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**Question 1**
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The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher

**Answer Details**

Let the age of the teacher be T. According to the problem, the mean age of the 10 pupils is 16. Therefore, the sum of their ages is 10 x 16 = 160. When the teacher's age is added to it, the sum becomes 160 + T. Also, it is given that the new mean age after adding the teacher's age is 19. So, we have: (160 + T) / 11 = 19 Multiplying both sides by 11, we get: 160 + T = 209 Subtracting 160 from both sides, we get: T = 49 Therefore, the age of the teacher is 49.

**Question 2**
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If the function f is defined by f(x + 2) = 2x2 + 7x = 5, find f(-1)

**Question 3**
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A binary operation ∗ $\ast $ is defined on a set of real numbers by x ∗ $\ast $ y = xy for all real values of x and y. If x ∗ $\ast $ 2 = x. Find the possible values of x

**Answer Details**

x ∗
$\ast $ y = xy

x ∗
$\ast $ 2 = x2

x ∗
$\ast $ 2 = x

∴ x2 - x = 0

x(x - 1) = 0

x = 0 or 1

**Question 4**
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The chances of three independent events X, Y, Z occurring are 12 $\frac{1}{2}$, 23 $\frac{2}{3}$, 14 $\frac{1}{4}$ respectively. What are the chances of Y and Z only occurring?

**Answer Details**

Chance of x = 12
$\frac{1}{2}$

Change of Y = 23
$\frac{2}{3}$

Chance of Z = 14
$\frac{1}{4}$

Chance of Y and Z only occurring

= Pr (Y ? Z ? Xc)

where Xc = 1 - Pr(X)

1 = 12
$\frac{1}{2}$ = 112
$\frac{1}{2}$

= Pr(Y) x Pr(Z) x Pr(Xc)

= 23
$\frac{2}{3}$ x 14
$\frac{1}{4}$ x 12
$\frac{1}{2}$

= 112

**Question 5**
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Simplify 1p $\frac{1}{p}$ - 1q $\frac{1}{q}$ + pq $\frac{p}{q}$ - qp

**Answer Details**

1p
$\frac{1}{p}$ - 1q
$\frac{1}{q}$ + pq
$\frac{p}{q}$ - qp
$\frac{q}{p}$ = q?ppq
$\frac{q?p}{pq}$ + p2?q2pq
$\frac{{p}^{2}?{q}^{2}}{pq}$

q?ppq
$\frac{q?p}{pq}$ x pqp2q2
$\frac{pq}{{p}^{2}{q}^{2}}$ = q?pp2q2
$\frac{q?p}{{p}^{2}{q}^{2}}$

?(p?q)(p+q)(p?q)
$\frac{?(p?q)}{(p+q)(p?q)}$

= ?1p+q

**Question 6**
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Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of 2212 $\frac{1}{2}$o at the centre of the circle

**Answer Details**

Diameter = 8cm

∴ Radius = 4cm

Length of arc = θ360
$\frac{\theta}{360}$ x 2 π
$\pi $r but Q = 2212
$\frac{1}{2}$

∴ Length 2212360
$\frac{22\frac{1}{2}}{360}$ x 2 x π
$\pi $ x 4

= 2212×8π360
$\frac{22\frac{1}{2}\times 8\pi}{360}$

= 180360
$\frac{180}{360}$

= π2

**Question 7**
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Make x the subject of the relation 1+ax1−ax $\frac{1+ax}{1-ax}$ = pq

**Answer Details**

To make x the subject of the given relation, we need to isolate x on one side of the equation. First, we can simplify the expression by finding the common denominator of the two fractions in the numerator: (1 + ax)/(1 - ax) = (1 - ax + ax + a^2x)/(1 - ax) = (1 + a^2x)/(1 - ax) Now we can cross-multiply to get rid of the denominator: (1 + a^2x) = pq - axpq Next, we can move the term with x to the left-hand side of the equation: a^2x + axpq = pq - 1 Finally, we can factor out x from the two terms on the left-hand side: x(a^2 + apq) = pq - 1 Therefore, we can isolate x by dividing both sides by (a^2 + apq): x = (pq - 1)/(a^2 + apq) So the correct option is: p-q / a(p+q) which is not among the given options.

**Question 8**
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If 2log3 y + log3 x2 = 4, then y is

**Answer Details**

The given equation is 2log3 y + log3 x2 = 4. We can simplify this equation using the laws of logarithms. First, we can use the power rule of logarithms to rewrite x2 as (x)2. Then, we can use the product rule of logarithms to combine the two logarithms on the left-hand side of the equation: 2log3 y + log3 (x)2 = log3 y2 + log3 (x)2 Now, we can use the sum rule of logarithms to combine the two logarithms on the right-hand side of the equation: log3 y2(x)2 = log3 (x2y2) We can now rewrite the original equation as: log3 (x2y2) = 4 Using the definition of logarithms, we know that log3 (x2y2) = 4 is equivalent to 3^4 = x2y2. Therefore, we have: x2y2 = 81 Taking the square root of both sides, we get: xy = ±9 Since y is a positive real number, we can discard the negative solution. Therefore, we have: xy = 9 Finally, we can solve for y: y = 9/x So, the answer is "±9/x".

**Question 9**
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The angle between latitudes 30oS and 13oN is

**Answer Details**

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.

∴ Total angle difference = (30 + 13) = 43o

**Question 10**
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Which of the following is a factor of 15 + 7x - 2x2

**Answer Details**

Factorize 15 + 7x - 2x2

(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0

∴ (5 - x)(3 + 2x) = 0

x = 5 or x = -32
$\frac{3}{2}$

Since 5 is a root, then (x - 5) is a factor

**Question 11**
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Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR

**Answer Details**

QPS - QRK = 31o

QRK + RKQ + KQR = 180

31 + 42 + KQR = 180o

KQR = 180 - 73 = 107o

**Question 12**
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Evaluate (x + 1x $\frac{1}{x}$ + 1)2 - (x + 1x $\frac{1}{x}$ + 1)2

**Answer Details**

(x + 1x
$\frac{1}{x}$ + 1)2 - (x + 1x
$\frac{1}{x}$ + 1)2

= (x + 1x
$\frac{1}{x}$ + 1 + x + −1x
$\frac{-1}{x}$ - 1)(x - 1x
$\frac{1}{x}$ + 1 - x + 1x
$\frac{1}{x}$ + 1)

= (2x) (2 + 2x
$\frac{2}{x}$) = 2x x 2(1 + 1x
$\frac{1}{x}$)

4x (1 + 1x
$\frac{1}{x}$) = 4x + 4

= 4(1 + x)

**Question 13**
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If sin θ $\theta $ = cos θ $\theta $, find θ $\theta $ between 0o and 360o

**Answer Details**

sin θ
$\theta $ = cos θ
$\theta $ 0 ≤
$\le $ θ
$\theta $ ≤
$\le $ 360o

The acute angle where sin θ
$\theta $ = cos θ
$\theta $ = 45o

But at the fourth Quadrant Cos θ
$\theta $ = +ve

at the 4th quadrant, value with respect to Q is

(360 - Q) where Q = acute angle

(360 - 45) = 315o

The two solution are 45o, 315o

**Question 14**
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If ₦225.00 yields ₦27.00 in x years simple interest at the rate of 4% per annum, find x

**Answer Details**

To solve this problem, we use the formula for simple interest: I = P*r*t Where: I = Interest earned P = Principal amount r = Rate of interest per year (as a decimal) t = Time period in years In this case, we are given: P = ₦225.00 (the principal amount) I = ₦27.00 (the interest earned) r = 4% = 0.04 (the rate of interest per year) Substituting these values into the formula, we get: 27 = 225 * 0.04 * t Simplifying the equation, we have: 27 = 9t Dividing both sides by 9, we get: t = 3 Therefore, the answer is 3 years. Explanation: The interest earned on the principal amount of ₦225.00 is ₦27.00. This means that for every year, the interest earned is 27/225 = 0.12 or 12% of the principal amount. We know that the rate of interest per year is 4%, which is much lower than the interest earned per year of 12%. This means that it will take more than 1 year for the interest earned to reach the rate of interest. In fact, it takes 3 years for the interest earned to reach the rate of interest of 4%, which is why the answer is 3.

**Question 15**
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ClassFrequency1−526−10411−15516−20221−25326−30231−35136−401 $\begin{array}{cc}Class& Frequency\\ 1-5& 2\\ 6-10& 4\\ 11-15& 5\\ 16-20& 2\\ 21-25& 3\\ 26-30& 2\\ 31-35& 1\\ 36-40& 1\end{array}$

Find the median of the observation in the table given.

**Answer Details**

Median = L1 + (Effm
$\frac{Ef}{fm}$) - fo

∑f2 $\frac{\sum f}{2}$

= 202 $\frac{20}{2}$

= 10, L1 = 10.5, fo = 6, fm = 5

Median = 10.5 + (10−6)5 $\frac{(10-6)}{5}$5

= 10.5 + 4

= 14.5

∑f2 $\frac{\sum f}{2}$

= 202 $\frac{20}{2}$

= 10, L1 = 10.5, fo = 6, fm = 5

Median = 10.5 + (10−6)5 $\frac{(10-6)}{5}$5

= 10.5 + 4

= 14.5

**Question 16**
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Solve the following equation (3x - 2)(5x - 4) = (3x - 2)2

**Answer Details**

(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6

= 9x2 = 12x + 4

6x2 - 10x + 4 = 0

6x2 - 6x - 4x + 4 = 0

6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0

x = 1 or 23

**Question 17**
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The following marks were obtained by twenty students in an examination: 53, 30, 70, 84, 59, 43, 90, 20, 78, 48, 44, 60, 81, 73, 50, 37, 67, 68, 64, 52. Find the numbers of students who scored at least 50 marks

**Answer Details**

To find the number of students who scored at least 50 marks, we need to count the number of marks that are greater than or equal to 50. Looking at the marks obtained by the students, we can see that there are 14 marks that are greater than or equal to 50. Therefore, the number of students who scored at least 50 marks is 14. We can count these marks by going through the list one by one, or by using a table or chart to organize the data. It's important to pay close attention to the wording of the question and to make sure we're answering the question being asked. In this case, the question is asking for the number of students, not the number of marks.

**Question 18**
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In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST

**Answer Details**

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = √167 $167$

= 12.92 = 12.9cm

**Question 19**
**Report**

In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = 32∘
$?$. Find PSQ

**Answer Details**

< PSO = 12 $\frac{1}{2}$ < SOQ = 12 $\frac{1}{2}$(180) = 90∘ $\circ $

< RTS = < PQS = 32∘ $\circ $ (Alternative angle)

< PSQ = 90 - < PSQ = 90∘ $\circ $ - 32∘ $\circ $

= 58∘

**Question 20**
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In the figure, the area of the square of the square PQRS is 100cm2. If the ratio of the area of the square TUYS to the area of the area of the square XQVU is 1 : 16, Find YR

**Answer Details**

Since area of square PQRS = 100cm2

each lenght = 10cm

Also TUYS : XQVU = 1 : 16

lengths are in ratio 1 : 4, hence TU : UV = 1: 4

Let TU = x

UV = 1: 4

hence TV = x + 4x = 5x = 10cm

x = 2cm

TU = 2cm

UV = 8cm

But TU = SY and UV = YR

YR = 8cm

**Question 21**
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Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0

**Answer Details**

x2 + y - 5 = 0.....(i)

y - 7x + 3 = 0.........(ii)

y = 7x - 3, substituting the value of y in equation (i)

x2 + (7x - 3) - 5 = 0

x2 + 7x + 3 = 0

(x + 8)(x - 1) = 0

x = -8 or 1

**Question 22**
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If sin x = cos 50o, then x equals

**Answer Details**

We know that sin x = cos 50°. To solve for x, we need to find the angle whose sine is equal to cos 50°. We can use the trigonometric identity that sin (90° - θ) = cos θ. So, sin x = cos 50° can be rewritten as sin x = sin (90° - 50°). Using the identity, we have: sin x = sin 40° Now, we need to find the angle x whose sine is equal to sin 40°. Since sine is a periodic function, there are multiple angles whose sine is equal to a given value. One such angle is x = 40°. However, sine is also negative in the third and fourth quadrants. In the third quadrant, x = 180° - 40° = 140° and in the fourth quadrant, x = 360° - 40° = 320° also satisfy the equation sin x = sin 40°. However, since x has to be between 0° and 360°, we can eliminate the30°. Therefore, the possible values of x are 40°, 140°, and 320°. However, since we know that sin x = cos 50° and cos is positive in the first quadrant, x cannot be in the third or fourth quadrants. Therefore, the only possible value of x is x = 40°. Hence, the answer is x = 40°.

**Question 23**
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If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

**Question 24**
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PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

**Question 25**
**Report**

The bar chart shows the distribution of marks in a class test. How many students took the test?

**Answer Details**

Number of students that took the test = ∑f
$\sum f$. Where f
$f$ is the frequencies

= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 0 + 1 + 0 + 2 = 17

**Question 26**
**Report**

simplify 1√3−2 $\frac{1}{\surd 3-2}$ - 1√3+2

**Answer Details**

1√3−2
$\frac{1}{\surd 3-2}$ - 1√3+2
$\frac{1}{\surd 3+2}$

L.C.M = (3- 2) (3 + 2)

∴ 1√3−2
$\frac{1}{\sqrt{3-2}}$ - 1√3−2
$\frac{1}{\sqrt{3-2}}$ = √3+2−√3−2√3−2+√3−2
$\frac{\sqrt{3+2}-\sqrt{3-2}}{\sqrt{3-2}+\sqrt{3-2}}$

√3+2−√3+23−2√3+2√3−4
$\frac{\surd 3+2-\surd 3+2}{3-2\surd 3+2\surd 3-4}$ = 43−2
$\frac{4}{3-2}$

= 4−1
$\frac{4}{-1}$

= -4

**Question 27**
**Report**

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

**Answer Details**

2x + 3 ≠
$\ne $ 2x - 3 for any value of x

∴ for the △
$\u25b3$ to be isosceles, either

2x - 3 = x + 3 or 2x + 3 = x + 3

solve the two equations we arrive at

x = 6 or x = 0

When x = 6, the sides are 9, 15, 9

When x = 0, the sides are 3, 4, -3 since lengths of a △
$\u25b3$can never be negative then the value of x = 6

**Question 28**
**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**

To solve this problem, we need to first determine the set of possible numbers that could be selected at random between 20 and 30. Since the problem states that both 20 and 30 are inclusive, the set of possible numbers is: {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} To find the probability that the selected number is prime, we need to determine the number of prime numbers in the set of possible numbers, and then divide that by the total number of possible numbers. Prime numbers are numbers that are only divisible by 1 and themselves. The prime numbers in the set of possible numbers are: { 23, 29 } Therefore, the probability that the selected number is prime is 2/11 or approximately 0.18.

**Question 29**
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A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

**Answer Details**