SIMPLE INTEREST 2

Anjali deposited the amount of 11,000/= in a bank which gives an interest rate of 5% for 2 years. Find the simple interest she got?

Solution

I = ?,  P = 11,000/=, T = 2 years, R = 5%

I  = PRT

      100

I  = 11,000 x 5 x 2

            100

I  = 11000 x 5 x 2

            100

I  = 110 x 5 x 2

          

I  = 1100/=

Hence the simple interest was Tsh 1100/=

TRY THIS………………..

Chudasama deposited the amount of 9000/= in a bank which gives an interest rate of 5% for 3 years. Find the simple interest she got?

COMPOUND INTEREST 1

Caroline invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 8,000 sh at the end of 3 years?

Solution

n=3, t=1, R=20%, A₃=8,000, P=?

A_n = P(1 + ^RT/100)ⁿ

A₃ = P(1 + ^(20x1)/100)³

A₃= P(1 + ²⁰/100)³

A₃ = P(¹⁰⁰/100 + ²⁰/100)³

A₃ = P(¹²⁰/100)³  But A₃=8,000,

8,000 = P(1.2)³  = P(1.2x1.2x1.2)  

8,000 = 1.728P

8000  =  1.728P

1.728      1.728

8000  =  P

1.728

P = 4629.63

Hence at the start she invested Tsh 4629.63

TRY THIS………………..

Caroline invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 8,000 sh at the end of 3 years?

SIMULTANEOUS EQUATIONS 1

Solve the following equations by substitution method.

4x - 2y = 6

x + y = 9

solution

4x - 2y = 6 ……………….(i)

x + y = 9……………….(ii)

1(4x - 2y = 6

 2(x + y = 9

By using elimination method,

   4x - 2y = 6

  2x + 2y = 18

    6x  + 0    = 24

6x = 24

6      6

x = 4

from equation (1),

x + y = 9

y = 9 – x

y = 9 - 4

y = 5

Hence x=4 and y=5.

TRY THIS........

Solve the following equations by substitution method.

4x - y = 8

2x + y = 34

SUM OF A GP 1

Find the sum of the 1st seven terms of the geometrical progression 2+6+18+54+…….

solution

G₁ = 2, r=3, n=7

S_n = G₁(rⁿ-1)/r-1

               

S₇ = 2(3⁷-1)/3-1

             

S₇ = ¹2(3⁷-1)/2₁

              

S₇ = (3⁷-1)

S₇ = 2187 - 1

S₇ = 2186

Hence the sum of the 1st seven terms is 2186.

TRY THIS………………

Find the sum of the 1st six terms of the geometrical progression 7+14+28+56+…..

DEGREES TO RADIAN 1

Change 540⁰ into radians.

Solution

s = ∏Ө/180

            

s = ∏ x 540

              180

s =  3∏ radians

       

Hence 540⁰ = 3∏ radians            1

TRY THIS…………………………

Change 380⁰ into radians

CIRCLES 2

Find the value of y in the figure below if O is the centre of the circle. [diagram not to scale]

Solution

Angle at the centre = 2 x (angle at the circumference)

= 2 x (y)

=2y

Angle 2y and 7y are opposite angles. They add up to 360⁰.

Therefore 

2y + 7y = 360⁰

9y = 360⁰

9y = 360⁰ 

9         9 

y = 40⁰          

Hence y = 40⁰          

TRY THIS………….

Find the value of y in the figure below if O is the centre of the circle. [diagram not to scale]

SIMPLE INTEREST 1

Dikshi deposited the amount of 7000/= in a bank which gives an interest rate of 3% for 2 years. Find the simple interest she got?

Solution

I = ?,  P = 7,000/=, T = 2 years, R = 3%

I  = PRT

      100

I  = 7000 x 3 x 2

            100

I  = 7000 x 3 x 2

            100

I  = 70 x 3 x 2

          

I  = 420/=

Hence the simple interest was Tsh 420/=

TRY THIS………………….

Edmund deposited the amount of 6000/= in a bank which gives an interest rate of 3% for 5 years. Find the simple interest she got?

TRANSPOSITION OF FORMULAE 1

If A = MPE make E the subject of the formula.

solution.

A = MPE

  A   =   MPE

MP      MP

  A   =   E

MN     

Hence   E =      A 

                       MP

TRY THIS………………….

If A = QDP make P the subject of the formula.

DIFFERENCE OF 2 SQUARES 1

Evaluate 176² – 166²

Solution

376² – 366² = (376 + 366)( 376 – 366)

                     = (732)( 10)

                     = 7320

376² – 366² = 7320

TRY THIS…………….

Evaluate 306² – 106²

INEQUALITIES 1

If 10x + 8 ≥ 7 + 3x ≥ 9x – 10; find x.

Solution

10x + 8 ≥ 7 + 3x and  7 + 3x ≥ 9x - 10

10x – 3x ≤ 7 - 8 and  7 +10 ≤ 9x - 3x

7x ≤ 1 and  17 ≤ 6x

7x ≤ 1  and  17 ≤ 6x

7      7           6      6

x ≤ ¹/7 and  ¹⁷/6 ≤ x

x ≤ ¹/7 and  x  ≥ ¹⁷/6

TRY THIS…………….

If 12x + 8 ≥ 7 + 3x ≥ 9x – 20; find x

PERCENTAGE PROFIT - 3

A man got a profit of 800/= after selling an item. Find the buying price if the percentage profit was 5%.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                         B.P

5% = 800   X  100   

          B.P.

B.P. x 5 = 80,000   

        

B.P. x 5¹ =   80,000   

  ₁5                   5

B.P. = 16,000

                      

Hence Buying Price was 16,000/=

TRY THIS…………………………..

A man got a profit of 1000/= after selling an item. Find the buying price if the percentage profit was 10%.

PERCENTAGE PROFIT - 2

A man got a profit of 100/= after selling an item for 2000/=. Find the percentage profit.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                         B.P

%’ge profit = 100   X  100   

                        2000

%’ge profit = 5%   

                       

Hence Percentage Profit is 5%

TRY THIS…………………………..

A man got a profit of 400/= after selling an item for 2000/=. Find the percentage profit.

LENGTH OF AN ARC - 1

Find the length of an arc if the radius of the circle is 10cm.

Solution

L = ∏r

      180⁰

L = ∏ x  10

             180⁰

L = ∏ cm

      18

Hence length of an arc is ^∏/18 cm

TRY THIS…………………………..

Find the length of an arc if the radius of the circle is 20cm.

RADIAN TO DEGREES - 1

Change ^5∏/4 radians into degrees.

Solution

Ө = 180⁰ x s                but s = ^5∏/4

       ∏

Ө = 180⁰  x  5∏              

       ∏        4

Ө =  ⁴⁵180  x  5∏             

          ₁∏      4₁

Ө = 45⁰ x 5

Ө = 225⁰

Hence ^5∏/4 radian is equal to 225⁰.

TRY THIS…………………………..

Change ^5∏/3 radians into degrees.

PERCENTAGE PROFIT 1

A man sold an item at 4000/=. Find the profit made if the percentage profit was 25%.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                         B.P

25%  =      P         X  100   

               4000

4000  x 25 = P x 100   

100,000 = 100P

100000 = 100P

  100       100

        

P = 1000

                      

Hence the profit was 1000/=

TRY THIS…………………………..

A man sold an item at 7000/=. Find the profit made if the percentage profit was 17%.

REPEATING DECIMALS-1

Learn Maths By Video     

Learn Maths By Video      

MATRIX - 1

Learn Maths By Video

PERFECT SQUARES-1

If 9x² – 12x + e is a perfect square, find e.

Solution

a = 9, b = -12, c = e.

b² = 4ac

(-12)² = 4 x 9 x e

144 = 36e

144 = 36e

 36      36

           

e = 4

Hence t = 4

TRY THIS………………………

If 9x² – 12x + e is a perfect square, find e

ALGEBRA-2

ARITHMETICS

Evaluate 13 x 237 + 463 x 13.

Solution

= 13 x 237 + 463 x 13

= 13 x (237 + 463)

= 13 x 700

= 9100

Hence 13 x 237 + 463 x 13 = 9100

TRY THIS………….

Evaluate 125 x 516 + 484 x 125.

RADICALS -1

CIRCLES-1

Find w in the figure below. [Diagram not to scale]

Solution

Angle inscribed in a semi-circle = 90⁰

<NMP = 90⁰

5w – 15 = 90⁰

5w – 15 = 90⁰  + 15

5w          = 105⁰  

5w          = 105⁰  

5                 5

w = 21⁰

TRY THIS………………..

Find a in the figure below. [Diagram not to scale]

PROBABILITY - 3

A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and then a number ten card?

solution

These are called independent events.

Probabilities:           

P(jack)  =       4

                     52

P(no. 10)      =          4                    

                               52                     

P(jack and no.10)  =         P(jack)           •         P(no.10)

                        =          4          •          4

                                    52                   52

             

             =           16                            

                       2704                         

             

             =           1                               

                       169

TRY THIS……………………………………

A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a king and then a number six card? 

LOGARITHMS -1

If Log_ax = 14, find Log_a(¹/x).

Solution

= Log_a(¹/x)

= Log_ax^-1

= -1 x Log_ax

= -1 x 14

= -14

TRY THIS………..

If Log_af= 43, find Log_a(¹/f)

ALGEBRA-1

Expand 8w(w – 3)

solution

= 8w(w – 3)

= (8w x w) – (8w x 3)

= 8w² – 24w     answer

TRY THIS………..

Expand 10p(p+ 5)

PROBABILITY - 2

A bag of candy contains 8 lemon flavored sour balls, and 5 lime flavored sour balls.  If Ana reaches in, takes one out and eats it, and then 20 minutes later selects another and eats that one as well, what is the probability that they were both lemon flavored candies?

solution

These are called independent events.

It is a ''Without replacement" question.

P(lemon 1st) = 8/13
P(lemon 2nd) = 7/12 

(remember...one is already eaten!)

Therefore:
P(lemon, lemon) = (8/13)(7/12) = 56/156 = 28/78=14/39

the probability that they were both lemon flavored candies is 14/39

TRY THIS………………….

A bag of candy contains 8 lemon flavored sour balls, and 5 lime flavored sour balls.  If Ana reaches in, takes one out and eats it, and then 20 minutes later selects another and eats that one as well, what is the probability that they were both lime flavored sour balls?

SETS - 3

If n(A)= 36 , n(B)= 50 and n(AuB)=65, find n(AnB).

Solution

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

       65 = 36 + 50 - n(AnB)

       65 = 86  - n(AnB)

       65 - 86 = - n(AnB)

      -21 = - n(AnB)

21 = n(AnB) (after dividing by -1 on both sides)

Hence n(AnB) = 21 answer

TRY THIS……………………

If n(A)= 35 , n(B)= 45 and n(AuB)=70, find n(AnB).

STATISTICS - 2

Distribution of length of nails in mm is as shown below. 

Length(mm)	15 - 21	22- 28	29 - 35	36 - 42	43 - 49	50 - 56	57 - 63
Frequency	4	6	8	16	8	5	3

Calculate the median.

Solution

First we prepare the frequency distribution table.

Class interval	Frequency(f)	Cumulative frequency
15-21	4	4
22-28	6	10
29-35	8	18
36-42	16	34
43-49	8	42
50-56	5	47
57-63	3	50
	∑f = 50

N = 50, N/2=25, 

Median class must fall in the cumulative frequency of 34. This has to be 36 – 42.

n_b = 18, n_w = 16, Upper boundary(U)= 42.5, Lower boundary(L) =35.5

i = Upper boundary – Lower boundary

i = 42.5 – 35.5

i = 7

Median = L + (N/2 – n_b)i/n_w

                              

Median = 35.5 + (50/2 – 18)7

                            16

Median = 35.5 + (25 – 20)7

                             16

Median = 35.5 + (5)7

                        16

  Median = 35.5 +  35

                           16

  Median = 35.5 +  2.1875

                               

 Median = 37.6875

Hence the median is 37.69 ( to 2d.p)                     

 

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

Distribution of Kiswahili test marks was given as hereunder. 

Length(mm)	10 - 19	20- 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79
Frequency	3	7	9	15	8	5	3

Calculate the median.