SEQUENCE AND SERIES A15

The first term of an AP is 5 and the last term is 57.If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 5, n=20, A_n = 57

S_n = n(A₁ + A_n)

       2

S₂₀ = 20(5 + 57)

         2

S₂₀ = 10 x 62

S₂₀ = 620

Hence the sum of all 23 terms is 620.

SIMPLE INTEREST A1

Rayan deposited the amount of 6,000/= in a bank for 5 years and got a profit of 1200/=. Find the interest rate?

Solution

I = 1200/=, P = 6,000/=, T = 5 years, R = ?

I  = PRT

      100

1200  = 6,000 x R x 5

                      100

1200  = 6,000 x R x 5

                      100

1200  = 60 x R x 5

                      

1200  = 300R

           

⁴1200  = 300R

   300       300

Hence the interest rate was  4%

SEQUENCE AND SERIES A14

Salma invested a certain amount of money in a bank which gives an interest rate of 10% compounded annually. How much did she invest at the start if she got 8000 sh at the end of 2 years?

solution

n=2, T=1, R=10%, A₂=8000, P=?

A_n = P(1 + ^RT/100)ⁿ

A₂ = P(1 + ^(10x1)/100)²

A₂ = P(1 + ¹⁰/100)²

A₂ = P(¹⁰⁰/100 + ¹⁰/100)²

A₂ = P(¹¹⁰/100)²  But A₂=8000,

8000 = P(1.1)²  

8000 = 1.21P

8000  =  1.21P

1.21       1.21

8000  =  P

1.21       

P = 6611.57

Hence at the start she invested Tsh 6611.67

STATISTICS A8

The masses of 10 cabbages in kilograms were recorded as follows:

5, 8, 4, 3, 6, 7, 2, 5, 1 and 9. Calculate the median of these data.

Solution

Arrange the data in order of magnitude starting with the smallest.

1, 2, 3, 4, 5 ,5,  6, 7, 8, 9

There are eight data so far. We get the two data in the middle which are 5 and 5.

Median can be calculated by finding the average of the two middle values.

So, median  = 5+ 5

                           2

                      = 5

Hence median is 5.

INEQUALITIES A1

2x – 1 ≤ x + 5 ≤ 6x

solution

2x – 1 ≤ x + 5 and  x + 5 ≤ 6x

2x – x ≤ 1+ 5 and  5 ≤ 6x - x

x ≤ 6 and  5 ≤ 5x (divide by 5 both sides)

x ≤ 6 and  1 ≤ x

x ≤ 6 and  x ≥ 1 

STATISTICS A7

In Biology test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59
No. of students	2	4	5	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 34.5 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	-20	2	29	-40
20-29	24.5	-10	4	98	-40
30-39	34.5	0	5	172.5	0
40-49	44.5	10	6	267	60
50-59	54.5	20	3	163.5	60
			∑f = 20	∑fx = 730	∑fd = 40

Mean = A +  ∑fd

                       ∑f

Mean = 34.5 +  40

                            20

Mean = 34.5 +  2

                           

         

Hence Mean =  36.5

Also you can prove your answer by using the normal formula of calculating mean, that is ∑fx/∑f and try to see if you get the same answer. 

CONGRUENCE OF SIMPLE POLYGONS A1

Prove that the following triangle is isosceles.

Solution

Given: triangle LMN with M produced to X.

Required to prove: ∆LMN is isosceles

Proof: LX = XN (given)

<LXM = <MXN = 90⁰ (given)

MX = XM (common line)

∆LXM ≡ ∆NXM by SAS rule

ML = MN (Equal corresponding sides)

Hence ∆LMN is isosceles (proved)

SEQUENCE AND SERIES A14

The first term of an AP is 10 and the last term is 80.If the arithmetic progression consists of 30 terms, calculate the sum of all the terms.

Solution

A₁ = 10, n=30, A_n = 80

S_n = n(A₁ + A_n)

       2

S₃₀ = 30(10 + 80)

         2

S₃₀ = 15 x 90

S₃₀ = 1350

Hence the sum of all 30 terms is 1350.

PROBABILITY A6

SEQUENCE AND SERIES A13

Find the number of terms in the following geometric progression 5+10+20+40+ … + 1280

Solution

G₁=5, G_n=1280, r=2.

G_n = G₁r^n-1;

1280 = 5 x 2^n-1;

1280 = 5 x 2^n-1;       (Divide by 5 on both sides)

   5         5

256 = 2^n-1;

256 = 2ⁿ x 2^-1;

256 = 2ⁿ x ¹/2;

256  =  2ⁿ;  (Divide by ¹/2 on both sides)

  ¹/2      

512 =  2ⁿ;     (512 = 2x2x2x2x2x2x2x2x2)

2⁹ =  2ⁿ;

9 = n

There are 9 terms in a Geometrical Progression.

STATISTICS A6

In Kiswahili test the following marks were recorded.

marks	10-19	20-29	30-39	40-49	50-59	60-69
No. of students	2	4	9	7	5	3

Calculate the mean.

Solution

Here you are required to produce the frequency distribution table.

Class interval	Class mark (x)	Frequency(f)	fx
10-19	14.5	2	29
20-29	24.5	4	98
30-39	34.5	9	310.5
40-49	44.5	7	311.5
50-59	54.5	5	272.5
60-69	64.5	3	193.5
		∑f = 30	∑fx = 1215

Mean = ∑fx

               ∑f

Mean = 1215

                 30

Mean =  40.5

         

Hence Mean =  40.5

SEQUENCE AND SERIES A12

The first term of an AP is 18 and the last term is 100.If the arithmetic progression consists of 20 terms, calculate the sum of all the terms.

Solution

A₁ = 18, n=20, A_n = 100

S_n = n(A₁ + A_n)

       2

S₂₀ = 20(18 + 100)

         2

S₂₀ = 10 x 118

S₂₀ = 1180

Hence the sum of all 20 terms is 1180.

PROBABILITY A5

RADIANS 7

SEQUENCE AND SERIES A11

The sum of the 1st six terms of a geometrical progression is 189. If the common ratio is 2 find the 1st term.

solution

S₆ = 189 G₁ = ?, r=2, n=6

S_n = G₁(rⁿ-1)/r-1

               

S₆ = G₁(r⁶-1)/ r-1

              

189= G₁(2⁶-1)/2-1

               

189= G₁(64-1)

                  1

189= G₁(63)

                 

189 = 63 G₁

³ 189   =  G₁

   63

G₁ = 3

Hence the 1st term is 3.

STATISTICS A5

The masses of 10 coconuts in kilograms were recorded as follows:

5, 8, 4, 3, 6, 7, 2, 5, 1,  and 1. Calculate the median of these data.

Solution

Arrange the data in order of magnitude starting with the smallest.

1, 1, 2, 3,  4, 5 ,5,  6, 7, 8

There are eight data so far. We get the two data in the middle which are 4 and 5.

Median can be calculated by finding the average of the two middle values.

So, median  = 4+ 5

                         2

                      = 4.5

Hence median is 4.5.

REPEATING DECIMAL A8

RADIANS 6

STATISTICS A4

In Mathematics test the following marks were recorded.

marks	31-40	41-50	51-60	61-70	71-80	81-90
No. of students	5	7	14	10	8	6

Calculate the mean.

Solution

Here you are required to produce the frequency distribution table.

Class interval	Class mark (x)	Frequency(f)	fx
31-40	35.5	5	177.5
41-50	45.5	7	318.5
51-60	55.5	14	777
61-70	65.5	10	655
71-80	75.5	8	604
81-90	85.5	6	513
		∑f = 50	∑fx = 3045

Mean = ∑fx

               ∑f

Mean = 3045

               50

Mean =  60.9

         

Hence Mean =  60.9