RADICALS B1

 

LOGARITHMS B5

 

Evaluate Log₂(1024 x 64 x 256).

 

Solution

 

= Log₂(1024 x 64 x 256)

 

= Log₂1024 +  Log₂64 +  Log₂256         (applying the product rule)

 

= Log₂2¹⁰+  Log₂2⁶    +  Log₂2⁸           (Since1024= 2¹⁰, 64=2⁶ ,32=2⁵)

 

= 10Log₂2 +  6Log₂2    +  8Log₂2         ( remember  Log_aa^c = cLog_aa )

 

= (10 x 1) +  (6 x 1) +  (8 x 1)                  ( remember  Log_aa = 1 )

 

= 10 + 6 + 8

 

= 24 answer

 

 

hence Log₂(1024 x 64 x 256) = 22

 

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

 

Evaluate Log₂(512 x 128 x 1024). 

ALGEBRA B4

Expand -10(y – 11)

 

Solution

 

= -10(y – 11)

 

= (-10 x y) – (-10 x 11)

 

= 10y – ^-110

 

= 10y + ⁺50

 

= 10y + 50 answer

 

TRY THIS………..

 

Expand ^-10(f– 7)

BODMAS B1

 

Evaluate 17 x 237 + 463 x 17.

 

Solution

 

= 17 x 237 + 463 x 17

 

= 17 x (237 + 463)

 

= 17 x 700

 

= 11900 answer

 

 

TRY THIS………….

 

 

Evaluate 125 x 516 + 484 x 125.

BANKING B1

Ann deposited the amount of 140,000/= in a bank which gives an interest rate of 5% for 5 years. Find the simple interest she got?

 

Solution

 

I = ?,  P = 140,000/=, T = 5 years, R = 5%

 

I  = PRT

      100

 

I  = 140000 x 5 x 5

            100

 

I  = 140000 x 5 x 5

            100

 

I  = 1400 x 5 x 5

          

I  = 350,000/=

 

Hence the simple interest was Tsh 350,000/=

 

TRY THIS………………..

 

Veda deposited the amount of 48000/= in a bank which gives an interest rate of 3% for 6years. Find the simple interest she got?

 

 

EXPONENTIALS B3

 

If (x⁵)(y²) = 7500; find x and y.

 

solution

 

(x⁵)(y²) = 7500  [7500=2x2x5x5x5x5x5 by prime factorization].

 

(x⁵)(y²) = (5⁵ )x (2²)     [ since 7500=5⁵ x 2²].

 

equating equal powers,

 

(x⁵) = (5⁵)     

 

x   =   5    [since equal powers cancel out]

 

also,

 

(y²) = (2²) [since equal powers cancel out]

 

 y    =   2

 

∴ x   =   5 and y    =   2

 

TRY THIS…………

 

If (x⁵)(y³) = 15,000; find x and y.

 

PERIMETERS B1

Find the perimeter of a regular polygon with 7 sides inscribed in a circle of radius 6cm.

Solution

P = 2nrSin(180⁰/n)

n= 7 sides and r=6cm

P = 2 x 7 x 6 x Sin(180⁰/7)

P = 84 x Sin25.71⁰

P = 84 x 0.4338

Hence P = 36.44cm

TRY THIS…………….. 

Find the perimeter of a regular polygon with 12 sides inscribed in a circle of radius 7cm.

MID POINT B1

Find the midpoint of a line from (13, 6) to (9, 10)

 

Solution

 

x1=13, x2=9, y1=6, y2=10.

 

Midpoint = (x1 + x2, y1 + y2)

                          2             2

 

Midpoint = (13 + 9, 6 + 10)

                          2          2

 

Midpoint = (22, 16)

                      2    2

 

Hence midpoint = (11, 6)

 

TRY THIS……………………..

 

 

Find the midpoint of a line from (22, -8) to (-10, 16) 

LOGARITHMS B4

 

If log 2= 0.3010; find the value of log 50,000 without using tables.

 

solution

 

=log50,000

 

=log(100,000÷ 2)

 

=log100,000 – log2

 

=log10⁵ – log2

 

=5log10 – log2

 

=(5x1) – (0.3010)

 

=5 – 0.3010)

  

=4.699

 

Hence log50,000=4.699

 

 

TRY THIS……………

 

If log 2= 0.3010; find the value of log 5,000 without using tables.

 

FUNCTIONS B3

 If F(x) = 5x + 10; Find F^-1(40).

 

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= 5x + 10

 

y – 10 = 5x

 

y – 10  = ¹5x

   5          ₁5

 

y – 10   = x

    5

 

x  =   y – 10     after rearranging

            5

 

F^-1(x) = x – 10     after interchanging x and y variables.

                5

Now we calculate F^-1(40) as hereunder;

 

F^-1(40) = 40 – 10     

                  5

F^-1(40) =  30      

                 5

F^-1(40) =  6      [after dividing 30 by 5]

               

Hence, F^-1(40) =  6     

                             

TRY THIS……………………………

 

If F(x) = 7x - 20; Find F^-1(6).

LOGARITHMS B4

 

If log₃(10x + 7)=3; find x.

 

Solution

 

log₃(10x + 7)=3

 

 (10x + 7)=3³         

                                            

 10x + 7=27

 

10x = 27 – 7

 

10x =  20

 

10x =  20²   

10      10

 

x = 2

 

Hence x = 2

 

TRY THIS…………….

 

If log₃(5x + 1)=4; find x

ANGLES B1

If x-15⁰ and 4x + 35⁰ are complementary angles, find the value of x.

 

solution

 

Complementary angles add up to 90⁰.

 

so,  x-15⁰ + 4x + 35⁰ = 90⁰.

so,  x + 4x + 35⁰ - 15⁰ = 90⁰.

so,  5x + 20⁰ = 90⁰.

so,  5x = 90⁰ -  20⁰

so,  5x = 70⁰

 

so,  5x = 70¹⁴

       5       5

 

Hence x = 14

 

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

 

 If 3x - 22⁰ and 2x + 87⁰ are complementary angles, find the value of x. 

GEOMETRY B2

 

An interior angle of a regular polygon is 62⁰ greater than an exterior angle. Find the interior angle.

 

Solution

 

Let i = interior angle, e = exterior angle.

 

Now i  + e=180⁰…………………(1)

 

But i = e+28⁰ …………………(2)

 

Substitute (2) in (1) above.

 

e+62⁰   + e=180⁰

 

e+ e+62⁰   =180⁰

 

2e+ 62⁰   =180⁰

 

2e=180⁰ - 62⁰   

 

2e=118⁰

 

2e=118⁰          dividing by 2 both sides.

2      2

 

e = 59⁰

 

But i  + e=180⁰…………………(1)

 

i  + 59⁰=180⁰.

 

i  =180⁰ - 59⁰

 

i = 121⁰

 

Hence i = 121⁰ 

  

TRY THIS………………………   

 

The interior angle of a regular polygon is 74⁰ greater than an exterior angle. Find the interior angle.

ALGEBRA B4

 

Factorize 225-9m²

 

Solution

 

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

 

225-9m² = 15² - 3²m²       

             = 15² - (3m)²     

             = (15 - 3m)(15 + 3m)

 

Hence 225 - 9m² = (15 - 3m)(15+ 3m)

 

TRY THIS…………….

 

Factorize 81 - 49c²

LOGARITHMS B3

 

If Log(30x-110/5+x) = 1. Find x.

Solution

Log(30x-110/5+x) = 1.

Log₁₀(30x-110/5+x) = 1.

 30x-110 = 10¹. After changing into exponential form

   5+x

 

30x-110 = 10    since [10¹=10]

   5+x

 

30x-110 = 10(5+x)    after cross multiply

  

30x-110 = 50+10x

 

30x-10x = 50+110 [collecting like terms]

 

20x = 160

 

x=8  [after dividing by 20 both sides]

Hence x=8.

TRY THIS……………

If Log(30x-50/7+x) = 1. Find x.

SETS B1

 

If n(A)= 83 , n(B)= 93 and n(AuB)= 130, find n(AnB).

Solution

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

130 = 83 + 93 - n(AnB)

130 = 176 - n(AnB)

130 - 176 = - n(AnB)

-46 = -n(AnB)

n(AnB) = 41       [after dividing by -1 both sides]

Hence n(AnB) = 41 answer

TRY THIS………….

If n(A)= 92 , n(B)= 98 and n(AuB)= 160, find n(AnB).

ALGEBRA B3

 

Simplify 21x-7(x-y)+5

 

Solution

 

=21x-7(x-y)+5

 

=21x-7x+7y+5        [since -7(x-y)= -7x+7y]

 

=14x+3y+5 answer

 

TRY THIS……..

 

Simplify 44x-9(x-y)+17

MAKING SUBJECT B1

 

QUADRATICS B2

 

What must be added to x²  + 8x to make the expression a perfect square?

 

Solution

 

a=1, b=8,c=?

 

We apply the formula: b² = 4ac

 

(8)² = 4 x 1 x c

 

64 = 4c

64 = 4c

4      4

 

16 = c

Hence number to be added is 16

 

TRY THIS……………………

What must be added to x²  + 12x to make the expression a perfect square?  

EXPONENTIALS B2

 

If 4¹⁰ = 32^a-2; find a

Solution

4¹⁰ = 32^a-2 

(2²)¹⁰ = (2⁵)^a-2 

2²⁰ = 2^5(a-2) 

2²⁰ = 2^5a-10

2²⁰ = 2^5a-10  [Same bases both sides cancels out]

20 = 5a – 10

20 + 10 = 5a

30 = 5a

30 = 5a

5      5

 

6 = a

Hence a=6

 

TRY THIS……………………….

 

If 4³ = 64^a-2; find a