ALGEBRA 1

Calculate the value of 4x + k + 80 + y , when x = 8, k = 10 and y =− 33 .

Solution

=4x + k + 80 + y

=4(8) + 10 + 80 + (-33)

=4(8) + 10 + 80 + (-33)

=32 + 10 + 80 – 33

= 58 – 33

= 25 answer

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

Calculate the value of 8x + k +- 30 + y, when x = 11, k = 17 and y =− 10 .

A. P. 2

The 1st term of an A.P. is 72 and the common difference is 80. Find the 10th term.

Solution

A₁= 72, d = 80

A_n = A₁ + (n-1)d

A₁₀ = A₁ + (10-1)d

A₁₀ = A₁ + 9d

A₁₀ = 72 + (9x80) [ after substituting A₁= 72, d = 80 as given above]

∴ A₁₀ = 72 + 720

A₁₀ = 792

Hence the 10th term is 792.

  

TRY THIS……………

The 1st term of an A.P. is 160 and the common difference is 219. Find the 5th term.

EXPONENTIALS 3

Simplify 9a⁶b⁹ × 6a⁵b¹⁶ .

Solution

= 9a⁶b⁹ × 6a⁵b¹⁶ .

= (9 x 6)( a⁶x a⁵)( b⁹ × b¹⁶)

= 54a⁶⁺⁵b⁹⁺¹⁶.

= 54a¹¹b²⁵.

TRY THIS……………………… 

 Simplify 11c⁹d¹⁷ × 5c⁷d¹³ .

LOG 1

If Log_ax = 4, find Log_a(¹/x¹⁵)

Solution

= Log_a(¹/x¹⁵)

= Log_ax^-15          [since 1/aⁿ = a^-n]

= -15 x Log_ax    [since Log_axⁿ   = nLog_ax ]

= -15 x 4

= -60 answer

TRY THIS………..

If Log_ab= 6, find Log_a(¹/b⁴⁰)

SIMULTANEOUS 1

Solve the following equations by elimination method.

4x - 2y = 6

3x - y = 7

Solution

4x - 2y = 6 ………..….(i)

3x - y = 7……………….(ii)

By using elimination method,

1(4x - 2y = 6

2(3x - y = 7

Then we add to eliminate y first by subtraction

   4x  - 2y = 6

   6x - 2y = 14

   -2x  + 0  = -8

-2x = -8                     [Dividing by 11 both sides] 

-2      -2

x = 4

From equation (1),

4x - 2y = 6

4(4) - 2y = 6

16 - 2y = 6

16-6 = 2y

10 =2y

10 =2y

2     2

y = 5 [after dividing by 2 both sides]

Hence x=4 and y=5.

TRY THIS........

Solve the following equations by elimination method.

5x + y = 20

2x + 3y = 21

SIMPLIFY 1

Simplify the expression r² + t²/ (r + t)

Solution

From difference of two squares p² - q² = (p - q)(p + q)

= r² + t²/ (r + t)

= (r + t)(r - t)

      r + t

= (r + t)(r - t)

      r + t

=  r - t    answer

      

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

Simplify the expression 

u² + y²/ (u + y)

EXPONENTIALS 2

If y⁵ = 11, find y¹⁰.

Solution

= y¹⁰

= (y⁵)²

= (11)²      remember y² = 11

= 11 x 11

= 121

Hence y¹⁰ = 121.

TRY THIS……………….

If  y⁹ = 6, find y²⁷.

GEOMETRY 1

If A and B are complementary angles such that A = 29° and B = x + 25° , find the value of x .

Solution

A + B = 90⁰. [Since complementary angles add up to 90⁰]

29⁰ + x + 25⁰ = 90⁰.

x + 54⁰ = 90⁰.

x = 90⁰ - 54⁰

x = 36⁰  answer.

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

If A and B are complementary angles such that A = 39° and B = x + 37° , find the value of x .

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?