Monday 13 October 2014

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  +  51
                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 1252/3        [Since 1/a-n = an]

=  5 x (1251/3)2        [Since 1251/3 = 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

=  m4x7

=  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.

Tuesday 7 October 2014

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  Logaac = cLogaa )

= (5 x 1) +  (4 x 1)          ( remember  Logaa = 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 Logax = 1/2  Loga16+ 1/3 Loga64 find x.

Solution

Logax = 1/2  Loga16+ 1/3 Loga64

Logax = Loga161/2  +  Loga641/3       (641/3  = 4 )   

Logax = Loga 4 +  Loga4

Logax = Loga (4 x 4)

Logax = Loga 16

x = 16

TRY THIS…………………

If Logax = 1/2  Loga900+ 1/3 Loga125 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

Monday 6 October 2014

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.