SETS - 2

If n(A)= 60 , n(AuB) = 160 and n(AnB)=10, find n(B)

Solution

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

  160     = 60 + n(B)  – 10

  160     = 60 -10 + n(B) 

  160     = 50 + n(B) 

  160 - 50        = n(B)

  110        = n(B)

Hence n(B) = 110 answer

TRY THIS…………………………….

If n(A)= 63 , n(AuB) = 125 and n(AnB) = 30, find n(B).

SIMPLE INTEREST-1

Priya deposited the amount of 90000/= in a bank which gives an interest rate of 3% for 4 years. Find the simple interest she got.

Solution

I = ?,  P = 90,000/=, T = 4 years, R = 3%

I  = PRT

      100

I  = 90,000 x 3 x 4

            100

I  = 90,000 x 3 x 4

            100

I  = 900 x 3 x 4

          

I  = 10,800/=

Hence the simple interest was Tsh 10,800/=

TRY THIS………………….

Edmund deposited the amount of 70,000/= in a bank which gives an interest rate of 5% for 4 years. Find the simple interest he got?

DIFFERENCE OF 2 SQUARES 3

Evaluate 2794² – 2784²

Solution

2794² – 2784² = (2794 + 2784)( 2794 – 2784)

                     = (5578)( 10)

                     = 55780

Hence 2794² – 2784² = 55780

TRY THIS…………….

Evaluate 931² – 831²

INEQUALITIES - 2

If 3x – 17 ≤ x + 5 ≤ 21x; find x

solution

3x – 17 ≤ x + 5 and  x + 5 ≤ 21x

3x – x ≤ 17+ 5 and  5 ≤ 21x - x

2x ≤ 22(divide by 2 both sides) and  5 ≤ 20x (divide by 20 both sides)

x ≤ 11 and  ¹/4 ≤ x

x ≤ 11 and  x ≥ ¹/4 

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

If 13x – 11 ≤ x + 25 ≤ 6x; find x

MID POINT 1

Find the midpoint of a line from (11, 4) to (7, 10)

Solution

x1=11, x2=7, y1=4, y2=10.

Mid point = (x1 + x2, y1 + y2)

                         2             2

Mid point = (11 + 7,  4 + 10)

                        2          2

Mid point = (18, 14)

                     2    2

Hence midpoint = (9, 7)

TRY THIS……………………..

Find the midpoint of a line from (14, 8) to (2, 16)

FACTORIZE-4

Factorize 2x² + 6x - 8

Solution

= 2x² + 6x - 8.  We split the middle term (+6x) to be (-2x + 8x).

= 2x² - 2x + 8x – 8                (-3x) = -14x + 5x.

= (2x² -2x) + (8x - 8)

= 2x(x - 1) + 8(x - 1)

= (2x + 8) (x -1)

Hence 2x² + 6x – 8 = (2x + 8) (x - 1) answer.

TRY THIS…..

Factorize 6x² + x – 2.

ARITHMETIC PROGRESSION - 4

The 1st term of an A.P. is 70 and the common difference is 34. Find the 10th term.

Solution

A₁= 70, d = 34

A_n = A₁ + (n-1)d

A₁₀ = A₁ + (10-1)d

A₁₀ = A₁ + 9d

A₁₀ = 70 + (9x34) [ after substituting A₁= 70, d = 34 as given above]

∴ A₁₀ = 70 + 306

A₁₀ = 376

Hence the 10th term is 376.

TRY THIS……………

The 1st term of an A.P. is 44 and the common difference is 33. Find the 7th term.

LOGARITHMIC EQUATIONS - 6

Find x if Log_x625 = 4

solution

Log_x625 = 4

625 = x⁴ .

5⁴ = x⁴ . [since 625 = 5⁴ by prime factorization]

5⁴ = x⁴ . [ powers cancel out]

x = 5

Hence x = 5

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

Find y if Log_y216 = 3

FUNCTIONS 7

If F(x) = 4x² + 12x - 420; find F(6) - F(2).

solution

PROCEDURE:

i) Find F(6)

ii) Find F(2)

iii) Do F(6) - F(2).

i) Find F(6)

F(x) = 4x² + 12x - 420

F(6) = 4(6)² + 12(6) - 420

F(6) = 4(6x 6) + (12 x 6) - 420

F(6) = 144 + 72 - 420

F(6) = 216 - 420

F(6) = -204.

ii) Find F(2)

F(x) = 4x² + 12x - 420

F(2) = 4(2)² + 12(2) - 420

F(2) = 4(2 x 2) + (12 x 2) - 420

F(2) = 32 + 24 - 420

F(2) = 56 - 420

F(2) = -364

iii) Do F(6) - F(2).

      F(6) - F(2) = -204 - (-364) = -204 + 364 = 160.

Hence F(6) - F(2) = 160.

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

If F(x) = 11x² + 2x - 10; find F(7) - F(4).

LOGARITHMIC EQUATIONS-5

If log₆(3x + 6)=2; find x

   Solution

log₆(3x + 6)=2

 (3x + 6)=6²   

                      

 3x + 6=36

3x = 36 – 6

3x =  30

3x =  30   

3        3

X = 10

Hence x = 10

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

If log₂(2x - 20)=4; find x