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**Question 1**
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The value of x + x ( xx) when x = 2 is

**Answer Details**

To solve this problem, we substitute x = 2 into the given expression: x + x(x*x) = 2 + 2(2*2) = 2 + 2(4) = 2 + 8 = 10. Therefore, the value of x + x(x*x) when x = 2 is 10. Here's a step-by-step breakdown of how we got this answer: 1. We start by substituting the value of x into the expression, which gives us: x + x(x*x) = 2 + 2(2*2) 2. We simplify the expression inside the parentheses first, which gives us: x + x(4) 3. Then we multiply x by 4, which gives us: x + 4x 4. We combine the two terms to get: 5x 5. Finally, we substitute x = 2 into this expression, which gives us: 5(2) = 10 Therefore, the value of x + x(x*x) when x = 2 is 10.

**Question 2**
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Given the quadrilateral RSTO inscribed in the circle with O as centre. Find the size angle x and given RST = 60o

**Answer Details**

To find the size of angle x, we can start by using the fact that the sum of the opposite angles in an inscribed quadrilateral is 180 degrees. In this case, angle RST is opposite to angle O, and angle OST is opposite to angle x. We are given that angle RST is 60 degrees, so we can use this to find angle O: angle RST + angle RTO + angle STO + angle OST = 360 degrees 60 + angle RTO + angle STO + angle OST = 360 angle RTO + angle STO + angle OST = 300 Since angle RTO and angle STO are both equal to angle O (because they are both subtended by the same arc), we can write: 3 x angle O = 300 angle O = 100 degrees Now that we know angle O, we can find angle x: angle OST = 180 - angle RST - angle O angle OST = 180 - 60 - 100 angle OST = 20 degrees Therefore, the size of angle x is 20 degrees, which corresponds to option D.

**Question 3**
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A cylindrical tank has a capacity of 3080 m3. What is the depth of the tank if the diameter of its base is 14 m?

(Take pi = 22/7)

**Answer Details**

We can use the formula for the volume of a cylinder to solve this problem. The formula is: Volume = πr^2h where π is the value of pi, r is the radius of the base, h is the height (or depth) of the cylinder. We are given that the diameter of the base of the cylinder is 14 m, which means the radius is 7 m (since radius is half the diameter). We are also given that the volume of the cylinder is 3080 m^3. Substituting these values into the formula, we get: 3080 = (22/7) * 7^2 * h Simplifying this expression, we get: 3080 = 22 * 7 * h 3080 = 154h h = 3080/154 h = 20 Therefore, the depth of the cylinder is 20 m, which means the answer is option C: 20 m.

**Question 6**
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What is the product of 2x2 − x + 1 and 3 − 2x

**Answer Details**

To find the product of (2x^2 - x + 1) and (3 - 2x), we need to multiply every term in the first expression with every term in the second expression, and then simplify the result. We can use the distributive property of multiplication to do this. First, we multiply 2x^2 by 3 and by -2x: 2x^2 * 3 = 6x^2 2x^2 * -2x = -4x^3 Then, we multiply -x by 3 and by -2x: -x * 3 = -3x -x * -2x = 2x^2 Finally, we multiply 1 by 3 and by -2x: 1 * 3 = 3 1 * -2x = -2x Now we can add up all these products: 6x^2 - 4x^3 - 3x + 2x^2 + 3 - 2x Simplifying this expression, we get: -4x^3 + 8x^2 - 5x + 3 Therefore, the answer is option B: -4x^3 + 8x^2 - 5x + 3.

**Question 7**
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Find the equation of the locus of a point P(x,y) such that PV = PW, where V = (1,1) and W = (3,5)

**Answer Details**

The locus of a point P(x,y) such that PV = PW where V = (1,1) and W = (3,5). This means that the point P moves so that its distance from V and W are equidistance.

PV = PW

√(x−1)2+(y−1)2=√(x−3)2+(y−5)2
$\sqrt{(x-1{)}^{2}+(y-1{)}^{2}}=\sqrt{(x-3{)}^{2}+(y-5{)}^{2}}$.

Squaring both sides of the equation,

(x-1)2 + (y-1)2 = (x-3)2 + (y-5)2.

x2-2x+1+y2-2y+1 = x2-6x+9+y2-10y+25

Collecting like terms and solving, x + 2y = 8.

**Question 8**
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If two graphs y = px2 + q and y = 2x2 -1 intersect at x = 2, find the value of p in terms q.

**Answer Details**

The value of p can be found by solving for p when the two equations intersect at x=2. To do this, we first need to find the y-values for both equations when x=2. When x=2, the first equation becomes: y = p * 2^2 + q = 4p + q And the second equation becomes: y = 2 * 2^2 - 1 = 7 Now we have two equations with two unknowns (p and q) and we can use substitution to solve for p. We substitute the value of y from the second equation into the first equation: 4p + q = 7 And then we can isolate p by subtracting q from both sides: 4p = 7 - q And finally, dividing both sides by 4: p = (7 - q) / 4 So the answer is: p = (7 - q) / 4.

**Question 9**
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In the diagram MN, PQ and RS are parallel lines. What is the value of the angle marked X?

**Answer Details**

MN || PQ || RS

MN = PQ = RS (parallel lines)

Label the angle in the lines

a = i (corresponding angles are equal)

b = x (corresponding angles are equal)

If |MN| = |RS|

If a = i

and a = 63 = i

a + b = 180 (Adjacent interior angles are supplementary i.e add to 180)

∴ i + x = 180

63 + x = 180

x = 180 - 63

x = 1170

**Question 10**
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The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ₦ 80 000. Find the amount allocated to fertilizer

**Answer Details**

Total angle at a point = 3600

∴ To get the angle occupied by fertilizer we have,

40 + 50 + 80 + 70 + 30 + fertilizer(x) = 360

270 + x = 360

x = 360 - 270

x = 90

Total amount allocated to the farm

= ₦ 80,000

∴Amount allocated to the fertilizer

= fertilizer (angle) × Total amounttotal angle
$\frac{\text{fertilizer (angle) \xd7 Total amount}}{\text{total angle}}$

= 90360
$\frac{90}{360}$ × 80,000

= ₦20,000

**Question 11**
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A man covered a distance of 50 miles on his first trip, on a later trip he traveled 300 miles while going 3 times as fast. His new time compared with the old distance was?

**Answer Details**

Let's denote the man's speed on his first trip as "s". On his first trip, he covered a distance of 50 miles, so his time would be: time = distance / speed = 50 / s On his later trip, he traveled 300 miles while going 3 times as fast, so his new speed would be: new speed = 3s His time for the later trip would then be: time = distance / speed = 300 / (3s) = 100 / s So the ratio of his new time to his old time would be: new time / old time = (100/s) / (50/s) = 100/50 = 2 Therefore, his new time compared to his old time is twice as much. In simpler terms, the man covered a longer distance on his second trip but also traveled faster. Even though he traveled faster on the second trip, the increased distance resulted in his new time being twice as much as his old time.

**Question 12**
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If 2√3−√2√3+2√2 $\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{3}+2\sqrt{2}}$ = m + n √ 6, find the values of m and n respectively.

**Answer Details**

2√3−√2√3+2√2
$\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{3}+2\sqrt{2}}$= m + n√6

2√3−√2√3+2√2
$\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{3}+2\sqrt{2}}$ x √3−2√2√3−√2
$\frac{\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

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

2×3−4√6−6+2×23−2√6+2√6−4×2
$\frac{2\times 3-4\sqrt{6}-6+2\times 2}{3-2\sqrt{6}+2\sqrt{6}-4\times 2}$

= 6−4√6−√6+43−8
$\frac{6-4\sqrt{6}-\sqrt{6}+4}{3-8}$

= 0−4√6−65
$\frac{0-4\sqrt{6}-6}{5}$

= 10−5√65
$\frac{10-5\sqrt{6}}{5}$

= − 2 + √6

∴ m + n√6
$6$ = − 2 + √6

m = − 2, n = 1

**Question 13**
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Divide 4x3-3x+1 by 2x-1

**Answer Details**

Dividing polynomials is a process similar to dividing numbers. To divide 4x^3 - 3x + 1 by 2x - 1, we need to find a polynomial that when multiplied by 2x - 1 gives us 4x^3 - 3x + 1. The polynomial that fits the bill is 2x^2 + x - 1. This can be verified by multiplying 2x^2 + x - 1 by 2x - 1, which gives us 4x^3 - 3x + 1. Therefore, the answer is 2x^2 + x - 1.

**Question 14**
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A car dealer bought a second-hand car for ₦250,000 and spent ₦70,000 refurbishing it. He then sold the car for ₦400,000. What is the percentage gain?

**Answer Details**

Total cost = N(250,000 + 70,000) = ₦320,000

Selling price = ₦400,000 (given)

Gain = SP - CP = N(400,000 - 320,000) = ₦80,000

Gain % = gain/CP x 100 = (80,000/320,000) x 100

Gain % = 25%

**Question 15**
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Make S the subject of the relation

p = s + sm2nr

**Answer Details**

p = s + sm2nr
$\frac{s{m}^{2}}{nr}$

p = s + ( 1 + m2nr $\frac{{m}^{2}}{nr}$)

p = s (1 + nr+m2nr $\frac{nr+{m}^{2}}{nr}$)

nr × p = s (nr + m2)

s = nrpnr+m2 $\frac{nrp}{nr+{m}^{2}}$

p = s + ( 1 + m2nr $\frac{{m}^{2}}{nr}$)

p = s (1 + nr+m2nr $\frac{nr+{m}^{2}}{nr}$)

nr × p = s (nr + m2)

s = nrpnr+m2 $\frac{nrp}{nr+{m}^{2}}$

$\frac{{\mathrm{}}^{}}{}$

**Question 16**
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In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon.

**Answer Details**

2x + x = 180°, => 3x = 180°, and thus x = 60°

Each exterior angle = 60° but size of ext. angle = 360°/n

Therefore 60° = 360°/n

n = 360°/60° = 6 sides

**Question 17**
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If α and β are the roots of the equation 3x2 + bx − 2 = 0. Find the value of 1α $\frac{1}{\alpha}$ + 1β

**Answer Details**

1α
$\frac{1}{\alpha}$ + 1β
$\frac{1}{\beta}$ = β−ααβ
$\frac{\beta -\alpha}{\alpha \beta}$

3x2 + 5x + 5x − 2 = 0.

Sum of root = α + β

Product of root = αβ

x2 + 5x3 $\frac{5x}{3}$ − 23 $\frac{2}{3}$ = 0

αβ = − −23 $\frac{-2}{3}$

α + β = 53 $\frac{5}{3}$

∴ α+βαβ $\frac{\alpha +\beta}{\alpha \beta}$ = − 5323 $\frac{{\textstyle \frac{5}{3}}}{{\textstyle \frac{2}{3}}}$

= − 23 $\frac{2}{3}$ × 33 $\frac{3}{3}$

= 52 $\frac{5}{2}$

3x2 + 5x + 5x − 2 = 0.

Sum of root = α + β

Product of root = αβ

x2 + 5x3 $\frac{5x}{3}$ − 23 $\frac{2}{3}$ = 0

αβ = − −23 $\frac{-2}{3}$

α + β = 53 $\frac{5}{3}$

∴ α+βαβ $\frac{\alpha +\beta}{\alpha \beta}$ = − 5323 $\frac{{\textstyle \frac{5}{3}}}{{\textstyle \frac{2}{3}}}$

= − 23 $\frac{2}{3}$ × 33 $\frac{3}{3}$

= 52 $\frac{5}{2}$

$\frac{\mathrm{}}{}$

**Question 18**
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Simplify (3√64a3)−1 $(\sqrt[3]{64}{a}^{3}{)}^{-1}$.

**Answer Details**

To simplify the expression (643a3)−1, we can first simplify the term under the cube root sign. The cube root of 64a^3 is equal to the cube root of 64 multiplied by the cube root of a^3. The cube root of 64 is 4 because 4 x 4 x 4 = 64, and the cube root of a^3 is a multiplied by the cube root of a. Therefore, the term simplifies to 4a(cuberoot(a)). So, now we have (3√(4a(cuberoot(a))))^-1. To simplify this further, we can use the rule that (a/b)^-1 = b/a. Applying this rule to the expression, we get: 1 / (3√(4a(cuberoot(a)))) = (3√(4a(cuberoot(a))))^(-1) = 1 / (4a(cuberoot(a))) Therefore, the simplified expression is 1 / (4a(cuberoot(a))) or equivalently (1/4)(cuberoot(a^-2)). So the correct answer is 1/4cuberoot(a^-2), which can also be written as 14a.

**Question 19**
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Simplify 3 n−1 $n-1$ × 27 n+181n

**Answer Details**

3n+1
$n+1$ × 27n+181n
$\frac{n+1}{{81}^{n}}$

= 3n+1
$n+1$ × 3 3(n+1)34n
$\frac{{3}^{(n+1)}}{{3}^{4n}}$

= 3n+1+3n+3−4n
$n+1+3n+3-4n$

= 34n−4n−1+3
$4n-4n-1+3$

= 32

= 9

**Question 20**
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The locus of a point which is equidistant from the line PQ forms a

**Answer Details**

The locus of a point that is equidistant from a line is a perpendicular bisector to the line. The perpendicular bisector of a line is a line that passes through the midpoint of the line and is perpendicular to it. In other words, the locus of a point that is equidistant from a line is a line that cuts the original line into two equal halves and is perpendicular to it.

**Question 21**
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Find the gradient of the line joining the points (3, 2) and (1, 4).

**Answer Details**

The gradient of a line is a measure of how steep the line is. It tells us the rate of change in the vertical direction for every unit change in the horizontal direction. To find the gradient of the line joining two points, we need to use the formula: Gradient = (Change in y-coordinate) / (Change in x-coordinate) = (y2 - y1) / (x2 - x1) Using this formula, the gradient of the line joining the points (3, 2) and (1, 4) is: Gradient = (4 - 2) / (1 - 3) = (2) / (-2) = -1 So, the gradient of the line joining the points (3, 2) and (1, 4) is -1.

**Question 22**
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Find ∫(x2 + 3x − 5)dx

**Answer Details**

The integral of x^2 + 3x - 5 with respect to x is: ∫(x^2 + 3x - 5) dx To solve this, we can use the power rule of integration, which states that the integral of x^n with respect to x is (x^(n+1))/(n+1) + C, where C is the constant of integration. Using this rule, we can integrate each term of the polynomial separately: ∫(x^2 + 3x - 5) dx = ∫x^2 dx + ∫3x dx - ∫5 dx = (x^3/3) + (3x^2/2) - (5x) + C Therefore, the antiderivative or indefinite integral of x^2 + 3x - 5 with respect to x is: (x^3/3) + (3x^2/2) - (5x) + C, where C is the constant of integration.

**Question 23**
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Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5

**Answer Details**

The range is the difference between the highest and lowest values in a set. To find the range of the given set of numbers, we need to first find the highest and lowest values. The lowest value in the set is 4, while the highest value is 10. Therefore, the range is 10 - 4 = 6. The mode is the value that appears most frequently in a set. To find the mode of the given set of numbers, we need to count how many times each value appears and find the value that appears most frequently. The value 10 appears 5 times in the set, which is more than any other value. Therefore, the mode of the set is 10. To find the sum of the range and the mode, we simply add the two values together. In this case, the range is 6 and the mode is 10, so the sum is 6 + 10 = 16. Therefore, the correct answer is: 16.

**Question 24**
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Simplify 4√27 $27$ + 5√12 $12$ − 3√75

**Answer Details**

Let's simplify each term first: - √27 can be simplified as √(3^3), which equals 3√3. - √12 can be simplified as √(2^2 × 3), which equals 2√3. - √75 can be simplified as √(3 × 5^2), which equals 5√3. Substituting these simplifications into the original expression, we get: 427(3√3) + 512(2√3) - 375(5√3) Simplifying the coefficients of √3, we get: 1281√3 + 1024√3 - 1875√3 Combining like terms, we get: 330√3 Therefore, the simplified expression is 330√3. Answer: 73" tabindex="0" class="mjx-chtml MathJax_CHTML" id="MathJax-Element-38-Frame">√3 divided by 3.

**Question 25**
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Evaluate 0.000002310.007 $\frac{0.00000231}{0.007}$ and leave the answer in standard form

**Answer Details**

The expression 0.00000231 divided by 0.007 can be evaluated by dividing the numerator and denominator. The answer is a decimal, which can be written in scientific notation, also known as standard form, for easier reading. Scientific notation is a way of writing very small or very large numbers by multiplying a number between 1 and 10 by 10 raised to a power. In scientific notation, the answer to 0.00000231 divided by 0.007 is 3.3 x 10^4.

**Question 26**
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In how many ways can the letters of the word ACCEPTANCE be arranged?

**Answer Details**

**Question 27**
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Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 10, 9, 7, 10, 6, 5

**Answer Details**

The sum of the range of a set of numbers is simply the difference between the largest and smallest number in the set. To find the mode, we need to look for the number that appears most frequently in the set. In this set of numbers, the smallest number is 4 and the largest number is 10, so the range would be 10 - 4 = 6. The mode of this set is the number 10, since it appears the most times (7 times). So, the sum of the range and the mode of the set is 6 + 10 = 16.

**Question 28**
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If a rod 10cm in length was measured as 10.5cm, calculate the percentage error

**Answer Details**

Percentage error is a measure of how accurately a measurement represents the true value of a quantity. It is calculated as the difference between the measured value and the true value, divided by the true value, and multiplied by 100 to get a percentage. In this case, the true value of the rod's length is 10 cm and the measured value is 10.5 cm. To calculate the percentage error, we use the formula: Percentage error = (|measured value - true value| / true value) * 100 = (|10.5 - 10| / 10) * 100 = (0.5 / 10) * 100 = 5% Therefore, the percentage error is 5%.

**Question 29**
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Find the range of the following set of numbers 0.4, −0.4, 0.3, 0.47, −0.53, 0.2 and −0.2

**Answer Details**

The range of a set of numbers is the difference between the largest and smallest values in the set. To find the range of the given set of numbers 0.4, -0.4, 0.3, 0.47, -0.53, 0.2, and -0.2, we need to find the largest and smallest values in the set. The largest value in the set is 0.47, and the smallest value in the set is -0.53. Therefore, the range of the set is: Range = largest value - smallest value = 0.47 - (-0.53) = 1.0 So the correct answer is: 1.0

**Question 30**
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y is inversely proportional to x and y and 6 when x = 7. Find the constant of the variation

**Answer Details**

When two quantities, such as x and y, are inversely proportional, it means that as one quantity increases, the other quantity decreases at a consistent rate. In mathematical terms, we can write the inverse proportionality as: y = k/x where k is a constant of variation that relates the two quantities. To find the constant of variation, we can use the given information that "y and 6 when x = 7". Substituting these values into the equation, we get: 6 = k/7 Multiplying both sides by 7, we obtain: k = 42 Therefore, the constant of variation is 42, and the answer is.

**Question 31**
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Find the principal which amounts to ₦5,500 at a simple interest in 5 years at 2% per annum.

**Answer Details**

Principal, P = Amount, A - Interest, I.

A = P + I

I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10

But A = P + I,

=> 5500 = P + (P/10)

=> 55000 = 10P + P

=> 55000 = 11P

Thus P = 55000/11 = ₦5,000

**Question 32**
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Find the sum to infinity of the series

14
$\frac{1}{4}$, 18
$\frac{1}{8}$, 116
$\frac{1}{16}$,..........

**Answer Details**

Sum to infinity

? = arn ? 1

= a1
$\frac{a}{1}$ ? r

a = 14
$\frac{1}{4}$

r = 18
$\frac{1}{8}$ ÷ 14
$\frac{1}{4}$

r =