Week 5 and 6 – Jss 3 First Term Mathematics Lesson Notes

WEEK 5
WORD PROBLEMS
Sum and Difference

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

SOLUTION
The number be
+ 30 = -18
= – 48

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

SOLUTION
Let the number be
– (-5) = 20
+ 5 = 20
– 15
If is less than -5
-5 – = 20
= -5 – 20
= -25

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

   SOLUTION
   

Let the numbers be n, n + r, and n + 2
n + (n + 1) + (n + 2) = 138
3n + 3 + 138
3n = 138
n = 45
n + 1 = 46
n + 2 = 47
The numbers are 45, 46, and 47
Exercise: Ex 2.1 No 6, 7, 8

 
 
 Product
Examples: (1) Find the product of 1, -0.8 and – 2 (2) what number must be multiplied by 25 to make ? (3) The product of 3 numbers is 3600. If two of the numbers are equal and the third number are 25. Find the two equal numbers.
SOLUTION
1.     () x         (2) Let the number be         (3) 25 x x = 3600
    =                     25 =                 25² = 3600
    = 3                 100x = 3                ² =
                         =                 ² = 144
                         = 0.03                 = 12                
Solving combine products with sum and difference
Examples: (1) Add the sum of the 20 and 30.5 to the position difference between 25 and 45. (2) From the sum of 8 ad 7 subtract the negative difference between 15 and 24 (3) find the product of 3 and
SOLUTION
1.    (20 + 30.5) + (45 – 25)        (2) -8 + 7 – 115 – 25    (3) Sum = +
    = 50.5 + 20                 = -8 + 7 + 9        
    = 70.5                     = 8                 =
Product = x
= = 1
Exercise: Ex. 2.3 No 1, 2, 3, page 17.
Word problems with fractions
Examples:

1. Find the three-fifth of the sum of 45 and -60. (45 – 60) = (-15) = -9
2. Divide the difference between 25 and 10 by the product of 6 and 5. = = 0.5
3. Find one-quarter of the sum of the product of 2 and 3 and the product of 1 and

 Solution: = = 1
Exercise: Ex 2.4 No 1, 2, 3, page 18
Problems leading to equations
Examples

1. When of a number is added to 30. The result is 20 added to the number find the number.
2. When the sum of 28 and a certain number is divided by 5. The result is equal to treble the original number. What is the number?

SOLUTION
(1) Let the number be (2) Let the number be
Hence + 30 = + 30                 = 3
Multiply each term by 5                15 = 28 +
3 + 150 = 5 + 100                     =
5 – 3 = 150 -100                     = 2
2 = 50
= 25
ASSIGNMENT
EXERCISE 2.6; NO 1, 2, AND 3 PAGE 20.

 WEEK 6

 CHANGE OF SUBJECT OF A FORMULA
Subject of formula: A formula is always written in terms of the subject e.g. V = ²h is a formula for volume of a cone, r is the radius, height (h) =.
Examples 1: Make r the subject of the formula in A = ² (2) Make T the subject of the formula in = K (3) Make x the subject of the formula in y =k^m/x
SOLUTION
a)    A = ²        b) = K         c)    y =
    ² =                PT = KV     xy = km
    ² =             T =          x =
Exercise

1. Make b the subject of the formula in A = (a + b) x
2. Make x the subject of the formula in a = b ( 1 – x)
3. Make w the subject of the formula in L =

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

1. Write in terms of A and r
2. Find the height of the cylinder of base radius 7cm and the total surface area of 396cm². = .

SOLUTION

1. A = 2(r+h)        b) h = – 7

   r + h =              h = 2m
   h = – r

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

1. if T = (a) Express m in terms of T an K (b) Find m when T = 20 and K = 50, =
2. make h the subject of the formula V = ²h
3. hence find the value of h when V = 256, = , r = 21