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Ajụjụ 1 Ripọtì
Udoh deposited ₦150.00 in the bank. At the end of 5 years the simple interest on the principal was ₦55.00. At what rate per annum was the interest paid?
Akọwa Nkọwa
.I = PTR100
R = 100PT
100×50150×6
= 223
= 713
%
Ajụjụ 2 Ripọtì
Find n if log24 + log27 - log2n = 1
Akọwa Nkọwa
log24 + log27 - log2n = 1
= log2(4 x 7) - log2n = 1
= log228 - log2n
= log282n
282n
= 21
= 2
2n = 28
∴ n = 14
Ajụjụ 3 Ripọtì
Factorize x2 + 2a + ax + 2x
Ajụjụ 4 Ripọtì
Find the smallest number by which 252 can be multiplied to obtain a perfect square
Akọwa Nkọwa
Let the smallest number be x and the perfect square be y 252x = y.
By trial and error method, 252 x 9 = 1764
Check if y = 1764
y2 = 42
x = 7
Ajụjụ 6 Ripọtì
In \( \triangle \)XYZ, determine the cosine of angle Z.
Ajụjụ 7 Ripọtì
make U the subject of the formula \(S = \sqrt{\frac{6}{u} - \frac{w}{2}}\)
Akọwa Nkọwa
S = √6u−w2
S = 12−uw2u
2us2 = 12 - uw
u(2s2 + w) = 12
u = 122s2+w
Ajụjụ 9 Ripọtì
Of the nine hundred students admitted in a university in 1979, the following was the distribution by state: Anambrs 185, Imo 135, Kaduna 90, Kwara 110, Ondo 155, Oyo 225. In a pie chart drawn to represent this distribution, The angle subtended at the centre by anambra is
Akọwa Nkọwa
Anambra = 185900
x 3601
= 74o
Ajụjụ 10 Ripọtì
Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ
Akọwa Nkọwa
In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)
Ajụjụ 11 Ripọtì
An arc of a circle of radius 6cm is 8cm long. Find the area of the sector
Akọwa Nkọwa
Radius of the circle r = 6cm, Length of the arc = 8cm
Area of sector = θ360
x 2π
r2........(i)
Length of arc = θ360
x 2π
r........(ii)
from eqn. (ii) θ
= 240π
, subt. for θ
in eqn (i)
Area x 2401
x 1360
x π61
= 24cm2
Ajụjụ 12 Ripọtì
An open rectangular box externally measures 4m x 3m x 4m. Find the total cost of painting the box externally if it costs ₦2.00 to paint one square meter
Akọwa Nkọwa
Total surface area(s) = 2(4 x 3) + 2(4 x 4)
= 2(12) + 2(16)
= 24 + 32
= 56cm2
1m2 costs ₦2.00
∴ 56m∴ will cost 56 x ₦2.00
= ₦112.00
Ajụjụ 13 Ripọtì
The table below gives the scores of a group of students in a Mathematical test.
| Scores | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 2 | 4 | 7 | 14 | 12 | 6 | 4 | 1 |
If the mode in m and the number of students who scored 4 or less is s. What is (s, m)?
Akọwa Nkọwa
M = mode = the number having the highest frequency = 4
S = Number of students with 4 or less marks
= 14 + 7 + 4 + 2
= 27
∴ (M,S) = (27, 4)
Ajụjụ 14 Ripọtì
If 5(x + 2y) = 5 and 4(x + 3y) = 16, find 3(x + y)
Akọwa Nkọwa
5(x + 2y) = 5
∴ x + 2y = 1.....(i)
4(x + 3y) = 16 = 42
x + 3y = 2 .....(ii)
x + 2y = 1.....(i)
x + 3y = 2......(ii)
y = 1
Substitute y = 1 into equation (i) = x + 2y = 1
∴ x + 2(1) = 1
x + 2 = 1
∴ x = 1
∴ 3x + y = 3-1 + 1
= 3 = 1
Ajụjụ 15 Ripọtì
Simplify \(\left(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\right)x\frac{1}{\sqrt{3}}\)
Akọwa Nkọwa
(1√5+√3−1√5−√3
x 1√3
)
1√5+√3−1√5−√3
= √5−√3−(√5+√3)(√5+√3)(5√3)
= −2√35−3
= −2√37
Ajụjụ 16 Ripọtì
Solve the equation 3x2 + 6x - 2 = 0
Akọwa Nkọwa
3x2 + 6x - 2 = 0
Using almighty formula i.e. x = b±√b2−4ac2a
a = 3, b = 6, c = -2
x = −6±√62−4(3)(−2)2(3)
x = 6±√36−246
x = 6±√606
x = −6±√4×156
x = −1±√153
Ajụjụ 17 Ripọtì
If \( y = \frac{x}{x - 3} + \frac{x}{x + 4} \) find y when \( x = -2 \)
Akọwa Nkọwa
y = xx−3
+ xx+4
when x = -2
y = −2−5
+ (−2)−2+4
= −25
+ −22
= 4+−1010
= −1410
= -75
Ajụjụ 18 Ripọtì
A man kept 6 black, 5 brown and 7 purple shirts in a drawer. What is the probability of his picking a purple shirt with his eyes closed?
Ajụjụ 19 Ripọtì
The ages of Tosan and Isa differ by 6 and the product of their ages is 187. Write their ages in the form (x, y), where x > y.
Akọwa Nkọwa
x - y = 6.......(i)
xy = 187.......(ii)
From equation (i), x(6 + y)
sub. for x in equation (ii) = y(6 + y)
= 187
y2 + 6y = 187
y2 + 6y - 187 = 0
(y + 17)(y - 11) = 0
y = -17 or y = 11
y cannot be negative, y = 11
Sub. for y in equation(i) = x - 11
= 16
x = 6 + 11
= 17
∴(x, y) = (17, 11)
Ajụjụ 21 Ripọtì
If x varies directly as y3 and x = 2 when y = 1, find x when y = 5
Akọwa Nkọwa
x α
y3
x = ky3
k = xy3
when x = 2, y = 1
k = 2
Thus x = 2y3 - equation of variation
= 2(5)3
= 250
Ajụjụ 22 Ripọtì
If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)
Akọwa Nkọwa
cosθ
= ab
, Sinθ
= √b2−a2a
Tanθ
= √b2−a2a2
, Tan 2 = √b2−a2a2
1 + tan2θ
= 1 + b2−a2a2
= a2+b2−a2a2
= b2a2
Ajụjụ 23 Ripọtì
A number of pencils were shared out among Bisi, Sola and Tunde in the ratio of 2 : 3 : 5 respectively. If Bisi got 5, how many were share out?
Akọwa Nkọwa
Let x r3epresent total number of pencils shared
B : S : T = 2 + 3 + 5 = 10
2 : 3 : 5
= 210
x y
= 5
2y =5
2y = 50
∴ y = 502
= 25
Ajụjụ 24 Ripọtì
Find the median of the numbers 89, 141, 130, 161, 120, 131, 131, 100, 108 and 119
Akọwa Nkọwa
Arrange in ascending order
89, 100, 108, 119, |120, 130|, 131, 131, 141, 161
Median = 120+1302
= 125
Ajụjụ 25 Ripọtì
Factorize (4a + 3)2 - (3a - 2)2
Akọwa Nkọwa
(4a + 3)2 - (3a - 2)2 = a2 - b2
= (a + b)(a - b)
= [(4a + 3) + (3a - 2)][(4a + 3) + (3a - 2)]
= [(4a + 3 + 3a - 2)][(4a + 3 - 3a + 2)]
= (7a + 1)(a + 5)
∴ (a + 5)(7a + 1)
Ajụjụ 26 Ripọtì
Find the reciprocal of \( \frac{\frac{2}{3}}{\frac{1}{2}+\frac{1}{3}} \)
Akọwa Nkọwa
23
= 23
= 23
x 65
= 45
reciprocal of 45
= 145
= 54
Ajụjụ 27 Ripọtì
Factorize completely 8a + 125ax3
Akọwa Nkọwa
8a + 125ax3 = 23a + 53ax3
= a(23 + 53x3)
∴a[23 + (5x)3]
a3 + b3 = (a + b)(a2 - ab + b2)
∴ a(23 + (5x)3)
= a(2 + 5x)(4 - 10x + 25x2)
Ajụjụ 28 Ripọtì
Divide the L.C.M of 48, 64, and 80 by their H.C.F
Akọwa Nkọwa
48 = 24 x 3, 64 = 26, 80 = 24 x 5
L.C.M = 26 x 3 x 5
H.C.F = 24
26×3×524
= 22 x 3 x 5
= 4 x 3 x 5
= 12 x 5
= 60
Ajụjụ 29 Ripọtì
The people in a city with a population of 0.9 million were grouped according to their ages. Use the diagram to determine the number of people in the 15 - 29 years group
Akọwa Nkọwa
15 - 29 years is represented by 104∘
Number of people in the group is 104360 x 0.9m
= 260000 = 26 x 104
Ajụjụ 30 Ripọtì
A girl walks 45 meters in the direction 050o from a point Q to a point X. She then walks 24 meters in the direction 140o from X to a point Y. How far is she then from Q?
Akọwa Nkọwa
QY = 452 + 242 = 2025 + 576
= 2601
QY = √2601
= 51
Ajụjụ 31 Ripọtì
Find the values of x which satisfy the equation 16x - 5 x 4x + 4 = 0
Akọwa Nkọwa
16x - 5 x 4x + 4 = 0
(4x)2 - 5(4x) + 4 = 0
let 4x = y
y2 - 5y + 4 = 0
(y - 4)(y - 1) = 0
y = 4 or 1
4x = 4
x = 1
4x = 1
i.e. 4x = 4o, x = 0
∴ x = 1 or 0
Ajụjụ 32 Ripọtì
If U and V are two distinct fixed points and W is a variable points such that UWV is a right angle, what is the locus of W?
Ajụjụ 33 Ripọtì
Three boys shared some oranges. The first received 1/3 of the oranges and the second received 2/3 of the remaining. If the third boy received the remaining 12 oranges, how many oranges did they share
Akọwa Nkọwa
Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
∴1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges
Ajụjụ 34 Ripọtì
Find the total surface area of solid cone of radius \(2\sqrt{3}\) cm and slanting side \(4\sqrt{3}\)
Akọwa Nkọwa
Total surface area of a solid cone
r = 2√3
= πr2
+ π
rH
H = 4√3
, π
r(r + H)
∴ Area = π
2√3
[2√3
+ 4√3
]
= π
2√3
(6√3
)
= 12π
x 3
= 36π
cm2
Ajụjụ 35 Ripọtì
If Musa scored 75 in biology instead of 57, his average mark in four subjects would have been 60. What was his total mark?
Akọwa Nkọwa
Let x represent Musa's total mark when he scores 57 in biology and Let Y represent Musa's total mark when he now scored 75 in biology, if he scored 75 in biology his new total mark will be Y4
= 60, y = 4 x 60 = 240
To get his total mark when he scored 57, subtract 57 from 75 to give 18, then subtract this 18 from the new total mark(ie. 240)
= 240 - 18
= 222
Ajụjụ 36 Ripọtì
A regular polygon of n sides has 160o as the size of each interior angle. Find n
Akọwa Nkọwa
Interior + exterior = 360
160 + exterior = 360
Exterior = 360 - 160
Exterior = 20
n = 360/exterior
n = 360/20
n = 18
Ajụjụ 37 Ripọtì
If \( \frac{a}{c} = \frac{c}{d} = k \), find the value of \( \frac{3a^2-ac+c^2}{3b^2-bd+d^2} \)
Akọwa Nkọwa
ac
= cd
= k
∴ ab
= bk
cd
= k
∴ c = dk
= 3a2−ac+c23b2−bd+d2
= 3(bk)2−(bk)(dk)+dk23b2−bd+a2
= 3b2k2−bk2d+dk23b2−bd+d2
k = 3b2k2−bk2d+dk23b2−bd+d2
Ajụjụ 39 Ripọtì
If P = 18, Q = 21, R = -6 and S = -4, Calculate \( \frac{(P-Q)^3+S^2}{R^3} + S^2 \)
Akọwa Nkọwa
(P−Q)3+S2R3
= (18−21)3+(−4)2(−6)3
= −27+16R3
= −11−216
= 11216
Ajụjụ 40 Ripọtì
Simplify \( \frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{3}}}{3^{-\frac{1}{6}} \times 3^{\frac{2}{3}}} \)
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
Find all real numbers \(x\) which satisfy the inequality \(\frac{1}{3}(x + 1) - 1 > \frac{1}{5}(x + 4)\)
Akọwa Nkọwa
13
(x + 1) - 1 > 15
(x + 4)
= x+13
- 1 > x+45
x+13
- x+45
- 1 > 0
= 5x+5−3x−1215
= 2x - 7 > 15
= 2x > 12
= x > 11
Ajụjụ 42 Ripọtì
Simplify \( \frac{1}{x-2} + \frac{1}{x+2} + \frac{2x}{x^2-4} \)
Akọwa Nkọwa
1x−2
+ 1x+2
+ 2xx2−4
= (x+2)+(x−2)+2x(x+2)(x−2)
= 4xx2−4
Ajụjụ 43 Ripọtì
In the figure, \( \triangle \)PQT is isosceles. PQ = QT, SRQ = \(35^\circ\), TPQ = \(20^\circ\) and PQR is a straight line.Calculate TSR
Akọwa Nkọwa
Given △ isosceles PQ = QT, SRQ = 35∘
TPQ = 20∘
PQR = is a straight line
Since PQ = QT, angle P = angle T = 20∘
Angle PQR = 180∘ - (20 + 20) = 140∘
TQR = 180∘ - 140∘ = 40∘ < on a straight line
QSR = 180∘ - (40 + 35)∘ = 105∘
TSR = 180∘ - 105∘
= 75∘
Ajụjụ 44 Ripọtì
Simplify \( \frac{1}{5x+5} + \frac{1}{7x+7} \)
Akọwa Nkọwa
15x+5
+ 17x+7
= 15(x+1)
+ 17(x+1)
= 7+535(x+1)
= 1235(x+1)
Ajụjụ 45 Ripọtì
In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.
Akọwa Nkọwa
PT x QT = TR x TS
24 x 8 = TR x 12
TR = 24×812
= = 16cm
Ajụjụ 46 Ripọtì
At what points does the straight line y = 2x + 1 intersect the curve y = 2x2 + 5x - 1?
Ajụjụ 47 Ripọtì
PQ and PR are tangents from P to a circle centre O as shown in the figure. If QRP = \(34^\circ\), find the angle marked x
Akọwa Nkọwa
From the circle centre 0, if PQ & PR are tangents from P and QRP = 34∘
Then the angle marked x i.e. QOP
34∘ x 2 = 68∘
Ajụjụ 48 Ripọtì
The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume
Akọwa Nkọwa
Volume of solid = cross section x H
Since the cross section is a trapezium
= 12(6+11)×12×8
= 6 x 17 x 8 = 816m3
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