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

A_n = 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