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**Question 1**
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Calculate the median of 14, 17, 10, 13, 18 and 10.

**Answer Details**

The median of a set of numbers is the middle value when the numbers are arranged in order. If there is an odd number of values, the median is the middle value, and if there is an even number of values, the median is the average of the two middle values. In this case, the set of numbers {14, 17, 10, 13, 18, 10} has 6 values, which is an even number. To find the median, we need to first arrange the values in order: {10, 10, 13, 14, 17, 18} The median is the average of the two middle values, which are 13 and 14, so: Median = (13 + 14) / 2 = 13.5 So, the median of the set of numbers {14, 17, 10, 13, 18, 10} is 13.5.

**Question 2**
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Given that r = 3vπh−−−√ make v the subject of the formula

**Answer Details**

square both sides to remove the big square root

→ r2 ${}^{2}$ = 3vπh $\frac{\mathrm{}}{}$

cross multiply

3v = r2 ${}^{2}$ * πh

v = πr2h3

**Question 3**
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The 10th term of an AP is 32. If the first term is 3/2, what is the 4th term?

**Answer Details**

The 10th term of an arithmetic progression (AP) is given as 32, and the first term is 3/2. We need to find the value of the 4th term. We can use the formula to find the nth term of an AP: nth term = a + (n-1)d where a is the first term, d is the common difference, and n is the term we want to find. We can first find the common difference, d, using the formula for the 10th term: 10th term = a + (10-1)d 32 = 3/2 + 9d 29.5 = 9d d = 29.5/9 Now, we can find the 4th term using the same formula: 4th term = 3/2 + (4-1)(29.5/9) 4th term = 3/2 + 3.2778 4th term = 35/6 Therefore, the answer is: 35/3.

**Question 4**
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If 8, p, q and 26 is an A.P. Find the values of p and q

**Answer Details**

The common difference is 6.

Just add 6 to get the next term

8, 14, 20 and 26

**Question 5**
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Factorize 4a\(^2\) - 9b\(^2\)

**Answer Details**

To factorize 4a\(^2\) - 9b\(^2\), we can use the difference of squares formula, which states that a\(^2\) - b\(^2\) = (a+b) (a-b). We just need to recognize that 4a\(^2\) is a\(^2\) multiplied by 4, and 9b\(^2\) is b\(^2\) multiplied by 9. Then we have: 4a\(^2\) - 9b\(^2\) = (2a)\(^2\) - (3b)\(^2\) = (2a + 3b) (2a - 3b) Therefore, the correct answer is (2a+3b) (2a-3b).

**Question 6**
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A rectangular pyramid has an area 24cm2 ${}^{2}$ and height 7.5cm. Find its volume?

**Answer Details**

Volume of a rectangular pyramid = length∗width∗height3or area∗height3 $\frac{}{}$

= 24∗7.53 $\frac{\mathrm{}}{}$ → 1803 $\frac{\mathrm{}}{}$

Volume of the rectangular pyramid = 60cm3

**Question 7**
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The locus of points equidistant from a fixed point.

**Answer Details**

The locus of points equidistant from a fixed point is a circle. A circle is a two-dimensional shape that is defined as a set of points that are all the same distance away from a fixed point, called the center. The distance between the center of the circle and any point on the circle is called the radius. So, when we talk about the locus of points equidistant from a fixed point, we are talking about all the points that are the same distance away from the fixed point. This creates a circular shape, with the fixed point as the center of the circle and the distance between the center and any point on the circle as the radius. Therefore, the correct answer is: circle.

**Question 8**
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Find x if the mean of 2x, 4x, 2x - 13 and 6x is 4.

**Answer Details**

To find the value of x, we need to use the formula for calculating the mean (also known as the average). The mean of a set of numbers is equal to the sum of the numbers divided by the total number of numbers in the set. In this problem, we have four numbers: 2x, 4x, 2x - 13, and 6x. To find their mean, we need to add them up and divide by 4 (since there are four numbers in the set). Therefore, we have: (2x + 4x + 2x - 13 + 6x) / 4 = 4 Simplifying the left-hand side of the equation, we get: 14x - 13 / 4 = 4 Multiplying both sides by 4, we get: 14x - 13 = 16 Adding 13 to both sides, we get: 14x = 29 Dividing both sides by 14, we get: x = 29/14 Therefore, x is approximately equal to 2.07 (rounded to two decimal places). So the answer is not one of the given options, but rather x = 29/14.

**Question 9**
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Find the length of a chord 3cm from the centre of a circle of radius 5cm

**Answer Details**

Using pythagoras formula:

Hyp2 ${}^{\mathrm{}}$ = adj2 ${}^{\mathrm{}}$ + opp2 ${}^{\mathrm{}}$

52 ${}^{\mathrm{}}$ = opp2 ${}^{\mathrm{}}$ + 32 ${}^{\mathrm{}}$

52 ${}^{\mathrm{}}$ - 32 ${}^{\mathrm{}}$ = adj2 ${}^{\mathrm{}}$

4 = adj

length of the chord = 2 * 4 = 8cm

**Question 10**
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Evaluate Log2 ${}_{2}$ 8√2

**Answer Details**

where Log2 ${}_{2}$ 8√2 → Log ${}_{2}$ √128

→ Log22

12812
${}^{\frac{1}{2}}$

=
$\frac{1}{2}$ * (Log2
${}_{2}$ 128) →

$\frac{1}{2}$ * (Log

${}_{2}$ 2
${}^{7}$)

= 7 * 12 * (Log

${}_{2}$ 2)

where (Log ${}_{2}$ 2) = 1

→ 7 *

$\frac{1}{2}$ * 1

=

$\frac{7}{2}$ or 3.5

**Question 11**
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A poultry farmer realized 200 eggs from his birds and sold 5 \(\frac{1}{2}\) crates of eggs. What percentages of eggs are left?

**Answer Details**

To determine the percentage of eggs left, we need to first calculate the total number of eggs produced by the poultry farmer. We know that the farmer produced 200 eggs, but we need to convert the 5 1/2 crates of eggs sold into a number of eggs. Let's assume that one crate contains 30 eggs (this may vary depending on the size of the crate, but for the purposes of this example, we'll use 30). 5 1/2 crates can be written as 5 + 1/2 crates, which is equivalent to 5 crates + 0.5 x 30 eggs = 5 crates + 15 eggs. So the farmer sold a total of 5 x 30 + 15 = 165 eggs. To find the percentage of eggs left, we can subtract the number of eggs sold (165) from the total number of eggs produced (200) and then divide the result by the total number of eggs and multiply by 100 to get the percentage: Percentage of eggs left = ((200 - 165) / 200) x 100 = 35 / 2 = 17.5% Therefore, the answer is option (C) 17.5%.

**Question 12**
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Solve for k in the equation 18k+2 = 1

**Answer Details**

(8?1)k+2 $({8}^{?1}{)}^{?+2}$ = (80) $({8}^{0})$

base 8 cancel out on both sides

-1(k+2) = 0

-k -2 = 0

: k = -2

**Question 13**
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find the limit of y = x3+6x−7x−1 $\frac{{}^{}}{}$ as x tends to 1

**Answer Details**

x3+6x−7x−1 $\frac{{}^{}}{}$

When numerator is differentiated → 3x2 ${}^{2}$ + 6

When denominator is differentiated → 1

: 3x2+61 $\frac{\mathrm{}}{}$

substitute x for 1

3∗12+61 $\frac{3\ast {1}^{2}+6}{1}$ = 3+61 $\frac{3+6}{1}$

= 91 $\frac{9}{1}$

= 9

**Question 14**
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The number line represented by the inequality

**Answer Details**

x ≥ 2 is represented by an arrow with an filled-in dot above the 2.

The arrow points in the direction of all the numbers that are greater than 2.

**Question 15**
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In the diagram above, XY = 8cm and OX = 5cm. Find Oz

**Answer Details**

hyp = 5cm, adj = 8cm2 $\frac{\mathrm{\mathrm{4cm,\; opp\; =\; xcm}}}{}$

Pythagoras theorem:

hyp22

= opp2
${}^{\mathrm{}}$ + adj2
${}^{\mathrm{}}$

52 ${}^{\mathrm{}}$ = x2 ${}^{\mathrm{}}$ + 42 ${}^{\mathrm{}}$

x

${}^{2}$ = 25 - 16

x = √9

x = 3cm

**Question 16**
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If y varies inversely as x and x = 3 when y =4. Find the value of x when y = 12

**Answer Details**

This problem involves an inverse variation, which means that as one variable increases, the other variable decreases, and vice versa. The relationship between the variables can be expressed as: y = k/x where k is the constant of variation. To solve for the value of k, we can use the given information that x = 3 when y = 4: 4 = k/3 Multiplying both sides by 3 gives: k = 12 Now that we know the constant of variation, we can use it to find the value of x when y = 12: 12 = 12/x Multiplying both sides by x gives: 12x = 12 Dividing both sides by 12 gives: x = 1 Therefore, when y = 12, x = 1. Option (B) is the correct answer.

**Question 17**
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If sin θ = - 35 and θ lies in the third quadrant, find cos θ

**Answer Details**

Where sin θ = opphyp →

$\frac{-3}{5}$

opp = -3, hyp = 5

using pythagoras formula

hyp2 ${}^{2}$ = adj2 ${}^{2}$ + opp2 ${}^{2}$

adj2 ${}^{2}$ = hyp2 ${}^{2}$ - opp2 ${}^{2}$

adj2 ${}^{2}$ = 52 ${}^{2}$ - 32 ${}^{2}$ → 25 - 9

adj2 ${}^{2}$ = 16

adj = 4

cos θ = adjhyp →

$\frac{4}{5}$

In third quadrant: cos θ is negative → - 45

**Question 18**
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If n(P) = 20 and n(Q) = 30 and n(PuQ) = 40, find the value n(PnQ)

**Answer Details**

The values n(P), n(Q), n(PuQ), and n(PnQ) represent the number of elements in the sets P, Q, the union of sets P and Q, and the intersection of sets P and Q, respectively. To find n(PnQ), we need to figure out how many elements are in both sets P and Q. This is because the intersection of two sets only includes the elements that are present in both sets. We can use the formula n(PuQ) = n(P) + n(Q) - n(PnQ) to find n(PnQ). Plugging in the given values, we get: 40 = 20 + 30 - n(PnQ) Solving for n(PnQ), we get: n(PnQ) = 20 + 30 - 40 = 10 So the answer is 10.

**Question 19**
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The equivalent of (10110.011)\(_2\) in base 10 is?

**Answer Details**

To convert a binary number to a decimal number, we need to understand the positional value of each digit in the binary number. In a binary number, each digit represents a power of 2, with the rightmost digit representing 2^0, the next digit to the left representing 2^1, and so on. So, for the binary number (10110.011)_2, we can break it down into its decimal equivalent as follows: 1 x 2^4 + 0 x 2^3 + 1 x 2^2 + 1 x 2^1 + 0 x 2^0 + 0 x 2^-1 + 1 x 2^-2 + 1 x 2^-3 = 16 + 0 + 4 + 2 + 0 + 0 + 0.25 + 0.125 = 22.375 Therefore, the equivalent of (10110.011)_2 in base 10 is 22.375.

**Question 20**
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Evaluate n2+1
${}^{{}^{+}}$Cn+5

if n = 3

**Answer Details**

32+1 ${}^{{\mathrm{}}^{}}$C3+5 ${}_{\mathrm{}}$

9+1 ${}^{\mathrm{}}$C3+5 ${}_{\mathrm{}}$

10
${}^{}$C

${}_{8}$ = 10!8!2!
$\frac{\mathrm{}}{}$

10∗9∗8!8!2! $\frac{\mathrm{}}{}$ = 10∗92 $\frac{\mathrm{}}{}$

= 45

**Question 21**
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The cost C of running a school is directly proportional to the number of students N, if 20 students cost #10,000, How many students can #1,000,000 cover?

**Answer Details**

To answer this question, we need to use the formula for direct proportion: C = k * N, where C is the cost, N is the number of students, and k is the proportionality constant. Since we know that 20 students cost #10,000, we can find the value of k by substituting the known values into the formula: 10,000 = k * 20 Solving for k: k = 10,000 / 20 = 500 Now that we have the value of k, we can use it to find the number of students that #1,000,000 would cover: C = k * N 1,000,000 = 500 * N N = 1,000,000 / 500 = 2000 So, #1,000,000 would cover 2000 students.

**Question 22**
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If the mean of 2, 5, (x+1), (x+2), 7 and 9 is 6. Find the median

**Answer Details**

Firstly; solving for x

6 = 2+5+x+1+x+2+7+96
$$

cross multiply to have:

6 * 6 = 2 + 5 + x+1 + x+2 + 7 + 9

36 = 2x + 26

36 - 26 = 2x

10 = 2x

x = 5

Median = 7+62 $\frac{7+6}{2}$

→ 6.5

**Question 23**
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If sec2 ${}^{2}$θ + tan2 ${}^{2}$θ = 3, then the angle θ is equal to?

**Answer Details**

We're given the equation sec²θ + tan²θ = 3, and we need to solve for θ. One way to approach this is to use the trigonometric identity: tan²θ + 1 = sec²θ Substituting this into the given equation, we get: tan²θ + 1 + tan²θ = 3 Simplifying this equation, we get: 2tan²θ = 2 Dividing both sides by 2, we get: tan²θ = 1 Taking the square root of both sides, we get: tanθ = ±1 This means that θ must be one of the angles whose tangent is ±1. These angles are 45º and 225º (or -135º), since tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants. However, we need to check whether these values of θ satisfy the original equation. Let's start with θ = 45º: sec²45º + tan²45º = 2 + 1 = 3 So this value of θ does satisfy the equation, and therefore it is the solution. Therefore, the angle θ is 45º. So the answer is (c) 45º.

**Question 24**
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Find the equation of a straight line parallel to the line 2x - y = 5 and having intercept of 5

**Answer Details**

To find the equation of a straight line parallel to 2x - y = 5, we need to determine the slope of the given line. The slope of a line is defined as the change in y divided by the change in x, which can be written as Δy/Δx. We can rewrite 2x - y = 5 as y = 2x - 5, which is in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. From this form, we can see that the slope of the line is 2. A line parallel to this line will have the same slope of 2. We also know that the new line has an intercept of 5, which means it passes through the point (0, 5). Using the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute m = 2 and (x1, y1) = (0, 5) to get the equation: y - 5 = 2(x - 0) Simplifying this equation gives: y = 2x + 5 Therefore, the equation of a straight line parallel to 2x - y = 5 and having intercept of 5 is 2x + y = 5. Option (a) is the correct answer.

**Question 25**
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In how many ways can the letter of ZOOLOGY be arranged?

**Answer Details**

Zoology has 7 letters in total, with O repeated thrice

\(\frac{7!}{3!}\) → \(\frac{7*6*5*4*3*2*1}{3*2*1}\)

= 840ways

**Question 26**
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Simplify 13−√2 $\frac{1}{3-\surd 2}$ in the form of p + q√2

**Answer Details**

Rationalization with conjugate 3+√2 $3+\surd 2$

$\frac{1}{3-\surd 2}$

→ 1∗[3+√2][3−√2][3+√2] $\frac{\mathrm{}}{}$

= 3+√29−3√2+3√2+√4

= 3+√29−2
$\frac{3+\surd 2}{9-2}$ →

=
$\frac{3}{7}$ + √27

**Question 27**
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If a dress is sold for #3800.00 at 20% discount. what is its original price?

**Answer Details**

The original price of the dress can be found by reversing the discount. If the dress was sold for #3800.00 at a 20% discount, then 20% of the original price was taken off to get #3800.00. To find the original price, we need to add back the 20% that was taken off. So if we take 100% and add 20% to it, we'll have the original price of the dress. Therefore, the original price of the dress was #3800.00 / (100% - 20%) = #3800.00 / 80% = #4750.00.