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**Question 1**
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If p + 1, 2P - 10, 1 - 4p2are three consecutive terms of an arithmetic progression, find the possible values of p

**Answer Details**

2p - 10 = p+1+1−4P22
$\frac{p+1+1-4{P}^{2}}{2}$ (Arithmetic mean)

= 2(2p - 100 = p + 2 - 4P2)

= 4p - 20 = p + 2 - 4p2

= 4p2 + 3p - 22 = 0

= (p - 2)(4p + 11) = 0

∴ p = 2 or -411

**Question 2**
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Make ax $\frac{a}{x}$ the subject of formula x+1x?a

**Answer Details**

x+ax?a
$\frac{x+a}{x?a}$ = m

x + a = mx - ma

a + ma = mx - x

a(m + 1) = x(m - 1)

ax
$\frac{a}{x}$ = m?1m+a

**Question 3**
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a cylindrical drum of diameter 56 cm contains 123.2 litres of oil when full. Find the height of the drum in centimeters

**Answer Details**

To solve the problem, we need to use the formula for the volume of a cylinder, which is V = πr²h, where V is the volume, r is the radius, and h is the height of the cylinder. We are given the diameter of the drum, which is 56 cm. To find the radius, we need to divide the diameter by 2: radius (r) = diameter / 2 = 56 cm / 2 = 28 cm We are also given that the drum contains 123.2 litres of oil when full. To convert litres to cubic centimeters, we need to multiply by 1000: 123.2 litres * 1000 = 123200 cubic centimeters Now we can use the formula for the volume of a cylinder to find the height (h): V = πr²h h = V / (πr²) h = 123200 / (π28²) h ≈ 123200 / 2463.47 h ≈ 50.00 cm Therefore, the height of the drum is approximately 50.00 cm.

**Question 4**
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If x, y can take values from the set (1, 2, 3, 4), find the probability that the product of x and y is not greater than 6

**Answer Details**

∣∣ ∣ ∣ ∣ ∣ ∣∣123411234224683369124481216∣∣ ∣ ∣ ∣ ∣ ∣∣
$\left|\overline{)\begin{array}{ccccc}& 1& 2& 3& 4\\ 1& 1& 2& 3& 4\\ 2& 2& 4& 6& 8\\ 3& 3& 6& 9& 12\\ 4& 4& 8& 12& 16\end{array}}\right|$

P (product of x and y not greater than 6) = 1016
$\frac{10}{16}$

= 58

**Question 5**
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Evaluate ∫π2 $2\pi $(sec2 x - tan2x)dx

**Answer Details**

∫π2
$2\pi $(sec2 x - tan2x)dx

∫π2
$2\pi $ dx = [X]π2
$2\pi $

= π
$\pi $ - 2 + c

when c is an arbitrary constant of integration

**Question 6**
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In a recent zonal championship games involving 10 teams, teams X and Y were given probabilities 25 $\frac{2}{5}$ and 13 $\frac{1}{3}$ respectively of winning the gold in the football event. What is the probability that either team will win the gold?

**Answer Details**

p(x) = 25
$\frac{2}{5}$ p(y) = 13
$\frac{1}{3}$

p(x or y) = p(x ∪ y)

= p(x) + p(y)

= 25
$\frac{2}{5}$ + 13
$\frac{1}{3}$

= 115

**Question 7**
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From the top of a vertical mast 150m high., two huts on the same ground level are observed. One due east and the other due west of the mast. Their angles of depression are 60o and 45o respectively. Find the distance between the huts

**Answer Details**

150Z
$\frac{150}{Z}$ = tan 60o,

Z = 150tan60o
$\frac{150}{tan{60}^{o}}$

= 1503
$\frac{150}{3}$

= 50√3
$3$cm

150XxZ
$\frac{150}{XxZ}$ = tan45o = 1

X + Z = 150

X = 150 - Z

= 150 - 50√3
$3$

= 50( √3
$3$ - √3
$3$)m

**Question 8**
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Two chords PQ and RS of a circle intersected at right angles at a point inside the circle. If ∠QPR = 35o,find ∠PQS

**Answer Details**

Since PQ and RS intersect inside the circle at right angles, then the line joining the point of intersection to the center of the circle will bisect both chords. Let O be the center of the circle, and let T be the point of intersection of the two chords. Then, angle QTR = 90 degrees and the angle subtended by chord PQ at the center O is twice angle QPR. Therefore, angle POQ = 2 * angle QPR = 70 degrees (since angle QPR = 35 degrees). Similarly, angle ROS = 70 degrees. Since PQ and RS are chords of a circle, then angle POQ = angle PTS and angle ROS = angle TQS. Thus, angle PTS + angle TQS = 140 degrees. Also, angle PTS + angle PTQ + angle QTS = 180 degrees (because they form a straight line). Therefore, angle TQS = 180 - 140 - 90 = 50 degrees. Since angle PQT = angle RQT (because they are opposite angles), then angle PQS = angle RQS = (180 - angle QTS)/2 = (180 - 50)/2 = 65 degrees. Therefore, the answer is 55 degrees.

**Question 10**
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The midpoint of the segment of the line y = 4x + 3 which lies between the x-ax 1 is and the y-ax 1 is

**Answer Details**

To find the midpoint of a line segment, we need to find the average of the endpoints. The x-intercept of the line y = 4x + 3 is found by setting y = 0 and solving for x: 0 = 4x + 3 x = -3/4 So the x-coordinate of the midpoint is the average of -3/4 and 0: x = (-3/4 + 0)/2 = -3/8 To find the y-coordinate of the midpoint, we plug in x = -3/8 to the equation of the line: y = 4(-3/8) + 3 = -3/2 + 3 = 3/2 So the midpoint is (-3/8, 3/2). Therefore, the answer is (-3/8, 3/2).

**Question 11**
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If m and n are the mean and median respectively of the set of numbers 2, 3, 9, 7, 6, 7, 8, 5, find m + 2n to the nearest whole number

**Answer Details**

To find the mean (m), you need to add up all the numbers in the set and then divide by the total number of numbers. In this case, the sum is 47 and the total number of numbers is 8, so the mean is 47/8 = 5.875. To find the median (n), you need to arrange the numbers in order from smallest to largest and then find the middle number. In this case, the numbers in order are: 2, 3, 5, 6, 7, 7, 8, 9. The middle number is 7, so the median is 7. To find m + 2n, you just need to substitute the values of m and n into the expression and solve. m + 2n = 5.875 + 2(7) = 19.875. To the nearest whole number, m + 2n is 20. Therefore, the answer is: 19.

**Question 12**
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Differentiate xcosx $\frac{x}{cosx}$ with respect to x

**Answer Details**

let y = xcosx
$\frac{x}{cosx}$ = x sec x

y = u(x) v (x0

dydx
$\frac{dy}{dx}$ = Udydx
$\frac{dy}{dx}$ + Vdudx
$\frac{du}{dx}$

dy x [secx tanx] + secx

x = x secx tanx + secx

**Question 13**
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In the diagram, QTR is a straight line and < PQT = 30?
$?$. find the sin of < PTR

**Answer Details**

10sin30o=15sinx=100.5=15sinx $\frac{10}{\mathrm{sin}?{30}^{o}}=\frac{15}{\mathrm{sin}?x}=\frac{10}{0.5}=\frac{15}{\mathrm{sin}?x}$

1520=sinx $\frac{15}{20}=\mathrm{sin}?x$

sin x = 1520=34 $\frac{15}{20}=\frac{3}{4}$

N.B x = < PRQ

**Question 14**
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In the figure, PQST is a parallelogram and TSR is a straight line. If the area of △
$\u25b3$QRS is 20cm2, find the area of the trapezium PQRT.

**Answer Details**

A△ $\u25b3$ = 12 $\frac{1}{2}$ x 8 x h = 20

= 12 $\frac{1}{2}$ x 8 x h = 4h

h = 204 $\frac{20}{4}$

= 5cm

A△ $\u25b3$(PQTS) = L x H

A△ $\u25b3$PQRT = A△ $\u25b3$QSR + A△ $\u25b3$PQTS

20 + 50 = 70cm2

ALTERNATIVE METHOD

A△ $\u25b3$PQRT = 12 $\frac{1}{2}$ x 5 x 28

= 70cm2

**Question 15**
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Find the variance of the numbers k, k+1, k+2.

**Answer Details**

To find the variance of the numbers k, k+1, k+2, we can use the formula for variance which is the average of the squared differences from the mean. First, we need to find the mean of the three numbers. Mean = (k + k + 1 + k + 2) / 3 = (3k + 3) / 3 = k + 1 So, the mean is k + 1. Next, we find the squared differences from the mean for each number: For k, the difference from the mean is k - (k+1) = -1. The squared difference is (-1)^2 = 1. For k+1, the difference from the mean is (k+1) - (k+1) = 0. The squared difference is 0^2 = 0. For k+2, the difference from the mean is (k+2) - (k+1) = 1. The squared difference is 1^2 = 1. Now we can find the variance: Variance = [(1^2 + 0^2 + 1^2) / 3] = 2/3 = 0.67 (rounded to two decimal places) Therefore, the answer is option (A) 2/3 or as a percentage approximately 66.7%.

**Question 16**
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The kinetic element with respect to the multiplication shown in the diagram below is

⊕pprsprprpqpqrsrrrrrsqsrq

**Answer Details**

**Question 17**
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Evaluate [10.03 $\frac{1}{0.03}$ ÷ $\xf7$ 10.024 $\frac{1}{0.024}$]-1 correct to 2 decimal places

**Answer Details**

[10.03
$\frac{1}{0.03}$ + 10.024
$\frac{1}{0.024}$]

= [10.03×0.024
$\frac{1}{0.03\times 0.024}$]-1

= [0.0240.003
$\frac{0.024}{0.003}$]-1

= 0.030.024
$\frac{0.03}{0.024}$

= 3024
$\frac{30}{24}$ = 1.25

**Question 18**
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In the diagram above; O is the centre of the circle and |BD| = |DC|. If ∠DCB = 35o, find ∠BAO.

**Answer Details**

**Question 19**
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If b3 = a-2 and c13 $\frac{1}{3}$ = a12 $\frac{1}{2}$b, express c in terms of a

**Answer Details**

c13
$\frac{1}{3}$ = a12
$\frac{1}{2}$b

= a12
$\frac{1}{2}$b x a-2

= a-32
$\frac{3}{2}$

= (c13
$\frac{1}{3}$)3

= (a-32
$\frac{3}{2}$)13
$\frac{1}{3}$

c = a-12

**Question 20**
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Find the positive value of x if the standard deviation of the numbers 1, x + 1 is 6

**Answer Details**

mean (x) = 1+x+1+2x+13
$\frac{1+x+1+2x+1}{3}$

= 3x+33
$\frac{3x+3}{3}$

= 1 + x

X(X−X)(X−X)21−xx2x+1002x+1xx22x2
$\begin{array}{ccc}X& (X-X)& (X-X{)}^{2}\\ 1& -x& {x}^{2}\\ x+1& 0& 0\\ 2x+1& x& {x}^{2}\\ & & 2{x}^{2}\end{array}$

S.D = √∑(x−7)2∑f
$\sqrt{\frac{\sum (x-7{)}^{2}}{\sum f}}$

= √(6)2
${\sqrt{(6)}}^{2}$

= 2x23
$\frac{2{x}^{2}}{3}$

= 2x2

= 18

x2 = 9

∴ x = ±
$\pm $ √9
$9$

= ±
$\pm $3

**Question 21**
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If 10112 + x7 = 2510, solve for X.

**Answer Details**

10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o

= 8 + 0 + 2 + 1

= 1110

x7 = 2510 - 1110

= 1410

71472R00R2
$\begin{array}{cc}7& 14\\ 7& 2R0\\ & 0R2\end{array}$

X = 207

**Question 22**
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The bar chart shows the distribution of marks scored by 60 pupils in a test in which the maximum score was 10. If the pass mark was 5, what percentage of the pupils failed the test?

**Answer Details**

x012345678910f194710879821
$\begin{array}{cccccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ f& 1& 9& 4& 7& 10& 8& 7& 9& 8& 2& 1\end{array}$

no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = 2560 $\frac{25}{60}$ x 100%

= 41.70%

**Question 23**
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The locus of all points at a distance 8cm from a point N passes through points T and S. If S is equidistant from T and N, find the area of triangle STN.

**Question 24**
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find the equation of the curve which passes through by 6x - 5

**Answer Details**

m = dydv
$\frac{dy}{dv}$ = 6x - 5

∫dy = ∫(6x - 5)dx

y = 3x2 - 5x + C

when x = 2, y = 5

∴ 5 = 3(2)2 - 5(2) +C

C = 3

∴ y = 3x2 - 5x + 3

**Question 25**
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A chord of a circle radius √3cm $\sqrt{3cm}$ subtends an angle of 60∘ $\circ $ on the circumference of he circle. Find the length of the chord

**Answer Details**

**Question 26**
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