PERFECT SQUARES 1

If 16x² + 40x + m is a perfect square, find m.

Solution

a = 16, b = 40, c = m.

b² = 4ac

(40)² = 4 x 16 x m

1600 = 64m

1600  =    64m

64          64

           

m = 25

Hence m = 25

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

If 16x² + 24x + Q is a perfect square, find Q.

SUM OF AN AP 1

The first term of an AP is 8 and the last term is 57.If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 8, n=20, A_n = 57

S_n = n(A₁ + A_n)

     2

S₂₀ = 20(8 + 57)

      2

S₂₀ = 10 x 65

S₂₀ = 650

Hence the sum of all 20 terms is 650.

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

The first term of an AP is 17 and the last term is 65. If the arithmetic progression consists of 40 terms, calculate the sum of all the terms. 

LOGARITHM 5

If Log_ax = ¹/₂  Log_a144+ ¹/₃ Log_a27 find x.

Solution

Log_ax = ¹/₂  Log_a144+ ¹/₃ Log_a27

Log_ax = Log_a144^1/2  +  Log_a27^1/3   [Square root of 144 is 12, Cube root of 125 is 5]

Log_ax = Log_a 12 +  Log_a3

Log_ax = Log_a (12 x 3)               [Since Log_a m +  Log_an = Log_a mn]

Log_ax = Log_a 36

x = 36

TRY THIS………………

If Log_ax = ¹/₃  Log_a216+ ¹/₄ Log_a16; find x.

LENGTH OF AN ARC 1

Find the length of an arc if the radius of the circle is 9cm.

Solution

L = ∏r

      180⁰

L = ∏ x  9

            180⁰

L = ∏ cm

     20

Hence length of an arc is ∏/20 cm.

TRY THIS……………

Find the length of an arc if the radius of the circle is 15cm.

FACTORIZATION 1

Factorize 1-r²

Solution

We use difference of two squares a² – b² = (a - b)(a + b)

1-r² = 1²-r²      

        = (1 - r)(1 + r)

Hence 1-r² = (1 - r)(1 + r)

TRY THIS…………….

Factorize 36 - c²

MID POINT 1

Find the midpoint of a line from (9, 4) to (5, 10)

Solution

x1=13, x2=5, y1=4, y2=10.

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

                        2             2

Mid point = (13 + 5,  4 + 10)

                        2          2

Mid point = (18, 14)

                     2    2

Hence midpoint = (9, 7)

TRY THIS……………………..

Find the midpoint of a line from (18, 8) to (2, 16)

FINDING MEDIAN 1

The masses of 11 cabbages in kilograms were recorded as follows:

12,8, 4, 3, 6,11, 7, 2, 5, 1 and 9. Calculate the median of these data.

Solution

Arrange the data in order of magnitude starting with the smallest.

1, 2, 3, 4,  5,  6, 7, 8, 9,11,12

There are eight data so far. The middle number is 6.

Hence median is 6.

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

The masses of 13 pineapples in kilograms were recorded as follows:

12, 8, 4, 3, 6,11, 7, 2,13,15, 5, 1 and 9. Calculate the median of these data.

ARITHMETIC PROGRESSION 1

The first term of an AP is 13 and the last term is 57.If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 5, n=20, A_n = 57

S_n = n(A₁ + A_n)

    2

S₂₀ = 20(13 + 57)

      2

S₂₀ = 10 x 70

S₂₀ = 700

Hence the sum of all 20 terms is 700.

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

The first term of an AP is 17 and the last term is 83.If the arithmetic progression consists of 40 terms, calculate the sum of all the terms. 

LOGARITHMS 4

If Log_ax = ¹/₂  Log_a144+ ¹/₃ Log_a125 find x.

Solution

Log_ax = ¹/₂  Log_a144+ ¹/₃ Log_a125

Log_ax = Log_a144^1/2  +  Log_a125^1/3   [Square root of 144 is 12, Cube root of 125 is 5]

Log_ax = Log_a 12 +  Log_a5

Log_ax = Log_a (12 x 5)     [Since Log_a m +  Log_an = Log_a mn]

Log_ax = Log_a 60

x = 60

TRY THIS………………

If Log_ax = ¹/₃  Log_a216+ ¹/₄ Log_a81; find x.

LOGARITHMS 3

Evaluate log₂1024⁸

Solution

= log₂1024⁸

= log₂(2¹⁰)⁸

= log₂2⁸⁰

= 80 x log₂2    [since log_aa = 1]

= 80 x 1

= 80

Hence log₂1024⁸ = 80

TRY THIS…………………..

Evaluate log₂2048¹⁰