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**Question 2**
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Given that x2 + y2 + z2 = 194, calculate z if x = 7 and √y
$y$ = 3

**Answer Details**

Given that x2 + y2 + z2 = 194, calculate z if x = 7 and √y
$y$ = 3

x = 7

∴ x2 = 49

√y
$y$ = 3

∴ y2 = 81 = x2 + y2 + z2 = 194

49 + 81 + z2 = 194

130 + z2 = 194

z2 = 194 - 130

= 64

z = √64
$64$

= 8

**Question 3**
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The H.C.F. of a2bx + ab2x and a2b - b2 is

**Answer Details**

To find the H.C.F. of the given terms, we need to factorize them first: a^{2}bx + ab^{2}x = abx(a + b) a^{2}b - b^{2} = b(a - b) The H.C.F. of two or more terms is the product of their common factors raised to the lowest power. Both the terms have a common factor of 'b', so the H.C.F. will have 'b' as a factor. Also, the first term has a common factor of 'abx', and the second term does not have any factor of 'abx'. So, the H.C.F. will not have any factor of 'abx'. Therefore, the H.C.F. of a^{2}bx + ab^{2}x and a^{2}b - b^{2} is 'b'. Hence, the correct answer is: b.

**Question 4**
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Below are the scores of a group of students in a test:

Scores123456No. of students1456x2 $\begin{array}{ccccccc}Scores& 1& 2& 3& 4& 5& 6\\ \text{No. of students}& 1& 4& 5& 6& x& 2\end{array}$

If the average scores is 3.5, find the value of x.

**Answer Details**

Scores123456No. of students1456x2
$\begin{array}{ccccccc}Scores& 1& 2& 3& 4& 5& 6\\ \text{No. of students}& 1& 4& 5& 6& x& 2\end{array}$

Average = 3.5

3.5 = (1×1)+(2×4)+(3×5)+(4×6)+5x+(6×2)1+4+5+6+x+2
$\frac{(1\times 1)+(2\times 4)+(3\times 5)+(4\times 6)+5x+(6\times 2)}{1+4+5+6+x+2}$

3.51
$\frac{3.5}{1}$ = 1+8+15+24+5x+1218+x
$\frac{1+8+15+24+5x+12}{18+x}$

3.51
$\frac{3.5}{1}$ = 60+5x18+x
$\frac{60+5x}{18+x}$

60 + 5x = 3.5(18 ÷
$\xf7$ x)

60 + 5x = = 63 + 1.5x

5x - 1.5x = 63 - 60

1.5x = 3

x = 31.5
$\frac{3}{1.5}$

3015
$\frac{30}{15}$ = 2

**Question 5**
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A car painter charges ₦40.00 per day for himself and ₦10.00 per day for his assistant. if a fleet of cars were painted for ₦2000.00 and the painter worked 10days more than his assistant, how much did the assistant receive?

**Answer Details**

To solve this problem, we can use a system of equations. Let's call the number of days the assistant worked "x". Then the number of days the car painter worked would be "x + 10" (since he worked 10 days more than the assistant). We can then set up the following equations: 40(x+10) + 10x = 2000 (this represents the total amount the car painter and his assistant earned for painting the cars) Simplifying this equation: 50x + 400 = 2000 50x = 1600 x = 32 So the assistant worked for 32 days. To find out how much he earned, we can plug this value back into one of the original equations: 10x = 10(32) = 320 Therefore, the assistant earned ₦320.00.

**Question 8**
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4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16.

**Answer Details**

Given that 4, 16, 30, 20, 10, 14 and 26

Adding up = 120

nos ≥
$\ge $ 16 are 16 + 30 + 20 + 26 = 92

The requires sum of angles = 92120
$\frac{92}{120}$ x 360o1
$\frac{{360}^{o}}{1}$

= 276o

**Question 9**
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Find the probability that a number selected at random from 40 to 50 is a prime

**Answer Details**

From 40 to 50 = 11 & number are prime i.e. 41, 43, 47

prob. of selecting a prime No. is 311

**Question 10**
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If a = 2, b = -2 and c = -12 $\frac{1}{2}$, evaluate (ab2 - bc2)(a2c - abc)

**Answer Details**

(ab2 - bc2)(a2c - abc)

[2(2)2 - (- 2x12
$\frac{1}{2}$)] [22(-12
$\frac{1}{2}$) - 2(-2)(-12
$\frac{1}{2}$)]

[8 = 12
$\frac{1}{2}$][-2 - 2] = 172
$\frac{17}{2}$ x 42

= -34

**Question 11**
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Find the sum of the first 18 terms of the progression 3, 6, 12......

**Answer Details**

3 + 6 + 12 + .....18thy term

1st term = 3, common ratio 63
$\frac{6}{3}$ = 2

n = 18, sum og GP is given by Sn = a(rn−1)r−1
$\frac{({r}^{n}-1)}{r-1}$

s18 = 3(218−1)2−1
$\frac{({2}^{18}-1)}{2-1}$

= 3(217 - 1)

**Question 12**
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Two numbers are removed at random from the numbers 1, 2, 3 and 4. What is the probability that the sum of the numbers removed is even?

**Answer Details**

12341(1,1)(1,2)(1,3)(1,4)2(2,1)(2,2)(2,3)(2,4)3(3,1)(3,2)(3,3)(3,4)4(4,1)(4,2)(4,3)(4,4)
$\begin{array}{cccc}1& 2& 3& 4\\ 1(1,1)& (1,2)& (1,3)& (1,4)\\ 2(2,1)& (2,2)& (2,3)& (2,4)\\ 3(3,1)& (3,2)& (3,3)& (3,4)\\ 4(4,1)& (4,2)& (4,3)& (4,4)\end{array}$

sample space = 16

sum of nos. removed are (2), 3, (4), 5

3, (4), 5, (6)

(4), 5, (6), 7

(5), 6, 7, (8)

Even nos. = 8 of them

Pr(even sum) = 816
$\frac{8}{16}$

= 12

**Question 13**
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Factorize 9(x + y)2 - 4(x - y)2

**Answer Details**

9(x + y)2 - 4(x - y)2

Using difference of two squares which says

a2 - b2 = (a + b)(a - b) = 9(x + y)2 - 4(x - y)2

= [3(x + y)]2 - [2(x - y)]-2

= [3(x + y) + 2(x - y) + 2(x - y)][3(x + y) - 2(x - y)]

= [3x +3y + 2x - 2y][3x + 3y - 2x + 2y]

= (5x + y)(x + 5y)

**Question 14**
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Simplify √160r2 $\sqrt{160{r}^{2}}$ + √71r4 $\sqrt{71{r}^{4}}$ + √100r2

**Answer Details**

√160r2+71r4+100r8
$\sqrt{160{r}^{2}+71{r}^{4}+100{r}^{8}}$

Simplifying from the innermost radical and progressing outwards we have the given expression

√160r2+71r4+100r8
$\sqrt{160{r}^{2}+71{r}^{4}+100{r}^{8}}$ = √160r2+81r4
$\sqrt{160{r}^{2}+81{r}^{4}}$

√160r2+9r2
$\sqrt{160{r}^{2}+9{r}^{2}}$ = √169r2
$\sqrt{169{r}^{2}}$

= 13r

**Question 15**
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The mean of ten positive numbers is 16. When another number is added, the mean becomes 18. Find the eleventh number

**Answer Details**

The mean of ten positive numbers is given to be 16. This means that the sum of these ten numbers is 10 times 16, which is 160. When an eleventh number is added, the mean becomes 18. Let's call this number "x". Now we have a total of eleven numbers, and their sum is 11 times 18, which is 198. We can find the eleventh number by subtracting the sum of the first ten numbers (160) from the sum of all eleven numbers (198): x = 198 - 160 = 38 Therefore, the eleventh number is 38, which is.

**Question 16**
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An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7]

**Answer Details**

We can use the formula for the length of an arc of a circle given by: L = rθ where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the center of the circle in radians. In this problem, we are given L = 22cm and r = 15cm, and we need to find θ. We are also given that π = 22/7. Using the formula above and substituting the given values, we can solve for θ: 22 = 15θ θ = 22/15 radians To convert this to degrees, we can multiply by 180/π: θ = (22/15) * (180/22/7) θ = 84 degrees (rounded to the nearest degree) Therefore, the answer is: - 84o

**Question 17**
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If x + 1x $\frac{1}{x}$ = 4, find x2 + 1x2

**Answer Details**

x + 1x
$\frac{1}{x}$ = 4, find x2 + 1x2
$\frac{1}{{x}^{2}}$

= (x + 1x
$\frac{1}{x}$)2 = x2 + 1x2
$\frac{1}{{x}^{2}}$ + 2

x2 + 1x2
$\frac{1}{{x}^{2}}$ = ( x + 1x2
$\frac{1}{{x}^{2}}$)2 - 2

= (4)2 - 2

= 16 - 2

= 14

**Question 18**
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A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]

**Answer Details**

The area of a sector of a circle is given by the formula A = (θ/360)πr^2, where A is the area of the sector, r is the radius of the circle, and θ is the central angle of the sector in degrees. We are given that the radius of the circle is 7cm and the area of the sector is 44cm^2, so we can plug these values into the formula: 44 = (θ/360)(22/7)(7)^2 Simplifying this expression, we get: 44 = (θ/360)(22)(7) 44 = (77θ/180) θ = (180/77)(44) θ ≈ 102.6 degrees Therefore, the angle of the sector is approximately 103 degrees to the nearest degree, which is.

**Question 19**
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Find the curved surface area of the frustrum in the figure.

**Answer Details**

x4=6+x6 $\frac{x}{4}=\frac{6+x}{6}$

6x = 4(6 + x) = 24 + 4x

x = 12 = c = πRL−πL $\pi RL-\pi L$

= π(6)√182+62−π×4×√122+42 $\pi (6)\sqrt{{18}^{2}}+{6}^{2}-\pi \times 4\times \sqrt{{12}^{2}}+{4}^{2}$

= 6π√360−4π√160 $\pi \sqrt{360}-4\pi \sqrt{160}$

= 36π√10−16π√10 $\pi \sqrt{10}-16\pi \sqrt{10}$

= 20π√10 $\pi \sqrt{10}$cm2

**Question 20**
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Each of the interior angles of a regular polygon is 140o. How many sides has the polygon?

**Answer Details**

To find out the number of sides of a regular polygon when given the measure of its interior angles, we can use the formula: Interior angle = (n-2) x 180 / n Where 'n' is the number of sides of the polygon. In this case, we know that each interior angle of the polygon measures 140 degrees. Substituting this value into the formula, we get: 140 = (n-2) x 180 / n Simplifying this equation, we get: 140n = (n-2) x 180 140n = 180n - 360 360 = 40n n = 9 Therefore, the regular polygon has 9 sides. Thus, the correct answer is 9.

**Question 21**
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If Cos θ $\theta $ = 1213 $\frac{12}{13}$. Find θ $\theta $ + cos2θ

**Answer Details**

Cos θ
$\theta $ = 1213
$\frac{12}{13}$

x2 + 122 = 132

x2 = 169- 144 = 25

x = 25

= 5

Hence, tanθ
$\theta $ = 512
$\frac{5}{12}$ and cosθ
$\theta $ = 1213
$\frac{12}{13}$

If cos2θ
$\theta $ = 1 + 1tan2θ
$\frac{1}{ta{n}^{2}\theta}$

= 1 + 1(5)212
$\frac{1}{\frac{(5{)}^{2}}{12}}$

= 1 + 125144
$\frac{1}{\frac{25}{144}}$

= 1 + 14425
$\frac{144}{25}$

= 25+14425
$\frac{25+144}{25}$

= 16925

**Question 22**
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Find the solution of the equation x - 8√x $x$ + 15 = 0

**Answer Details**

x - 8√x
$x$ + 15 = 0

x + 15 = 8√x
$x$

square both sides = (x + 15)2 = (8 √x
$x$2

x2 + 225 + 30x = 64x

x2 + 225 + 30x - 64x = 0

x2 - 34x + 225 = 0

(x - 9)(x - 25) = 0

x = 9 or 25

**Question 23**
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Simplify (1x?1+1y?1 $\frac{1}{{x}^{?1}}+\frac{1}{{y}^{?1}}$)-1

**Answer Details**

Simplify (1x?1+1y?1
$\frac{1}{{x}^{?1}}+\frac{1}{{y}^{?1}}$)-1 = (1x?1+1y?1
$\frac{1}{{x}^{?1}}+\frac{1}{{y}^{?1}}$)-1

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

= xy $\frac{x}{y}$

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

= xy $\frac{x}{y}$

**Question 24**
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Three children shared a basket of mangoes in such a way that the first child took 14 $\frac{1}{4}$ of the mangoes and the second 34 $\frac{3}{4}$ of the remainder. What fraction of the mangoes did the third child take?

**Answer Details**

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket.

If the first child takes 14
$\frac{1}{4}$ it will remain 1 - 14
$\frac{1}{4}$ = 34
$\frac{3}{4}$

Next, the second child takes 34
$\frac{3}{4}$ of the remainder

which is 34
$\frac{3}{4}$ i.e. find 34
$\frac{3}{4}$ of 34
$\frac{3}{4}$

= 34
$\frac{3}{4}$ x 34
$\frac{3}{4}$

= 916
$\frac{9}{16}$

the fraction remaining now = 34
$\frac{3}{4}$ - 916
$\frac{9}{16}$

= 12?916
$\frac{12?9}{16}$

= 316

**Question 25**
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A cylinder pipe, made of metal is 3cm thick.If the internal radius of the pope is 10cm.Find the volume of metal used in making 3m of the pipe.

**Answer Details**

To find the volume of metal used to make 3 meters of the pipe, we first need to find the volume of one meter of the pipe, and then multiply it by 3. The pipe is a cylinder with an internal radius of 10cm and a thickness of 3cm, which means that the external radius of the pipe is 13cm (10cm + 3cm). To find the volume of one meter of the pipe, we need to find the volume of the outer cylinder (with radius 13cm) and subtract the volume of the inner cylinder (with radius 10cm): Volume of outer cylinder = π x radius^2 x height = π x 13^2 x 100 = 16,900π cm^3 Volume of inner cylinder = π x radius^2 x height = π x 10^2 x 100 = 10,000π cm^3 Volume of metal in one meter of the pipe = Volume of outer cylinder - Volume of inner cylinder = 16,900π - 10,000π = 6,900π cm^3 Therefore, the volume of metal used to make 3 meters of the pipe is: Volume of metal in 3 meters of the pipe = Volume of metal in 1 meter x 3 = 6,900π x 3 = 20,700π cm^3 So the answer is 20,700π cubic cm.

**Question 26**
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Simplify 1x−2 $\frac{1}{x-2}$ + 1x+2 $\frac{1}{x+2}$ + 2xx2−4

**Answer Details**

1x−2
$\frac{1}{x-2}$ + 1x+2
$\frac{1}{x+2}$ + 2xx2−4
$\frac{2x}{{x}^{2}-4}$

= (x+2)+(x−2)+2x(x+2)(x−2)
$\frac{(x+2)+(x-2)+2x}{(x+2)(x-2)}$

= 4xx2−4

**Question 27**
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If a2 + b2 = 16 and 2ab = 7.Find all the possible values of (a - b)

**Answer Details**

a2 + b2 = 16 and 2ab = 7

To find all possible values = (a - b)2 + b2 - 2ab

Substituting the given values = (a - b)2

= 16 - 7

= 9

(a - b) = ±
$\pm $9

= ±
$\pm $3

OR a - b = 3, -3

**Question 28**
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If f(x - 4) = x2 + 2x + 3, Find, f(2)

**Answer Details**

f(x - 4) = x2 + 2x + 3

To find f(2) = f(x - 4)

= f(2)

x - 4 = 2

x = 6

f(2) = 62 + 2(6) + 3

= 36 + 12 + 3

= 51

**Question 29**
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At what value of x is the function x2 + x + 1 minimum?

**Answer Details**

To find the value of `x` that minimizes the function `x^2 + x + 1`, we need to find the vertex of the parabola represented by this function. The vertex of a parabola of the form `ax^2 + bx + c` is given by `(-b/2a, f(-b/2a))`. In this case, `a = 1`, `b = 1`, and `c = 1`, so the vertex is located at `(-1/2, f(-1/2))`. To find `f(-1/2)`, we substitute `-1/2` for `x` in the function: `f(-1/2) = (-1/2)^2 + (-1/2) + 1 = 3/4` Therefore, the vertex of the parabola is `(-1/2, 3/4)`. Since the parabola opens upwards (the coefficient of `x^2` is positive), the value of `x` that minimizes the function is the x-coordinate of the vertex, which is `-1/2`. Therefore, the answer is `1`.

**Question 30**
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