Wednesday 28 December 2016

EXPONENTIALS 4


Simplify the following by writing in power form:
w90
w10

Solution

= w90 - 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, An = 97

Sn = n(A1 + An)
       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 + 512 + 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.


Tuesday 27 December 2016

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

An = 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 Logax = 0.2, find Loga(1/x)

Solution

= Loga(1/x)

= Logax-1

= -1 x Logax

= -1 x 0.2

= -0.2


TRY THIS………..


If Logab= 0.6, find Loga(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.

Saturday 24 December 2016

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 Logay = Loga26 +  Loga10 ; find y.

solution

Logay = Loga26 +  Loga10

Logay = Loga(26x 10)         [since Logam +  Logan = Loga(m x n)] 

Logay = Loga260

Hence y = 260                 [since Loga cancels out]

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

NECTA 1997 QN 13a



If Logax = Loga5 +  Loga3 ; 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).