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**Question 1**
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P, Q and R are subsets of the universal set U. The Venn diagram showing the relationship (P∩Q)∪R is

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**Question 2**
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The locus of the points which is equidistant from the line PQ forms a

**Answer Details**

The locus of points which are equidistant from the line PQ is a line that is perpendicular to PQ and passes through the midpoint of PQ. This is because a point is equidistant from a line if and only if it is on the perpendicular bisector of the line segment connecting any two points on the line. So, the correct option is "perpendicular line to PQ". The points on this line are equidistant from the line PQ, and the line passes through the midpoint of PQ, which is equidistant from both P and Q. A circle with center P or Q would not include all the points equidistant from PQ, and a pair of parallel lines would not be equidistant from PQ.

**Question 3**
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Integrate 1+xx3dx

**Answer Details**

∫1+xx3dx $\int \frac{1+x}{{x}^{3}}\mathrm{d}x$

= ∫(1x3+xx3)dx $\int (\frac{1}{{x}^{3}}+\frac{x}{{x}^{3}})\mathrm{d}x$

= ∫(x−3+x−2)dx $\int ({x}^{-3}+{x}^{-2})\mathrm{d}x$

= −12x2−1x+k

**Question 4**
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3 girls share a number of apples in the ration 5:3:2. If the highest share is 40 apples, find the smallest share

**Answer Details**

Let's call the shares of the three girls "5x", "3x", and "2x", where "x" is a constant factor that will allow us to determine the actual number of apples in each share. We are told that the highest share is 40 apples, which corresponds to the "5x" share. Therefore, we can set up an equation: 5x = 40 Solving for "x", we get: x = 8 Now we can determine the actual number of apples in each share by multiplying the corresponding factor by "x": - The largest share ("5x") has 5 x 8 = 40 apples - The second-largest share ("3x") has 3 x 8 = 24 apples - The smallest share ("2x") has 2 x 8 = 16 apples Therefore, the smallest share has 16 apples, which is option (C).

**Question 5**
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If log104 = 0.6021, evaluate log1041/3

**Question 6**
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P varies jointly as m and u, and varies inversely as q. Given that p = 4, m = 3 and u = 2 and q = 1, find the value of p when m = 6, u = 4 and q =85

**Answer Details**

P ∝
$\propto $ mu, p ∝1q
$\propto \frac{1}{q}$

p = muk ................ (1)

p = 1qk
$\frac{1}{q}k$.... (2)

Combining (1) and (2), we get

P = muqk
$\frac{mu}{q}k$

4 = m×u1k
$\frac{m\times u}{1}k$

giving k = 46=23
$\frac{4}{6}=\frac{2}{3}$

So, P = muq×23=2mu3q
$\frac{mu}{q}\times \frac{2}{3}=\frac{2mu}{3q}$

Hence, P = 2×6×43×85
$\frac{2\times 6\times 4}{3\times \frac{8}{5}}$

P = 2×6×4×53×8
$\frac{2\times 6\times 4\times 5}{3\times 8}$

p = 10

**Question 7**
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In how many ways can a student select 2 subjects from 5 subjects?

**Answer Details**

To calculate the number of ways a student can select 2 subjects from 5 subjects, we can use the formula for combinations: C(n, k) = n! / (k! (n - k)!) where: n = number of subjects (5) k = number of subjects to be selected (2) So, substituting the values in the formula, we get: C(5, 2) = 5! / (2! (5 - 2)!) C(5, 2) = 120 / (2 * 3!) C(5, 2) = 120 / (2 * 6) C(5, 2) = 120 / 12 C(5, 2) = 10 So, a student can select 2 subjects from 5 subjects in 10 different ways.

**Question 8**
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A basket contains 9 apples, 8 bananas and 7 oranges. A fruit is picked from the basket, find the probability that it is neither an apple nor an orange.

**Answer Details**

The probability of picking a fruit that is neither an apple nor an orange is the same as the probability of picking a banana, since bananas are the only remaining fruit in the basket. To calculate the probability, we need to divide the number of bananas in the basket by the total number of fruits in the basket. The total number of fruits in the basket is: 9 (apples) + 8 (bananas) + 7 (oranges) = 24 So the probability of picking a banana is: 8 (bananas) / 24 (total fruits) = 1/3 Therefore, the answer is the third option: 3.

**Question 9**
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The nth term of the progression 42 $\frac{4}{2}$, 73 $\frac{7}{3}$, 104 $\frac{10}{4}$, 135 $\frac{13}{5}$ is ..

**Answer Details**

Using Tn = 3n+1n+1
$\frac{3n+1}{n+1}$,

T1 = 3(1)+1(1)+1=42
$\frac{3(1)+1}{(1)+1}=\frac{4}{2}$

T2 = 3(2)+1(2)+1=73
$\frac{3(2)+1}{(2)+1}=\frac{7}{3}$

T3 = 3(3)+1(3)+1=104

**Question 10**
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If the variance of 3+x, 6, 4, x and 7-x is 4 and the mean is 5, find the standard deviation

**Answer Details**

To find the standard deviation of the given set of numbers, we need to find the variance first, using the formula: Variance = (sum of squares of deviations from the mean) / (number of observations) Let's start by finding the mean of the given set of numbers: Mean = (3 + x + 6 + 4 + 7 - x) / 5 = 20 / 5 = 4 Next, we can substitute this mean into the formula for variance, which gives us: 4 = [(3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2] / 5 Simplifying this equation, we get: 20 = (3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2 20 = 1 + 4 + 0 + (x - 4)^2 + (3 - x)^2 Simplifying further, we get: 15 = 2(x^2 - 8x + 13) Expanding and simplifying this equation, we get: 2x^2 - 16x + 26 = 15 2x^2 - 16x + 11 = 0 We can solve this quadratic equation using the quadratic formula: x = [16 ± sqrt(16^2 - 4(2)(11))] / (2(2)) x = [16 ± sqrt(176)] / 4 x = [16 ± 4sqrt(11)] / 4 x = 4 ± sqrt(11) Now that we have found the possible values of x, we can calculate the standard deviation using the formula: Standard deviation = sqrt(variance) We already know the variance is 4, so: Standard deviation = sqrt(4) = 2 Therefore, the answer is (B) 2.

**Question 11**
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The bar chart above shows the allotment of time(in minutes) per week for selected subjects in a certain school. What is the total time allocated to the six subjects per week?

**Answer Details**

80 + 160 + 200 + 80 = 128 + 72 = 720minutes

**Question 12**
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If tanθ=34 $\mathrm{tan}\theta =\frac{3}{4}$, find the value of sinθ+cosθ $\mathrm{sin}\theta +\mathrm{cos}\theta $.

**Answer Details**

tanθ=oppadj=34 $\mathrm{tan}\theta =\frac{opp}{adj}=\frac{3}{4}$

hyp2=opp2+adj2 $hy{p}^{2}=op{p}^{2}+ad{j}^{2}$

hyp=32+42−−−−−−√ $hyp=\sqrt{{3}^{2}+{4}^{2}}$

= 5

sinθ=35;cosθ=45 $\mathrm{sin}\theta =\frac{3}{5};\mathrm{cos}\theta =\frac{4}{5}$

sinθ+cosθ=35+45 $\mathrm{sin}\theta +\mathrm{cos}\theta =\frac{3}{5}+\frac{4}{5}$

= 75=125

**Question 13**
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If the midpoint of the line PQ is (2,3) and the point P is (-2, 1), find the coordinate of the point Q.

**Answer Details**

The midpoint formula states that the midpoint of a line segment with endpoints (x1,y1) and (x2,y2) is: ((x1+x2)/2, (y1+y2)/2) In this problem, we are given the midpoint of the line segment PQ, which is (2,3), and one endpoint, P, which is (-2,1). Let Q have coordinates (x,y), then we can use the midpoint formula to find the coordinates of Q: ((x + (-2))/2, (y + 1)/2) = (2,3) Simplifying this expression, we get: (x-2, y+1) = (4,6) Now we can solve for x and y by equating the x-coordinates and y-coordinates separately: x - 2 = 4 => x = 6 y + 1 = 6 => y = 5 Therefore, the coordinates of point Q are (6,5). So, the correct answer is (6,5).

**Question 14**
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Convert 2710 to another number in base three

**Answer Details**

To convert the number 27 from base 10 to base 3, we need to divide 27 by 3 repeatedly until the quotient becomes zero. The remainders of each division, read from bottom to top, will be the digits of the number in base 3. 27 divided by 3 is 9 with a remainder of 0. 9 divided by 3 is 3 with a remainder of 0. 3 divided by 3 is 1 with a remainder of 0. 1 divided by 3 is 0 with a remainder of 1. Therefore, the number 27 in base 3 is 1000.

**Question 15**
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Evaluate 3(x + 2) > 6(x + 3)

**Answer Details**

3(x + 2) > 6(x + 3)

3x + 6 > 6x + 18

3x - 6x > 18 - 6

-3x > 12

x < -4

**Question 16**
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What is the probability that an integer x (1≤x≤25) $(1\le x\le 25)$ chosen at random is divisible by both 2 and 3?

**Answer Details**

To find the probability that an integer between 1 and 25 is divisible by both 2 and 3, we need to count the number of integers that satisfy this condition, and divide by the total number of integers between 1 and 25. A number is divisible by both 2 and 3 if and only if it is divisible by their product, which is 6. Therefore, we need to count the number of integers between 1 and 25 that are divisible by 6. To do this, we can start by finding the smallest multiple of 6 that is greater than or equal to 1, which is 6 itself. The next multiples of 6 are 12, 18, and 24, all of which are less than or equal to 25. Therefore, there are 4 integers between 1 and 25 that are divisible by both 2 and 3. The total number of integers between 1 and 25 is 25, so the probability that an integer chosen at random from this range is divisible by both 2 and 3 is 4/25. Therefore, among the given options, the answer is the one represented by the fraction 4/25.

**Question 17**
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The mean of seven numbers is 10. If six of the numbers are 2, 4, 8, 14, 16 and 18, find the mode.

**Answer Details**

To find the mode of the seven numbers, we need to determine the number that appears most frequently among the seven numbers. We are given that the mean (average) of the seven numbers is 10. To find the sum of the seven numbers, we can multiply the mean by 7: mean = (sum of seven numbers) / 7 Rearranging this formula, we get: (sum of seven numbers) = mean x 7 Substituting the given mean of 10, we get: (sum of seven numbers) = 10 x 7 = 70 We are also given that six of the seven numbers are 2, 4, 8, 14, 16, and 18. To find the seventh number, we can subtract the sum of these six numbers from the sum of the seven numbers: (seventh number) = (sum of seven numbers) - (sum of six numbers) Substituting the values we know, we get: (seventh number) = 70 - (2 + 4 + 8 + 14 + 16 + 18) (seventh number) = 8 So the seven numbers are: 2, 4, 8, 14, 16, 18, and 8. To find the mode, we need to determine which number appears most frequently among these seven numbers. We can see that the number 8 appears twice, while all the other numbers appear only once. Therefore, the mode of these seven numbers is 8.

**Question 18**
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Calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest

**Answer Details**

To calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest, we can use the formula: time = (interest / principal) / rate where: interest = ₦600 principal = ₦3000 rate = 8% So, substituting the values in the formula, we get: time = (600 / 3000) / 0.08 time = (0.2) / 0.08 time = 2.5 years So, it takes approximately 2.5 years for ₦3000 to earn ₦600 if invested at 8% simple interest.

**Question 19**
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Score345678910Frequency10752311 $\begin{array}{ccccccccc}\text{Score}& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{Frequency}& 1& 0& 7& 5& 2& 3& 1& 1\end{array}$

The table above shows the scores of 20 students in further mathematics test. What is the range of the distribution?

**Answer Details**

The range of a distribution is the difference between the highest and lowest values in the distribution. In this case, the highest score is 10 and the lowest score is 3. Therefore, the range of the distribution is: 10 - 3 = 7 So the correct option is 7.

**Question 20**
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Simplify 3−5n91−n×27n+1

**Answer Details**

3−5n91−n×27n+1
$\frac{{3}^{-5n}}{{9}^{1-n}}\times {27}^{n+1}$

3−5n32(1−n)×33(n+1)
$\frac{{3}^{-5n}}{{3}^{2(1-n)}}\times {3}^{3(n+1)}$

3−5n÷32(1−n)×33(n+1)
${3}^{-5n}\xf7{3}^{2(1-n)}\times {3}^{3(n+1)}$

3−5n−2(1−n)+3(n+1) ${3}^{-5n-2(1-n)+3(n+1)}$

3−5n−2+2n+3n+3
${3}^{-5n-2+2n+3n+3}$

3−5n+5n+3−2
${3}^{-5n+5n+3-2}$

31
${3}^{1}$

= 3

**Question 21**
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A chord of a circle subtends an angle of 120° at the centre of a circle of diameter 43–√cm $4\sqrt{3}cm$. Calculate the area of the major sector.

**Answer Details**

Angle of major sector = 360° - 120° = 240°

Area of major sector : θ360×πr2 $\frac{\theta}{360}\times \pi {r}^{2}$

r = 43√2=23–√cm $\frac{4\sqrt{3}}{2}=2\sqrt{3}cm$

Area : 240360×π×(23–√)2 $\frac{240}{360}\times \pi \times (2\sqrt{3}{)}^{2}$

= 8πcm2

**Question 22**
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The remainder when 6p3 - p2 - 47p + 30 is divided by p - 3 is

**Answer Details**

Let f(p) = 6p3 - p2 - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

**Question 23**
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The graph is shown is correctly represented by

**Answer Details**

The graph represented by the given equation is y = x² - x - 2. To understand this, we first need to understand what a quadratic equation is. A quadratic equation is a polynomial equation of the second degree, meaning it has an exponent of 2. In other words, it's an equation where the highest exponent of the variable is 2. The equation y = x² - x - 2 is a quadratic equation, where y is the dependent variable and x is the independent variable. The graph of a quadratic equation is a parabola, which is a symmetrical U-shaped curve. Looking at the given equation, we can see that the coefficient of x² is positive, which means that the parabola will open upwards. The coefficient of x is negative, which means that the vertex of the parabola will be to the right of the y-axis. The constant term is -2, which means that the vertex will be two units below the y-axis. So, the graph represented by the given equation is a parabola that opens upwards, with its vertex at (0.5, -2.25) and its axis of symmetry being the line x = 0.5. Therefore, the correct option is the one that represents the equation y = x² - x - 2.

**Question 24**
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Find the length of a chord which subtends an angle of 90° at the centre of a circle whose radius is 8 cm.

**Answer Details**

Length of chord = 2rsin(θ2) $2r\mathrm{sin}(\frac{\theta}{2})$

= 2×8×sin(902) $2\times 8\times \mathrm{sin}(\frac{90}{2})$

= 16×2√2 $16\times \frac{\sqrt{2}}{2}$

= 82–√cm

**Question 25**
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Find the inverse ∣∣∣5364