Tuesday 31 December 2013

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, An = 57

Sn = n(A1 + An)
       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=?

An = 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

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, An = 80

Sn = n(A1 + An)
       2

S30 = 30(10 + 80)
         2

S30 = 15 x 90

S30 = 1350


Hence the sum of all 30 terms is 1350.

Saturday 21 December 2013

PROBABILITY A6




SEQUENCE AND SERIES A13

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

Solution

G1=5, Gn=1280, r=2.

Gn = G1rn-1;

1280 = 5 x 2n-1;

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

256 = 2n-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, An = 100

Sn = n(A1 + An)
       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

Sn = 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