SETS 2

If n(AnB)=45, n(B)= 96 and n(AuB)= 146, find n(A)

Solution

n(AuB) = n(A) + n(B) - n(AnB)

146 = n(A)  + 96 - 45

146 = n(A)  +  51

146 - 51 = n(A) 

95 = n(A) 

Hence n(A)  = 95 answer

TRY THIS………….

If n(AnB)=44, n(B)= 97 and n(AuB)= 145, find n(A)

FUNCTIONS 2

Given that F(x) = 3x  +  11. Find F(-4)

Solution

F(x) = 3x  +  11

F(-4) = 3(-4)  +  11

F(-4) = -12  +  11

F(-4) = -1

Hence  F(-4) = -1

TRY THIS………………..

Given that F(x) = 14x  +  38. Find F(-4)

LOGARITHMS 2

Simplify Log₂256 - Log₃243

Solution

= Log₂1024 - Log₃243

= Log₂2¹⁰ - Log₃3⁵    [Since 256=2⁸ and 243=3⁵].

= 10Log₂2 - 5Log₃3    [since Log_aa = 1]

= (10 x 1) - (5 x 1)

= 10 - 5

= 5

Hence Log₂1024 - Log₃243 = 5

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

Simplify Log₂2048 – Log₅625

PROBABILITY 2

	A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and then an eight? solution
	P(jack)  =   4  52 P(8)  =   4  52 P(jack and 8)  =  P(jack)  ·  P(8)  =   4   ·   4  52 52  =    16   2704  =    1   169

Hence probability of choosing a jack and then an eight is ¹/69

TRY THIS………….

A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a king and then a three?

FUNCTIONS 1

If F(x) = 2x + 20; Find F^-1(x).

Solution

HINT: F^-1(x) means inverse.

PROCEDURE:

Make x the subject and then interchange x and y variables.

Let y=F(x)

So,  y= 2x + 20

y – 20 = 2x

y – 20  = 2x

   2          2

y – 20   = x

   2

x  =   y – 20     after rearranging

           2

 y^-1 = x – 20     after interchanging x and y variables.

             2

F^-1(x) = x – 20     after interchanging x and y variables.

                2

Hence, F^-1(x) = x – 20     

                             2

TRY THIS……………………………

If F(x) = 2x - 27; Find F^-1(x).

LOGARITHMS 1

Evaluate Log100,000 -  log 0.001 -  log 0.0001

Solution

= Log100,000  -  log 0.001 -  log 0.0001

= Log10⁵ -  log 10^-3 – log 10^-4

= 5Log10 - (-3 log 10) – (-4log 10)

= (5x1) - (-3x1) - (-8x1)

=5 - (-3) – (-4)

= 5 + 3 + 4

=12

  

Hence Log100,000 -  log 0.001-  log 0.0001 = 12

TRY THIS……………………… 

Evaluate Log100,000 -  log 0.000001 -  log 0.00000001

SUM A.P. 1

The first term of an AP is 41 and the last term is 107.If the arithmetic progression consists of 30 terms, calculate the sum of all the terms. 

Solution

A₁ = 41, n=30, A_n = 107, S₃₀ =?

S_n = n(A₁ + A_n)

       2

S₃₀ = 30(41 + 107)

         2

S₃₀ = 15 x 148

S₃₀ = 2220

Hence the sum of all 20 terms is 1480.

TRYTHIS………………….

The first term of an AP is 15 and the last term is 113. If the arithmetic progression consists of 30 terms, calculate the sum of all the terms. 

SETS 1

In a certain school, 160 students take either Chemistry or biology. 130 take chemistry and 100 take both subjects. Find those who take biology only.

Solution

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

Let Chemistry = n(C), Biology = n(B).

n(C)= 130 ,

n(B)= ?

n(CuB) = 160,

n(CnB)= 100

Solution

n(CuB) = n(C) + n(B) - n(CnB)

160  =  130 + n(B)  – 100

160  =  n(B) + 30

160 - 30  =  n(B)

n(B)= 130

physics only = n(B) – n(CnB)

                    = 130 – 100

                    = 30 

Hence those taking only biology are 30.    

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 only.

AREA 1

Area of rectangle is 48m². Find its perimeter if width is 2m. 

Solution

A = L x w

48 = L x 4

48 = 2L

2      2

24 = L

Now; P =2(L + w)

= 2(24 + 4)

= 2 x 28

= 56m

Hence perimeter is 56m

TRY THIS………..

Area of rectangle is 84m². Find its perimeter if width is 4m.  

PROBABILITY 2

	A school survey found that 4 out of 10 students like pizza. If three students are chosen at random with replacement, what is the probability that all three students like pizza? Solution THIS IS AN INDEPENDENT EVENT. THE 1ST CHOICE HAS NO ANY EFECT FOR THE 2ND OR 3RD.
	P(student 1 likes pizza)  =   4  10 P(student 2 likes pizza)  =   4  10 P(student 3 likes pizza)  =   4  10 P(student 1 and student 2 and student 3 like pizza)  =   4   ·   4   ·   4   =   8  10 10 10 125

Hence probability that all three students like pizza is ⁸/125 .

TRY THIS………..

A school survey found that 3 out of 5 students like coffee. If three students are chosen at random with replacement, what is the probability that all three students like coffee?