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**Question 1**
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A binary operation * is defined by x * y = xy. If x * 2 = 12 - x, find the possible values of x

**Answer Details**

x * y = xy

x * 2 = 12 - x

Thus by comparison,

x = x, y = 2

But x * y = x * 2

xy = 12 - x

x2 = 12 - x

x2 + x - 12 = 0

x2 + 4x - 3x - 12 = 0

x(x + 4) - 3(x + 4) = 0

(x - 3)(x + 4) = 0

x - 3 = 0 or x + 4 = 0

So x = 3 or x = -4

**Question 2**
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Find the equation of the straight line through (-2, 3) and perpendicular to 4x + 3y - 5 = 0

**Answer Details**

4x + 3y - 5 = 0 (given)

The equation of the line perpendicular to the given line takes the form 3x - 4y = k

Thus, substitution x = -2 and y = 3 in 3x - 4y = k gives;

3(-2) - 4(3) = k

-6 - 12 = k

k = -18

Hence the required equation is 3x - 4y = -18

3x - 4y + 18 = 0

**Question 3**
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A man donates 10% of his monthly net earnings to his church. If it amounts to ₦4,500, what is his net monthly income?

**Answer Details**

We can begin by setting up an equation to represent the given situation. Let x be the man's net monthly income. Then, we know that he donates 10% of his net monthly income to his church, which amounts to ₦4,500. Mathematically, we can express this as: 10% of x = ₦4,500 To solve for x, we need to isolate the variable on one side of the equation. We can do this by dividing both sides of the equation by 10%, which is equivalent to multiplying both sides by 10/100 or 0.1: 10% of x ÷ 10% = ₦4,500 ÷ 10% x = ₦4,500 ÷ 0.1 x = ₦45,000 Therefore, the man's net monthly income is ₦45,000. Answer: ₦45,000

**Question 4**
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The gradient of a line joining (x,4) and (1,2) is 12 $\frac{1}{2}$. Find the value of x

**Answer Details**

Gradient m=y2−y1x2−x1
$\text{Gradient m}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

12=2−41−x $\frac{1}{2}=\frac{2-4}{1-x}$

1 - x = 2(2 - 4)

1 - x = 4 - 8

1 - x = -4

-x = -4 - 1

x = 5

12=2−41−x $\frac{1}{2}=\frac{2-4}{1-x}$

1 - x = 2(2 - 4)

1 - x = 4 - 8

1 - x = -4

-x = -4 - 1

x = 5

**Question 5**
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If log7.5 = 0.8751, evaluate 2 log75 + log750

**Answer Details**

If log 7.5 = 0.8751

Then 2log75 + log750

= 2(1.8751) + 2.8751

= 3.7502 + 2.8751

= 6.6253

**Question 6**
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What is the solution of x-5/x+3<-1?

**Question 7**
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Find the value of x in the figure above.

**Answer Details**

The figure above is a right triangle with sides x, 3 cm, and x√3 cm. Since it is a right triangle, we can use the Pythagorean theorem to find the value of x. The theorem states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the longest side, which is the hypotenuse. So, we have: x^2 + (3)^2 = (x√3)^2 Expanding the square on the right side: x^2 + 9 = x^2 * 3 Simplifying the equation: x^2 + 9 = 3x^2 Solving for x: 2x^2 = 9 x^2 = 4.5 Taking the square root of both sides: x = ±√(4.5) Since x has to be positive, we choose the positive square root: x = √(4.5) = 2√3 Finally, multiplying by 10: x = 10√3 So, the value of x in the figure is 10√3 cm.

**Question 8**
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The pie chart above shows the monthly distribution of a man's salary on food items. If he spent ₦8,000 on rice, how much did he spent on yam?

**Answer Details**

The man's salary was divided into four food items: Rice, Yam, Beans, and Others. The chart shows that rice takes up 20% of his salary, and yam takes up 40% of his salary. Since rice takes up 20% of his salary, and he spent ₦8,000 on it, we can calculate the total salary of the man by dividing his spend on rice by 20%. ₦8,000 / 20% = ₦40,000 Since the total salary is ₦40,000 and yam takes up 40% of the salary, we can calculate how much he spent on yam by multiplying the total salary by 40%. ₦40,000 * 40% = ₦16,000 Therefore, the man spent ₦16,000 on yam.

**Question 9**
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*y* varies directly as *w2*. When y = 8, w = 2. Find y when w = 3

**Answer Details**

The given statement "y varies directly as w^2" can be written as an equation: y = k w^2 where k is a constant of proportionality. We are also given that when y = 8, w = 2. We can use this information to solve for k: 8 = k (2^2) 8 = 4k k = 2 Now that we know the value of k, we can use the equation to find y when w = 3: y = 2 (3^2) y = 18 Therefore, when w = 3, y is equal to 18. In other words, the problem is asking us to find the value of y when the value of w is changed from 2 to 3, given that y varies directly with w^2. We can use the equation y = k w^2 and the given information to solve for the constant of proportionality k. Once we have found k, we can use the equation to find y when w = 3.

**Question 10**
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How many sides has a regular polygon whose interior angle is 135o

**Answer Details**

The formula to find the interior angle of a regular polygon is: Interior angle = (n - 2) × 180° / n Where "n" is the number of sides of the polygon. We are given that the interior angle of the regular polygon is 135°, so we can substitute this value into the formula and solve for "n": 135 = (n - 2) × 180° / n Multiplying both sides by "n": 135n = (n - 2) × 180° Distributing on the right-hand side: 135n = 180n - 360° Subtracting 135n from both sides: 0 = 45n - 360° Adding 360° to both sides: 360° = 45n Dividing both sides by 45: 8 = n So the regular polygon has 8 sides. Looking at the given answer options, we see that the answer is (D) 8.

**Question 11**
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The 4th term of an A.P is 13 while the 10th term is 31. Find the 21st term

**Answer Details**

Let's begin by recalling the formula for the nth term of an arithmetic progression (A.P): a_n = a_1 + (n - 1)d where a_n is the nth term of the A.P, a_1 is the first term, n is the number of the term, and d is the common difference between consecutive terms. We are given that the 4th term of the A.P is 13, so we can substitute these values into the formula to get: a_4 = a_1 + (4 - 1)d = 13 Simplifying this equation, we get: a_1 + 3d = 13 ---(1) We are also given that the 10th term of the A.P is 31, so we can use the formula again to get: a_10 = a_1 + (10 - 1)d = 31 Simplifying this equation, we get: a_1 + 9d = 31 ---(2) Now we need to solve for a_1 and d. We can do this by subtracting equation (1) from equation (2) to eliminate a_1: 6d = 18 d = 3 Substituting this value of d into equation (1), we get: a_1 + 3(3) = 13 a_1 = 4 So, the first term of the A.P is 4 and the common difference is 3. Now we can use the formula again to find the 21st term of the A.P: a_21 = a_1 + (21 - 1)d Substituting the values we found earlier, we get: a_21 = 4 + (20)(3) = 64 Therefore, the 21st term of the A.P is 64, and the correct answer is option (C).

**Question 12**
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ind the value of k if y - 1 is a factor of y3 + 4y2 + ky - 6

**Answer Details**

if y - 1 is a factor of y3 + 4y2 + ky - 6, then

f(1) = (1)3 + 4(1)2 + k(1) - 6 = 0 (factor theorem)

1 + 4 + k - 6 = 0

5 - 6 + k = 0

-1 + k = 0

k = 1

**Question 13**
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Solve for x in 8x-2 = 2/25

**Answer Details**

8x-2 = 2/25

x-2 = 2/25 x 1/8

x-2 = 2/200

x-2 = 1/100

1/x2 = 1/100

x2 = 100

x = 10

**Question 14**
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If cos(x + 40)o = 0.0872, what is the value of x?

**Answer Details**

We are given that cos(x + 40) = 0.0872. To find the value of x, we need to use the inverse cosine function, also known as arccosine or cos^-1. Taking the inverse cosine of both sides, we get: arccos(cos(x + 40)) = arccos(0.0872) The inverse cosine and cosine functions are inverses of each other, so they "cancel out" on the left-hand side, leaving us with: x + 40 = arccos(0.0872) Using a calculator or a table of trigonometric values, we can find that arccos(0.0872) is approximately 84.74 degrees. Subtracting 40 from both sides, we get: x = 84.74 - 40 x = 44.74 So the value of x is approximately 44.74 degrees. None of the given options is an exact match, but the closest one is 45 degrees.

**Question 15**
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Simplify 2√2−√3√2+√3

**Answer Details**

=2√2−√3√2+√3×√2−√3√2+√3
$=\frac{2\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$

=2√2(√2)+(2√2)(−√3)−√3(√2)−√3(−√3)(√2)2−(√3)2
$=\frac{2\sqrt{2}(\sqrt{2})+(2\sqrt{2})(-\sqrt{3})-\sqrt{3}(\sqrt{2})-\sqrt{3}(-\sqrt{3})}{(\sqrt{2}{)}^{2}-(\sqrt{3}{)}^{2}}$

=2×2−2√6−√6+32−3
$=\frac{2\times 2-2\sqrt{6}-\sqrt{6}+3}{2-3}$

=4−3√6+3−1
$=\frac{4-3\sqrt{6}+3}{-1}$

=7−3√6−1
$=\frac{7-3\sqrt{6}}{-1}$

=7−1−3√6−1
$=\frac{7}{-1}-\frac{3\sqrt{6}}{-1}$

=−7+3√6
$=-7+3\sqrt{6}$

=3√6−7

**Question 16**
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If ∣∣∣−x12−14∣∣∣=−12, find x

**Answer Details**

∣∣∣−x12−14∣∣∣=−12
$\left|\begin{array}{cc}-x& 12\\ -1& 4\end{array}\right|=-12$

-4x - (-1)12 = -12

-4x + 12 = -12

-4x = -12 - 12

-4x = - 24

x = 6

-4x - (-1)12 = -12

-4x + 12 = -12

-4x = -12 - 12

-4x = - 24

x = 6

**Question 17**
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A number is chosen at random from 10 to 30 both inclusive. What is the probability that the number is divisible by 3?

**Answer Details**

Sample space S = {10, 11, 12, ... 30}

Let E denote the event of choosing a number divisible by 3

Then E = {12, 15, 18, 21, 24, 27, 30} and n(E) = 7

Prob (E) = n(E)n(E)
$\frac{n(E)}{n(E)}$

Prob (E) = 721
$\frac{7}{21}$

Prob (E) = 13

**Question 18**
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Evaluate ∫sin2xdx

**Answer Details**

The value of the integral of sin(2x)dx is -(1/2)cos(2x) + k, where k is an arbitrary constant of integration. The integral of sin(2x) can be found using substitution or by recognizing that sin(2x) is the derivative of -(1/2)cos(2x). The constant of integration k is added to account for the fact that there are infinitely many functions that have the same derivative as sin(2x). The constant can take any value and is introduced to reflect the inherent uncertainty in finding an antiderivative.

**Question 19**
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Evaluate Log28 + Log216 - Log24

**Answer Details**

=log28×164
$=lo{g}_{2}\frac{8\times 16}{4}$

=log232
$=lo{g}_{2}32$

=log225
$=lo{g}_{2}{2}^{5}$

=5log22
$=5lo{g}_{2}2$

=5×1
$=5\times 1$

=5

**Question 20**
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Values01234Frequency12219 $\begin{array}{cccccc}\text{Values}& 0& 1& 2& 3& 4\\ \text{Frequency}& 1& 2& 2& 1& 9\end{array}$

Find the mode of the distribution above

**Answer Details**

To find the mode of the distribution, we look for the value that appears most frequently in the dataset. From the given frequency table, we can see that the value "4" appears 9 times, which is more than any other value. Therefore, the mode of this distribution is "4". In other words, the mode is the value that occurs most frequently in the data set. It is a measure of central tendency that can be useful in describing a dataset.

**Question 21**
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The locus of a dog tethered to a pole with a rope of 4m is a

**Answer Details**

The locus of a dog tethered to a pole with a rope of 4m is a circle with radius 4m. When a dog is tethered to a pole with a rope, it can move around the pole within the radius of the rope. Therefore, the dog's possible positions form a circle centered at the pole, with the radius equal to the length of the rope, which in this case is 4 meters. Since the circle has a fixed radius of 4m, it is not a semi-circle, but a full circle. Therefore, the correct answer is "circle with radius 4m."

**Question 22**
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Calculate the mid point of the line segment y - 4x + 3 = 0, which lies between the x-axis and y-axis.

**Answer Details**

y - 4x + 3 = 0

When y = 0, 0 - 4x + 3 = 0

Then -4x = -3

x = 3/4

So the line cuts the x-axis at point (3/4, 0).

When x = 0, y - 4(0) + 3 = 0

Then y + 3 = 0

y = -3

So the line cuts the y-axis at the point (0, 3)

Hence the midpoint of the line y - 4x + 3 = 0, which lies between the x-axis and the y-axis is;

[12(x1+x2),12(y1+y2)]
$[\frac{1}{2}({x}_{1}+{x}_{2}),\frac{1}{2}({y}_{1}+{y}_{2})]$

[12(34+0),12(0+−3)]
$[\frac{1}{2}(\frac{3}{4}+0),\frac{1}{2}(0+-3)]$

[12(34),12(−3)]
$[\frac{1}{2}(\frac{3}{4}),\frac{1}{2}(-3)]$

[38,−32]

**Question 23**
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A woman bought a grinder for ₦60,000. She sold it at a loss of 15%. How much did she sell it?

**Answer Details**

The woman bought the grinder for ₦60,000 and sold it at a loss of 15%. This means that she sold it for 100% - 15% = 85% of its original price. To find out how much she sold it for, we can calculate 85% of ₦60,000: 85% of ₦60,000 = 0.85 x ₦60,000 = ₦51,000 Therefore, the woman sold the grinder for ₦51,000. The answer is option C.

**Question 24**
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If P = {1,2,3,4,5} and P ∪ $\cup $ Q = {1,2,3,4,5,6,7}, list the elements in Q

**Question 25**
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Find y, if (5−62−7)(52)=(7−11)

**Answer Details**

(5−62−7)(52)=(7−11)
$\left(\begin{array}{cc}5& -6\\ 2& -7\end{array}\right)\left(\begin{array}{c}5\\ 2\end{array}\right)=\left(\begin{array}{c}7\\ -11\end{array}\right)$

By matrices multiplication;

5x - 6y = 7 ........(1)

2x - 7y = -11 ......(2)

2 x (1): 10x - 12y = 14 .......(3)

5 x (2): 10x - 35y = -55 ......(4)

(3) - (4): 23y = 69

y = 69/23 = 3

**Question 26**
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Find the median of 5,9,1,10,3,8,9,2,4,5,5,5,7,3 and 6

**Answer Details**

To find the median of a set of numbers, we need to arrange the numbers in order from smallest to largest and then find the middle number. Arranging the given numbers in order from smallest to largest, we get: 1, 2, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 9, 9, 10 There are 15 numbers in this set, so the median will be the average of the two middle numbers: the 7th and 8th numbers. The 7th number is 5, and the 8th number is also 5, so the median of this set of numbers is: (median) = (5 + 5) / 2 = 10 / 2 = 5 So the median of the given set of numbers is 5. Looking at the given answer options, we see that the answer is (B) 5.

**Question 27**
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Find the mid point of S(-5, 4) and T(-3, -2)

**Answer Details**

To find the midpoint of the line segment between two points, we need to average the x-coordinates and the y-coordinates of the two points separately. So, to find the midpoint of S(-5, 4) and T(-3, -2), we take the average of their x-coordinates and the average of their y-coordinates: Midpoint x-coordinate = (S x-coordinate + T x-coordinate) / 2 = (-5 + (-3)) / 2 = -4 Midpoint y-coordinate = (S y-coordinate + T y-coordinate) / 2 = (4 + (-2)) / 2 = 1 Therefore, the midpoint of S(-5, 4) and T(-3, -2) is (-4, 1).

**Question 28**
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From the Venn diagram above, the shaded parts represent

**Answer Details**

The Venn diagram above represents three sets: P, Q, and R. The shaded parts represent the elements that are in both P and Q (the intersection of P and Q), and the elements that are in both P and R (the intersection of P and R). Therefore, the shaded parts represent the set (P∩Q) and the set (P∩R). Option (A) (P∩Q)∪(P∩R) is the correct answer.

**Question 29**
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If sinθ=1213 $\mathrm{sin}\theta =\frac{12}{13}$, find the value of 1+cosθ

**Question 30**
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What is the common ratio of the G.P. (√10+√5)+(√10+2√5)+...
$(\sqrt{10}+\sqrt{5})+(\sqrt{10}+2\sqrt{5})+...$?

**Answer Details**

Common ratio r of the G.P is

r=Tn+1Tn=T2T1
$r=\frac{{T}_{n}+1}{{T}_{n}}=\frac{{T}_{2}}{{T}_{1}}$

r=√10+2√5√10+√5
$r=\frac{\sqrt{10}+2\sqrt{5}}{\sqrt{10}+\sqrt{5}}$

r=√10+2√5√10+√5×√10−√5√10−√5
$r=\frac{\sqrt{10}+2\sqrt{5}}{\sqrt{10}+\sqrt{5}}\times \frac{\sqrt{10}-\sqrt{5}}{\sqrt{10}-\sqrt{5}}$

=(√10)(√10)+(√10)(−√5)+(2√5)(√10)+(2√5)(−√5)(√10)2−(√5)2
$=\frac{(\sqrt{10})(\sqrt{10})+(\sqrt{10})(-\sqrt{5})+(2\sqrt{5})(\sqrt{10})+(2\sqrt{5})(-\sqrt{5})}{(\sqrt{10}{)}^{2}-(\sqrt{5}{)}^{2}}$

10−√50+2√50−1010−5
$\frac{10-\sqrt{50}+2\sqrt{50}-10}{10-5}$

√505
$\frac{\sqrt{50}}{5}$

√25×25
$\frac{\sqrt{25\times 2}}{5}$

5√25
$\frac{5\sqrt{2}}{5}$

√2

**Question 31**
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If gt2 - k - w = 0, make g the subject of the formula

**Answer Details**

We can solve for g by manipulating the given equation, gt² - k - w = 0, to isolate g on one side of the equation. First, we can add k and w to both sides of the equation to obtain: gt² = k + w Next, we can divide both sides of the equation by t² to solve for g: g = (k + w)/t² Therefore, the solution is: g = (k + w)/t² Hence, the answer is: (k + w)/t²

**Question 32**
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In how many ways can a team of 3 girls be selected from 7 girls?

**Answer Details**

A team of 2 girls can be selected from 7 girls in 7C3
${}^{7}{C}_{3}$

=7!(7?3)!3!
$=\frac{7!}{(7?3)!3!}$

=7!4!3!ways

**Question 33**
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Evaluate ∫(2x+3)12δx