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Question 1 Report
Find the standard derivation of the following data -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Answer Details
x = ∑xN
$\frac{\sum x}{N}$
= 011
$\frac{0}{11}$
= 0
x(x−x)(x−x)2−5−525−4−416−3−39−2−24−1−1100011122433944165525110
$\begin{array}{ccc}x& (x-x)& (x-x{)}^{2}\\ -5& -5& 25\\ -4& -4& 16\\ -3& -3& 9\\ -2& -2& 4\\ -1& -1& 1\\ 0& 0& 0\\ 1& 1& 1\\ 2& 2& 4\\ 3& 3& 9\\ 4& 4& 16\\ 5& 5& 25\\ & & 110\end{array}$
S.D = √∑(x−x)2∑f
$\sqrt{\frac{\sum (x-x{)}^{2}}{\sum f}}$
= √11011
$\sqrt{\frac{110}{11}}$
= √10
Question 2 Report
The angle of elevation of a building from a measuring instrument placed on the ground is 30o. If the building is 40m high, how far is the instrument from the foot of the building?
Answer Details
40x
$\frac{40}{x}$ = tan 30o
x = 40tan36
$\frac{40}{tan36}$
= 401√3
$\frac{40}{1\sqrt{3}}$
= 40√3m
Question 3 Report
Make F the subject of the formula t = √v1f+1g
Answer Details
t = √v1f+1g
$\sqrt{\frac{v}{\frac{1}{f}+\frac{1}{g}}}$
t2 = v1f+1g
$\frac{v}{\frac{1}{f}+\frac{1}{g}}$
= vfgftg
$\frac{vfg}{ftg}$
1f+1g
$\frac{1}{f}+\frac{1}{g}$ = vt2
$\frac{v}{{t}^{2}}$
= (g + f)t2 = vfg
gt2 = vfg - ft2
gt2 = f(vg - t2)
f = gt2gv−t2
Question 4 Report
Each of the base angles of a isosceles triangle is 59o and the verticles of the triangle lie on a circle. Determine rhe angle which the base of the triangle subtends at the centre of the circle.
Question 5 Report
Find the value of k if 5+2r(r+1)(r−2) $\frac{5+2r}{(r+1)(r-2)}$ expressed in partial fraction is kr−2 $\frac{k}{r-2}$ + Lr+1 $\frac{L}{r+1}$ where K and L are constants
Answer Details
5 + 2r = k(r + 1) + L(r - 2)
but r - 2 = 0 and r = 2
9 = 3k
k = 3
Question 6 Report
In the figure, XYZ is a triangle with XY = 5cm, XZ = 2cm and XZ is produced to E making the angle < YZE = 150∘ $\circ $. If the angle XYZ = θ $\theta $, calculate the value of sinθ
Answer Details
5sin36o=2sinθ=512=2sinθ $\frac{5}{\mathrm{sin}{36}^{o}}=\frac{2}{\mathrm{sin}\theta}=\frac{5}{\frac{1}{2}}=\frac{2}{\mathrm{sin}\theta}$
10 = 2sinθ $\frac{2}{\mathrm{sin}\theta}$
sinθ $\theta $ = 210 $\frac{2}{10}$
= 15
Question 7 Report
Two binary operations ∗ $\ast $ and ⊕ $\oplus $ are defines as m ∗ $\ast $ n = mn - n - 1 and m ⊕ $\oplus $ n = mn + n - 2 for all real numbers m, n.
Find the value of 3 ⊕ $\oplus $ (4 ∗ $\ast $ 50)
Answer Details
m ∗
$\ast $ n = mn - n - 1, m ⊕
$\oplus $ n = mn + n - 2
3 ⊕
$\oplus $ (4 ∗
$\ast $ 5) = 3 ⊕
$\oplus $ (4 x 5 - 5 - 1) = 3 ⊕
$\oplus $ 14
3 ⊕
$\oplus $ 14 = 3 x 14 + 14 - 2
= 54
Question 8 Report
The angle between the positive horizontal axis and a given line is 135o. Find the equation of the line if it passes through the point (2,3)
Answer Details
The problem requires finding the equation of a line that passes through a given point (2,3) and makes an angle of 135 degrees with the positive x-axis. To solve this problem, we need to first find the slope of the line. Since the given angle is measured from the positive x-axis and is 135 degrees, we know that the angle made with the negative x-axis is 45 degrees. Therefore, the slope of the line is the tangent of 45 degrees, which is 1. Now that we have the slope, we can use the point-slope form of the equation of a line to find the equation of the line. The point-slope form is: y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point on the line. Plugging in the values we have, we get: y - 3 = 1(x - 2) Simplifying this equation gives us: y - x + 3 = 0 This equation is in the form of y = mx + b, where m is the slope and b is the y-intercept. We can see that the slope is 1, which we found earlier, and the y-intercept is 3. Therefore, the equation of the line that passes through the point (2,3) and makes an angle of 135 degrees with the positive x-axis is: y - x + 3 = 0 which is equivalent to: y = x - 3 So, the answer is neither (a) nor (b), but is (c) x + y = 5.
Question 9 Report
Given that 1/2 log10 P = 1, find the value Of P
Question 10 Report
An arc of a circle subtends an angle 70o at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.(π $\pi $ = 227 $\frac{22}{7}$)
Question 11 Report
Find the simple interest rate percent per annum at which ₦1,000 accumulates to ₦1,240 in 3 years
Answer Details
The formula for simple interest is: I = P * r * t Where: I = Interest P = Principal (initial amount borrowed or invested) r = Interest rate per annum (as a decimal) t = Time in years From the given information: P = ₦1,000 I = ₦1,240 - ₦1,000 = ₦240 t = 3 years Substituting these values into the formula, we get: 240 = 1000 * r * 3 Simplifying the equation, we get: r = 240 / (1000 * 3) = 0.08 Converting to a percentage, we get: r = 0.08 * 100% = 8% Therefore, the simple interest rate percent per annum is 8%. Answer: 8%
Question 12 Report
In an examination, the result of a certain school is as shown in the histogram above. How many candidates did the school present?
Answer Details
3 + 5 + 8 + 1 + 2 = 19
Question 14 Report
A point P moves so that is equidistant from point L and M. If LM is 6cm, find the distance of P from LM when P is 10cm from L
Answer Details
p from LM = √102
$\sqrt{{10}^{2}}$ - 82
= √36
$36$ = 6cm
Question 15 Report
Given that loga2 = 0.693 and loga3 = 1.097, find loga 13.5
Answer Details
loga 13.5 = loga 272
$\frac{27}{2}$
= 3loga 3 - log2a
= 3 x 1.097 - 0.693
= 2.598
Question 17 Report
Age202530354045Number of people351123 $\begin{array}{ccccccc}\text{Age}& 20& 25& 30& 35& 40& 45\\ \text{Number of people}& 3& 5& 1& 1& 2& 3\end{array}$
Find the median age of the frequency distribution in the table above.
Answer Details
To find the median age, we need to arrange the ages in ascending order and then find the middle value. However, since the table only gives us frequency counts for each age, we first need to construct a cumulative frequency distribution. Age | Number of People | Cumulative Frequency --- | --- | --- 20 | 3 | 3 25 | 5 | 8 30 | 1 | 9 35 | 1 | 10 40 | 2 | 12 45 | 3 | 15 The total number of people is 15, which is an odd number, so the median is simply the value that corresponds to the middle person. Since the cumulative frequency for 8 people is at age 25 and the cumulative frequency for 9 people is at age 30, the median age is 25. Therefore, the answer is: - 25
Question 18 Report
A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime
Answer Details
Possible outcomes are 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30. Prime numbers has only two factors
itself and 1
The prime numbers among the group are 23, 29. Probability of choosing a prime number
= Number of primeNo. of total Possible Outcomes
$\frac{\text{Number of prime}}{\text{No. of total Possible Outcomes}}$
= 211
Question 19 Report
Find the distance between the point Q (4,3) and the point common to the lines 2x - y = 4 and x + y = 2
Answer Details
To find the distance between point Q and the point common to the lines 2x - y = 4 and x + y = 2, we need to first find the coordinates of the point of intersection of the two lines. We can solve the system of equations: 2x - y = 4 x + y = 2 by either substitution or elimination to get the coordinates of the point of intersection, which is (1, 1). Then, we can use the distance formula to find the distance between point Q (4, 3) and (1, 1): d = √[(4 - 1)² + (3 - 1)²] = √[9 + 4] = √13. Therefore, the answer is √13.
Question 20 Report
Integrate 1x $\frac{1}{x}$ + cos x with respect to x
Question 22 Report
Simplify 2?3+3?53?5?2?3
Answer Details
(2?3+3?5)(3?5+2?3)(3?5?2?3)(3?5?2?3)
$\frac{(2\sqrt{3}+3\sqrt{5})(3\sqrt{5}+2\sqrt{3})}{(3\sqrt{5}?2\sqrt{3})(3\sqrt{5}?2\sqrt{3})}$
= 5+12?1533
$\frac{5+12\sqrt{15}}{33}$
=19+4?1511
Question 23 Report
A chord of a circle of a diameter 42cm subtends an angle of 60o at the centre of the circle. Find the length of the mirror arc
Question 24 Report
Find the difference between the range and the variance of the following set of numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7 where ∑d2 $\sum {d}^{2}$ = 60
Question 25 Report
If U = (s, p, i, e, n, d, o, u, r), X = (s, p, e, n, d) Y = (s, e, n, o, u), Z = (p, n, o, u, r) find X ∩ Y ∪ Z
Question 26 Report
The figure shows circles of radii 3cm and 2cm with centres at X and Y respectively. The circles have a transverse common tangent of length 25cm. Calculate XY.
Question 27 Report
The table above shows that the scores of a group of students in a test. If the average score is 3.5, find the value of x
Answer Details
mean = | 60 + 5x |
18 + x | |
3.5 = | 60 + 5x |
18 + x | |
7 = | 60 + 5x |
2 | 18 + x |
7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2
Question 28 Report
A cone with the sector angle of 45o is cut out of a circle of radius of the cone.
Question 29 Report
Find the minimum value of X2 - 3x + 2 for all real values of x
Answer Details
y = X2 - 3x + 2, dydx
$\frac{dy}{dx}$ = 2x - 3
at turning pt, dydx
$\frac{dy}{dx}$ = 0
∴ 2x - 3 = 0
∴ x = 32
$\frac{3}{2}$
d2ydx2
$\frac{{d}^{2}y}{d{x}^{2}}$ = ddx
$\frac{d}{dx}$(ddx
$\frac{d}{dx}$)
= 270
∴ ymin = 232
$\frac{3}{2}$ - 332
$\frac{3}{2}$ + 2
= 94
$\frac{9}{4}$ - 92
$\frac{9}{2}$ + 2
= -14
Question 30 Report
If X ∗ $\ast $ Y = X + Y - XY, find x when (x ∗ $\ast $ 2) + (x ∗ $\ast $ 3) = 68
Answer Details
x ∗
$\ast $ y = x + y - xy
(x ∗
$\ast $ 2) + (x ∗
$\ast $ 3) = 68
= x + 2 - 2x + x + 3 - 3x
= 86
3x = 63
x = -21
Question 31 Report
Differentiate 6x3−5x2+13x2 $\frac{6{x}^{3}-5{x}^{2}+1}{3{x}^{2}}$ with respect to x
Answer Details
6x3−5x2+13x2
$\frac{6{x}^{3}-5{x}^{2}+1}{3{x}^{2}}$
let y = 3x2
y = 6x33x2
$\frac{6{x}^{3}}{3{x}^{2}}$ - 6x23x2
$\frac{6{x}^{2}}{3{x}^{2}}$ + 13x2
$\frac{1}{3{x}^{2}}$
Y = 2x - 53
$\frac{5}{3}$ + 13x2
$\frac{1}{3{x}^{2}}$
dydx
$\frac{dy}{dx}$ = 2 + 13
$\frac{1}{3}$(-2)x-3
= 2 - 23x3
Question 32 Report
Sn is the sum of the first n terms of a series given by Sn = n2n - 1. Find the nth term
Answer Details
The given series is: 1 + 3 + 5 + 7 + ... + (2n - 1) Notice that each term in the series is an odd number, and the difference between consecutive terms is 2. To find the nth term, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)d where a1 is the first term, d is the common difference, and n is the term number. In this case, a1 = 1 and d = 2, so we have: an = 1 + (n - 1)2 an = 2n - 1 Therefore, the answer is (D) 2n - 1.
Question 33 Report
Find the value of (0.006)3 + (0.004)3 in standard form