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**Question 1**
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The gradient of the straight line joining the points P(5, -7) and Q(-2, -3) is

**Answer Details**

PQ = y1−y0x1−x0 $\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}$ = −3−(−7)−2−5 $\frac{-3-(-7)}{-2-5}$ = −3+7−2−5 $\frac{-3+7}{-2-5}$ = 4−7

**Question 2**
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If log3x2 = -8, what is x?

**Answer Details**

We can start by using the definition of logarithms to rewrite the equation: log3(x^2) = -8 This means that 3 raised to the power of -8 is equal to x^2: 3^(-8) = x^2 To solve for x, we can take the square root of both sides: sqrt(3^(-8)) = sqrt(x^2) On the left side, we can simplify the expression using the rule that says sqrt(a^b) = a^(b/2): 3^(-8/2) = x Simplifying the exponent, we get: 3^(-4) = x Recall that a negative exponent means the reciprocal of the corresponding positive exponent. So: 3^(-4) = 1/3^4 Using the exponent rule that says a^b = a*a*a*...*a (b times), we get: 1/81 = x Therefore, the correct answer is option (D) 181.

**Question 3**
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Simply 223×112445

**Answer Details**

223×112445
$\frac{2\frac{2}{3}\times 1\frac{1}{2}}{4\frac{4}{5}}$

83×32÷245
$\frac{8}{3}\times \frac{3}{2}\xf7\frac{24}{5}$

83×32×524
$\frac{8}{3}\times \frac{3}{2}\times \frac{5}{24}$

56

**Question 4**
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The bar chart above shows the distribution of marks in a class test. If the pass mark is 5, what percentage of students failed the test?

**Question 5**
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If ∣∣∣53x2∣∣∣ $\left|\begin{array}{cc}5& 3\\ x& 2\end{array}\right|$ = ∣∣∣3545∣∣∣ $\left|\begin{array}{cc}3& 5\\ 4& 5\end{array}\right|$, find the value of x

**Answer Details**

∣∣∣53x2∣∣∣
$\left|\begin{array}{cc}5& 3\\ x& 2\end{array}\right|$ = ∣∣∣3545∣∣∣
$\left|\begin{array}{cc}3& 5\\ 4& 5\end{array}\right|$

10 - 3x = 15 - 20

-3x = 15 - 20 - 10

-3x = -15

x = 5

**Question 6**
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Given that I3 is a unit matrix of order 3, find |I3|

**Answer Details**

A unit matrix is a square matrix in which all the diagonal elements are equal to 1, and all the other elements are equal to 0. The symbol I3 represents the unit matrix of order 3, which is a 3x3 matrix. So, the matrix I3 can be written as: |1 0 0| |0 1 0| |0 0 1| To find the determinant of I3, we can use the formula for the determinant of a 3x3 matrix: |a b c| |d e f| |g h i| = a(ei - fh) - b(di - fg) + c(dh - eg) Applying this formula to I3, we get: |1 0 0| |0 1 0| |0 0 1| = 1(1*1 - 0*0) - 0(0*1 - 0*0) + 0(0*1 - 0*0) = 1 Therefore, the determinant of I3 is 1, and the correct answer is option (C) 1. Options (A) -1, (B) 0, and (D) 2 are incorrect.

**Question 7**
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The mean of seven numbers is 96. If an eighth number is added, the mean becomes 112. Find the eighth number.

**Answer Details**

The mean of seven numbers is the sum of all seven numbers divided by 7. If the mean is 96, then the sum of the seven numbers is 96 * 7 = 672. When an eighth number is added, the mean becomes 112, which means that the sum of all eight numbers is 112 * 8 = 896. To find the eighth number, we can subtract the sum of the first seven numbers from the sum of all eight numbers: Eighth number = 896 - 672 = 224. So, the eighth number is 224.

**Question 8**
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Solve for x and y in the equations below

x2 - y2 = 4

x + y = 2

**Answer Details**

We can solve for x and y by using substitution and elimination methods. Starting with the first equation: x^2 - y^2 = 4 We can rewrite this as: x^2 = 4 + y^2 Next, we substitute this expression for x^2 into the second equation: x + y = 2 x^2 + 2xy + y^2 = 4 + y^2 + 2xy + y^2 x^2 + 2xy + y^2 = 4 + 2xy + 2y^2 x^2 = 4 x = ±2 Since x cannot be negative, we can conclude that x = 2. Finally, we can substitute x = 2 into the second equation to find y: 2 + y = 2 y = 0 So, the solution is x = 2, y = 0

**Question 9**
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In a class of 46 students, 22 play football and 26 play volleyball. If 3 students play both games, how many play neither?

**Answer Details**

To find the number of students who play neither football nor volleyball, we need to subtract the total number of students who play either football or volleyball (including those who play both) from the total number of students in the class. First, we add the number of students who play football and the number of students who play volleyball, then subtract the number of students who play both (since they were counted twice). 22 + 26 - 3 = 45 So there are 45 students who play either football or volleyball or both. To find the number of students who play neither, we subtract this from the total number of students in the class: 46 - 45 = 1 Therefore, only 1 student plays neither football nor volleyball. The correct answer is 1.

**Question 11**
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The binary operation * is defined on the set of integers such that p * q = pq + p - q. Find 2 * (3 * 4)

**Answer Details**

To find 2 * (3 * 4), we need to evaluate the expression in parentheses first, i.e., 3 * 4. Using the given binary operation *, we have: 3 * 4 = 3(4) + 3 - 4 = 11 Now we can substitute this value into the original expression: 2 * (3 * 4) = 2 * 11 = 2(11) + 2 - 11 = 13 Therefore, the answer is 13.

**Question 12**
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Evaluate 219 $\frac{21}{9}$ to 3 significant figures

**Answer Details**

To evaluate 21/9 to 3 significant figures, we need to round the result to the nearest thousandth. The result of 21 divided by 9 is 2.33333333333... The third digit after the decimal point is 3, which is greater than or equal to 5. Therefore, we need to round up the second digit after the decimal point, which is 3. Therefore, rounding 2.33333333333... to 3 significant figures gives us 2.33. Therefore, the correct option is 2.33.

**Question 13**
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If y = x2 - 1x $\frac{1}{x}$, find δyδx

**Answer Details**

To find δyδx (the derivative of y with respect to x), we need to apply the power rule of differentiation, which states that if y = x^n, then δyδx = n*x^(n-1). Applying this rule to y = x^2 - 1/x, we get: δyδx = 2x + 1/x^2 Therefore, the answer is not one of the options provided. Option A (2x - 1/x^2) is close but has a minus sign instead of a plus sign before the second term. Option B (2x + x^2) and option C (2x - x^2) are incorrect because they don't take into account the derivative of the second term (-1/x). Option D (2x + 1/x^2) is the correct answer based on the power rule of differentiation.

**Question 14**
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The sum to infinity of a geometric progression is −110
$-\frac{1}{10}$ and the first term is −18
$-\frac{1}{8}$. Find the common ratio of the progression.

**Answer Details**

Sr = a1−r
$\frac{a}{1-r}$

−110
$-\frac{1}{10}$ = 18×11−r
$\frac{1}{8}\times \frac{1}{1-r}$

−110
$-\frac{1}{10}$ = 18(1−r)
$\frac{1}{8(1-r)}$

−110
$-\frac{1}{10}$ = 18−8r
$\frac{1}{8-8r}$

cross multiply...

-1(8 - 8r) = -10

-8 + 8r = -10

8r = -2

r = -1/4

**Question 16**
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Find the median of 2,3,7,3,4,5,8,9,9,4,5,3,4,2,4 and 5

**Answer Details**

Arrange all the values in ascending order,

2,2,3,3,3,4,4,__4,4__,5,5,5,7,8,9,9

**Question 17**
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The binary operation on the set of real numbers is defined by m*n = mn2
$\frac{mn}{2}$ for all m, n ∈
$\in $ R. If the identity element is 2, find the inverse of -5

**Answer Details**

m ∝
$\propto $ n = mn2−a
$\frac{mn}{2}-a$

Identify = e = 2

a ∝
$\propto $ a-1 = e

a ∝
$\propto $ a-1 = 2

-5 ∝
$\propto $ a-1 = 2

−5×a−12=2
$\frac{-5\times {a}^{-1}}{2}=2$

a−1=2×2−5
${a}^{-1}=\frac{2\times 2}{-5}$

a−1=−45

**Question 18**
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Make 'n' the subject of the formula if w = v(2+cn)1-cn

**Answer Details**

w = v(2+cn)1−cn
$\frac{v(2+cn)}{1-cn}$

2v + cnv = w(1 - cn)

2v + cnv = w - cnw

2v - w = -cnv - cnw

Multiply through by negative sign

-2v + w = cnv + cnw

-2v + w = n(cv + cw)

n = −2v+wcv+cw
$\frac{-2v+w}{cv+cw}$

n = 1c−2v+wv+w
$\frac{1}{c}\frac{-2v+w}{v+w}$

Re-arrange...

n = 1cw−2vv+w

**Question 19**
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The probabilities that a man and his wife live for 80 years are 23 $\frac{2}{3}$ and 35 $\frac{3}{5}$ respectively. Find the probability that at least one of them will live up to 80 years

**Answer Details**

The probability that the man will not live for 80 years is 1 - 2/3 = 1/3. Similarly, the probability that the woman will not live for 80 years is 1 - 3/5 = 2/5. The probability that both of them will not live for 80 years is the product of their individual probabilities, which is (1/3) * (2/5) = 2/15. Therefore, the probability that at least one of them will live up to 80 years is 1 - 2/15 = 13/15. So, the answer is 13/15.

**Question 20**
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The locus of a point equidistant from the intersection of lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a

**Answer Details**

The locus of a point equidistant from the intersection of two lines is the perpendicular bisector of the line segment joining the intersection point of the lines. Therefore, we need to find the intersection point of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0, and then find the perpendicular bisector passing through that point. To find the intersection point of the two lines, we can solve their simultaneous equations. Multiplying the first equation by 4 and the second equation by 3, we get: 12x - 28y + 28 = 0 12x - 18y + 3 = 0 Subtracting the second equation from the first, we get: -10y + 25 = 0 Solving for y, we get: y = 5/2 Substituting this value of y in either of the original equations, we get: 3x - 7(5/2) + 7 = 0 3x = 17/2 x = 17/6 So, the intersection point of the lines is (17/6, 5/2). Now, we need to find the perpendicular bisector of the line segment joining this point with any other point equidistant from the intersection point. Since the question does not specify any other point, we can assume that we are looking for the perpendicular bisector of the line segment joining the intersection point with itself. This means we need to find the equation of the line passing through the point (17/6, 5/2) and perpendicular to the line joining the intersection point with itself, which is the x-axis. The line passing through (17/6, 5/2) and perpendicular to the x-axis has a slope of 0, so its equation is of the form y = b, where b is the y-coordinate of the point (17/6, 5/2). Therefore, the equation of the perpendicular bisector is: y = 5/2 This is the equation of a line parallel to the x-axis and passing through the point (0, 5/2). Therefore, the locus of a point equidistant from the intersection of the lines 3x - 7y + 7 = 0 and 4x - 6y + 1 = 0 is a line parallel to 7x + 13y + 8 = 0, which is.

**Question 21**
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In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.

**Answer Details**

Join PR

QRP = QPR

= 180 - 80 = 100/20 = 50o

SRP = SPR

= 180 - 30 = 150/2 = 75o

∴ SPQ = SPR - QPR

= 75 - 50 = 25o

**Question 22**
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Simplify 3x+152y+6xy

**Question 23**
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Find the standard deviation of 2,3,8,10 and 12

**Answer Details**

x(x−ϰ)(x−ϰ)22−5253−41681110391252576
$\begin{array}{ccc}x& (x-\varkappa )& (x-\varkappa {)}^{2}\\ 2& -5& 25\\ 3& -4& 16\\ 8& 1& 1\\ 10& 3& 9\\ 12& 5& 25\\ & & 76\end{array}$

S.D = √(x−ϰ)2n
$\sqrt{\frac{(x-\varkappa {)}^{2}}{n}}$

S.D = √765
$\sqrt{\frac{76}{5}}$

S.D = 3.9

**Question 24**
**Report**

Evaluate n+1Cn-2 If n =15

**Answer Details**

n+1(n−2)(n+1)!
$\frac{n+1(n-2)}{(n+1)!}$

(n+1)+(n−2)!(n−2)!(n+1)!
$\frac{(n+1)+(n-2)!(n-2)!}{(n+1)!}$

(n+1)(n+1−1)(n+1−2)(n+1−3)!3!(n−2)!
$\frac{(n+1)(n+1-1)(n+1-2)(n+1-3)!}{3!(n-2)!}$

(n+1)(n)(n−1)(n−2)!3!(n−2)!
$\frac{(n+1)(n)(n-1)(n-2)!}{3!(n-2)!}$

(n+1)(n)(n−1)3!
$\frac{(n+1)(n)(n-1)}{3!}$

Since n = 15

(15+1)(15)(15−1)3!
$\frac{(15+1)(15)(15-1)}{3!}$

16×15×143×2×1
$\frac{16\times 15\times 14}{3\times 2\times 1}$

= 560

**Question 25**
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Find the range of 4,9,6,3,2,8,10 and 11

**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 set {4, 9, 6, 3, 2, 8, 10, 11}, we need to find the largest and smallest values in the set and then subtract the smallest value from the largest value. The smallest value in the set is 2 and the largest value is 11. So, the range is 11 - 2 = 9. So, the range of the set {4, 9, 6, 3, 2, 8, 10, 11} is 9.

**Question 26**
**Report**

The grades of 36 students in a test are shown in the pie chart above. How many students had excellent?

**Answer Details**

Angle of Excellent

= 360 - (120+80+90)

= 360 - 290

= 70∘
$\circ $

If 360∘
$\circ $ represents 36 students

1∘
$\circ $ will represent 36/360

50∘
$\circ $ will represent 36/360 * 70/1

= 7

**Question 27**
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U is inversely proportional to the cube of V and U = 81 when V = 2. Find U when V = 3

**Answer Details**

When two quantities, U and V, are inversely proportional, it means that as one of them increases, the other decreases, and vice versa, in such a way that their product remains constant. In mathematical terms, we can express this relationship as U*V^3 = k, where k is a constant. In this problem, we are told that U is inversely proportional to the cube of V. Therefore, we can write U*V^3 = k, where k is a constant of proportionality that we need to find. We are also given that U equals 81 when V equals 2. Substituting these values into the equation, we get: 81*2^3 = k k = 648 Now that we have the constant of proportionality, we can use the equation U*V^3 = 648 to find U when V equals 3: U*3^3 = 648 U*27 = 648 U = 648/27 Simplifying this expression, we get U = 24. Therefore, the answer is 24.

**Question 28**
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Evaluate ∫31(X2−1)dx

**Answer Details**

∫31(x2−1)dx=[13x2−x]31=(9−3)−(13−1)=6−(−23)=6+23=623

**Question 29**
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In how many ways can the letters of the word TOTALITY be arranged?

**Answer Details**

8!3!
$\frac{8!}{3!}$

8×7×6×5×4×3×2×13×2×1
$\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1}$

==> 8 x 7 x 6 x 5 x 4 = 6720

**Question 30**
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If y varies directly as √n $n$ and y = 4 when n = 4, find y when n = 179

**Answer Details**

y α√