ALGEBRA EXPRESSIONS C2a

 

BANKING 1

 

Rayan deposited the amount of 20,000/= in a bank for 5 years and got a profit of 1500/=. Find the interest rate?

 

Solution

 

I = 1500/=, P = 20,000/=, T = 5 years, R = ?

 

I  = PRT

      100

 

1500  = 20000 x R x 5

                   100

 

1500  = 20000 x R x 5

                  100

 

1500  = 200 x R x 5

                      

1500  = 1000R

           

 ^1.51500  =   1000R

   1000       1000

 

 

Hence the interest rate was  1.5%

 

TRY THIS………….

 

Rayan deposited the amount of 6,000/= in a bank for 5 years and got a profit of 4200/=. Find the interest rate?

PROFIT AND LOSS K5

 

A man got a profit of 7400/= 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% = 7400   X  100   

            B.P.

 

B.P. x 10% = 740,000   x B.P.  [multiplying by B.P. both sides]

                       B.P.

 

 

B.P. x 10 = 740,000   

 

        

B.P. x 10¹ =   740,000   

  ₁10                   10

 

B.P. =74,000

                      

Hence Buying Price was 74,000/=

 

TRY THIS………………

 

A man got a profit of 90,500/= after selling an item. Find the buying price if the percentage profit was 20%.

 

EXPONENTIALS C2a

 

9^x =3^x +72; find x

 

Solution

 

9^x =3^x +72

9^x - 3^x = 72    taking 3x on the left side

9^x - 3^x – 72 = 0.  Taking 72 on the left side as well

3^2x - 3^x – 72 = 0.   Rewriting 9^x as 3^2x.

(3^x)² – (3^x) – 72 = 0.   Factoring 3^x out.

      Let u = 3^x  ……… (i)

(u)² – (u) – 72 = 0.  

u² – 9u + 8u – 72 = 0   [Factorizing by splitting the middle term].

 

(u² – 9u) + (8u – 72) = 0  

u(u – 9) + 8(u – 9) =0

(u – 9)(u + 8) =0

 u – 9 = 0 or  u + 8  =0

u = 9 or  u =-8

 

We neglect the –ve answer and we stick with the positive one.

 

 Now from (i) u = 3^x 

9 = 3^x

3² = 3^x   equal bases they cancel out.

 

x = 2

 

TRY THIS………….

 

9^x =3^x + 11¹/₄; find x

4           

 

LOGARITHMS K8

 Log₃27 + Logx =8. Find x

 

Solution

 

Log₃27 + Logx =8

 

Log₃3³ + Logx =8

 

3Log₃3 + Logx =8

 

(3 x 1) + Logx =8

 

3 + Logx =8

 

Logx =8 - 3

 

Logx =5

 

Log₁₀x =5 [Since any log without a base it is a base 10 logarithm]

 

x = 10⁵  [after changing logarithmic form into exponential form]

 

x = 100,000 answer. 

TRY THIS………….

 

Log₅625 + Logu =9. Find u

SETS K3

 There are 37 villagers at the meeting. 12 are farmers and 18 are workers and 8 are both farmers and workers. How many villagers are

i)                 Either farmers or workers.

ii)             Neither farmers nor workers

 

Solution

 

n(U)=37

n(F)=12

n(W)=18

n(FnW)=8

n(FuW)=?

n(FuW)’=?

 

i)  n(FuW) = n(F) + n(W) - n(FnW)

n(FuW) = 12 + 18 - 8

n(FuW) = 30 – 8

n(FuW) = 22

 

---------------------------------------- 

n(U) = n(FuW) + n(FuW)’

37 = 22+ n(FuW)’

37 - 22=n(FuW)’

15=n(FuW)’

 

Hence those who are neither farmers nor workers are 15

 TRY THIS.....................

 There are 40 villagers at the meeting. 12 are farmers and 18 are workers and 6 are both farmers and workers. How many villagers are

i)                 either farmers or workers.

ii)             neither farmers nor workers

========================= 

ALGEBRA K9

 

Expand y(y – 4)(y – 10)

 

Solution

 

= y(y – 4) (y – 10)

 

= y[y (y – 10) – 4 (y – 10)]

 

= y[y² – 10y – 4y + 40] 

 

= y³ – 14y² + 44y

 

= y³ – 14y² + 44y    answer

 

TRY THIS………..

 

Expand w(w – 9) (w – 10)

PROFIT AND LOSS K4

 

Jemson bought a radio at 50,000/= and after one year he sold it at a loss of 14%. Find the price of a radio after a year.

 

Solution

 

= 50,000 - (14% of 50,000)

= 50,000 - (0.14 of 50,000)  after dividing 14% by 100

= 50,000 - 7,000

= 43000/= 

 

Hence the new price is 43000/=    

 

TRY THIS………….

 

Patson bought a radio at 70,000/= and after one year he sold it at a loss of 24%. Find the price of a radio after a year.

LOGARITHMS K6

 

If Log_ax = 0.82, find Log_a(¹/x⁷)

 

Solution

 

= Log_a(¹/x⁷)

 

= Log_ax^-7

 

= -7 x Log_ax

 

= -7 x 0.82

 

= -5.74

 

TRY THIS………..

 

If Log_ac= 60, find Log_a(¹/c⁵)

ALGEBRA K8

 

Expand (y – 4)(y – 10)

 

Solution

 

= (y – 4) (y – 10)

 

= y (y – 10) – 4 (y – 10)

 

= y² – 10y – 4y + 40     [Since -4 x -40 = +40]

 

= y² – 14y + 44 answer

 

TRY THIS………..

 

Expand (w – 9) (w – 10)