PRIME FACTORS 1

Write 3872 as a product of prime factors.

Solution

2	3872
2	1936
2	968
2	484
2	242
2	121
11	121
11	11
	1

 Then, 3872 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Hence   3872 = 2⁶ x 11².

TRY THIS........

Write 1056 as a product of prime factors.

EXPONENTIALS 1

Simplify (7a)² + 11a². 

  

Solution

= (7a)² + 11a²    

= 7².a² + 11a²              Since (mn)² =m² . n²

= 49a² + 11a²

= 60a²

TRY THIS………………

Simplify (6a)² + 20a².   

VARIATION 1

x is inversely proportional to y. x=18 while y=5. Find y when x is 6.

Solution

x⍺k

    y

x = k

      y

18 = k

        5

k = 18 x 5

k = 90

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k = 90, y= ?, x = 6.

x = k

      y

6 = 90

         y

y x 6 = 90  x  y   (multiply by y on both sides)

                y

6y = 90

6       6

y = 15 .

TRY THIS…………….

x is inversely proportional to y. x=48 while y=4. Find y when x is 8.    

BODMAS 1

Evaluate 190 + 50 x 30 – 160 ÷ 4

Solution

We apply BODMAS.

= 190 + 50 x 30 – 160 ÷ 4

= 190 + 50 x 30 – 40    [After dividing]

= 190 + 1500 – 40    [After multiplying]

= 1690 – 40    [After adding]

= 1650       [After subtracting]

Hence 190 + 50 x 30 – 160 ÷ 4= 1650      

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

Evaluate 900 + 40 x 30 – 200 ÷ 5

PROBABILITY 1

A number is chosen at random from 1 – 15 inclusive. Find the probability that it is a multiple of five or an even number.

Solution

THIS IS A NON-MUTUALLY EXCLUSIVE EVENT.

Let n(S) represent sample space

P(E) = probability of even number

P(M) = probability of a multiple of 5

n(S) = {1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12, 13, 14, 15} = 15

n(E) = {2, 4, 6, 8, 10, 12, 14} = 7

n(M) = {5, 10, 15 } = 3

But 14 appears on both categories.

P(E) = ⁷/15

P(M) = ³/15

P(EnM) = ¹/15

………………………….

P(EuM) = P(E) + P(M) - P(EnM) 

P(EuM) = ⁷/15 + ³/15 -¹/15

            = ⁹/15

            = ³/5

∴P(EuM) = ³/5

TRY THIS………………

A number is chosen at random from 4 – 30 inclusive. Find the probability that it is a multiple of 5 or an even number.