APPROXIMATIONS 4

Estimate the value of 1143  ÷ 52.

Solution

= 1123     

     52

= 1000      [ since 1143 ≈ 1000   and   52 ≈ 50 ]

     50

= 20    [After dividing 1000 by 20]

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

Estimate the value of 7312  ÷ 18

PERIMETER 2

Area of rectangle is 60m². Find its perimeter if width is 4m. 

Solution

A = L x w

60 = L x 4

60 = 4L

4      4

15 = L

Now; P =2(L + w)

= 2(15 + 4)

= 2 x 19

= 38m

Hence perimeter is 38m

TRY THIS………..

Area of rectangle is 85m². Find its perimeter if width is 5m. 

UNITS OF VOLUME 1

Change 4.5 liters of water into milliliters

Solution

1L  =  1000mL

4.5L = ?

___________________________  

 = 4.5 x  1000

         1

= 4.5 x 1000

4500mL.

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

Change 6.3 liters of milk into milliliters 

ALGEBRA 11

Simplify 6n + 18m + 2n – 5m

Solution

= 6n + 18m + 2n – 5m

= 6n + 2n + 18m– 5m         [Grouping like terms together]

= 8n + 13m answer

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

Simplify 6n + 18m + 2n – 5m

LOGARITHMS 18

Evaluate       Log0.0000001

                       

Solution

=    Log0.0000001

      

We write 0.0000001 as an exponent. So 0.0000001=10^-7

=    Log10^-7

      

=    Log₁₀10^-7

            

=    -7Log₁₀10       [Since Log_ccⁿ = nLog_cc]

      

=    -7 x 1             [Since Log_cc = 1]

Hence           Log0.0000001         =  -7

TRY THIS………..

Evaluate    Log0.0000000001

APPROXIMATIONS 3

Estimate the value of 6.3  x  0.064

solution

= 6.3  x  0.064

= 6.0 x 0.06      [ 6.3 ≈ 6.0 to ones and 0.064 ≈ 0.06 to hundredths]

= 0.36

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

Estimate the value of 4.2  x  0.078

PROBABILITY 13

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a spade and an ace?

Solution

(This is a problem with replacement)

n(E) = 13, n(S) = 52

P(E) = n(E)

            n(S)

1st pick(spade) = 13/52

2nd pick(an ace) = 4/52.

P(a spade and an ace) =   13      x    4

                                         52           52

P(a spade and an ace) =  ¹13      x    4 ¹

                                        ₄52           52 ₁₃

P(a spade and an ace) =  1      x    1

                                        4           13

P(a spade and an ace) =     1      

                                         52     

Therefore Probability of drawing a spade and an ace is 1/52 

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

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a number 2 and a joker?

VARIATION 5

x is inversely proportional to y. x=18 while y=5. Find y when x is 12.

Solution

x ⍺ k

      y

x = k

      y

18 = k

        5

k = 18 x 5

k = 90

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

k = 90, y= ?, x = 12.

x = k

      y

12 = 90

         y

y x 12 = 90  x  y   (multiply by y on both sides)

                y

12y = 90

12      12

y = 7.5  .

TRY THIS…………….

x is inversely proportional to y. x=24 while y=4. Find y when x is 8.

EXPONENTIALS 10

Simplify the following by writing in power form:

 4w⁷⁰

  w²⁰

Solution

= 4w^{70 - 20} 

= 4w⁵⁰

TRY THIS……………….

Simplify the following by writing in power form:

8 m⁵⁰

   m³⁰    

BODMAS 1

Evaluate 18 x (45 – 41) ÷ (101-99)

Solution

We apply BODMAS.

= 18 x (45 – 41) ÷ (101-99)

= 18 x 4 ÷ (101-99)     [After dealing with 1^st brackets]

= 18 x 4 ÷ 2     [After dealing with 2^nd brackets]

= 18 x  2     [After dividing]

= 36     [After multiplying]

Hence  18 x (45 – 41) ÷ (101-99) = 36

TRY THIS……………….

Evaluate 8 x (56 – 42) ÷ (88-86)