5x – 13 = 0
Solution
X -3 -2 -1 0 1 2 3 4 5
2x2 18 8 2 0 2 8 18 32 50
-3x 9 6 3 0 -3 -6 -9 -12 -15
-10 -10 -10 -10 -10 -10 -10 -10 -10 -10
y 17 4 -5 -10 -11 -8 -1 10 25
(Ci) ROOTS OF THE EQUATION Y = 2X2 -3X -10 are found at the point where the curve intersects with the x axis. These values are -1.6 or 3.2
ii. y is minimum at -11. Therefore the corresponding value of x = 1
iii. Finding the roots of the equation 2x2 -3x -6= 0implies that the original equation 2x2 – 3x – 10 = -4, hence, draw a parallel line to the x-axis to pass through y =-4
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
TEST
The incomplete table is for the relation y = 2x2 – 5x + 1
x -3 -2 -1 0 1 2 3 4
y 8 1 -1 26
(i) Copy and complete the table
(ii) Using a scale of 2cm on the axis and 2cm to 10units on the y-axis draw the graph of the relation y = 2x2 – 5x + 1 for -3≤ x ≤ 5
(iii) On the same axes, draw the graph of y = x + 6
(iv) From the graphs, estimate the least value of y and the corresponding value of x
(v) Find also from the graphs the solution of the equation 2x2 – 5x +1 = x + 6
ASSIGNMENT;
Copy and complete the table below for y = 5sin3θ – 3 cos 2θ
θ 0 30 60 90 120 150 180
y -3 1.5
(b) using a scale of 2cm to 300 on θ axis and 2cm to 1 unit on the y-axis, draw the graph of the relation y = 5sin3θ -3cos 2θ for 00≤ θ ≤ 1800
(c) Use your graph to find the (i) solution of y = 5sinθ – 3cos2θ(ii) values of θ for which y is maximum.
WEEK 9
REVISION QUESTIONS
1. Factorise 16x2y – 24x3y3 (a) 8x2y(2 – 3xy2) (b) 4x2(4y – 6xy3) (c) 2xy(8x – 12x2y2) (d) 8(2x2y – 3x3y3)
2. Calculate the length of the diagonal of a square whose area is pcm2 (a) p (b) 2 p (c) p 2 (d) 2p
3. If 3log a – 6log a = log 64 what is a? (a) 4 (b) 6 (c.) 8 (d) 16
4. If 5 times a certain integer is subtracted from twice the square of the integer, the result is 63. Find the integer. (a) 21 (b) 9 (c.) 7 (d) 4
5. Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height 8cm (Take ∏ 22/7) (a) 126.5cm2 (b) 165cm2 (c) 176cm2 (d) 214.5cm2.
6. 1/3 of 4/5 of a certain number is 3. Find the number correct to 3 significant figures.
(a) 3.75 (b) 3.80 (c.) 11.25 (d)11.3
7. Lena bought 400 Alpha Company shares at #1.50 each and sold them at #2.05 each. What was her gain? (a) #0.55 (b) 220.00 (c) #200.00 (d) #420.00
8. In an A.P., the first term is 2, and the sum of the 1st and 6th terms is 16 ½. What is the 4th term?
(a) 12 (b) 9 ½ (c)8 (d) 7
9. A trader sold a pair of shoes for C 2,800.00 making a loss of 20% on his cost price. Find his loss as a percentage of his selling price (a) 16 2/3% (b) 20% (c.) 25% (d) 75%
10. The first term of a G.P is 6. If its common ratio is 2, find the 6th term (a) 60 (b) 72 (c) 96 (d) 192
11. A money lender collects s200 simple interest on a capital after 2years at 5%. Calculate the capital invested (a) s 1,000.00 (b) s2,000.00 (c) s3,000.00 (d) s4,000.00
12. Ladi sold a car for #84,000.00 at a loss of 4%. How much did Ladi buy the car?. (a) #80,500.00 (b) #80,640.00 (c) #87,360.00 (d) #87,500.00
13. Correct 0.04945 to two significant figures (a) 0.040 (b) 0.049 (c) 0.050 (d) 0.049
14. The total surface area of the walls of a room, 7m long, 5m wide and xm high is 96m2. Find the value of x (a) 2/3 (b) 2 (c) 4 (d) 8
15. Simplify (0.3 x 105) ÷ (0.4 x 107), leaving your answer in the standard form (a) 7.5 x 10-4
(b) 7.5 x 10-3 (c) 7.5 x 10-2 (d) 7.5 x 10-1
16. A trader bought 100 tubers of yam at 5 for #350.00. She sold them in sets of 4 for #290.00. Find her gain percent (a) 3.6% (b) 3.5% (c) 3.4% (d) 2.5%
17. Express the square root of 0.000144 in standard form (a) 1.2 x 10-4 (b) 1.2 x10-3 (c) 1.2 x 10-2 (d) 1.2 x 10-1
18. The nth term of a sequence is represented by 3 x2(2-n). Write down the first three terms of the sequence (a) 3/2, 3, 6 (b) 6,3,3/2 (c) 3/2, 3 ,1/3 (d) 2/3 , 3, 8/3
19. A trader makes a loss of 15% when selling an article. Find the ratio, selling price: cost price(a) 3:20 (b) 3:17 (c) 17:20 (d) 20:23
20. A boy estimated his transport fare for a journey as #190 instead of #200. Find the percentage error in this estimate (a) 95% (b) 47.5% (c) 5.26% (d) 5%
THEORY
ANSWER ANY TWO QUESTIONS, Number 3 IS COMPULSORY AND ANY OTHER ONE
1a. A man underestimated his expenses by 6.5%, but actually spent #400.00 . What was his estimate?
b. Correct the following to 2 significant figures
(i) 0.0000086439 (ii) 426009
c.. Simplify 17. 36 x 4.65
2a. Evaluate by using logarithm tables 16.82 x 0.00635 correct to three significant figures.
0.04822
b
Using logarithm table, evaluate 3 1.376 correct to three significant figure.
5 0.007
3ai.The 14th term of an A.P is 96 while 25th term is 173. Find the
(a) 19th term
(b) Sum of 13th and 56th terms
(c) Product of 6th and 13th terms
aii. Find the sum of the first 25 terms of the sequence 11,15,19, 23, 27.....
3bi. The 3rd and 9th terms od a G.P are 54 and 39,366 respectively. Find the
(a) 6th
(b) Sum of the 4th and 7th terms
(c) Product of 2nd and 5th terms
Bii The sum of infinity of geometric progression is 18. If the first term is 1/3, find the common ratio.
1. Factorise 16x2y – 24x3y3 (a) 8x2y(2 – 3xy2) (b) 4x2(4y – 6xy3) (c) 2xy(8x – 12x2y2) (d) 8(2x2y – 3x3y3)
2. Calculate the length of the diagonal of a square whose area is pcm2 (a) p (b) 2 p (c) p 2 (d) 2p
3. If 3log a – 6log a = log 64 what is a? (a) 4 (b) 6 (c.) 8 (d) 16
4. If 5 times a certain integer is subtracted from twice the square of the integer, the result is 63. Find the integer. (a) 21 (b) 9 (c.) 7 (d) 4
5. Calculate the surface area of a hollow cylinder which is closed at one end, if the base radius is 3.5cm and the height 8cm (Take ∏ 22/7) (a) 126.5cm2 (b) 165cm2 (c) 176cm2 (d) 214.5cm2.
6. 1/3 of 4/5 of a certain number is 3. Find the number correct to 3 significant figures.
(a) 3.75 (b) 3.80 (c.) 11.25 (d)11.3
7. Lena bought 400 Alpha Company shares at #1.50 each and sold them at #2.05 each. What was her gain? (a) #0.55 (b) 220.00 (c) #200.00 (d) #420.00
8. In an A.P., the first term is 2, and the sum of the 1st and 6th terms is 16 ½. What is the 4th term?
(a) 12 (b) 9 ½ (c)8 (d) 7
9. A trader sold a pair of shoes for C 2,800.00 making a loss of 20% on his cost price. Find his loss as a percentage of his selling price (a) 16 2/3% (b) 20% (c.) 25% (d) 75%
10. The first term of a G.P is 6. If its common ratio is 2, find the 6th term (a) 60 (b) 72 (c) 96 (d) 192
11. A money lender collects s200 simple interest on a capital after 2years at 5%. Calculate the capital invested (a) s 1,000.00 (b) s2,000.00 (c) s3,000.00 (d) s4,000.00
12. Ladi sold a car for #84,000.00 at a loss of 4%. How much did Ladi buy the car?. (a) #80,500.00 (b) #80,640.00 (c) #87,360.00 (d) #87,500.00
13. Correct 0.04945 to two significant figures (a) 0.040 (b) 0.049 (c) 0.050 (d) 0.049
14. The total surface area of the walls of a room, 7m long, 5m wide and xm high is 96m2. Find the value of x (a) 2/3 (b) 2 (c) 4 (d) 8
15. Simplify (0.3 x 105) ÷ (0.4 x 107), leaving your answer in the standard form (a) 7.5 x 10-4
(b) 7.5 x 10-3 (c) 7.5 x 10-2 (d) 7.5 x 10-1
16. A trader bought 100 tubers of yam at 5 for #350.00. She sold them in sets of 4 for #290.00. Find her gain percent (a) 3.6% (b) 3.5% (c) 3.4% (d) 2.5%
17. Express the square root of 0.000144 in standard form (a) 1.2 x 10-4 (b) 1.2 x10-3 (c) 1.2 x 10-2 (d) 1.2 x 10-1
18. The nth term of a sequence is represented by 3 x2(2-n). Write down the first three terms of the sequence (a) 3/2, 3, 6 (b) 6,3,3/2 (c) 3/2, 3 ,1/3 (d) 2/3 , 3, 8/3
19. A trader makes a loss of 15% when selling an article. Find the ratio, selling price: cost price(a) 3:20 (b) 3:17 (c) 17:20 (d) 20:23
20. A boy estimated his transport fare for a journey as #190 instead of #200. Find the percentage error in this estimate (a) 95% (b) 47.5% (c) 5.26% (d) 5%
THEORY
ANSWER ANY TWO QUESTIONS, Number 3 IS COMPULSORY AND ANY OTHER ONE
1a. A man underestimated his expenses by 6.5%, but actually spent #400.00 . What was his estimate?
b. Correct the following to 2 significant figures
(i) 0.0000086439 (ii) 426009
c.. Simplify 17. 36 x 4.65
2a. Evaluate by using logarithm tables 16.82 x 0.00635 correct to three significant figures.
0.04822
b
Using logarithm table, evaluate 3 1.376 correct to three significant figure.
5 0.007
3ai.The 14th term of an A.P is 96 while 25th term is 173. Find the
(a) 19th term
(b) Sum of 13th and 56th terms
(c) Product of 6th and 13th terms
aii. Find the sum of the first 25 terms of the sequence 11,15,19, 23, 27.....
3bi. The 3rd and 9th terms od a G.P are 54 and 39,366 respectively. Find the
(a) 6th
(b) Sum of the 4th and 7th terms
(c) Product of 2nd and 5th terms
Bii The sum of infinity of geometric progression is 18. If the first term is 1/3, find the common ratio.
