PROBABILITY K2

 

If a number is chosen at random from the integers 1, 2, 3, 4, …………….., 16. What will be the probability that it is a multiple of three?

 

Solution

 

n(s)={1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}

n(s)=16

 

n(E) = {3,6,9,12,15}

n(E) =5

p(E) = n(E)

          n(s)

 

p(E) = 5

        16

 

probability that it is a multiple of three is ⁵/16.

 

TRY THIS………

 

If a number is chosen at random from the integers, 3, 4, …………….., 22. What will be the probability that it is a multiple of four?

PERCENTAGES K1

 Kitakuli earns 320,000/= per month. Her boss has promised to offer her a salary increase of 13% per month. Calculate the amount to be added to her in a year.

 

Solution

 

Increase per one month

= 13% x 320,000/=

= ¹³/₁₀₀ x 320,000/=

= 13 x 3200

= 41,600/=

 

Increase for the whole year

        =increase per one month x 12

        =41,600 x 12

        =499200

Hence for the whole year she will get 499,200/=

    

TRY THIS………

 

Kahotipoti earns 270,000/= per month. Her boss has promised to offer her a salary increase of 11% per month. Calculate the amount to be added to her in a year.

LOGARITHMS K5

 

If nLog₅125= Log₂1024 ; Find n

 

Solution

 

nLog₅125= Log₂1024

 

nLog₅5³= Log₂2¹⁰  [Since 125=5³  and 1024=2¹⁰]

 

3nLog₅5= 10Log₂2

 

3n x 1= 10 x 1   [since Log_aa=1]

 

3n= 10

 

n = 10/3 answer

 

TRY THIS………

 

If uLog₅125= Log₂256⁷ ; Find u

ALGEBRA K7

 Factorize the expression  Cos⁴m – Sin⁴m

 

Solution

 

= Cos⁴m – Sin⁴m

 

= (Cos²m)² – (Sin²m)².

 

= (Cos²m – Sin²m) (Cos²m – Sin²m).[Since a² – b² =(a-b)(a+b)]

 

= (Cos²m – Sin²m) (Cos²m – Sin²m) answer

 

TRY THIS………….

 

Factorize the expression  Cos⁸c – Sin⁸c

EXPONENTIALS K8

 

Evaluate the following giving your answer in standard form.

1982.7 x 10^-9

   5 x 10^-30

 

Solution

 

=  1982.7 x 10^-9

       5 x 10^-30

 

=  1982.7  x   10^-9

       5         10^-30

 

=  396.54 x 10^{-9 –(-30)}

 

=  396.54x 10^{-9 + 30}    

 

=  396.54x 10²¹      [then we change 396.54 into standard form as well]

 

= 3.9654x 10² x 10²¹   [when we have exponents with same base, we add  the powers.]

 

= 3.9654 x 10²³ [Since 10² x 10²¹

 

= 3.97 x 10²³ [correct to 2 d.p.]

 

Hence   1982.7 x 10^-9   = 3.97 x 10²³

                5 x 10^-20

 

 

TRY THIS………………..

 

Evaluate the following giving your answer in standard form.

21765 x 10^-9

   2 x 10^-30

ALGEBRA K6

Expand (y – 4)(y – 10)

 

Solution

 

= (y – 4) (y – 10)

 

= y (y – 10) – 4 (y – 10)

 

= y² – 10y – 4y + 40    [Since -4 x -40 = +40]

 

= y² – 14y + 44 answer

 

TRY THIS………..

 

Expand (w – 9) (w – 10)

EXPONENTIALS K7

 

EXPONENTIALS K6

 

Simplify 500

                64^-2/3

 

Solution

 

= 500

    64^-2/3

 

=  500 x 1

             64^-2/3

 

= 500 x 64^2/3        [Since 1/a^-n= aⁿ]

 

= 500 x (64^1/3) ²        [Since 64^1/3= cube root of 64 = 4]

 

= 500 x (4)²                

 

= 500 x 16

 

= 8000        

         

Hence    500       = 8000

              64^-2/3        

 

 

TRY THIS…………………………

 

 

Simplify 200           
             27^-2/3        

 

ALGEBRA K5

 The sum of four consecutive numbers which are multiples of 6 is 204. Find the 4th number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+6	n+12	n+18	204

 

Then, n + (n+6) + (n+12) + (n+18) = 204

 

4n + 6+12+18= 204

 

4n + 36 = 204

 

4n = 204 – 36

 

4n = 168 

 

4n = 168

4      4

 

n = 42

 

4th number = n + 18

 

                   = 42 + 18

 

                   = 60

 

Hence the 4th number is 60

 

TRY THIS………………….. 

 

The sum of four consecutive numbers which are multiples of 7 is 406. Find the 4th number

SUM OF AP K1

The first term of an AP is 13 and the last term is 57.If the arithmetic progression consists of 40 terms, calculate the sum of all the terms. 

 

Solution

 

A₁ = 13, n=40, A_n= 57

 

Sn = n(A₁ + A_n)

      2

 

S₂₀ = 40(13 + 57)

       2

 

S₂₀ = 20 x 70

 

S₂₀ = 1400

 

 Hence the sum of all 20 terms is 1400.

 

TRY THIS………….

 

The first term of an AP is 11 and the last term is 163.If the arithmetic progression consists of 60 terms, calculate the sum of all the terms. 

ALGEBRA K4

  

If 48 + 24 =56; Find a.

            a

 

Solution

 

48 + 88 =56

        a

 

   88 = 56 - 48

    a

 

   88 =  8

    a

 

 

88 = 8a   after multiplying by a both sides.

 

88  = 8a

8       8

 

11 = a

 

Hence a=11.

 

 

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

 

 

22 + 72 = 30    ; find c.

         c

 

 

 

ALGEBRA K3

 

The sum of four consecutive numbers is 606. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+1	n+2	n+3	1606

 

Then, n + (n+1) + (n+2) + (n+3) = 1606

 

4n + 1+2+3= 1606

 

4n + 6 = 1606

 

4n = 1606 – 6

 

4n = 1600 

 

4n = 1600

 4        4

 

n = 400

 

3rd number = n +2

 

                   = 400 + 2

 

                   = 402

 

Hence the 3rd number is 402

 

TRY THIS………………….. 

 

The sum of five consecutive numbers is 610. Find the largest number.

 

EXPONENTIALS K5

 Simplify (7a)² + 31a². 

  

Solution

 

= (7a)² + 31a²    

 

= 7².a² + 31a²              Since (mn)² =m² . n²

 

= 49a² + 31a²

 

= 80a²

 

 

TRY THIS………………

 

  

Simplify (6a)² + 50a².   

EXPONENTIALS K4

 

Find x if 128^x = 0.125; find x.

 

Solution

 

128^x = 0.125

 

(2⁷)^x = 0.125          [since 128=2⁷ ]

 

2^7x = 0.125      

    

2^7x = ¹²⁵/₁₀₀₀       [converting 0.125 into fraction]  

     

2^7x = ¹/₈             [after simplifying]

      

2^7x = 8^-1             [since ¹/_a=a^-1]

 

2^7x = (2³)^-1

 

2^7x = 2^{3 x-1}   

        

2^7x = 2^-3            [since (a^c)^d =a^cd]

 

2^7x = 2^-3            [same bases cancel out]

 

7x = -3         

 

7x = -3

7       7

 

x= ^-3/7

 

Hence x= ^-3/7  .

 

TRY THIS…………………….

 

 

Find x if 64^4m = 0. 5; find m.

SETS K2

 

In a certain school, 140 students take either Chemistry or physics. 160 take chemistry and 100 take both subjects. Find those who take physics.

 

Solution

 

In most cases, OR stands for union whereas AND/BOTH, stands for intersection.

 

Let Chemistry = n(C),

      physics = n(P).

 

n(C)=160 ,

n(P)=?

 

n(CuP)= 140,

n(CnP)=100

 

Now,the formula

 

n(CuP)= n(C) + n(P) – n(CnP)

 

140 =  160 + n(P)  – 100

 

140 =  160 – 100 + n(P) 

 

140 =  60+ n(P) 

 

140– 60  =  n(P)

 

n(P)=80

 

Hence those taking  physics are 80.  

 

TRY THIS………….

 

In a certain school, 185 students take either Chemistry or French. 160 take chemistry and 130 take both subjects. Find those who take French.

VARIATION K3

 x is directly proportional to y. x=32 while y=4. Find y when x is 152.

 

Solution

 

x ⍺ y

x = ky

 

32 = k x 4

 

32 = 4k    [dividing by 4 both sides]

4      4

 

k = 8

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

 

k = 8, y= ? x = 152.

 

x = ky

 

152 = 8 x y

 

152 = 8y

8       8

 

y = 19  .

 

TRY THIS…………….

 

x is directly proportional to y. x=40 while y=4. Find y when x is 180.