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Question 2 : The 9th term of an arithmetic sequence is 12 and the 17th term is 28, find the 4th term of the arithmetic sequence.
 19.77
 25.4
 28.2
 16.5

We can solve this system   where a1 is the first term and d is the common differeence

12  =  a1  + 8d

28  =  a1  + 16d             subtract the first equation from the second

16   =  8d                      divide both sides by  8

2   = d

And using either equation, we can find the first term as

12  =   a1  +  8(2)

12  =  a1  + 16

12 - 16  = a1

-4 = a1

So  the  4th term is

-4 + 2(3)   =     2

Question 3 : A survey conducted for 50 members of staff of Cadbury plc shows that 30 of them have degree while 35 possess H.N.D. if 8 of the employees possess neither of the certificate how many have both degree and H.N.D

It's obvious that    50 - 8  =  44 have certificates

So we have the following equation

N( Degree U HND)  =  N(Degree)  + N(HND) -  N(Degree ∩ HND)

42    =   30   +  35  -     N(Degree ∩ HND)

42  =  65  - N(Degree ∩ HND)

N(Degree ∩ HND)  =  65  - 42    =    23

Question 5 : Using factorization method solve for x, 5x²+ 12x +7 = 0

(5x  + 7) ( x + 1)   = 0

Setting both factors to 0  and solving for x we have that

x = -7/5   =   - 1  2/5        or  x   =  -1

Question 6 : If A= {3,5,7,9} B = {2,3,4,8,7,9} then AuB is
 

A U B    =  { 2,3,4,5,7,8, 9 }

Question 7 : The sum of eight times a number and 15 is less than thrice the same number minus 10, find the number.

8x + 15  <  3x - 10        add 10 to both sides, subtract 8x from both sides

25 < -5x            divide both sides by -5  and reverse the inequality sign

-5 > x     →    x < -5

Question 8 : Determine the value of x and y simultaneously given that x + y = 10 and xy = 21

x + y  = 10     subtract x from both sides

y  = 10 - x  

Sub this into the other equation

x ( 10 - x)  =  21

10x - x^2  =  21         rearrange as

x^2 - 10x  +  21   =  0   factor

(x - 7) ( x - 3)  =  0

Setting each factor to 0   and solving for x  we have that

x = 7   and   x  = 3

Question 9 : Given that A{prime number less than 20}, B {even numbers less than 15}, C {Multiples of 4 less than 15} then AuBnC is

A  =  { 2, 3, 5, 7, 11, 13, 17 , 19 }

B =  { 2, 4, 6 , 8 , 10, 12, 14 }

C =  { 4, 8, 12 }

So  A U B ∩ C  

{ 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13 14, 17, 19}  ∩ { 4, 8, 12}   =

{ 4, 8, 12 }

Question 13 : Find the sum of 28th term of the AP 3+10+17+-----

The 28th  term is given by

3 + 7(27)   =  192

So.....the sum of the first 28 terms is  

(first term + last term)(number of terms) / 2  =

(3 + 192) (28)  / 2  =

(195) * (14)  =

2730

Question 15 : Given Ƹ = {1,2,3,4,5,.6,7,8}, A= {1,2,4,8}, B= {1,3,5,7}, C = {2,4,6,8}, find A n B¹ u C¹.

B '  = { 2, 4, 6, 8 }

C' = { 1, 3, 5, 7 }

So

A ∩ B' U C'  =

{ 2, 4, 8 }  U { 1, 3, 5, 7 }  =

{1, 2, 3, 4, 5, 7, 8 }

Question 16 : Find the value of K in the perfect square 12x² - 6x + K

Write  12x^2 - 6x

Factor  out  the 12

12  ( x^2  -  (1/2)x  )

Take (1/2)  of (1/2)  = 1/4.....square this  =  1/16

So  we  have

12 (x ^2  - (1/2)x  + 1/16)

12x^2  - 6x  + 12/16

12x^2 - 6x  +  3/4

So....  K  =  3/4

Question 17 : Given equation 4x² - 7x + 9 = 0, find the value of x.

Because the disriminant < 0,  this has no real roots

14.

 All the element are not the same