MID POINT 1

Find a and b if the midpoint of a line from (a, 6) to (8, b) is

(30, 7)

Solution

x1=a, x2=8, y1=6, y2=b.

(30, 7)= (x1 + x2, y1 + y2)

                   2             2

(30, 7)= (a + 8,  6 + b)

                  2          2

Equating equal values of x;

30= a + 8

         2      

2 x 30= (a + 8)  x 2 ¹

                2 ₁     

60 = a + 8

60-8 = a

a = 52.

Equating equal values of y;

7 = 6 + b

        2

2 x 7= (6 + b)  x 2 ¹  multiplying by 2 both sides.

                2 ₁     

14 = 6 + b

14 – 6 = b

b = 8

Hence a = 52 and b=8

TRY THIS……………………..

Find m and n if the midpoint of a line from (m, 19) to (13, n) is (9, 11)

POLYGONS 1

Find the exterior angle of a regular polygon with 18 sides.

Solution

n = 6

e =   360

         n

e =   360

        18

e =   360 ²⁰

        18₁

e = 20⁰

Hence the exterior angle is 20⁰

TRY THIS………………………………………..

Find the exterior angle of a regular polygon with 12 sides.

PRIME FACTORS 1

Write 560 as a product of prime factors.

Solution

2	560
2	280
2	140
2	70
5	35
7	7
	1

 Then, 480 = 2 x 2 x 2 x 2 x 5 X 7

Hence   480 = 2⁴ x 5 x 7.

TRY THIS........

Write 1000 as a product of prime factors.

SETS 1

If n(B)= 60 , n(AuB) = 170 and n(AnB)=10, find n(A)

Solution

n(AuB) = n(A) + n(B) - n(AnB)

  170    = n(A)  + 60– 10

  170     = 60 -10 + n(A) 

  170     = 50 + n(A) 

  170 - 50        = n(A)

  120        = n(A)

Hence n(A) = 120 answer

TRY THIS…………………………….

If n(B)= 55 , n(AuB) = 169 and n(AnB)=16, find n(A)

ANGLES 1

If A and B are supplementary angles such that A = 27° and B = x + 55° , find the value of x .

Solution

A + B = 180⁰. [Since supplementary angles add up to 180⁰]

27⁰ + x + 55⁰ = 180⁰.

x + 82⁰ = 180⁰.

x = 180⁰ - 82⁰

x = 98⁰  answer.

TRY THIS.............................. 

If A and B are supplementary angles such that A = 53° and B = x + 42° , find the value of x .

EXPONENTIALS 1

If a² = 3; find the value of a¹⁰ + a⁸.

solution

=a¹⁰ + a⁸.

=(a²)⁵ + (a²)⁴.

=(3)⁵  +  (3)⁴

=243 + 81

=324 answer  

  

TRY THIS.........................

If a² = 5; find the value of a¹⁰ - a⁶.

ALGEBRA 2

The sum of five consecutive numbers is 525. Find the 4th number.

Solution

Let the numbers be as shown in the table below

1st number	2nd number	3rd number	4th number	5th number	TOTAL
n	n+1	n+2	n+3	n+4	525

Then, n + (n+1) + (n+2) + (n+3) + (n+4) = 525

5n + 1+2+3+4= 525

5n + 10= 525

5n = 525- 10

5n = 515 

5n = 515

5       5

n = 103

4th number = n + 3

                   = 103 + 3

                   = 106

Hence the 4th number is 106

TRY THIS………………….. 

The sum of five consecutive numbers is 1815. Find the largest number.  

FUNCTIONS 1

FACTORIZE 1

Factorize 1-25m²

Solution

We use difference of two squares a² – b² = (a - b)(a + b)

1-25m² = 1²-5²m²       

            = 1² - (5m)²       

            = (1 - 5m)(1 + 5m)

Hence 1-25m² = (1 - 5m)(1 + 5m)  answer

TRY THIS…………….

Factorize 1 – 49y²

LOGARITHMS 1

If Log_ax = -20, find Log_a(¹/x⁶)

Solution

= Log_a(¹/x⁶)                        

= Log_ax^-6                   [since ¹/cⁿ =c^-n]

= -6 x Log_ax

= -6 x -20

= 120

TRY THIS………..

If Log_aW= -70, find Log_a(¹/W⁴)

ALGEBRA 1

Express as a single fraction m - 3 + m+5

                                               6         2

solution

= m - 3 + m+5

      6          2

= m - 3 + 3(m+5)

            6         

= m - 3 + 3m+15

            6         

= m + 3m + 15- 3

            6         

= 4m +  12

         6         

= 4(m +  3)

         6         

= 2(m +  3)

         3         

= 2m +  6       answer

         3         

TRY THIS...................

Express as a single fraction 4n - 1 + n+1

                                                3          2