The 1st term of an A.P. is 10 and the common difference is
58. Find the 16th term.
Solution
A1=
10, d = 58
An =
A1 + (n-1)d [ formula for n terms ]
A16 =
A1 + (16-1)d
A16 =
A1 + 15d [ formula for 16 terms ]
A16 =
10 + (16 x 58)
A16 =
10 + 928
A16 =
938
Hence
the 16th term is 938.
TRY THIS……………..
The
1st term of an A.P. is 23 and the common difference is 58. Find the 22nd term.
Given
that F(x) = 13x - 55. Find F(8)
Solution
F(x) =
13x - 55
F(8) =
13(8) - 55
F(8) =
104 - 55
F(8) =
49
Hence
F(8) = 49
TRY THIS……………..
Given
that F(x) = 55x - 20. Find F(3)
In Ntoma
district the number of people who speak Kiswahili or Lingala is 410. 200 of
them speak Kiswahili and 320 of them speak Lingala. How many speak both
languages?
solution
In
most cases, OR stands for union whereas AND/BOTH, stands for intersection.
Let
Kiswahili=n(K), Lingala= n(L).
n(K)=
200 ,
n(L)=
320,
n(KuL)
= 410,
n(KnL)=?
n(KuL)
= n(K) + n(L) - n(KnL)
410
= 200 + 320 – n(KnL)
410
= 520 – n(KnL)
n(KnL)
= 520 – 410
n(KnL)
= 110
Hence
n(KnL)=110 answer
TRY THIS……………..
In
Bibanja district the number of people who speak Spanish or German is 210. 170
of them speak Spanish and 180 of them speak German. How many speak both
languages?
Find the slope of a line which passes through (-15, -2) and
(6,-9)
Solution
x1 =
-15, y1 =-2, x2 = 6, y2 =
-9
m
= y2 –y1
x2 –
x1
m =
-9 –(-2)
6 –(-15)
m =
-9 + 2
6 + 15
m
= - 7
21
Hence
the slope is -7/21
TRY THIS……………..
Find
the slope of a line which passes through (-15, -6) and (1,-9)
Solve
for x if 2x – 1 ≤ x + 50 ≤ 6x.
solution
2x –
1 ≤ x + 50 and x + 50 ≤ 6x
2x –
x ≤ 1+ 50 and 50 ≤ 6x - x
x ≤ 51
and 50 ≤ 5x (having divided by
5 both sides)
x ≤ 6
and 10 ≤ x
x ≤ 6 and x ≥ 10
TRY
THIS………………….
Solve
for x if 2x – 8 ≤ x + 40 ≤ 11x
The first term of an AP
is 5 and the last term is 67.If the arithmetic progression consists of 20
terms, calculate the sum of all the terms.
Solution
A1 = 5,
n=20, An = 67
Sn = n(A1 +
An)
2
S20 = 20(5
+ 67)
2
S20 = 10
x 72
S20 = 720
Hence
the sum of all 20 terms is 720.
TRY
THIS………………….
The first term of an AP is 7 and the last term is 75. If the arithmetic progression consists of 24 terms, calculate the sum of all the terms
Dikshi
deposited the amount of 90,000/= in a bank which gives an interest rate of 3%
for 2 years. Find the simple interest she got.
Solution
I =
?, P = 90,000/=, T = 2 years, R = 3%
I
= PRT
100
I
= 90000 x 3 x 2
100
I
= 90000 x 3 x 2
100
I
= 900 x 3 x 2
I
= 5400/=
Hence
the simple interest was Tsh 5400/=
TRY
THIS………………….
Edmund
deposited the amount of 40,000/= in a bank which gives an interest rate of 6%
for 5 years. What was her simple interest?
If n(A)=
85 , n(B)= 96 and n(AuB)= 119, find n(AnB).
Solution
n(AuB) = n(A) + n(B) -
n(AnB)
119 = 85 + 96 - n(AnB)
119 = 181 - n(AnB)
119 - 181 = - n(AnB)
-62 = -n(AnB)
n(AnB) = 62 [after
dividing by -1 both sides]
Hence n(AnB) = 62 answer
TRY THIS………….
If n(A)= 91 , n(B)= 95 and
n(AuB)= 127, find n(AnB).
Simplify the expression c2 - t2/
(c + t)
Solution
from difference of two squares p2 - q2 =
(p - q)(p + q)
= c2 - t2/ (c + t)
= (c + t)(c - t)
c + t
= (c + t)(c - t)
c + t
= c - t answer
TRY THIS.........................
Simplify the expression
u2 - y2/ (u + y)
Rayan deposited the
amount of 8,000/= in a bank for 5 years and got a profit of 2800/=. Find the
interest rate?
Solution
I = 2800/=, P =
8,000/=, T = 5 years, R = ?
I = PRT
100
Where P=Principal amount
I = simple Interest
T= Time
R=Rate
2800 = 8,000
x R x 5
100
2800 = 8,000 x
R x 5
100
2800 = 80 x R
x 5
2800 = 400R
72800 = 400R
400 400
Hence the interest
rate was Tsh 7%
TRY THIS……………….
Kibonde deposited
the amount of 15,000/= in a bank for 4 years and got a profit of 3000/=. Find
the interest rate?
Expand 8w(5w + 7 – w2)
Solution
= 8w(5w + 7 – w2)
= (8w x 5w) + (8w x 7) - (8w
x w2)
= 40w2 + 56w - 8w3 answer
