INEQUALITIES B1

If 7x – 1 ≤ 2x + 5 ≤ 9x -1; Find x.

solution

7x – 1 ≤ 2x + 5 and  2x + 5 ≤ 9x -1

7x – 2x ≤ 1+ 5 and  5 +1 ≤ 9x - 2x

5x ≤ 6 and  6 ≤ 7x

5x ≤ 6  and  6 ≤ 7x

5       5           7     7

x ≤ ⁶/5 and  ⁶/7 ≤ x

x ≤ ⁶/5 and  x  ≥ ⁶/7

DIFFERENCE OF 2 SQUARES B1

Evaluate 176² – 166²

Solution

176² – 166² = (176 + 166)( 176 – 166)

                     = (342)( 10)

                     = 3420

Hence 176² – 166² = 3420

TRANSPOSITION OF FORMULAE B1

if A = MNE make E the subject of the formula

solution

A = MNE

  A   =   MNE

MN      MN

  A   =   E

MN     

Hence   E =      A  

                         MN      

SEQUENCE AND SERIES B2

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 6,000 sh at the end of 3 years?

Solution

n=3, t=1, R=20%, A₃=6,000, 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=6,000,

6,000 = P(1.2)³  = P(1.2x1.2x1.2)  

6,000 = 1.728P

6000  =  1.728P

1.728      1.728

6000  =  P

1.728

P = 3472.22

Hence at the start she invested Tsh 3472.22

SIMPLE INTEREST B2

Sonali deposited the amount of 30,000/= in a bank for 3 years and got a profit of 2700/=. Find the interest rate?

Solution

I = 2700/=, P = 30,000/=, T = 3 years, R = ?

I  = PRT

      100

2700  = 30,000 x R x 3

                      100

2700  = 30,000 x R x 3

                      100

2700  = 300 x R x 3

                     

2700  = 900R

           

³2700  = 900R

   900       900

Hence the interest rate was 3%

RADIANS TO DEGREES

Change ^11∏/20 radians into degrees.

Solution

Ө = 180⁰ x s                but s = ^11∏/20

       ∏

Ө = 180⁰  x  11∏              

       ∏        20

Ө =  ⁹180  x  11∏             

          ₁∏     20₁

Ө = 9⁰ x 11

Ө = 99⁰

Hence ^11∏/20 radian is equal to 99⁰.

STATISTICS B2

In French test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89
No. of students	4	5	6	7	11	8	6	3

Calculate the mean.

Solution

In this question we are going to apply the method of assumed mean.

Here you are required to produce the frequency distribution table.

Let us take 64.5 as our assumed mean (A). Take away the assumed mean from each class mark.

Class interval	Class mark (x)	d=x-A	Frequency (f)	fx	fd
10-19	14.5	-50	4	58	-200
20-29	24.5	-40	5	122.5	-200
30-39	34.5	-30	6	207	-180
40-49	44.5	-20	7	311.5	-140
50-59	54.5	-10	11	599.5	-110
60-69	64.5	0	8	516	0
70-79	74.5	10	6	447	60
80-89	84.5	20	3	223.5	60
			∑f = 50	∑fx = 2458	∑fd = -710

Mean = A +  ∑fd

                       ∑f

Mean = 64.5 +  -710

                              50

Mean = 64.5 +  (-14.2)

                    

Mean = 50.3

         

Hence Mean =  50.3

PERCENTAGES B1

Aniket sold an item at 3000/=. Find the profit made if the percentage profit was 8¹/3%.

Solution

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

                         B.P

8¹/3% =      P         X  100   

                 3000

3000  x 8¹/3 = P x 100   

¹⁰⁰⁰3000  x  25 = P x 100

                     3₁

1000 x 25 = 100P

25000 = 100P

25000 = 100P

  100       100

        

P = 250

                      

Hence the profit was 250/=

SIMPLE INTEREST B1

Sonam deposited the amount of 30,000/= in a bank which gives an interest rate of 4% and got a profit of 4800/=. For how long did she deposit the money in a bank?

Solution

I = 4,800/=, P = 30,000/=, T = ?, R = 4%

I  = PRT

      100

4,800  = 30,000 x 4 x T

                      100

4,800  = 30,000 x 4 x T

                      100

4,800  = 1,200 x  T

                     

4,800  = 1,200T

           

⁴4,800  =  1,200T

  1,200      1,200

T = 4

Hence the time was  4 years.

STATISTICS B1

In French test the following marks were recorded.

marks	30-34	35-39	40-44	45-49	50-54	55-59	60-64	65-69
No. of students	4	5	6	7	10	9	5	4

Calculate the mean.

Solution

In this question we are going to apply the method of assumed mean.

Here you are required to produce the frequency distribution table.

Let us take 42 as our assumed mean (A). Take away the assumed mean from each class mark.

Class interval	Class mark (x)	d=x-A	Frequency(f)	fx	fd
30-34	32	-10	4	128	-40
35-39	37	-5	5	185	-25
40-44	42	0	6	252	0
45-49	47	5	7	329	35
50-54	52	10	10	520	100
55-59	57	15	9	513	135
60-64	62	20	5	310	100
65-69	67	25	4	248	100
			∑f = 50	∑fx = 2485	∑fd = 405

Mean = A +  ∑fd

                       ∑f

Mean = 42 +  405

                          50

Mean = 42 +  (8.1)

                    

Mean = 50.1

         

Hence Mean =  50.1