VARIATIONS F1

 

x is inversely proportional to y. x=22 while y=5. Find y when x is 10.

 

Solution

 

x ⍺ k

      y

 

x = k

      y

 

22 = k

        5

 

k = 22 x 5

 

k = 110

 

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

 

k = 110, y= ?, x = 10.

 

x = k

      y

 

10  = 110

          y

 

y x 10 = 110  x  y   (multiply by y on both sides)

               y

 

 

10y = 110

10      10

 

Hence y = 11.

 

 

TRY THIS…………….

 

 

x is inversely proportional to y. x=80 while y=4. Find y when x is 8. 

LOGARITHMS F8

 

Simplify Log₂256 - Log₃729

 

Solution

 

= Log₂256 - Log₃729

 

= Log₂2⁸ - Log₃3⁵    [Since 256=2⁸ and 729=3⁶].

 

= 8Log₂2 - 6Log₃3    [since Log_aa = 1]

 

= (8 x 1) - (6 x 1)

 

= 8 - 6

 

= 2

 

Hence Log₂256 -Log₃729 = 2

 

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

 

Simplify Log₂2048 - Log₅625 

LENGTH OF AN ARC F1

 

Find the length of an arc if the radius of the circle is 70cm.

 

Solution

 

L = ∏r

      180⁰

 

L = ∏ x  70

             180⁰

 

L = 7∏ cm

       18

 

Hence length of an arc is ^7∏/18 cm

 

TRY THIS…………………………..

 

 

Find the length of an arc if the radius of the circle is 30cm.

COMPOUND INTEREST F1

Mwakajonga invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 20,000 sh at the end of 3 years?

 

Solution

 

n=3, t=1, R=20%, A₃=8,500, P=?

A_n = P(1 + ^RT/100)ⁿ

A₃ = P(1 + ^(20x1)/100)³

A₃= P(1 + ²⁰/100)³

A₃ = P(¹⁰⁰/100 + ²⁰/100)³

A₃ = P(¹²⁰/100)³  But A₃=20,000.

20,000 = P(1.2)³  = P(1.2x1.2x1.2)  

 

20,000 = 1.728P

 

20000  =  1.728P

1.728       1.728

 

20000  =  P

1.728

 

P ≈ 11574/=

 

Hence at the start she invested Tsh 11574/=

 

TRY THIS………………..

 

 

Tunsubilege invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 16,000 sh at the end of 5 years?

EXPAND F1

 

Expand 11w(5w + 3 - 11w)

 

Solution

 

= 11w(5w + 3 - 11w)

 

= (11w x 5w) + (11w x 3) - (11w x 11w)

 

= 55w² + 33w - 121w²

 

= [55w² - 121w²] + 33w collecting like terms

 

= - 66w² + 33w

 

= 33w - 66w² answer

 

 

TRY THIS………..

  

Expand 6a(5a+ 12 - 9a)

LOGARITHMS F7

 If log_x 2401 – log₃729 = -2; find x

 

solution

 

log_x 2401 – log₃729 = -2

 

log_x 2401 – log₃3⁶ = -2

 

log_x 2401 – 6log₃3 = -2

 

log_x 2401 – 6 x 1 = -2

 

log_x 2401 – 6 = -2

 

log_x 2401 = -2 + 6

 

log_x 2401 = 4 

 

2401 = x⁴        2401=7x7x7x7=7⁴ by prime factors

 

7⁴ = x⁴      Powers cancel out

 

x = 7

 

Hence x = 7.

 

SETS F2

 If n(A)= 55 , n(B)= 90 and n(AuB)= 136, find n(AnB).

 

Solution

 

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

 

136 = 55 + 90 - n(AnB)

 

136 = 145 - n(AnB)

 

136 - 145 = - n(AnB) [after transferring 145 on the left hand side]

 

-9 = -n(AnB)

 

n(AnB) = 9 [after dividing by -1 both sides]

 

Hence n(AnB) = 9 answer

MATRICES F1

 

SETS F1

 

In a certain school, 160 students take either Chemistry or biology. 140 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)= 140 ,

n(B)= ?

n(CuB) = 160,

n(CnB)= 100

 

 

Solution

 

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

 

160  =  140 + n(B)  – 100

 

160  =  n(B) + 40

 

160 - 40  =  n(B)

 

n(B)= 120

 

physics only = n(B) – n(CnB)

 

                    = 140 – 100

 

                    = 40 

 

 

Hence those taking only biology are 40.    

 

TRY THIS………….

 

In a certain school, 196 students take either Chemistry or French. 150 take chemistry and 110 take both subjects. Find those who take French only.

FUNCTIONS F3

 If F(x) = 2x + 50; Find F^-1(200)

 

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 + 50

 

y – 50 = 2x

 

y – 50  = 2x    dividing by 2 on both sides

   2          2

 

y – 50   = x

   2

 

x  =   y – 50     after rearranging

           2

 

 y^-1 = x – 50     after interchanging x and y variables.

             2

 

F^-1(x) = x – 50     after interchanging x and y variables.

                2

 

Now we find F^-1(200)

 

F^-1(200) = 200– 50     

                     2

 

F^-1(200) = 150     

                   2

 

F^-1(200) = 75   

 

TRY THIS……………………………

 

If F(x) = 2x + 80; Find F^-1(20)