COORDINATES A1

 

Find n if a line which passes through (-11, -8) and (6,n) has a slope of -¹/17

 

Solution

 

x₁ = -11,  y₁ =-8,  x₂ = 6,  y₂ = n

 

m =  y2 –y1

          x2 – x1

 

 

-1 =  y2 –y1

 17     x2 – x1

 

-1  =   n –(-8)

17      6 –(-11)

 

-1 =   n + 8

 17    6 + 11

 

-1 =   n + 8          

 17      17

 

17(n+8) = -1 x 17

 

7n + 136 = -17

 

7n = -17 - 136

 

7n = -153

 

7n = -153

7         7

 

n = -9

 

 

TRY THIS……………………… 

 

 

Find t if a line which passes through (6, -2) and (-3,t) has a slope of  ¹/9

LOGARITHMS K12

 Evaluate Log60 – Log0.003 + Log500.

 

Solution

 

= Log60 – Log0.003 + Log500.

 

= Log(60 x 500)

              0.003

 

= Log(30000)

            0.003

 

= Log(30,000,000)

                 3

 

= Log 10,000,000

             

= Log₁₀10,000,000

 

= Log₁₀10⁷

            

= 7Log₁₀10

 

= 7 x 1

 

= 7

 

∴ Log600 – Log0.003 + Log50 = 7

 

TRY THIS……………………….

 

Evaluate Log60,000 – Log4.8 + Log800,0000

QUADRATICS A1

 What must be added to x²  + 14x to make the expression a perfect square?

 

Solution

 

a=1, b=14,c=?

 

b² = 4ac

 

(14)² = 4 x 1 x c

 

196 = 4c

 

196 = 4c      [dividing by 4 both sides]

4        4

 

49 = c

 

Hence number to be added is 49

 

TRY THIS……………………

 

What must be added to x²  + 20x to make the expression a perfect square?

  

AREAS A1

 The areas of the two circles are in the ratio of 49:4 . Calculate the radius of the smaller circle when the radius of the larger circle is 40 cm. 

 

Solution

 

let R = radius of larger circle

let r = radius of smaller circle

R= 40cm, r = ? Ratio = 49:4.

 

 

For larger circle

A1 = ∏R² …………(i)

 

For smaller circle

A2 = ∏r² …………(ii)

 

A1 = ∏R²

A2    ∏r²

 

A1 = R²

A2     r²

 

 

A1 = (R/r)² 

A2

 

√(^A1/A2) =  (R/r)

 

√(⁴⁹/4) =  (63/r)

 

7 = 63                 We cross multiply

2     r

 

(2 x 63) = (7 x r)   after cross multiply

 

126 = 7r

 

126 = 7r

7       7

 

r = 18

 

∴ radius of smaller circle is 18cm.

 

TRY THIS………..

 

NECTA 1997 QN 5(c)

The areas of two circles are in the ratio of 16:9 . Calculate the radius of the smaller circle when the radius of larger circle is 24 cm.

 

WORD PROBLEMS A2

 Karungi earns 350,000/= per month. Her boss has promised to offer her a salary increase of 17% per month. Calculate the amount to be added to her in 14 months.

 

Solution

 

Increase per one month

= 17% x 350,000/=

= ¹⁷/₁₀₀ x 350,000/=

= 17 x 3,500/=

= 59,500/=

 

Increase for the whole year

        =increase per one month x 12

        =59,500x 14

        =833,000

Hence for the whole year she will get 833,000/=

    

TRY THIS………

 

Mihayo earns 270,000/= per month. Her boss has promised to offer her a salary increase of 11% per month. Calculate the amount to be added to her in a year.

LOGARITHMS K11

 If Log_ax = 0.8, find Log_a(¹/x⁹)

 

Solution

 

= Log_a(¹/x⁹)

 

= Log_ax^-9

 

= -9 x Log_ax

 

= -9 x 0.8

 

= -7.2

 

TRY THIS………..

 

If Log_aw= 60, find Log_a(¹/w⁷)

ALGEBRA C2C

 

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)

GEOMETRIC PROGRESSION K1

 

In a G.P. the ratio of 8^th term and 6^th term is 4. If the sum of 2nd term and 4^th term is 240, Find the

(a)         Common ratio

(b)        The 1^st term.

 

Solution

 

G_n =G₁r^n-1.

 

G₈ =G₁r^8-1

 

G₈ =G₁r⁷

G₆ =G₁r^6-1.

G₆ =G₁r⁵.

 

^G6/_G8 =4

G₁r⁷/ G₁r⁵.=4

r⁷/r⁵=4       [G1 cancels out]

r^7-5=4

r²=4

√r²=√4   root sign

r = 2.

 

Hence the common ratio is 2

 

Now the 2^nd  term plus 4^th term is 240

G_n =G₁r^n-1.

 

G₉ =G₁r^9-1

G₉ =G₁r⁸

 

G₁₀ =G₁r^10-1

G₁₀ =G₁r⁹

 

G₁r + G₁r³ = 240

G₁(r  +  r³) = 240

G₁(2 +  2³) = 240       [since r=2]

G₁(2 +  8) = 240    

10G₁= 240    

Now G₁= ²⁴⁰/₁₀ = 24    

 

Hence the st term is 24.     

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

In a G.P. the ratio of 7^th term and 4^th term is 27. If the sum of 3rd term and 5^th term is 180, Find the

(a)         Common ratio

(b)        The 1^st term.

(c)          The 8^th term

SETS K4

 

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 50 villagers at the meeting. 11 are farmers and 20 are workers and 8 are both farmers and workers. How many villagers are

i)                 Either farmers or workers.

ii)             Neither farmers nor workers

LOGARITHMS K10

 

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

EXPONENTIALS K9

 

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            

WORD PROBLEMS A1

 

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………….

 

Jemson 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.

FACTORIZE K3

 

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)

FACTORIZE K2

 

Factorize completely 2t – 288t³

 

Solution

 

= 2t – 288t³ 

 

= 2t(1 – 144t²)

 

= 2t(1 – 12²t²)

 

= 2t[(1)² – (12t)²]

 

From difference of two squares, a² – b² = (a-b)(a+b)

 

= 2t[(1 – 12t)(1 + 12t)]

 

= 2t(1 – 12t)(1 + 12t) answer

 

TRY THIS……………..

 

Factorize completely 3m – 75m³

VARIATION K2

 

x is inversely proportional to y. x=54 while y=7. Find y when x is 9.

 

Solution

 

x ⍺ k

      y

 

x = k

      y

 

54 = k

        7

 

k = 54 x 7

 

k = 378

 

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

 

k = 378, y= ?, x = 9.

 

x = k

      y

 

9  = 378

        y

 

y x 9 = 378  x  y   (multiply by y on both sides)

             y

 

9y= 378

 

9y =378           [Dividing by 10 both sides]

9      9

 

Hence y = 42.

 

 

TRY THIS…………….

 

 

x is inversely proportional to y. x=80 while y=4. Find y when x is 10. 

LOGARITHMS K9

 

If Log_ax = 7, find Log_a(¹/x⁴⁶)

Solution

= Log_a(¹/x⁴⁶)

= Log_ax^-46          [since 1/aⁿ = a^-n]

= -46 x Log_ax    [since Log_axⁿ   = nLog_ax ]

= -46 x 7

= - 322 answer

TRY THIS………..

If Log_ab= 16, find Log_a(¹/b³⁰)