CIRCLES-1

Find w in the figure below. [Diagram not to scale]

Solution

Angle inscribed in a semi-circle = 90⁰

<NMP = 90⁰

5w – 15 = 90⁰

5w – 15 = 90⁰  + 15

5w          = 105⁰  

5w          = 105⁰  

5                 5

w = 21⁰

TRY THIS………………..

Find a in the figure below. [Diagram not to scale]

PROBABILITY - 3

A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and then a number ten card?

solution

These are called independent events.

Probabilities:           

P(jack)  =       4

                     52

P(no. 10)      =          4                    

                               52                     

P(jack and no.10)  =         P(jack)           •         P(no.10)

                        =          4          •          4

                                    52                   52

             

             =           16                            

                       2704                         

             

             =           1                               

                       169

TRY THIS……………………………………

A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a king and then a number six card? 

LOGARITHMS -1

If Log_ax = 14, find Log_a(¹/x).

Solution

= Log_a(¹/x)

= Log_ax^-1

= -1 x Log_ax

= -1 x 14

= -14

TRY THIS………..

If Log_af= 43, find Log_a(¹/f)

ALGEBRA-1

Expand 8w(w – 3)

solution

= 8w(w – 3)

= (8w x w) – (8w x 3)

= 8w² – 24w     answer

TRY THIS………..

Expand 10p(p+ 5)

PROBABILITY - 2

A bag of candy contains 8 lemon flavored sour balls, and 5 lime flavored sour balls.  If Ana reaches in, takes one out and eats it, and then 20 minutes later selects another and eats that one as well, what is the probability that they were both lemon flavored candies?

solution

These are called independent events.

It is a ''Without replacement" question.

P(lemon 1st) = 8/13
P(lemon 2nd) = 7/12 

(remember...one is already eaten!)

Therefore:
P(lemon, lemon) = (8/13)(7/12) = 56/156 = 28/78=14/39

the probability that they were both lemon flavored candies is 14/39

TRY THIS………………….

A bag of candy contains 8 lemon flavored sour balls, and 5 lime flavored sour balls.  If Ana reaches in, takes one out and eats it, and then 20 minutes later selects another and eats that one as well, what is the probability that they were both lime flavored sour balls?

SETS - 3

If n(A)= 36 , n(B)= 50 and n(AuB)=65, find n(AnB).

Solution

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

       65 = 36 + 50 - n(AnB)

       65 = 86  - n(AnB)

       65 - 86 = - n(AnB)

      -21 = - n(AnB)

21 = n(AnB) (after dividing by -1 on both sides)

Hence n(AnB) = 21 answer

TRY THIS……………………

If n(A)= 35 , n(B)= 45 and n(AuB)=70, find n(AnB).

STATISTICS - 2

Distribution of length of nails in mm is as shown below. 

Length(mm)	15 - 21	22- 28	29 - 35	36 - 42	43 - 49	50 - 56	57 - 63
Frequency	4	6	8	16	8	5	3

Calculate the median.

Solution

First we prepare the frequency distribution table.

Class interval	Frequency(f)	Cumulative frequency
15-21	4	4
22-28	6	10
29-35	8	18
36-42	16	34
43-49	8	42
50-56	5	47
57-63	3	50
	∑f = 50

N = 50, N/2=25, 

Median class must fall in the cumulative frequency of 34. This has to be 36 – 42.

n_b = 18, n_w = 16, Upper boundary(U)= 42.5, Lower boundary(L) =35.5

i = Upper boundary – Lower boundary

i = 42.5 – 35.5

i = 7

Median = L + (N/2 – n_b)i/n_w

                              

Median = 35.5 + (50/2 – 18)7

                            16

Median = 35.5 + (25 – 20)7

                             16

Median = 35.5 + (5)7

                        16

  Median = 35.5 +  35

                           16

  Median = 35.5 +  2.1875

                               

 Median = 37.6875

Hence the median is 37.69 ( to 2d.p)                     

 

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

Distribution of Kiswahili test marks was given as hereunder. 

Length(mm)	10 - 19	20- 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79
Frequency	3	7	9	15	8	5	3

Calculate the median.

SETS - 2

In a certain school, 196 students take either Chemistry or physics. 130 take chemistry and 100 take both subjects. Find those who take physics only.

Solution

In most cases, OR stands for union whereas AND/BOTH, stands for intersection.

Let Chemistry = n(C), physics = n(P).

n(C)= 130 ,

n(P)= ?

n(CuP) = 196,

n(CnP)= 100

n(CuP) = n(C) + n(P) - n(CnP)

196  =  130 + n(P)  – 100

196  =  n(P) + 30

196 - 30  =  n(P)

n(P)= 166

physics only = n(P) – n(CnP)

                    = 166 – 100

                    = 66 

Hence those taking only physics are 66.      

TRY THIS…..

In a certain school, 220 students take either Chemistry or physics. 140 take chemistry and 110 take both subjects. Find those who take physics only.

FACTORIZATION - 1

Factorize 2x² - x - 15.

Solution

= 2x² - x – 15.  We split the middle term (-x) to be (-6x + 5x).

= 2x² - 6x + 5x – 15     (-7x = -6x + 5x).

= (2x² - 6x) + (5x - 15)

= 2x(x - 3) + 5(x - 3)

= (2x + 5) (x - 3)

Hence 2x² - x – 15 = (2x + 5) (x - 3) answer.

TRY THIS…..

Factorize 3x² + x – 14.

SETS - 1

If n(A)= 62 , n(B)= 100 and n(AnB)=20, find n(AuB).

Solution

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

             = 62 + 100 – 20

             = 162– 20

             = 142

Hence n(AuB) = 142 answer

TRY THIS………….

If n(A)= 110 , n(B)= 178 and n(AnB)=83, find n(AuB).