EXPONENTIALS 4

Simplify the following by writing in power form:

w⁹⁰

w¹⁰

Solution

= w^{90 - 10} 

=w⁸⁰

TRY THIS……………….

Simplify the following by writing in power form:

h⁵⁷

h²² 

ARITHMETIC PROGRESSION 4

The first term of an AP is 31 and the last term is 97.If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 31, n=20, A_n = 97

S_n = n(A₁ + A_n)

       2

S₂₀ = 20(31 + 97)

         2

S₂₀ = 10 x 128

S₂₀ = 1280

Hence the sum of all 20 terms is 1280.

TRYTHIS………………….

The first term of an AP is 11 and the last term is 33.If the arithmetic progression consists of 40 terms, calculate the sum of all the terms. 

MID-POINT 3

Find the midpoint of a line from (13, 12) to (5, 20)

Solution

x1=13, x2=5, y1=12,y2=20.

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

                         2           2

Mid point = (13 + 5,  12 + 20)

                         2          2

Mid point = (18, 32)

                     2     2

Hence midpoint = (9, 16)

TRY THIS……………………..

Find the midpoint of a line from (-4, 12) to (6, -2)

ARITHMETICS 1

Evaluate 696² – 686²

Solution

We apply difference of two squares: a² - b² = (a+b)(a-b)

696² – 686²  = (696 + 686)( 696 – 686)

                   = (1382)( 10)

                   = 13820   [only add a zero at 1382]

696² – 686² = 13820

TRY THIS…………….

Evaluate 808² – 708²

VECTORS 1

If u = 16i + 2j and v = 3i + 10j find 5u - 2v.

solution

= 5u - 3v

= 5(16i + 2j) - 2(3i + 10j)

= 80i + 10j - 6i - 20j

= (80i - 6i) + (10j - 20j)    [grouping like terms]

= 74i - 10j

Hence 5u - 3v = 74i - 10j .

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

If u = 8i - 3j and v = 5i + 10j;  find 5u + 3v.

LOGARITHMS 15

If log₉(2x + 19)=2; find x.

   Solution

Log₉(2x + 19)=2

(2x + 19)=9²           

                                            

2x + 19=81

2x =81 – 19

2x = 62

2x =   62   

2        2

x = 31

Hence x = 31

TRY THIS………………

If log₇(2t - 25)=2; find t 

STANDARD FORM - 1

Evaluate the following giving your answer in standard form.

882.7 x 10^-9

   5 x 10^-50

Solution

=  882.7 x 10^-9

       5 x 10^-50

=  882.7  x   10^-9

       5         10^-50

=  176.54 x 10^{-9 –(-50)}

=  176.54 x 10^{-9 + 50}    

=  176.54 x 10⁴¹   [then we change 176.54 into standard form as well]

= 1.7654 x 10² x 10⁴¹   [when we have exponents with same base, we add 

   the powers.]

 

= 1.7654 x 10⁴³

= 1.77 x 10⁴³ [correct to 2 d.p.]

Hence   782.7 x 10^-9   = 1.77 x 10⁴³

                5 x 10^-50

TRY THIS………………..

Evaluate the following giving your answer in standard form.

0.001768 x 10^-9

   4 x 10^-60

LOGARITHMS 14

Evaluate Log100 +  log 0.001 -  log 0.00000001

Solution

= Log100 +  log 0.001 -  log 0.00000001

= Log10² +  log 10^-3 – log 10^-8

= 2Log10 + (-3 log 10) – (-8log 10)

= (2x1) + (-3x1) - (-8x1)

=2 + (-3) – (-8)

= 2 – 3 + 8

=7

  

Hence Log100 +  log 0.001-  log 0.00000001 = 7

TRY THIS……………………… 

Evaluate Log1000 +  log 0.00001 -  log 0.00000001

SLOPE OR GRADIENT 2

Find the slope of a line which passes through (-1, -2) and (6,-9)

Solution

x₁ = -1,  y₁ =-2,  x₂ = 6,  y₂ = -9

m = y2 –y1

         x2 – x1

m =   -9 –(-2)

         6 –(-1)

m =   -9 + 2

         6 + 1

m =    - 7

           7

m =    - 1

Hence the slope is -1

TRY THIS……………………… 

Find the slope of a line which passes through (-8, -2) and (3,-12)  

BINARY OPERATIONS 2

If p*k = 4pk + p – 2k; find 7*2.

Solution

p*k = (4 x p x k) + p – (2xk)   [rewriting the given expression more clearly]

7*2 = (4 x 7 x 2) + 7 – (2x2)   [substituting 7 for p and 2 for k]

7*2 = 56 + 7 – 4

7*2 = 63 – 4   [we add first before subtracting]

7*2 = 59.

Hence   7*2 = 59.

  

 TRY THIS........

  

If p*k = 2p - k – 5pk; find 3*-3.

PERCENTAGE PROFIT 3

Dikshi got a profit of 700/= after selling an item for 14000/=. Find the percentage profit.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                         B.P

%’ge profit = 700   X  100   

                      14000

%’ge profit = 700   

                       140

%’ge profit = 5%   

                       

Hence Percentage Profit was 5%   

TRY THIS………………..

Janhwi got a profit of 3500/= after selling an item for 21,000/=. Find the percentage profit.

SETS 5

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)

-9 = -n(AnB)

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

Hence n(AnB) = 9 answer

TRY THIS………….

If n(A)= 89 , n(B)= 92 and n(AuB)= 137, find n(AnB).

INEQUALITIES 2

If 10x + 21 ≥ 7 + 3x ≥ 9x – 10; find x. 

Solution

10x + 42 ≥ 7 + 3x and  7 + 3x ≥ 9x - 29

10x – 3x ≤ 7 - 42 and  7 +29 ≤ 9x - 3x

7x ≤ -35 and  36 ≤ 6x 

7x ≤ -35  and  36 ≤ 6x

7        7            6      6

x ≤ -5 and  6 ≤ x

x ≤ -5 and  x  ≥ 6

TRY THIS…………….

If 10x - 63 ≥ 7 + 3x ≥ 9x – 65; find x

PERCENTAGE PROFIT 2

A man got a profit of 1800/= after selling an item. Find the buying price if the percentage profit was 10%.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                        B.P

10% = 1800   X  100   

            B.P.

10% = 180,000   

              B.P.

B.P. x 10% = 180,000   x B.P.

                       B.P.

B.P. x 10 = 180,000   

        

B.P. x 10¹ =   180,000    [dividing by 10 both sides]

  ₁10                  10

B.P. = 18,000

                      

Hence Buying Price was 18,000/=

TRY THIS………………

A man got a profit of 2000/= after selling an item. Find the buying price if the percentage profit was 30%.

VARIATIONS 3

x is inversely proportional to y. x=12 while y=5. Find y when x is 15.

Solution

x ⍺ k

      y

x = k

      y

12 = k

        5

k = 12 x 5

k = 60

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

k = 60, y= ?, x = 15.

x = k

      y

15 = 60

         y

y x 15 = 60  x  y   (multiply by y on both sides)

                y

15y = 60

15      15

Hence y = 4  .

TRY THIS…………….

x is inversely proportional to y. x=40 while y=4. Find y when x is 10.  

LOGARITHMS 13

If log 2= 0.3010; find the value of log 3,200,000 without using tables.

solution

log3,200,000=log(32 x 100,000)

=log32 + log100,000

=log2⁵ + log10⁵

=5log2 + 5log10

=5(0.3010) + (5 x 1)

=(1.5050) +  5

=6.5050

Hence log3,200,000=6.5050

TRY THIS……………

If log 2= 0.3010; find the value of log 160,000 without using tables. 

ARITHMETIC PROGRESSION 3

The 1st term of an A.P. is 83 and the common difference is 34. Find the 10th term.

Solution

A₁= 83, d = 34

A_n = A₁ + (n-1)d

A₁₀ = A₁ + (10-1)d

A₁₀ = A₁ + 9d

A₁₀ = 83 + (9x34)   [after substituting A₁= 83, d = 34 as given above]

∴ A₁₀ = 83 + 306

A₁₀ = 389

Hence the 10th term is 389.

TRY THIS……………

The 1st term of an A.P. is 18 and the common difference is 35. Find the 17th term.

LOGARITHMS 12

If Log_ax = 0.2, find Log_a(¹/x)

Solution

= Log_a(¹/x)

= Log_ax^-1

= -1 x Log_ax

= -1 x 0.2

= -0.2

TRY THIS………..

If Log_ab= 0.6, find Log_a(¹/b)

FUNCTIONS 7

Find the maximum value of the quadratic equation:5-2t-8t².

Solution

a=-8, b=-2, c=5

Maximum = 4ac-b²

                       4a

Maximum = (4 x -8 x 5) - (-2)²

                          4(-8)

Maximum = (-160)- 4

                         32

Maximum= -164

                      32

Maximum = -41

                      8

Hence maximum value is -41/8

TRY THIS………………………………..

NECTA  2003 QN. 10c

Find the maximum value of the quadratic equation:3+30t-5t².

UNITS OF DISTANCE 3

Convert 2335m into km.

Solution

1km  = 1000m

  ?     = 2335m

After cross-multiplying;

= 1 x 2335

      1000

=   2335

     1000

= 2.335 km

Hence 2335m = 2.335km

TRY THIS........

Convert 5041m into km.

FACTORS 2

Write 484 as a product of prime factors.

Solution

2	968
2	484
2	242
2	121
11	121
11	11
	1

 Then, 968 = 2 x 2 x 2 x 2 x 11 x 11

Hence   968 = 2⁴ x 11².

LOGARITHMS 11

If Log_ay = Log_a26 +  Log_a10 ; find y.

solution

Log_ay = Log_a26 +  Log_a10

Log_ay = Log_a(26x 10)         [since Log_am +  Log_an = Log_a(m x n)] 

Log_ay = Log_a260

Hence y = 260                 [since Log_a cancels out]

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

NECTA 1997 QN 13a

If Log_ax = Log_a5 +  Log_a3 ; find x.

SETS 4

If n(A)= 60 , n(AuB) = 140 and n(AnB)=40, find n(B)

Solution

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

  140     = 60 + n(B)  – 40

  140     = 60 -40 + n(B) 

  140     = 20 + n(B) 

  140 - 20        = n(B)

  120        = n(B)

Hence n(B) = 120 answer

TRY THIS…………………………….

If n(A)= 64 , n(AuB) = 136 and n(AnB) = 31, find n(B).