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).