PERCENTAGE PROFIT 5

A man got a profit of 100/= after selling an item for 2500/=. Find the percentage profit.

Solution

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

                        B.P

%’ge profit = 100   X  100   

                     2500

%’ge profit = 4%   

    

Hence Percentage Profit is 4%

TRY THIS…………………………..

A man got a profit of 200/= after selling an item for 3000/=. Find the percentage profit.

ARC LENGTH 2

Find the length of an arc if the radius of the circle is 40cm.

Solution

L = ∏r

      180⁰

L = ∏ x  40

             180⁰

L = 2∏ cm

       9

Hence length of an arc is ^2∏/9 cm

TRY THIS…………………………..

 Find the length of an arc if the radius of the circle is 25cm.

SETS 1

If n(A)= 82 , n(B)= 100 and n(AnB)=20, find n(AuB).

Solution

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

             = 82 + 100 – 20

             = 182– 20

             = 162

Hence n(AuB) = 162 answer

TRY THIS………….

If n(A)= 115 , n(B)= 128 and n(AnB)=89, find n(AuB).

PRIME FACTORS 2

Using a factor tree give the prime factors of 60.

Solution

Hence the prime factors are 2  x  2  x  3  x  5

TRY THIS………….

Using a factor tree give the prime factors of 132.

FACTORIZING 10

Solve by Factorization 2x² - 3x - 20

Solution

2x² - 3x – 20=0  We split the middle term (-3x) to be (-8x + 5x).

2x² - 8x + 5x – 20=0                (-3x) = -8x + 5x.

(2x² -8x) + (5x - 20)=0

2x(x - 4) + 5(x - 4)=0

(2x + 5) (x -4) =0

2x + 5=0 or x -4=0

2x = -5  or  x = 4

2       2

x = ^-5/2  or  x = 4

Hence x= ^-5/2 or x=4 answer.

TRY THIS…..

 Solve by factorization; 2x² - x – 28.

GEOMETRIC PROGRESSION 4

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

solution

G₁ = 2, r=3, n=8

S_n = G₁(rⁿ-1)/r-1

               

S₈ = 2(3⁸-1)/3-1

             

S₈ = ¹2(3⁸-1)/2₁

              

S₈ = (3⁸-1)

S₇ = 6561 – 1               [Since  3⁸  =6561]

S₇ = 6560

Hence the sum of the 1st seven terms is 6560.

TRY THIS………………

Find the sum of the 1st six terms of the geometrical progression 5+10+20+40+…..

SIMPLE INTEREST 4

Anjali deposited the amount of 13,000/= in a bank which gives an interest rate of 5% for 2 years. Find the simple interest she got?

Solution

I = ?,  P = 13,000/=, T = 2 years, R = 5%

I  = PRT

      100

I  = 13,000 x 5 x 2

            100

I  = 13000 x 5 x 2

            100

I  = 130 x 5 x 2

          

I  = 1300/=

Hence the simple interest was Tsh 1300/=

TRY THIS………………..

Chudasama deposited the amount of 90,000/= in a bank which gives an interest rate of 3% for 5 years. Find the simple interest she got?

LOGARITHMS 20

If log_x3125 – log₃729 = -1; find x

solution

log_x3125 – log₃729 = -1

log_x3125 – log₃3⁶ = -1

log_x3125 – 6log₃3 = -1

log_x3125 – 6 x 1 = -1

log_x3125 – 6 = -1

log_x3125 = -1 + 6

log_x3125 = 5

3125 = x⁵

5⁵ = x⁵      Powers cancel out

x = 5

  

Hence x = 5.

TRY THIS………….

If log_k1024 – log₇343 = 7; find k

UNITS OF DISTANCE 1

Change 73980m into Km.

Solution

1Km = 1000m

?  =  73980m

= 1  x  73980

       1000

=     73980

       1000

= 73.98Km

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

Change 78010m into Km.

COMPOUND INTEREST 2

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 8,500 sh at the end of 3 years?

Solution

n=3, t=1, R=20%, A₃=8,500, 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=8,500

8,500 = P(1.2)³  = P(1.2x1.2x1.2)  

8,500 = 1.728P

8500  =  1.728P

1.728      1.728

8500  =  P

1.728

P = 4919/=

Hence at the start she invested Tsh 4919

TRY THIS………………..

Tushabe 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 18,000 sh at the end of 4 years?

UNITS OF WEIGHT 1

Change 23980g into Kg.

Solution

1Kg = 1000g

?  =  23980g

= 1  x  23980

       1000

=     23980

       1000

= 23.98Kg

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

Change 45080g into Kg.

LOGARITHMS 19

Evaluate    Log46 +  Log5000  -  Log23

                       

Solution

=    Log46 +  Log5000  -  Log23

= Log(46x5000)

                    23

= Log(¹230000)

                    123

= Log₁₀10000

= Log₁₀10⁴

= 4Log₁₀10

= 4 x 1

= 4

Hence Log46 +  Log5000  -  Log23 = 4

TRY THIS………..

Evaluate  Log3400 +  Log600  -  Log0.204

PROBABILITY 12

A number is chosen at random from 1 – 15 inclusive. Find the probability that it is a multiple of seven or an even number.

Solution

THIS IS A NON-MUTUALLY EXCLUSIVE EVENT.

Let n(S) represent sample space

P(E) = probability of even number

P(M) = probability of a multiple of 7

n(S) = {1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12, 13, 14, 15} = 15

n(E) = {2, 4, 6, 8, 10, 12, 14} = 7

n(M) = {7, 14 } = 2

But 14 appears on both categories.

P(E) = ⁷/15

P(M) = ²/15

P(EnM) = ¹/15

………………………….

P(EuM) = P(E) + P(M) - P(EnM) 

P(EuM) = ⁷/15 + ²/15 -¹/15

              = ⁸/15

              = ⁸/15

∴ P(EuM) = ⁸/15

TRY THIS………………

A number is chosen at random from 4 – 30 inclusive. Find the probability that it is a multiple of 3 or an even number.

ALGEBRA 12

Simplify 60 - 18m + 12 + 5m

Solution

= 60 - 18m + 12 + 5m

= 60 + 12 + 5m - 18m  [Grouping like terms together]

= 72 - 13m   [since 5m - 18m  = -13m]

= 72 - 13m answer

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

Simplify 6 - 18m + 12 + 5m. 

WEEK AND DAYS 1

Change 875 days into weeks.

Answer

1 week = 7 days

?  =  875 days

= 875 x 1

      7

= 875

     7

= 125weeks

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

Change 1456 days into weeks.

BODMAS 2

Evaluate 90 + 50 x 30 – 80 ÷ 4

Solution

We apply BODMAS.

= 90 + 50 x 30 – 80 ÷ 4

= 90 + 50 x 30 – 20    [After dividing]

= 90 + 1500 – 20    [After multiplying]

= 1590 – 20    [After adding]

= 1570       [After subtracting]

Hence 90 + 50 x 30 – 80 ÷ 4 = 1570      

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

Evaluate 90 + 50 x 30 – 80 ÷ 4

SIMULTANEOUS EQN 1

Solve the following equations by substitution method.

x + y = 10

2x + y = 16

Solution

x + y = 10 ----------- (i)

2x + y = 16 ---------- (ii)

from equation (i)

x + y = 10

x = 10 – y ----------(iii)

substitute equation (iii) in (ii) above,

2(10 – y) + y = 16

20–2y  + y = 16  

20 – y = 16           [since –2y+y = -y]

20 –16 = y   

4 = y          

      

So; y = 4

From equation (iii)

x = 10 – 4 ----------(iii)

x = 6

Hence x=6 and y=4.

TRY THIS………………………..

Solve the following equations by substitution method.

5x – y = 11

x  +  y = 7

Answer: x=3;   y=4

FACTORIZING 9

Factorize 9-w²

Solution

We use difference of two squares a² – b² = (a - b)(a + b)

9-w² = 3²-w²       

        = (3 - w)(3 + w)

Hence 9-w² = (3 - w)(3 + w)

TRY THIS…………….

Factorize 49 - c²

INEQUALITIES 4

If 11x + 8 ≥ 7 + 3x ≥ 9x – 13; find x

Solution

11x + 8 ≥ 7 + 3x and  7 + 3x ≥ 9x - 13

11x – 3x ≤ 7 - 8 and  7 +13 ≤ 9x - 3x

8x ≤ -1 and  20 ≤ 6x

8x ≤ -1  and  20 ≤ 6x

8       8          6       6

x ≤ -¹/8 and  ¹⁰/3 ≤ x

x ≤ -¹/8 and  x  ≥ ¹⁰/3

TRY THIS…………….

  

If 14x + 8 ≥ 7 + 5x ≥ 11x – 20; find x

PRIME FACTORS 1

Using a factor tree give the prime factors of 48.

Solution

Hence the prime factors of 48 are 2  x  2  x  2  x  2  x  3

TRY THIS………….

 Using a factor tree give the prime factors of 68.

APPROXIMATIONS 4

Estimate the value of 1143  ÷ 52.

Solution

= 1123     

     52

= 1000      [ since 1143 ≈ 1000   and   52 ≈ 50 ]

     50

= 20    [After dividing 1000 by 20]

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

Estimate the value of 7312  ÷ 18

PERIMETER 2

Area of rectangle is 60m². Find its perimeter if width is 4m. 

Solution

A = L x w

60 = L x 4

60 = 4L

4      4

15 = L

Now; P =2(L + w)

= 2(15 + 4)

= 2 x 19

= 38m

Hence perimeter is 38m

TRY THIS………..

Area of rectangle is 85m². Find its perimeter if width is 5m. 

UNITS OF VOLUME 1

Change 4.5 liters of water into milliliters

Solution

1L  =  1000mL

4.5L = ?

___________________________  

 = 4.5 x  1000

         1

= 4.5 x 1000

4500mL.

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

Change 6.3 liters of milk into milliliters 

ALGEBRA 11

Simplify 6n + 18m + 2n – 5m

Solution

= 6n + 18m + 2n – 5m

= 6n + 2n + 18m– 5m         [Grouping like terms together]

= 8n + 13m answer

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

Simplify 6n + 18m + 2n – 5m

LOGARITHMS 18

Evaluate       Log0.0000001

                       

Solution

=    Log0.0000001

      

We write 0.0000001 as an exponent. So 0.0000001=10^-7

=    Log10^-7

      

=    Log₁₀10^-7

            

=    -7Log₁₀10       [Since Log_ccⁿ = nLog_cc]

      

=    -7 x 1             [Since Log_cc = 1]

Hence           Log0.0000001         =  -7

TRY THIS………..

Evaluate    Log0.0000000001

APPROXIMATIONS 3

Estimate the value of 6.3  x  0.064

solution

= 6.3  x  0.064

= 6.0 x 0.06      [ 6.3 ≈ 6.0 to ones and 0.064 ≈ 0.06 to hundredths]

= 0.36

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

Estimate the value of 4.2  x  0.078

PROBABILITY 13

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a spade and an ace?

Solution

(This is a problem with replacement)

n(E) = 13, n(S) = 52

P(E) = n(E)

            n(S)

1st pick(spade) = 13/52

2nd pick(an ace) = 4/52.

P(a spade and an ace) =   13      x    4

                                         52           52

P(a spade and an ace) =  ¹13      x    4 ¹

                                        ₄52           52 ₁₃

P(a spade and an ace) =  1      x    1

                                        4           13

P(a spade and an ace) =     1      

                                         52     

Therefore Probability of drawing a spade and an ace is 1/52 

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

A certain man takes 2 cards from a standard deck of 52 cards.

What will be the probability of getting a number 2 and a joker?

VARIATION 5

x is inversely proportional to y. x=18 while y=5. Find y when x is 12.

Solution

x ⍺ k

      y

x = k

      y

18 = k

        5

k = 18 x 5

k = 90

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

k = 90, y= ?, x = 12.

x = k

      y

12 = 90

         y

y x 12 = 90  x  y   (multiply by y on both sides)

                y

12y = 90

12      12

y = 7.5  .

TRY THIS…………….

x is inversely proportional to y. x=24 while y=4. Find y when x is 8.

EXPONENTIALS 10

Simplify the following by writing in power form:

 4w⁷⁰

  w²⁰

Solution

= 4w^{70 - 20} 

= 4w⁵⁰

TRY THIS……………….

Simplify the following by writing in power form:

8 m⁵⁰

   m³⁰    

BODMAS 1

Evaluate 18 x (45 – 41) ÷ (101-99)

Solution

We apply BODMAS.

= 18 x (45 – 41) ÷ (101-99)

= 18 x 4 ÷ (101-99)     [After dealing with 1^st brackets]

= 18 x 4 ÷ 2     [After dealing with 2^nd brackets]

= 18 x  2     [After dividing]

= 36     [After multiplying]

Hence  18 x (45 – 41) ÷ (101-99) = 36

TRY THIS……………….

Evaluate 8 x (56 – 42) ÷ (88-86)