Monday 23 March 2015

ARITHMETIC PROGRESSION 1


The 1st term of an A.P. is 40 and the common difference is 58. Find the 12th term.

Solution

A1= 40, d = 58

An = A1 + (n-1)d   [ formula for n terms ]

A12 = A1 + (12-1)d

A12 = A1 + 11d    [ formula for 12 terms ]

A12 = 40 + (11 x 58)

A12 = 40 + 638

A12 = 678


Hence the 12th term is 678.

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


The 1st term of an A.P. is 456 and the common difference is 70. Find the 12th term.

FUNCTIONS 1


Given that F(x) = 15x  +  65. Find F(7)

Solution

F(x) = 15x  +  65

F(7) = 15(7)  +  65

F(7) = 105  +  65

F(7) = 170


Hence  F(7) = 170

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

Given that F(x) = 50x  -  70. Find F(3)

PERFECT SQUARES 2


If zx2 - 28x + 49 = 0 is a perfect square, find the value of z .

Solution

In zx2 - 28x + 49 = 0;   a=z, b=-28 and c=49.


For the perfect square, b2 = 4ac


(-28)2 = 4 x z x 49


784 = 196z
  
 784 =     196z
196    196


4 = z


Hence the value of z is 4

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


If zx2 - 24x + 16 = 0 is a perfect square, find the value of z .


WORD PROBLEMS 1


The sum of five consecutive numbers is 475. Find the 3rd number.

Solution

Let the numbers be as shown in the table below


1st number
2nd number
3rd number
4th number
5th
number
TOTAL
n
n+1
n+2
n+3
n+4
475


Then, n + (n+1) + (n+2) + (n+3)+ (n+4) = 475


5n + 1+2+3 +4= 475


5n + 10= 475


5n = 475 -10


5n = 465


5n = 465
5        5

n = 93


3rd number = n + 2
                  = 93 + 2
                  = 95



Hence the 3rd number is 95

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


The sum of five consecutive numbers is 510. Find the 3rd number.


LOGARITHMS 6


Evaluate Log100000 +  log 0.0001 -  log 0.00000001
  
Solution


= Log100000 +  log 0.0001 -  log 0.00000001

= Log105 +  log 10-4 – log 10-8

= 5Log10 + (-4 log 10) – (-8log 10)

= (5x1) + (-4x1) - (-8x1)

=5 + (-4) – (-8)

= 5 – 4 + 8

=9


Hence Log100000 +  log 0.0001-  log 0.00000001 = 9

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


Evaluate Log10,000,000 +  log 0.0001 -  log 0.00000001


SETS 1


In Bibanja district the number of people who speak Kiswahili or Lingala is 400. 250 of them speak Kiswahili and 350 of them speak Lingala. How many speak both languages?

solution

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

Let Kiswahili=n(K), Lingala= n(L).

n(K)= 250 ,
n(L)= 350,
n(KuL) = 400,
n(KnL)=?

n(KuL) = n(K) + n(L) - n(KnL)

400  =  250 + 350 – n(KnL)

400  =  600 – n(KnL)

n(KnL) =  600 – 400 [– n(KnL) goes to the left it will be +ve

n(EnF) =  200

Hence n(EnF)=200 answer

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


In Bibanja district the number of people who speak English or Lingala is 500. 270 of them speak English and 380 of them speak Lingala. How many speak both languages?

GRADIENT 2


Find the slope of a line which passes through (-5, -2) and (6,-1)

Solution

x1 = -5,  y1 =-2,  x2 = 6,  y2 = -1

m = y2 –y1
      x2 – x1

m =   -1 –(-2)
        6 –(-5)

m =   -1 + 2
        6 + 5

m =    1
        11


Hence the slope is 1/11

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


Find the slope of a line which passes through (-5, -11) and (8,-1)

PROBABILITY 1


Given the interval of 1-15 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 = 15

Let P(E)= probability of even numbers = 7/15
Let P(R)= probability of prime  numbers = 6/15
P(EnR) = probability of having both even and prime numbers= 1/15 [since even 2 is both even and prime]

 P(EuR) = P(E) + P(R)  - P(EnR).

 P(EuR) =  7  +  6  1
                15    15   15

P(EuR) =    13   -   1 
                  15      15  

P(EuR) =    12  or   4
                  15        5 

Hence probability of an even number or a prime number is 4/5.

TRY THIS........



Given the interval of 1 - 23 inclusive, find the probability of having an even number or a prime number.


GRADIENT 1


Find the slope of a line which passes through (-10, -2) and (6,-9)

Solution

x1 = -10,  y1 =-2,  x2 = 6,  y2 = -9

m = y2 –y1
     x2 – x1

m =   -9 –(-2)
        6 –(-10)

m =   -9 + 2
        6 + 10

m =    - 7
         16


Hence the slope is -7/16

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


Find the slope of a line which passes through (-1, -11) and (8,-9)

FINDING MEAN 1


In Mathematics test the following marks were recorded.

marks
31-40
41-50
51-60
61-70
71-80
81-90
No. of students
5
7
14
10
9
5


Calculate the mean.

Solution

Here you are required to produce the frequency distribution table.


Class interval
Class mark (x)
Frequency(f)
fx
31-40
35.5
5
177.5
41-50
45.5
7
318.5
51-60
55.5
14
777
61-70
65.5
10
655
71-80
75.5
9
679.5
81-90
85.5
5
427.5

∑f = 50
∑fx = 3035



Mean = ∑fx
               ∑f

Mean = 3035
               50

Mean =  60.7
         

Hence Mean =  60.7


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

In biology test the following marks were recorded.


interval
20-24
25-29
30-34
35-39
40-44
45-49
No. of students
5
7
16
8
9
5