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).

MATRIX-4

ARITHMETIC PROGRESSION-1

Find the sum of the 1st 16 terms of the following arithmetic progression: 5+11+17+…………………

Solution

d=6, A₁ = 5,n=16

S_n = n[2A₁ + (n-1)d]

       2

S₁₆ = 16[2(5) + (16-1)d]

         2

S₁₆ = 8[10 + 15d]

 

S₁₆ = 8[10 + 15(6)]       since d=6

S₁₆ = 8[10 + 90]

           

S₁₆ = 8 x 100  

         

S₁₆ = 800

Therefore the sum of 1st 16 terms is 800.

TRY THIS......

Find the sum of the 1st 16 terms of the following arithmetic progression: 6+14+22+…………………

MATRIX - 3

STATISTICS 1

In Kiswahili test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59	60-69
No. of students	2	5	9	6	5	3

Calculate the mean.

Solution

Here you are required to produce the frequency distribution table.

Class interval	Class mark (x)	Frequency(f)	fx
10-19	14.5	2	29
20-29	24.5	5	122.5
30-39	34.5	9	310.5
40-49	44.5	6	267
50-59	54.5	5	272.5
60-69	64.5	3	193.5
		∑f = 30	∑fx = 1195

Mean = ∑fx

               ∑f

Mean = 1195

                 30

Mean =  39.83

         

Hence Mean =  39.83

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

In a Biology test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59	60-69
No. of students	4	7	12	8	5	4

Calculate the mean.

MATRIX 2

PERFECT SQUARES 1

If 16x² + 24x + t is a perfect square, find t.

Solution

a = 16, b = 24, c = t.

b² = 4ac

(-24)² = 4 x 16 x t

576 = 64t

576 = 64t

 64      64

           

t = 9

Hence t = 9

TRY THIS………………..

If 9x² + 24x + f is a perfect square, find f.

MATRIX 1