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**Question 1**
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In the diagram above, /PQ/ = /PS/ and /QR/ = /SR/. Which of the following is/are true? i. the line PR bisects ?QRS ii. The line PR is the perpendicular bisector of the line segment QS iii. Every point on PR is equidistant from SP and QP

**Answer Details**

Given that /PQ/ = /PS/ and /QR/ = /SR/, we can see that triangle QPR and SPR are congruent by the Side-Side-Side (SSS) criterion. Hence, we have ∠QPR = ∠SPR and ∠PQR = ∠PSR. i. From the above statement, we can say that the line PR bisects ∠QRS and ∠QSR, but it does not necessarily bisect the whole angle ∠QSRQ. Hence, statement (i) is true. ii. As triangle QPR and SPR are congruent, we can say that /QP/ = /SP/ and /QR/ = /SR/. Hence, the line PR is the perpendicular bisector of the line segment QS. Therefore, statement (ii) is also true. iii. Every point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. As we have already proved that line PR is the perpendicular bisector of the line segment QS, every point on PR is equidistant from SP and QP. Hence, statement (iii) is true. Therefore, all the three statements are true, and the correct answer is (v) I, II and III.

**Question 2**
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If 3p - q = 6 and 2p + 3q = 4, find q

**Answer Details**

To find the value of **q**, we need to solve the given system of equations:

3**p** - **q** = 6 ...(1)

2**p** + 3**q** = 4 ...(2)

One way to solve this system is to use the method of elimination, where we eliminate one of the variables by adding or subtracting the equations.

Multiplying equation (1) by 3, we get:

9**p** - 3**q** = 18 ...(3)

Now, we can eliminate **q** by adding equations (2) and (3):

2**p** + 3**q** = 4

9**p** - 3**q** = 18

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11**p** = 22

Dividing both sides by 11, we get:

**p** = 2

Substituting this value of **p** in equation (1), we get:

3(2) - **q** = 6

Simplifying this, we get:

6 - **q** = 6

Subtracting 6 from both sides, we get:

-**q** = 0

Dividing both sides by -1, we get:

**q** = 0

Therefore, **q** = 0 is the solution of the given system of equations.

In summary, to find the value of **q**, we used the method of elimination by multiplying equation (1) by 3 and adding it to equation (2) to eliminate **q**. This resulted in finding the value of **p** as 2. Substituting this value of **p** in equation (1), we found the value of **q** as 0.

**Question 3**
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Each interior angle of a regular nonagon(nine sided polygon) is equal to

**Answer Details**

A nonagon has nine sides and nine interior angles. In a regular nonagon, all the sides and angles are equal. To find the measure of each interior angle of a regular nonagon, we can use the formula: Interior angle = (n - 2) x 180° / n where n is the number of sides of the polygon. Substituting n = 9 in the formula, we get: Interior angle = (9 - 2) x 180° / 9 = 7 x 20° = 140° Therefore, each interior angle of a regular nonagon is equal to 140°. So the correct answer is option (C) 140^{o}.

**Question 4**
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Simplufy Log_{10}4 + Log_{10}25

**Answer Details**

Using the logarithmic identity Log_{b}(MN) = Log_{b}M + Log_{b}N, we can simplify Log_{10}4 + Log_{10}25 as follows: Log_{10}4 + Log_{10}25 = Log_{10}(4 x 25) = Log_{10}100 Since 100 is 10 raised to the power of 2, we can rewrite this as: Log_{10}100 = Log_{10}(10^{2}) = 2 Therefore, the simplified expression is 2.

**Question 5**
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In the diagram, ?PMN = ?PRQ and ?PNM = ?PQR. If /Pm/ = 3cm, /MQ/ = 7cm and /PN/ = 5cm, find /NR/

**Question 6**
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The 6th term of a G.P is -2 and its first term is 18. What is the common ratio?

**Question 7**
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In the diagram above, O is the center of the circle of radius 3.5cm, ∠POQ = 60°. What is the area of the minor sector POQ?

[Take π = 22/7].

**Question 8**
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A boy measured the length and breath of a rectangular lawn as 59.6m and 40.3m respectively instead of 60m and 40m. What is the percentage error in his calculation of the perimeter of the lawn?

**Answer Details**

The correct perimeter of the rectangular lawn can be calculated by adding twice the length and twice the breadth, i.e., 2 × length + 2 × breadth = 2 × 60m + 2 × 40m = 120m + 80m = 200m The perimeter calculated by the boy would be: 2 × 59.6m + 2 × 40.3m = 119.2m + 80.6m = 199.8m The difference between the correct perimeter and the calculated perimeter is: 200m - 199.8m = 0.2m To find the percentage error, we divide the difference by the correct perimeter and multiply by 100: (0.2m / 200m) × 100% = 0.1% Therefore, the percentage error in the boy's calculation of the perimeter of the lawn is 0.1%. The correct option is: 0.1%.

**Question 9**
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Find, correct to 1 decimal place, the volume of cylinder of height 8cm and base radius 3cm. [Take π = 3.142]

**Answer Details**

The formula for the volume of a cylinder is V = πr^{2}h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Substituting the given values in the formula, we have: V = π(3cm)^{2}(8cm) V = 72π cm^{3} V ≈ 226.1cm^{3} (rounded to 1 decimal place, using π ≈ 3.142) Therefore, the correct option is (D) 226.2cm^{3}.

**Question 10**
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Convert the decimal number 89 to a binary number

**Answer Details**

To convert a decimal number to binary, we need to continuously divide the decimal number by 2, until the quotient becomes 0. The binary number is obtained by writing the remainders (0 or 1) obtained in reverse order. Here's the step-by-step process: - Divide 89 by 2. Quotient is 44 and remainder is 1. - Divide 44 by 2. Quotient is 22 and remainder is 0. - Divide 22 by 2. Quotient is 11 and remainder is 0. - Divide 11 by 2. Quotient is 5 and remainder is 1. - Divide 5 by 2. Quotient is 2 and remainder is 1. - Divide 2 by 2. Quotient is 1 and remainder is 0. - Divide 1 by 2. Quotient is 0 and remainder is 1. So the remainders obtained in reverse order are: 1 0 1 1 0 0 0 Therefore, the binary representation of 89 is 1011001.

**Question 11**
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two towns X and Y both on latitude 60^{o}S have longitude 27^{o}E and 33^{o}W respectively. Find to the nearest kilometers, the distance between X and Y measured along the parallel of latitude. [Take 2πR = 4 x 10^{4}km, where R is the Radius of the earth]

**Question 12**
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A town P is 150km from a town Q in the direction 050^{o}. What is the bearing of Q from P?

**Question 13**
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The marks obtained by pupils of class are grouped as shown below; 0 - 4, 5 - 9,10 -14, 15 -19. Which of the following is/are not true? l. The mid values of the grouped marks are 2, 7, 12 and 17. II The class interval is 4. III The class boundaries are 0.5, 4.5, 9.5, 14.5 and 19.5

**Answer Details**

The mid values of the grouped marks can be found by taking the average of the upper and lower limits of each class interval. For example, the mid value of the first class interval (0 - 4) is (0+4)/2 = 2. Similarly, the mid values of the other class intervals can be calculated as 7, 12, and 17. The class interval is the difference between the upper limit of a class interval and the lower limit of the previous class interval. For example, the class interval between 0-4 and 5-9 is 5-4 = 1. The class interval in this case is not 4, but rather it is 5-0 = 5. The class boundaries are the values that separate one class interval from another. They are found by adding and subtracting half of the class interval from the upper and lower limits of each class interval. For example, the lower class boundary of the first class interval (0 - 4) is 0 - 0.5 = -0.5, and the upper class boundary of the second class interval (5 - 9) is 9 + 0.5 = 9.5. Therefore, the class boundaries for the given grouped marks are -0.5, 4.5, 9.5, 14.5, and 19.5. Therefore, the statement that is not true is II only, since the class interval is not 4. The correct class interval is 5, as explained above. So the answer is (B) II only.

**Question 14**
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In the diagram above, O is the center of the circle of radius 3.5cm, ?POQ = 60^{o}

Use the information to answer the question below [Take ? = 22/7]

What is the length of the arc PXQ?

**Answer Details**

The length of an arc is given by the formula L = 2πr(θ/360), where r is the radius of the circle and θ is the central angle of the arc in degrees. In this case, the radius of the circle is 3.5cm and the central angle of the arc PXQ is 60 degrees. Substituting these values into the formula, we have: L = 2 x (22/7) x 3.5 x (60/360) L = 11cm Therefore, the length of the arc PXQ is 11cm. Answer (C)

**Question 15**
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In the diagram above, TRQ is a straight line. Find p, if p = 1/3(a + b + c)

**Question 16**
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Make p the subject if formula y =

a + p/a - p

**Question 17**
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In the diagram above, find x correct to the nearest degree

**Question 18**
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If the hypotenuse of a right-angle isosceles triangle is 2, what is the length of each of the other side?

**Answer Details**

In a right-angle isosceles triangle, the two legs are congruent to each other. Let x be the length of each leg. By the Pythagorean theorem, we know that: x² + x² = 2² Simplifying the equation gives: 2x² = 4 Dividing both sides by 2, we have: x² = 2 Taking the square root of both sides gives: x = √2 Therefore, the length of each leg is √2, which is approximately 1.41. So, the correct answer is not in the options given, but it is approximately equal to, √2.

**Question 19**
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In the diagram above, KLMN is a cyclic quadrilateral. /KL/ = /KN/, ?NKM = 55^{o} and ?KML = 40^{o}. Find ?LKM

**Question 20**
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The angle of a sector of a circle of radius 35cm is 288^{o}. Find the perimeter of the sector. [Take π = 22/7]

**Question 21**
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The roots of a quadratic equation are -^{1}/_{4} and 3. The quadratic equation is

**Answer Details**

We know that if a quadratic equation has roots α and β, then the equation can be written as: (x - α)(x - β) = 0 Expanding the above expression, we get: x^{2} - (α + β)x + αβ = 0 Here, the roots of the quadratic equation are -^{1}/_{4} and 3. Therefore, α = -^{1}/_{4} and β = 3. Substituting these values in the above equation, we get: x^{2} - (α + β)x + αβ = 0 x^{2} - (-^{1}/_{4} + 3)x + (-^{1}/_{4} × 3) = 0 x^{2} - ^{11}/_{4} x - ^{3}/_{4} = 0 Hence, the quadratic equation is 4x^{2} - 11x - 3 = 0. Therefore, the correct option is: 4x^{2} - 11x - 3 = 0

**Question 22**
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The table gives the distribution of outcomes obtained when a die was rolled 100 times.

What is the experimental probability that it shows at most 4 when rolled again?
**Question 23**
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Simplify: \(\frac{5}{x - y} - \frac{4}{y - x}\)

**Answer Details**

To simplify \(\frac{5}{x-y}-\frac{4}{y-x}\), we first notice that \(y-x=-(x-y)\). Thus, we can rewrite the expression as \(\frac{5}{x-y}+\frac{4}{x-y}\). Now, we can combine the fractions by finding a common denominator, which is \(x-y\). Thus, we have \[\frac{5}{x-y}+\frac{4}{x-y}=\frac{5+4}{x-y}=\frac{9}{x-y}.\] Therefore, the simplified expression is \(\boxed{\frac{9}{x-y}}\).

**Question 24**
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The trapezium PQRS parallel to SR,?PQS = 34^{o} and ?SPQ = 2 ?SRQ. Find the size of, SQR

**Question 25**
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If tan x = \(2\frac{2}{5}\), find the value of sin x; 0 \(\leq\) x \(\leq\) 90^{o}

**Question 26**
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Simplify 56x\(^{-4}\) \(\div\) 14x\(^{-8}\)

**Answer Details**

To simplify the expression, we can use the rule of dividing powers with the same base. We divide the coefficients and subtract the exponents: 56x^{-4} ÷ 14x^{-8} = (56/14)x^{(-4)-(-8)} = 4x^{-4+8} = 4x^{4} Therefore, the simplified expression is 4x^{4}.

**Question 27**
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The number of goals scored by a football team in 20 matches is shown in the table above

What is the mean goal scored?

**Question 28**
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The number of goals scored by a football team in 20 matches is shown in the table above

What is the modal goal scored?

**Answer Details**

The modal score is the score that occurs most frequently. Looking at the table, we can see that the score of 2 occurs the most frequently, appearing 6 times. Therefore, the modal goal scored is 2.

**Question 29**
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One side of a rectangle is 8cm and the diagonal is 10cm. What is the area of the rectangle?

**Answer Details**

We know that the diagonal of a rectangle divides it into two right triangles with the diagonal as the hypotenuse, and the sides of the rectangle as the legs of the right triangles. Let's call the other side of the rectangle "x". Using the Pythagorean theorem, we have: 10^{2} = 8^{2} + x^{2} Simplifying and solving for x, we get: x = √(10^{2} - 8^{2}) = √36 = 6 Therefore, the area of the rectangle is: 8 x 6 = 48 cm^{2} Hence, the answer is 48cm^{2}.

**Question 30**
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Factorize 32x^{3} - 8xy^{2}

**Answer Details**

We can factorize the given expression by first finding the greatest common factor (GCF) of the two terms, which is 8x. We can factor out the GCF from the given expression as: 32x^{3} - 8xy^{2} = 8x(4x^{2} - y^{2}) We can then use the identity a^{2} - b^{2} = (a + b)(a - b) to factorize the expression further: 8x(4x^{2} - y^{2}) = 8x(2x + y)(2x - y) Therefore, the fully factorized form of 32x^{3} - 8xy^{2} is 8x(2x + y)(2x - y). So, the correct option is: - 8x(2x + y)(2x - y)

**Question 32**
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