FUNCTIONS 3

PROBABILITY 3

A jar contains 10 blue balls and 11 red balls. Two balls are drawn without replacement. What is the probability of getting two red balls?

Solution

Total number of balls = 10 + 11 = 21

Let P(A) = Probability of getting first red ball and
      P(B) = Probability of getting second red ball

Therefore:

P(A) = 11/21

After first withdraw we are left with 20 balls.

So P(B) = 10/20

Probability of 2 red balls = P(A) x P(B)

                                            = 11/21 × 10/20
                                            = 110/420

                                            = 11/42

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

A jar contains 4 green balls and 8 red balls. Two balls are drawn without replacement. What is the probability of getting two green balls?

FUNCTIONS 2

A P 1

The first term of an AP is 13 and the last term is 57.If the arithmetic progression consists of 40 terms, calculate the sum of all the terms. 

Solution

A₁ = 13, n=40, A_n = 57

S_n = n(A₁ + A_n)

       2

S₂₀ = 40(13 + 57)

         2

S₂₀ = 20 x 70

S₂₀ = 1400

Hence the sum of all 20 terms is 1400.

TRY THIS………….

The first term of an AP is 11 and the last term is 163.If the arithmetic progression consists of 60 terms, calculate the sum of all the terms. 

SETS 1

In a certain school, 140 students take either Chemistry or physics. 160 take chemistry and 100 take both subjects. Find those who take physics.

Solution

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

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

n(C)= 160 ,

n(P)= ?

n(CuP) = 140,

n(CnP)= 100

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

140  =  160 + n(P)  – 100

140  =  160 – 100 + n(P) 

140  =  60+ n(P) 

140 - 60  =  n(P)

n(P)= 80

Hence those taking  physics are 80.    

TRY THIS………….

In a certain school, 185 students take either Chemistry or French. 160 take chemistry and 130 take both subjects. Find those who take French.

PERIMETER 1

Area of rectangle is 60m². Find its perimeter if width is 4m. 

Solution

A = L x w

60 = L x 4

60 = 4L                                                

4      4

15 = L

Now; P =2(L + w)

= 2(15 + 4)

= 2 x 19

= 38m

Hence perimeter is 38m

TRY THIS………..

Area of rectangle is 80m². Find its perimeter if width is 4m. 

PROBABILITY 2

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of Kings?

Solution

(This is a problem with replacement)

n(E) = 4, n(S) = 52

P(E) = n(E)

            n(S)

1st pick = 4/52

2nd pick = 4/52 as well.

P(2 Kings) =   4      x    4

                        52           52

P(2 Kings) =    16    =      1 

                       2704        169  

Therefore Probability of drawing 2 spade cards is 1/169     

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

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a pair of diamonds?  

PRIME FACTORS 1

Write 3872 as a product of prime factors.

Solution

2	3872
2	1936
2	968
2	484
2	242
2	121
11	121
11	11
	1

 Then, 3872 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Hence   3872 = 2⁶ x 11².

TRY THIS........

Write 1056 as a product of prime factors.

EXPONENTIALS 1

Simplify (7a)² + 11a². 

  

Solution

= (7a)² + 11a²    

= 7².a² + 11a²              Since (mn)² =m² . n²

= 49a² + 11a²

= 60a²

TRY THIS………………

Simplify (6a)² + 20a².   

VARIATION 1

x is inversely proportional to y. x=18 while y=5. Find y when x is 6.

Solution

x⍺k

    y

x = k

      y

18 = k

        5

k = 18 x 5

k = 90

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k = 90, y= ?, x = 6.

x = k

      y

6 = 90

         y

y x 6 = 90  x  y   (multiply by y on both sides)

                y

6y = 90

6       6

y = 15 .

TRY THIS…………….

x is inversely proportional to y. x=48 while y=4. Find y when x is 8.