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**Question 1**
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Marks | 1 | 2 | 3 | 4 | 5 |

Frequency | 2y - 2 | y - 1 | 3y - 4 | 3 - y | 6 - 2y |

The table above is the distribution of data with mean equals to 3. Find the value of y.

**Answer Details**

Mean = ∑fx∑f
$\frac{\sum fx}{\sum f}$

3=26−y3y+2
$3=\frac{26-y}{3y+2}$

3(3y+2)=26−y
$3(3y+2)=26-y$

9y+6=26−y
$9y+6=26-y$

9y+y=26−6
$9y+y=26-6$

10y=20⟹y=2

**Question 2**
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Each of the interior angles of a regular polygon is 140°. Calculate the sum of all the interior angles of the polygon.

**Answer Details**

Since each interior angle = 140°;

Each exterior angle = 180° - 140° = 40°

Number of sides of the polygon = 360°40°
$\frac{360\xb0}{40\xb0}$

= 9

Sum of angles in the polygon = 140° x 9

= 1260°

**Question 3**
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If sinx=45
$\mathrm{sin}x=\frac{4}{5}$, find 1+cot2xcsc2x−1
$\frac{1+{\mathrm{cot}}^{2}x}{{\mathrm{csc}}^{2}x-1}$.

**Answer Details**

sinx=oppHyp=45
$\mathrm{sin}x=\frac{opp}{Hyp}=\frac{4}{5}$

52
$2$ = 42
$2$ + adj2
$2$

adj2
$2$ = 25 - 16 = 9

adj = √9
$9$ = 3

tanx=43
$\mathrm{tan}x=\frac{4}{3}$

cotx=143=34
$\mathrm{cot}x=\frac{1}{\frac{4}{3}}=\frac{3}{4}$

cot2x=(34)2=916
${\mathrm{cot}}^{2}x=(\frac{3}{4}{)}^{2}=\frac{9}{16}$

cscx=1sinx
$\mathrm{csc}x=\frac{1}{\mathrm{sin}x}$

= 145=54
$\frac{1}{\frac{4}{5}}=\frac{5}{4}$

csc2x=(54)2=2516
${\mathrm{csc}}^{2}x=(\frac{5}{4}{)}^{2}=\frac{25}{16}$

∴1+cot2xcsc2x−1=1+9162516−1
$\therefore \frac{1+{\mathrm{cot}}^{2}x}{{\mathrm{csc}}^{2}x-1}=\frac{1+\frac{9}{16}}{\frac{25}{16}-1}$

= 2516÷916
$\frac{25}{16}\xf7\frac{9}{16}$

= 259

**Question 4**
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Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30

**Answer Details**

To solve this system of equations, we can use substitution or elimination. Here's how to use substitution: From the first equation, we can rearrange it to solve for y in terms of x: 3x + y = 21 y = 21 - 3x Then we can substitute this expression for y into the second equation: xy = 30 x(21 - 3x) = 30 Expanding the left side: 21x - 3x^2 = 30 Rearranging and factoring: 3x^2 - 21x + 30 = 0 Dividing by 3: x^2 - 7x + 10 = 0 This quadratic equation can be factored as: (x - 2)(x - 5) = 0 So the possible values of x are 2 and 5. To find the corresponding values of y, we can substitute each value of x into the expression we found for y: When x = 2: y = 21 - 3(2) = 15 When x = 5: y = 21 - 3(5) = 6 Therefore, the solution to the system of equations is: x = 2 or 5, y = 15 or 6 So the correct option is: - x = 2 or 5, y = 15 or 6

**Question 5**
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If 251−x×5x+2÷(1125)x=625−1 ${25}^{1-x}\times {5}^{x+2}\xf7(\frac{1}{125}{)}^{x}={625}^{-1}$, find the value of x.

**Answer Details**

251−x×5x+2÷(1125)x=625−1
${25}^{1-x}\times {5}^{x+2}\xf7(\frac{1}{125}{)}^{x}={625}^{-1}$

(52)(1−x)×5(x+2)÷(5−3)x=(54)−1
$({5}^{2}{)}^{(1-x)}\times {5}^{(x+2)}\xf7({5}^{-3}{)}^{x}=({5}^{4}{)}^{-1}$

52−2x×5x+2÷5−3x=5−4
${5}^{2-2x}\times {5}^{x+2}\xf7{5}^{-3x}={5}^{-4}$

5(2−2x)+(x+2)−(−3x)=5−4
${5}^{(2-2x)+(x+2)-(-3x)}={5}^{-4}$

Equating bases, we have

2−2x+x+2+3x=−4
$2-2x+x+2+3x=-4$

4+2x=−4⟹2x=−4−4
$4+2x=-4\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2x=-4-4$

2x=−8
$2x=-8$

x=−4

**Question 6**
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If M varies directly as N and inversely as the root of P. Given that M = 3, N = 5 and P = 25. Find the value of P when M = 2 and N = 6.

**Answer Details**

The problem tells us that "M varies directly as N and inversely as the root of P". This means that M is directly proportional to N and inversely proportional to the square root of P, which can be written mathematically as: M = k(N / sqrt(P)) where k is a constant of proportionality. We can solve for k by using the given values of M, N, and P. When M = 3, N = 5, and P = 25, we can plug these values into the equation above and solve for k: 3 = k(5 / sqrt(25)) 3 = k(5 / 5) 3 = k So, we have found that k = 3. Now, we can use this value of k to find P when M = 2 and N = 6. We can rearrange the equation above to solve for P: M = k(N / sqrt(P)) sqrt(P) = k(N / M) P = (k(N / M))^2 Plugging in k = 3, N = 6, and M = 2, we get: P = (3(6 / 2))^2 P = 9(3)^2 P = 81 Therefore, the value of P when M = 2 and N = 6 is 81. So, option D is the correct answer.

**Question 7**
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If 2x2 $2$ + x - 3 divides x - 2, find the remainder.

**Answer Details**

When you divide a polynomial p(x) by (x - a), the remainder = p(a)

i.e. In the case of 2x2
$2$ + x - 3 ÷
$\xf7$ (x - 2), the remainder = p(2).

= 2(2)2
$2$ + 2 - 3

= 8 + 2 - 3

= 7.

**Question 8**
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A bricklayer charges ₦1,500 per day for himself and ₦500 per day for his assistant. If a two bedroom flat was built for ₦95,000 and the bricklayer worked 10 days more than his assistant, how much did the assistant receive?

**Answer Details**

Let the number of days worked by the assistant = t

∴
$\therefore $ The bricklayer worked (t + 10) days.

1500(t + 10) + 500(t) = N 95,000

1500t + 15,000 + 500t = N 95,000

2000t = N 95,000 - N 15,000

2000t = N 80,000

t = 40 days

∴
$\therefore $ The assistant worked for 40 days and received N (500 x 40)

= N 20,000

**Question 9**
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If the volume of a frustrum is given as V=πh3(R2+Rr+r2) $V=\frac{\pi h}{3}({R}^{2}+Rr+{r}^{2})$, find dVdR $\frac{\mathrm{d}V}{\mathrm{d}R}$.

**Answer Details**

To find dV/dR, we need to take the derivative of V with respect to R, while treating all other variables as constants. Using the product rule of differentiation, we get: dV/dR = πh/3 [2R + r] Therefore, the answer is πh/3 [2R + r]. To understand this, we need to remember that the derivative of a function gives us the rate of change of the function. In this case, the volume of a frustrum is a function of its height (h) and radii (R and r). By taking the derivative of the volume with respect to one of the radii (R), we get the rate of change of the volume with respect to that radius. In other words, dV/dR tells us how much the volume changes for a small change in R, while holding h and r constant.

**Question 10**
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Find the equation of a line perpendicular to the line 4y = 7x + 3 which passes through (-3, 1)

**Answer Details**

Equation: 4y = 7x + 3

⟹y=74x+34
$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y=\frac{7}{4}x+\frac{3}{4}$

Slope = coefficient of x = 74
$\frac{7}{4}$

Slope of perpendicular line = −174
$\frac{-1}{\frac{7}{4}}$

= −47
$\frac{-4}{7}$

The perpendicular line passes (-3, 1)

∴
$\therefore $ Using the equation of line y=mx+b
$y=mx+b$

m = slope and b = intercept.

y=−47x+b
$y=\frac{-4}{7}x+b$

To find the intercept, substitute y = 1 and x = -3 in the equation.

1=−47(−3)+b
$1=\frac{-4}{7}(-3)+b$

1=127+b
$1=\frac{12}{7}+b$

b=−57
$b=\frac{-5}{7}$

∴y=−47x−57
$\therefore y=\frac{-4}{7}x-\frac{5}{7}$

7y+4x+5=0

**Question 11**
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Find the distance between the points C(2, 2) and D(5, 6).

**Answer Details**

To find the distance between two points, we can use the distance formula which is: distance = √((x2 - x1)^2 + (y2 - y1)^2) Where (x1, y1) and (x2, y2) are the coordinates of the two points. Using this formula, let's find the distance between points C(2, 2) and D(5, 6). distance = √((5 - 2)^2 + (6 - 2)^2) distance = √(3^2 + 4^2) distance = √(9 + 16) distance = √25 distance = 5 units Therefore, the distance between points C and D is 5 units.

**Question 12**
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If ∣∣ ∣∣2−53x14032∣∣ ∣∣=132 $\left|\begin{array}{ccc}2& -5& 3\\ x& 1& 4\\ 0& 3& 2\end{array}\right|=132$, find the value of x.

**Answer Details**

∣∣ ∣∣2−53x14032∣∣ ∣∣=132
$\left|\begin{array}{ccc}2& -5& 3\\ x& 1& 4\\ 0& 3& 2\end{array}\right|=132$

⟹2∣∣∣1432∣∣∣−(−5)∣∣∣x402∣∣∣+3∣∣∣x103∣∣∣=132 $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2\left|\begin{array}{cc}1& 4\\ 3& 2\end{array}\right|-(-5)\left|\begin{array}{cc}x& 4\\ 0& 2\end{array}\right|+3\left|\begin{array}{cc}x& 1\\ 0& 3\end{array}\right|=132$

2(2−12)+5(2x)+3(3x)=132 $2(2-12)+5(2x)+3(3x)=132$

−20+10x+9x=132 $-20+10x+9x=132$

19x=152 $19x=152$

x=8

⟹2∣∣∣1432∣∣∣−(−5)∣∣∣x402∣∣∣+3∣∣∣x103∣∣∣=132 $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2\left|\begin{array}{cc}1& 4\\ 3& 2\end{array}\right|-(-5)\left|\begin{array}{cc}x& 4\\ 0& 2\end{array}\right|+3\left|\begin{array}{cc}x& 1\\ 0& 3\end{array}\right|=132$

2(2−12)+5(2x)+3(3x)=132 $2(2-12)+5(2x)+3(3x)=132$

−20+10x+9x=132 $-20+10x+9x=132$

19x=152 $19x=152$

x=8

**Question 13**
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If the universal set μ = {x : 1 ≤ x ≤ 20} and

A = {y : multiple of 3}

B = |z : odd numbers}

Find A ∩ B

**Answer Details**

To find A ∩ B, we need to find the elements that are common to both A and B. A = {y : multiple of 3} contains all the numbers that are multiples of 3 between 1 and 20. So, A = {3, 6, 9, 12, 15, 18}. B = {z : odd numbers} contains all the odd numbers between 1 and 20. So, B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}. To find A ∩ B, we need to find the common elements of A and B. The common elements of A and B are 3, 9, and 15. Therefore, A ∩ B = {3, 9, 15}. So, option C - {3, 9, 15} is the correct answer.

**Question 14**
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Evaluate 2sin30+5tan60sin60 $\frac{2\mathrm{sin}?30+5\mathrm{tan}?60}{\mathrm{sin}?60}$, leaving your answer in surd form.

**Answer Details**

2sin30+5tan60sin60
$\frac{2\mathrm{sin}30+5\mathrm{tan}60}{\mathrm{sin}60}$

sin30=12;tan60=√3;sin60=√32
$\mathrm{sin}30=\frac{1}{2};\mathrm{tan}60=\sqrt{3};\mathrm{sin}60=\frac{\sqrt{3}}{2}$

∴2sin30+5tan60sin60=2(12)+5(√3)√32
$\therefore \frac{2\mathrm{sin}30+5\mathrm{tan}60}{\mathrm{sin}60}=\frac{2(\frac{1}{2})+5(\sqrt{3})}{\frac{\sqrt{3}}{2}}$

= 1+5√3√32
$\frac{1+5\sqrt{3}}{\frac{\sqrt{3}}{2}}$

= 2(1+5√3)√3
$\frac{2(1+5\sqrt{3})}{\sqrt{3}}$

= 2+10√3√3
$\frac{2+10\sqrt{3}}{\sqrt{3}}$

Rationalizing, we get

= 2√3+303
$\frac{2\sqrt{3}+30}{3}$

= 23√3+10

**Question 15**
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This table below gives the scores of a group of students in a Further Mathematics Test.

Score | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Frequency | 4 | 6 | 8 | 4 | 10 | 6 | 2 |

Calculate the mean deviation for the distribution

**Answer Details**

Score(x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Total |

Frequency (f) | 4 | 6 | 8 | 4 | 10 | 6 | 2 | 40 |

fx | 4 | 12 | 24 | 16 | 50 | 36 | 14 | 156 |

x – ¯x $\stackrel{\xaf}{x}$ | -2.9 | -1.9 | -0.9 | 0.1 | 1.1 | 2.1 | 3.1 | |

|x – ¯x $\stackrel{\xaf}{x}$| | 2.9 | 1.9 | 0.9 | 0.1 | 1.1 | 2.1 | 3.1 | |

f|x – ¯x $\stackrel{\xaf}{x}$| | 11.6 | 11.4 | 7.2 | 0.4 | 11 | 12.6 | 6.2 | 60.4 |

Mean = ∑fx∑f $\frac{\sum fx}{\sum f}$

= 15640 $\frac{156}{40}$

= 3.9

M.D = ∑f|x–¯x|∑f $\frac{\sum f|x\u2013\stackrel{\xaf}{x}|}{\sum f}$

= 60.440 $\frac{60.4}{40}$

= 1.51

**Question 16**
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Determine the values for which x2−7x+10≤0

**Answer Details**

To determine the values of x that satisfy x² - 7x + 10 ≤ 0, we need to find the roots of the quadratic equation x² - 7x + 10 = 0 and then analyze the behavior of the quadratic function. To find the roots, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a In this case, a = 1, b = -7, and c = 10, so: x = (-(-7) ± √((-7)² - 4(1)(10))) / 2(1) x = (7 ± √9) / 2 x1 = 5 and x2 = 2 The roots of the equation are x = 5 and x = 2. Now, we need to analyze the behavior of the quadratic function in the intervals between the roots and to the left and right of the roots. We can do this by creating a sign chart: Interval | x² - 7x + 10 --------------------------------- (-∞, 2) | + (2, 5) | - (5, +∞) | + In the interval (-∞, 2), the quadratic function is positive because all the factors are positive. In the interval (2, 5), the function is negative because x - 2 is negative and x - 5 is positive. In the interval (5, +∞), the function is positive again because both factors are positive. Therefore, the solution to the inequality x² - 7x + 10 ≤ 0 is the interval [2, 5], because the function is non-positive in that interval and positive elsewhere. In interval notation, we can write: 2 ≤ x ≤ 5 So the correct answer is: 2 ≤ x ≤ 5.

**Question 17**
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Find the value of x for 2+2x3−2≥4x−65

**Answer Details**

2+2x3−2≥4x−65
$\frac{2+2x}{3}-2\ge \frac{4x-6}{5}$

2+2x−63≥4x−65
$\frac{2+2x-6}{3}\ge \frac{4x-6}{5}$

2x−43≥4x−65
$\frac{2x-4}{3}\ge \frac{4x-6}{5}$

5(2x−4)≥3(4x−6)
$5(2x-4)\ge 3(4x-6)$

10x−20≥12x−18
$10x-20\ge 12x-18$

10x−12x≥−18+20
$10x-12x\ge -18+20$

−2x≥2
$-2x\ge 2$

x≤−1

**Question 18**
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A binary operation Δ is defined by a Δ b = a + 3b + 2.

Find (3 Δ 2) Δ 5

**Answer Details**

The binary operation Δ is defined as a Δ b = a + 3b + 2. To find (3 Δ 2) Δ 5, we need to evaluate the expression step by step. First, we need to find 3 Δ 2: 3 Δ 2 = 3 + 3(2) + 2 = 11 Next, we need to find (3 Δ 2) Δ 5: (3 Δ 2) Δ 5 = 11 Δ 5 = 11 + 3(5) + 2 = 28 Therefore, the answer is 28.

**Question 19**
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In the diagram above, O is the centre of the circle ABC, < ABO = 26° and < BOC = 130°. Calculate < AOC.

**Answer Details**

< BAC = 1302
$\frac{130}{2}$ (angle subtended at the centre)

< BAC = 65°

Also, x = 26° (theorem)

y = 65° - 26° = 39°

< AOC = 180° - (39° + 39°)

= 102°

**Question 20**
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Calculate the volume of the regular three-dimensional figure drawn above, where

**Answer Details**

|AC| = |DF| = 13 cm

Using Pythagoras theorem,

|AC|2
$2$ = |AB|2
$2$ + |BC|2
$2$

132
$2$ = 122
$2$ + |BC|2
$2$

|BC|2
$2$ = 169 - 144 = 25

|BC| = √25
$25$

= 5 cm

Volume of triangular prism = 12×base×length×height
$\frac{1}{2}\times base\times length\times height$

= 12×5×12×18
$\frac{1}{2}\times 5\times 12\times 18$

= 540 cm3

**Question 21**
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Differentiate 2xsinx $\frac{2x}{\mathrm{sin}?x}$ with respect to x.

**Answer Details**

To differentiate 2xsinx with respect to x, we use the product rule of differentiation, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function. Applying this rule, we get: (2xsinx)' = 2(x)'sinx + 2x(sin x)' Simplifying this expression, we have: (2xsinx)' = 2sinx + 2xcosx Therefore, the answer is option (D) 2cscx(1−xcotx). None of the other options match the simplified expression we obtained using the product rule.

**Question 22**
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Find the value of k in the equation: √28+√112−√k=√175

**Answer Details**

To find the value of k, we need to isolate the variable on one side of the equation. Let's start by simplifying each of the square roots: √28 = √4 × √7 = 2√7 √112 = √16 × √7 = 4√7 √175 is already simplified, so we leave it as is. Now we can substitute these values into the original equation: 2√7 + 4√7 - √k = √175 Simplifying the left side, we get: 6√7 - √k = √175 Next, we'll isolate the radical term by moving the non-radical term to the other side: 6√7 = √k + √175 Squaring both sides to eliminate the radicals: (6√7)² = (√k + √175)² 36 × 7 = k + 2√k × √175 + 175 252 = k + 5√7√35 252 = k + 5√(7 × 35) 252 = k + 5√(5² × 7) 252 = k + 5 × 5√7 252 = k + 25√7 k = 252 - 25√7 Therefore, the value of k is 252 - 25√7.

**Question 23**
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A man bought a car newly for ₦1,250,000. He had a crash with the car and later sold it at the rate of ₦1,085,000. What is the percentage gain or loss of the man?