PROBABILITY 2

Given the interval of 1-12 inclusive, Find the probability of having an even number or a prime number.

Solution

This is a non-mutually exclusive event.

n(S) = number of sample space = 12

Let P(E)= probability of even numbers = ⁶/12

Let P(R)= probability of prime  numbers = ⁵/12

P(EnR) = probability of having both even and prime numbers= ¹/12

∴ P(EuR) = P(E) + P(R)  - P(EnR).

 P(EuR) =  6  +  5 – 1

                12    12   12

P(EuR) =  11  -   1 

                12     12  

P(EuR) =  10 

                12      

Hence probability of an even number or a prime number is ^5/6.

TRY THIS........

Given the interval of 1-15 inclusive, find the probability of having an even number or a prime number.

EXPONENTS 2

Simplify 5

            125^-2/3

Solution

=  5

  125^-2/3

=  5 x 1

         125^-2/3

=  5 x 125^2/3        [Since 1/a^-n = aⁿ]

=  5 x (125^1/3)²        [Since 125^1/3 = cube root of 125 = 5]

=  5 x (5)²        

=  5 x 25

= 125        

         

Hence   5        = 125

            125^-2/3

TRY THIS…………………………

simplify 5

             27^-2/3

DEGREES TO RADIAN

Change 1200⁰ into radians.

Solution     

s = ∏Ө/180

            

s = ∏ x 1200

              180

s =  20∏ radians

        3

Hence 500⁰ = 20∏ radians            .

                           3                               

TRY THIS…………….

Change 1500⁰ into radians

ALGEBRA 4

Expand 2x(6x – 5)

Solution

= 2x(6x – 5)

= (2x x 6x) – (2x x 5)

= 12x² – 10x

Hence 2x(6x – 5) = 12x² – 10x

TRY THIS……………

Expand 11a(7a – 4)

EXPONENTS 1

Simplify (m⁴)⁷

solution

= (m⁴)⁷

=  m^4x7

=  m²⁸

TRY THIS........

Simplify  (h¹⁸)⁵

ARITHMETICS 2

Evaluate  20 + 6 – 5 – 7 – 9 + 10.

Solution

= 20 + 6 – 5 – 7 – 9 + 10

= 20 + 6 + 10 – 5 – 7 – 9

= 20 + 6 + 10 – ( 5 + 7 + 9)

= 36 – 21

= 15

TRY THIS

Evaluate  35 + 6 – 4 – 7 – 11 + 8

ARITHMETICS 1

Evaluate 144 x 545 + 455 x 144

Solution

= 144 x 545 + 455 x 144

= 144 x (545 + 455)

= 144 x 1000

= 144000

Hence 144 x 545 + 455 x 144 = 144000

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

Evaluate 132 x 376 + 624 x 132

FACTORIZATION 7

Factorize 3x² - 8x - 35.

Solution

= 3x² - 8x - 35. We split the middle term (-8x) to be (+7x-15x).

= 3x² + 7x - 15x - 35.

= (3x² + 7x) – (15x – 35).

= x(3x + 7) – 5(3x + 7).

= (x - 5) (3x + 7)

Hence 3x² - 8x – 35 = (x - 5) (3x + 7) answer.

TRY THIS…..

2x² - x + 15.

MATRIX 2

LOGARITHMS 9

Evaluate Log₂(32 x 16).

Solution

= Log₂(32 x 16)

= Log₂32 +  Log₂16          (applying the product rule)

= Log₂2⁵ +  Log₂2⁴             ( 32= 2⁵ and 16=2⁴ )

= 5Log₂2 +  4Log₂2       ( remember  Log_aa^c = cLog_aa )

= (5 x 1) +  (4 x 1)          ( remember  Log_aa = 1 )

= 5 + 4

= 11

hence Log₂(32 x 16) = 11

TRY THIS…………………

Evaluate Log₂(8 x 32).

MATRIX 1

SETS 4

If n(A)= 42 , n(B)= 98 and n(AnB)=30, find n(AuB).

Solution

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

             = 42 + 98 – 30

             = 140 – 30

             = 110

Hence n(AuB) = 110 answer

TRY THIS…………………

If n(A)= 43 , n(B)= 97 and n(AnB)=50, find n(AuB).

LOGARITHMS 8

If Log_ax = ¹/₂  Log_a16+ ¹/₃ Log_a64 find x.

Solution

Log_ax = ¹/₂  Log_a16+ ¹/₃ Log_a64

Log_ax = Log_a16^1/2  +  Log_a64^1/3       (64^1/3  = 4 )   

Log_ax = Log_a 4 +  Log_a4

Log_ax = Log_a (4 x 4)

Log_ax = Log_a 16

x = 16

TRY THIS…………………

If Log_ax = ¹/₂  Log_a900+ ¹/₃ Log_a125 find x.

ALGEBRA 3

Expand 3x(4x – 5)

Solution

= 3x(4x – 5)

= (3x x 4x) – (3x x 5)

= 12x² – 15x

Hence 3x(4x – 5) = 12x² – 15x

TRY THIS……………

Expand 7a(7a – 4)

SETS 3

If n(B)= 90 , n(AuB)=110, n(AnB)=50, find n(A).

Solution

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

     110 = n(A)  + 90 - 50

     110 = n(A)  + 40      

     110 - 40 = n(A) 

      70 = n(A) 

Hence n(A) = 70 answer

TRY THIS…………………

If n(B)= 88 , n(AuB)=107, n(AnB)=52, find n(A).

PERFECT SQUARES 2

If 16x² + 8x + h is a perfect square, find h.

Solution

a = 16, b = 8, c = h.

b² = 4ac

(8)² = 4 x 16 x h

64 = 64h

 64  =   64h

 64       64

           

h = 1

Hence h = 1

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

If 25x² + 10x + w is a perfect square, find w.

DIFFERENCE OF 2 SQUARES 2

Evaluate 740² – 640²

Solution

740² – 640² = (740 + 640)( 740 – 640)

                    = (1380)( 100)

                    = 138,000

Hence 740² – 640² = 138,000

TRY THIS……………………..

Evaluate 770² – 670²

CIRCLES 3

Find y in the figure below;

Solution

From circle geometry,

(QR)(QS) = (PQ)²

y(y + 21) = 10²

y² + 21y = 100

y² + 21y – 100 = 0

We factorize by splitting the middle term: 21y = (25y – 4y).

y² + 25y – 4y – 100 = 0

(y² + 25y) – (4y – 100) = 0

y(y + 25) – 4(y + 25) = 0

(y - 4)( y + 25) = 0                              

y - 4 = 0  OR  y + 25 = 0

y = 4 OR y = -25 

Since we don’t have negative length, we take y = 4cm

TRY THIS……..

Find y in the figure below;

GRADIENT 2

Find the slope of a line which passes through (4, 13) and (14, 8)

Solution

x₁ = 4,  y₁ =13,  x₂ = 14,  y₂ = 8

m = y2 –y1

      x2 – x1

m =   8 –13

        14 –4

m =   -5

         10

m =   -1                           (After simplification)

          2

Hence the slope is -1/2

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

Find the slope of a line which passes through (4, 1) and (16, 8)

FACTORIZATION 6

Factorize 4x² + 13x + 3

Solution

4x² + 13x + 3

4x² + 12x + x + 3

(4x² + 12x) + (x + 3)

4x(x + 3) + 1(x + 3)

(4x + 1) (x + 3)

Hence 4x² + 13x + 3 = (4x + 1) (x + 3)

TRY THIS……………..

Factorize 6x² + 10x + 5

CIRCLES 1

Find y in the figure below

From circle geometry,

(QR)(QS) = (PQ)²

y(y + 16) = 6²

y² + 16y = 36

y² + 16y – 36 = 0

We factorize by splitting the middle term. 16y = (– 2y + 18y).

y² + 18y – 2y – 36 = 0

(y² + 18y) – (2y – 36) = 0

y(y + 18) – 2(y + 18) = 0

(y - 2)( y + 18) = 0                              

y - 2 = 0  OR  y + 18 = 0

y = 3 OR y = -18 

Since we don’t have negative length, we take y=2cm

TRY THIS……..

Find y in the figure below;

PERFECT SQUARES 1

If 4x² + 12x + t is a perfect square, find t.

Solution

a = 4, b = 12, c = t.

b² = 4ac

(12)² = 4 x 4 x t

144 = 16t

144 = 16t

 16      16

             

t = 9

Hence t = 9

TRY THIS……..

If 16x² + mx + 1 is a perfect square, find m.

GRADIENT 1

Find the slope of a line which passes through (7, 13) and (6, 8)

Solution

x₁ = 7,  y₁ =13,  x₂ = 6,  y₂ = 8

m = y2 –y1

       x2 – x1

m =   8 –13

         6 –7

m =   -5

         -1

  

Hence the slope is 5

TRY THIS……..

Find the slope of a line which passes through (1, 13) and (6, 8)

INEQUALITIES 1

Find x if 7x – 21 ≤ x + 15 ≤ 6x.

Solution

7x – 21 ≤ x + 15 and  x + 15 ≤ 6x

7x – x ≤ 21+ 15 and  15 ≤ 6x - x

6x ≤ 36 (divide by 6 both sides) and  15 ≤ 5x (divide by 5 both sides)

x ≤ 6 and  3 ≤ x

x ≤ 6 and  x ≥ 3

TRY THIS……..

Find x if 4x – 31 ≤ x + 15 ≤ 6x.