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Tambaya 1 Rahoto
In the diagram, find the size of the angle marked ao
Bayanin Amsa
2 x s = 280o(Angle at centre = 2 x < at circum)
S = 280o2
= 140
< O = 360 - 280 = 80o
60 + 80 + 140 + a = 360o
(< in a quad); 280 = a = 360
a = 360 - 280
a = 80o
Tambaya 3 Rahoto
The solution of the quadratic inequality (x2 + x - 12) ≥ 0 is
Bayanin Amsa
(x2 + x - 12) ≥
0 , (x - 3)(x + 4) ≥
0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3 ≥
0 and x + 4 ≥
0
or x - 3≤
0 and x + 4 ≤
0
when x ≥
3, the condition is satisfied
when x ≥
-4, the condition is not satisfied.
when x ≤
3, the condition is not satisfied
when x ≤
-4 , the condition is not satisfied. Thus, the solution of the inequality is x ≥
3 or x ≤
-4 ,
Tambaya 4 Rahoto
calculate the simple interest on N6,500 for 8 years at 5% per annum.
Tambaya 5 Rahoto
What is the mean of the data t, 2t-1, t-2, 2t-1, 4t and 2t+2?
Bayanin Amsa
Tambaya 6 Rahoto
Find the range of values of x which satisfy the inequalities 4x - 7 ≤ 3x and 3x - 4 ≤ 4x
Bayanin Amsa
4X - 7 ≤
3X and 3X - 4 ≤
4X
4X - 3X ≤
7 and 3X - 4X ≤
4
X ≤
7 and -X ≤
4 = X ≥
-4
Range -4 ≤
x ≤
7
Tambaya 7 Rahoto
If x > 0, find the range of number x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x and 3x+1
Bayanin Amsa
x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x, 3x+1
Range = 3x+2 - (x-3)
= 3x+2 - x - 3
= 2x + 5
Tambaya 9 Rahoto
If tan θ = 54 find sin2θ - cos2θ
Bayanin Amsa
(tan θ
= oppadj
)
|AB|2 = 52 + 42 →
|AB|2 = 41
→
AB = √41
sin2θ
- cos2θ
→
52√41
- (4√412
) = 2541
- 1641
= (941
)
Tambaya 11 Rahoto
Find the median of 4, 1, 4, 1, 0, 4, 4, 2 and 0
Bayanin Amsa
Tambaya 12 Rahoto
Find the gradient of a line which is perpendicular to the line with the equation 3x + 2y + 1 = 0
Bayanin Amsa
3X + 2Y + 1 = 0
2Y = -3X - 1
−32X−12
Gradient of 3X + 2Y +1 = 0 is -3/2
Gradient of a line perpendicular to 3X + 2Y + 1 = 0
=−1÷32=−1×−23=23
Tambaya 13 Rahoto
The bar chart above shows the number of times the word a, and , in, it, the , to appear in a paragraph in a book. What is the ratio of the least frequent word?
Bayanin Amsa
Ratio of least to most = 3:12
= 3/12
= 1/4
Tambaya 14 Rahoto
Add 11012,101112 and 1112
Tambaya 15 Rahoto
The locus of a point equidistant from two points p(6,2) and R(4,2) is a perpendicular bisector of PR passing through
Bayanin Amsa
Tambaya 16 Rahoto
A book seller sells Mathematics and English books. If 30 customers buy Mathematics books, 20 customers buy English books and 10 customers buy the two books, How many customers has he altogether.
Bayanin Amsa
n(M) only = 30-10 = 20
n(E) only = 20-10 = 10
n(M∩E) = 10
∴M∪E = 20+10+10
= 40
Tambaya 17 Rahoto
Find the capacity in liters of a cylindrical well of radius 1 meter and depth 14 meters
[π = 22/7]
Bayanin Amsa
V = πr2h
1m = 100cm
14cm = 1400cm
∴V=227×100x100x14001000=44,000liters
Tambaya 18 Rahoto
Express 1223456 to 3 significant figures
Tambaya 19 Rahoto
A binary operation on the real set of numbers excluding -1 is such that for all m, n ∈ R, mΔn = m+n+mn. Find the identity element of the operation.
Bayanin Amsa
mΔn = m+n+mn
Let e be the identity element
∴mΔe = eΔm = m
m+e+me = m
e+me = m-m
e+me = 0
e(1+m) = 0
e = 0 / (1+m)
e = 0
Tambaya 20 Rahoto
If p = varies inversely as the square of q and p=8 when q=4, find when p=32
Bayanin Amsa
P ∝ 1/q
P = k/q
K = q2P
= 428
∴P = 128/q
32 = 128/q
q2 = 128/32
q2 = 4
q = √4 = +/-2
Tambaya 21 Rahoto
In the diagram < OPQ is
Tambaya 22 Rahoto
If X = {n2 + 1:n = 0,2,3} and Y = {n+1:n=2,3,5}, find X∩Y.
Tambaya 24 Rahoto
Find the area of the figure above
[π = 22/7]
Bayanin Amsa
Area of the figure = Area of rect + area of semi circle
=L×h+12πr25×15+12×227×(52)2=75+(22×25)2×7=75+92328=84.8cm
Tambaya 25 Rahoto
The cost of kerosine per liter increase from N60 to N85. What is the percentage rate of increase?
Bayanin Amsa
N85 - N60 = N25 increase
∴ percentage increase =2560×1001=1253=41.67%=42%
Tambaya 26 Rahoto
In how many ways can the letters of the word ACCEPTANCE be arranged?
Bayanin Amsa
ACCEPTANCE = 10 Letters
A = 2 letters
C = 3 letters
E = 2 letters
Can be arranged in 10! / (2!3!2!) ways
Tambaya 27 Rahoto
The result of rolling a fair die 150 times is ass summarized in the table above. What is the probability of obtaining a 5
Bayanin Amsa
Total possible outcome
12+18+x+30+2x+45 = 105+3x
∴105+3x = 150
3x = 150-105
3x = 45
x = 15
P(obtaining 5) =2x(105+3x)Butx=15=2(15)(105+3(15))=30(105+45)=30150=15
Tambaya 29 Rahoto
Differentiate sin x - x cos x
Bayanin Amsa
sin x - x cos x
dy/dx = cos x - [1.cos x + x -sin x]
= co x - [cos x - x sin x]
= cos x - cos x + x sin x
= x sin x
Tambaya 30 Rahoto
If 125x = 2010 find x
Bayanin Amsa
125x = 20
1xX2 + 2xX1 + 5xX0 = 20
X2 + 2X + 5 = 20
X2 + 2X - 15 = 0
(X + 5)( X - 3) = 0
X + 5 implies X = -5
X - 3 implies X = 3
But X cannot be negative
∴X = 3
Tambaya 31 Rahoto
Find the derivative of y=x7−x5x4
Bayanin Amsa
Tambaya 32 Rahoto
The fifth term of an A.P is 24 and the eleventh term is 96. Find the first term.
Bayanin Amsa
U5 = 24, n = 5 and U11 = 96, n = 11
Un = a + (n-1)d
24 = a + (5-1)d imply 24 = a+4d .....eqn1
96 = a + (11-1)d imply 96 = a+10d ...eqn2
eqn1 - eqn2 -72 = -6d
d = 72/6 = 12
but 24 = a+4d
24 = a + 4(12)
24 = a + 48
a = 24-48
a = -24
Tambaya 33 Rahoto
The result of rolling a fair die 150 times is as summarized in the table given.
Number123456Frequency1218x302x45
What is the probability of obtaining a 5?
Bayanin Amsa
Number123456Frequency1218x302x45
12 + 18 + x + 30 + 2x + 45 = 150
3x + 105 = 150
3x = 150 - 105
3x = 45
x = 453
x = 15
probability of 5 = 30150
= 15
Tambaya 34 Rahoto
A binary operation * is defined on the set of positive integers is such x*y = 2x-3y+2 for all positive integers x and y. The binary operation is
Bayanin Amsa
X * Y = 2X - 3Y + 2
2*3 = 2(2) - 3(3) + 2
=4-9+2
= -3
But -3 does not belong to positive integer
Tambaya 35 Rahoto
If 2x2 - kx - 12 is divisible by x-4, Find the value of k.
Bayanin Amsa
2x2 - kx - 12 is divisible by x-4
implies x is a factor ∴ x = 4
f(4) implies 2(4)2 - k(4) - 12 = 0
32 - 4k - 12 = 0
-4k + 20 = 0
-4k = -20
k = 5
Tambaya 36 Rahoto
Solve the quadratic inequalities x2 - 5x + 6 ≥0
Bayanin Amsa
x2 - 5x + 6 = 0
(X-2)(X-3) = 0
X-2 = 0 implies X = 2
X-3 = 0 implies X = 3
∴ x ≤ 2, x ≥ 3
Tambaya 37 Rahoto
In the diagram above ∠OPQ is
Bayanin Amsa
a = a(base ∠s of Iss Δ)
∴ a+a+74 = 180
2a + 74 = 180
2a = 180-74
2a = 106
a = 53
∴∠OPQ = 53∘
Tambaya 38 Rahoto
Find the number of ways of selecting 6 out of 10 subjects for an examination
Bayanin Amsa
Tambaya 39 Rahoto
Find the mean deviation of 2, 4, 5, and 9
Tambaya 40 Rahoto
Which of the following angles is an exterior angle of a regular polygon?
Tambaya 41 Rahoto
If logx1/264 = 3, find the value of x
Bayanin Amsa
If logx1/264 = 3
(X 1/2)3 = 64
(X 1/2)3 = 4 3
X 1/2 = 4
X = 42
X = 16
Tambaya 42 Rahoto
Evaluate (38÷12+12)(18×23+13)
Bayanin Amsa
Tambaya 43 Rahoto
Factorize complete;y (4x+3y)2 - (3x-2y)2
Bayanin Amsa
(4x+3y)2 - (3x-2y)2
(4x+3y+3x-2y)(4x+3y-(3x-2y))
(4x+3y+3x-2y)(4x+3y-3x+2y)
(x+5y)(7x+y)
Tambaya 44 Rahoto
The probability of picking a letter T fr4om the word OBSTRUCTION is
Bayanin Amsa
OBSTRUCTION
Total possible outcome = 11
Number of chance of getting T = 2
P(picking T) = 2/11
Tambaya 45 Rahoto
simplify 16−12×4−12×2713
Bayanin Amsa
Tambaya 46 Rahoto
If 1+√21−√2
is expressed in the form of x+y√2 find the values of x and y
Bayanin Amsa
Tambaya 47 Rahoto
Make Q the subject of formula when L=43M√PQ
Bayanin Amsa
Tambaya 48 Rahoto
Evaluate ∫π2−π2cosxdx
Bayanin Amsa
∫π2−π2cosxdx=[sinx]π2−π2=sinπ2−sin−π2
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2
Tambaya 49 Rahoto
Find the area of the figure given
Bayanin Amsa
Area of semicircle + Area of rectangle
A = 12πr2 + LB
A = 12×2277×(52)2+(15×5)
= 12×227×254+75
A = 27528+751
275+210028=237528
A = 84.8cm2
Tambaya 50 Rahoto
Evaluate ∫21(6x2−2x)dx
Bayanin Amsa
∫21(6x2−2x)dx=[6x33−2x22]21=[2x3−x2]21
= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11
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