Thursday 31 March 2016

APPROXIMATIONS-1


Estimate the value of 87 x 71.

solution

round off to tens;
87 90  [approximating to 90]
71 70  [approximating to 70]

 = 90 x 70
 = 6300

Hence 87 x 71 6300

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


Estimate the value of 61 x 69.[ans. 4200]

ARITHMETIC PROGRESSION-1


The 1st term of an A.P. is 78 and the common difference is 16. Find the 10th term.

Solution

A1= 78, d = 20

An = A1 + (n-1)d

A10 = A1 + (10-1)d

A10 = A1 + 9d

A10 = 78 + (9x16) [ after substituting A1= 78, d = 16 as given above]

A10 = 78 + 144

A10 = 222

Hence the 10th term is 222.



TRY THIS……………


The 1st term of an A.P. is 30 and the common difference is 43. Find the 9th term.

ALGEBRA-1


Divide 68ax + 22ay - 10az by 2a.

solution

= 68ax + 22ay - 10az
              2a

= 3468ax  +  1122ay  -  510az       [cancelling by 2a throughout]
     2a             2a           2a

= 34x + 11y - 5z

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


Divide 80am - 32an + 52ap by 4a.

FUNCTIONS-1


If F(x) = 2x + 6; Find F-1(x).

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 + 6

y – 6 = 2x

y – 6  = 2x
  2        2

y – 6   = x
   2

x  =   y – 6     after rearranging
            2

 y-1 = x – 6     after interchanging x and y variables.
            2

F-1(x) = x – 6     after interchanging x and y variables.
               2

Hence, F-1(x) = x – 6     
                            2


TRY THIS……………………………

If F(x) = 3x + 10; Find F-1(x).


LOGARITHMS-1


If log 2= 0.3010; find the value of log 50,000,000 without using tables.

solution

log 50,000,000 = log(100,000,000÷ 2)

=log100,000,000- log2

=log108 – log2

=8log10 –log2

=(8x1) – (0.3010)

=8– (0.3010)

= 7.699

Hence log5,000,000 = 7.699


TRY THIS……………


If log 2= 0.3010; find the value of log 500,000without using tables.

SETS - 1


If n(A)= 176 , n(B)= 164 and n(AnB)=150, find n(AuB).

Solution

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

             = 176 + 164 – 150

             = 340 – 150

             = 190

Hence n(AuB) = 190 answer



TRY THIS…………………………….




If n(A)= 116 , n(B)= 124 and n(AnB)=80, find n(AuB).


Wednesday 30 March 2016

WORD PROBLEMS-1


Pedro sells 1 litre of milk for sh700/=. How many litres of milk should Pedro sell to get 42,700/=?

Solution

1litre = 700/=
  ?      = 42,700/=

after cross multiplication;

= 42,700x  1
         700

= 42700x  1
         700

= 427
     7

= 61

Hence Pedro should sell 61 litres.

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


Iniesta sells 1 litre of milk for sh200. How many litres of milk should Iniesta sell to get 72,600/= ?



BINARY OPERATIONS-1


If p*k = pk + p – 2k; find 4*2.

Solution

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

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

4*2 = 8 + 4 – 4

4*2 = 12 – 4   [we add first before subtracting]

4*2 = 8.

Hence   4*2 = 8.

 TRY THIS........


If p*k = p + k – pk; find 6*2.


PRIME FACTORS-1


Write 440 as a product of prime factors.

Solution

2
440
2
220
2
110
5
55
11
11

1

 Then, 440 = 2 x 2 x 2 x 5 x 11

Hence   440 = 23 x 5 x 11.

TRY THIS........


Write 560 as a product of prime factors.

RATIOS-1


Divide 300 in the ratio 7:3.

solution

1st add the ratios. 7 +3 = 10

then;   7  x 300 = 210
           10

also;   3  x 300 = 90
          10

Hence the ratio parts are 210 and 90.

TRY THIS........


Divide 3200 in the ratio 5:3.

EXPONENTIALS - 1


If 2(6m-7) x 3(n+9) = (344) x (293); Find m and n.

solution

equating equal bases;

2(6m-7) =  293

2(6m-7)293

6m - 7 = 93
6m  = 93 + 7
6m  = 100
16m  = 10050
16           63

m = 50/3

again for n we equate equal bases.

3(n+9) = 344

3(n+9) = 344

n + 9 = 44

n = 44 - 9

n = 35

Hence m = 20 and n = 35

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


If 8(2m-9) x 11(n+40) = (1160) x (823); Find m and n.