Nkojọpọ....
Tẹ mọ́ & Dì mú láti fà yíká. |
|||
Tẹ ibi lati pa |
Ibeere 1 Ìròyìn
T varies inversely as the cube of R. When R = 3, T = 281 , find T when R = 2
Awọn alaye Idahun
T α1R3
T = kR3
k = TR3
= 281
x 33
= 281
x 27
dividing 81 by 27
k = 22
therefore, T = 23
x 1R3
When R = 2
T = 23
x 123
= 23
x 18
= 112
Ibeere 2 Ìròyìn
Simplify 323×56×231115×34×227
Awọn alaye Idahun
323×56×231115×34×227
113×56×231115×34×227
11054÷661620
50
Ibeere 3 Ìròyìn
Solve the inequality -6(x + 3) ≤ 4(x - 2)
Awọn alaye Idahun
-6(x + 3) ≤
4(x - 2)
-6(x +3) ≤
4(x - 2)
-6x -18 ≤
4x - 8
-18 + 8 ≤
4x +6x
-10x ≤
10x
10x ≤
-10
x ≤
1
Ibeere 4 Ìròyìn
Simplify (√2+1√3)(√2−1√3 )
Awọn alaye Idahun
(√2+1√3)(√2−1√3
)
√4−√2√3+√2√3−1√9
= 2 - 13
= 16−13
= 53
Ibeere 5 Ìròyìn
Find the equation of a line perpendicular to line 2y = 5x + 4 which passes through (4, 2).
Awọn alaye Idahun
2y = 5x + 4 (4, 2)
y = 5x2
+ 4 comparing with
y = mx + e
m = 52
Since they are perpendicular
m1m2 = -1
m2 = −1m1
= -1
52
= -1 x 25
The equator of the line is thus
y = mn + c (4, 2)
2 = -25
(4) + c
21
+ 85
= c
c = 185
10+55
= c
y = -25
x + 185
5y = -2x + 18
or 5y + 2x - 18 = 0
Ibeere 6 Ìròyìn
A chord of circle of radius 7cm is 5cm from the centre of the maximum possible area of the square?
Awọn alaye Idahun
From Pythagoras theorem
|OA|2 = |AN|2 + |ON|2
72 = |AN|2 + (5)2
49 = |AN|2 + 25
|AN|2 = 49 - 25 = 24
|AN| = √24
= √4×6
= 2√6 cm
|AN| = |NB| (A line drawn from the centre of a circle to a chord, divides the chord into two equal parts)
|AN| + |NB| = |AB|
2√6 + 2√6 = |AB|
|AB| = 4√6 cm
Ibeere 8 Ìròyìn
The perpendicular bisector of a line XY is the locus of a point B. whose distance from Y is always twice its distance from X. C
Awọn alaye Idahun
Ibeere 9 Ìròyìn
A binary operation ⊕ om real numbers is defined by x ⊕ y = xy + x + y for two real numbers x and y. Find the value of 3 ⊕ - 23 .
Awọn alaye Idahun
N + Y = XY + X + Y
3 + -23
= 3(- 23
) + 3 + (- 23
)
= -2 + 3 -23
= 1−21−3
= 13
Ibeere 10 Ìròyìn
If log318 + log33 - log3x = 3, Find x.
Awọn alaye Idahun
log183
+ log33
- logx3
= 3
log183
+ log33
- logx3
= 3log33
log183
+ log33
- logx3
= log333
log3(18×3X
) = log333
18×3X
= 33
18 x 3 = 27 x X
x = 18×327
= 2
Ibeere 11 Ìròyìn
I how many was can the letters of the word ELATION be arranged?
Awọn alaye Idahun
ELATION
Since there are 7 letters. The first letter can be arranged in 7 ways, , the second letter in 6 ways, the third letter in 5 ways, the 4th letter in four ways, the 3rd letter in three ways, the 2nd letter in 2 ways and the last in one way.
therefore, 7 x 6 x 5 x 4 x 3 x 2 x 1 = 7! ways
Ibeere 12 Ìròyìn
Rationalize 2−√53−√5
Awọn alaye Idahun
2−√53−√5
x 3+√53+√5
(2−√5)(3+√5)(3−√5)(3+√5)
= 6+2√5−3√5−√259+3√5−3√5−√25
= 6−√5−59−5
= 1−√54
Ibeere 13 Ìròyìn
Find ∫10 cos4 x dx
Awọn alaye Idahun
∫10
cos4 x dx
let u = 4x
dydx
= 4
dx = dy4
∫10
cos u. dy4
= 14
∫
cos u du
= 14
sin u + k
= 14
sin4x + k
Ibeere 14 Ìròyìn
Find the value of x at the minimum point of the curve y = x3 + x2 - x + 1
Awọn alaye Idahun
y = x3 + x2 - x + 1
dydx
= d(x3)dx
+ d(x2)dx
- d(x)dx
+ d(1)dx
dydx
= 3x2 + 2x - 1 = 0
dydx
= 3x2 + 2x - 1
At the maximum point dydx
= 0
3x2 + 2x - 1 = 0
(3x2 + 3x) - (x - 1) = 0
3x(x + 1) -1(x + 1) = 0
(3x - 1)(x + 1) = 0
therefore x = 13
or -1
For the maximum point
d2ydx2
< 0
d2ydx2
6x + 2
when x = 13
dx2dx2
= 6(13
) + 2
= 2 + 2 = 4
d2ydx2
> o which is the minimum point
when x = -1
d2ydx2
= 6(-1) + 2
= -6 + 2 = -4
-4 < 0
therefore, d2ydx2 < 0
the maximum point is -1
Ibeere 15 Ìròyìn
In a right angled triangle, if tan θ
= 34
. What is cosθ
- sinθ
?
Awọn alaye Idahun
tanθ
= 34
from Pythagoras tippet, the hypotenus is T
i.e. 3, 4, 5.
then sin θ
= 35
and cosθ
= 43
cosθ
- sinθ
45
- 35
= 15
Ibeere 16 Ìròyìn
A man walks 100 m due West from a point X to Y, he then walks 100 m due North to a point Z. Find the bearing of X from Z.
Awọn alaye Idahun
tanθ
= 100100
= 1
θ
= tan-1(1) = 45o
The bearing of x from z is ₦45oE or 135o
Ibeere 17 Ìròyìn
Find the derivative of sinθcosθ
Awọn alaye Idahun
sinθcosθ
cosθd(sinθ)dθ−sinθd(cosθ)dθcos2θ
cosθ.cosθ−sinθ(−sinθ)cos2θ
cos2θ+sin2θcos2θ
Recall that sin2 θ
+ cos2 θ
= 1
1cos2θ
= sec2 θ
Ibeere 18 Ìròyìn
The seconds term of a geometric series is 4 while the fourth term is 16. Find the sum of the first five terms
Awọn alaye Idahun
T2 = 4, T4 = 16
Tx = arn-1
T2 = ar2-1 = 4 i.e. ar3 = 16, i.e. ar = 4
T4 = ar4-1
therefore, T4Tr
= ar3ar
= 164
r2 = 4 and r = 2
but ar = 4
a = 4r
= 42
a = 2
Sn = a(rn−1)r−1
S5 = 2(25−1)2−1
= 2(32−1)2−1
= 2(31)
= 62
Ibeere 19 Ìròyìn
Find the probability that a number picked at random from the set(43, 44, 45, ..., 60) is a prime number.
Awọn alaye Idahun
Ibeere 20 Ìròyìn
Solve for x and y respectively in the simultaneous equations -2x - 5y = 3, x + 3y = 0
Awọn alaye Idahun
-2x -5y = 3
x + 3y = 0
x = -3y
-2 (-3y) - 5y = -3
6y - 5y = 3
y = 3
but, x = -3y
x = -3(3)
x = -9
therefore, x = -9, y = 3
Ibeere 21 Ìròyìn
The inverse of matrix N = ∣∣∣2314∣∣∣
is
Awọn alaye Idahun
N = [2 3]
N-1 = adjN|N|
adj N = ∣∣∣4−3−12∣∣∣
|N| = (2 x4) - (1 x 3)
= 8 - 3
=5
N-1 = 15
∣∣∣4−3−12∣∣∣
Ibeere 22 Ìròyìn
The sum of four consecutive integers is 34. Find the least of these numbers
Awọn alaye Idahun
Let the numbers be a, a + 1, a + 2, a + 3
a + a + 1 + a + 2 + a + 3 = 34
4a = 34 - 6
4a = 28
a = 284
= 7
The least of these numbers is a = 7
Ibeere 23 Ìròyìn
The derivatives of (2x + 1)(3x + 1) is
Awọn alaye Idahun
(2x + 1)(3x + 1) IS
2x + 1 d(3x+1)d
+ (3x + 1) d(2x+1)d
2x + 1 (3) + (3x + 1) (2)
6x + 3 + 6x + 2 = 12x + 5
Ibeere 24 Ìròyìn
Class Intervals0−23−56−89−11Frequency3253
Find the mode of the above distribution.
Awọn alaye Idahun
Mode = L1 + (D1D1+D2
)C
D1 = frequency of modal class - frequency of the class before it
D1 = 5 - 2 = 3
D2 = frequency of modal class - frequency of the class that offers it
D2 = 5 - 3 = 2
L1 = lower class boundary of the modal class
L1 = 5 - 5
C is the class width = 8 - 5.5 = 3
Mode = L1 + (D1D1+D2
)C
= 5.5 + 32+3
C
= 5.5 + 35
x 3
= 5.5 + 95
= 5.5 + 1.8
= 7.3 ≈
= 7
Ibeere 25 Ìròyìn
The midpoint of P(x, y) and Q(8, 6). Find x and y. midpoint = (5, 8)
Awọn alaye Idahun
P(x, y) Q(8, 6)
midpoint = (5, 8)
x + 8 = 5
y+62
= 8
x + 8 = 10
x = 10 - 8 = 2
y + 6 = 16
y + 16 - 6 = 10
therefore, P(2, 10)
Ibeere 27 Ìròyìn
In how many ways can five people sit round a circular table?
Awọn alaye Idahun
The first person will sit down and the remaining will join.
i.e. (n - 1)!
= (5 - 1)! = 4!
= 24 ways
Ibeere 28 Ìròyìn
Class Interval3−56−89−11Frequency222 .
Find the standard deviation of the above distribution.
Awọn alaye Idahun
Class Interval3−36−89−11x4710f222f−x81420|x−¯x|2909|x−¯x|218018
¯x
= ∑fx∑f
= 8+14+202+2+2
= 426
¯x
= 7
S.D = √∑f(x−¯x)2∑f
= √18+0+186
= √366
= √6
Ibeere 29 Ìròyìn
A solid metal cube of side 3 cm is placed in a rectangular tank of dimension 3, 4 and 5 cm. What volume of water can the tank now hold
Awọn alaye Idahun
Volume of cube = L3
33 = 27cm3
volume of rectangular tank = L x B X h
= 3 x 4 x 5
= 60cm3
volume of H2O the tank can now hold
= volume of rectangular tank - volume of cube
= 60 - 27
= 33cm3
Ibeere 30 Ìròyìn
A circle of perimeter 28cm is opened to form a square. What is the maximum possible area of the square?
Awọn alaye Idahun
Perimeter of circle = Perimeter of square
28cm = 4L
L = 284
= 7cm
Area of square = L2
= 72
= 49cm2
Ibeere 32 Ìròyìn
Make R the subject of the formula if T = KR2+M3
Awọn alaye Idahun
Ibeere 33 Ìròyìn
Find the remainder when X3 - 2X2 + 3X - 3 is divided by X2 + 1
Awọn alaye Idahun
X2 + 1 X−2√X3−2X2+3n−3
= −6X3+n−2X2+2X−3
= (−2X2−2)2X−1
Remainder is 2X - 1
Ibeere 34 Ìròyìn
If 2q35 = 778, find q
Awọn alaye Idahun
2q35 = 778
2 x 52 + q x 51 + 3 x 50 = 7 x 81 + 7 x 80
2 x 25 + q x 5 + 3 x 1 = 7 x 8 + 7 x 1
50 + 5q + 3 = 56 + 7
5q = 63 - 53
q = 105
q = 2
Ibeere 35 Ìròyìn
Simplify (1681)14÷(916)−12
Awọn alaye Idahun
(1681)14÷(916)−12
(1681)14÷(169)12
(2434)14÷(4232)12
24×1434×14÷42×1232×12
23÷43
23×34
24
12
Ibeere 36 Ìròyìn
A man invested ₦5,000 for 9 months at 4%. What is the simple interest?
Awọn alaye Idahun
S.I. = P×R×T100
If T = 9 months, it is equivalent to 912
years
S.I. = 5000×4×9100×12
S.I. = ₦150
Ibeere 37 Ìròyìn
Find the sum of the first 18 terms of the series 3, 6, 9,..., 36.
Awọn alaye Idahun
3, 6, 9,..., 36.
a = 3, d = 3, i = 36, n = 18
Sn = n2
[2a + (n - 1)d
S18 = 182
[2 x 3 + (18 - 1)3]
= 9[6 + (17 x 3)]
= 9 [6 + 51] = 9(57)
= 513
Ibeere 38 Ìròyìn
Raial has 7 different posters to be hanged in her bedroom, living room and kitchen. Assuming she has plans to place at least a poster in each of the 3 rooms, how many choices does she have?
Awọn alaye Idahun
The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.
= 7 x 6 x 5
= 210 ways
or 7P3
= 7!(7−3)!
= 7!4!
= 7×6×5×4!4!
= 210 ways
Ibeere 40 Ìròyìn
Which of these angles can be constructed using ruler and a pair of compasses only?
Awọn alaye Idahun
Ibeere 42 Ìròyìn
The bar chart above shows the distribution of SS2 students in a school.
Find the total number of students
Awọn alaye Idahun
Ibeere 43 Ìròyìn
No012345Frequency143825 .
From the table above, find the median and range of the data respectively.
Awọn alaye Idahun
Ibeere 44 Ìròyìn
The pie chart shows the distribution of courses offered by students. What percentage of the students offer English?
Awọn alaye Idahun
90360×100=14×100
=25%
Ibeere 45 Ìròyìn
Factorize completely 9y2 - 16X2
Awọn alaye Idahun
9y2 - 16x2
= 32y2 - 42x2
= (3y - 4x)(3y +4x)
Ibeere 46 Ìròyìn
If | 2 3 | = | 4 1 |. find the value of y. 7
Awọn alaye Idahun
∣∣∣2353x∣∣∣
= ∣∣∣4132x∣∣∣
(2 x 3x) - (5 x 3) = (4 x 2x) - (3 x 1)
6x - 15 = 8x - 3
6x - 8x = 15 - 3
-2x = 12
x = 12−2
= -6
Ibeere 47 Ìròyìn
In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW
Awọn alaye Idahun
< TUW = 110∘
= 180∘
(< s on a straight line)
< TUW = 180∘
- 110∘
= 70∘
In △
XTU, < XUT + < TXU = 180∘
i.e. < YTS + 70∘
= 180
< XTU = 180 - 110∘
= 70∘
Also < YTS + < XTU = 180 (< s on a straight line)
i.e. < YTS + < XTU - 180(< s on straight line)
i.e. < YTS + 70∘
= 180
< YTS = 180 - 70 = 110∘
in △
SYT + < YST + < YTS = 180∘
(Sum of interior < s)
SYT + 40 + 110 = 180
< SYT = 180 - 150 = 30
< SYT = < XYW (vertically opposite < s)
Also < SYX = < TYW (vertically opposite < s)
but < SYT + < XYW + < SYX + < TYW = 360
i.e. 30 + 30 + < SYX + TYW = 360
but < SYX = < TYW
60 + 2(< TYW) = 360
2(< TYW) = 360∘
- 60
2(< TYW) = 300∘
TYW = 3002
= 150∘
< SYT
Ibeere 48 Ìròyìn
Solve the inequality x2 + 2x > 15.
Awọn alaye Idahun
x2 + 2x > 15
x2 + 2x - 15 > 0
(x2 + 5x) - (3x - 15) > 0
x(x + 5) - 3(x + 5) >0
(x - 3)(x + 5) > 0
therefore, x = 3 or -5
then x < -5 or x > 3
i.e. x< 3 or x < -5
Ibeere 49 Ìròyìn
What is the size of each interior angle of a 12-sided regular polygon?
Awọn alaye Idahun
Interior angle = (n - 2)180
but, n = 12
= (12 -2)180
= 10 x 180
= 1800
let each interior angle = x
x = (n−2)180n
x = 180012
= 150o
Ṣe o fẹ tẹsiwaju pẹlu iṣe yii?