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**Question 1**
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Find the tax on an income of N20,000 if no tax is paid on the first N10,000 and tax is paid at N50 in N1000 on the next N5000 and at N55 in N1000 on the remainder

**Answer Details**

To calculate the tax on an income of N20,000 using the given tax rules, we need to break down the income into three parts: 1. The first N10,000: No tax is paid on this amount. 2. The next N5,000: Tax is paid at N50 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N50 = N250. 3. The remainder of N5,000: Tax is paid at N55 for every N1000, so the tax on N5,000 would be (N5,000/N1,000) x N55 = N275. Therefore, the total tax on an income of N20,000 would be N0 + N250 + N275 = N525. In summary, the tax on an income of N20,000 using the given tax rules is N525, which is option (D).

**Question 2**
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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

**Question 3**
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In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find angle PQR.

**Answer Details**

Q1 = 50(alternate angle)

Q2 = 180o - 110o (straight line angle)

Q2 = 70

PQR = Q1 + Q2

= 50o + 70o

= 120o

**Question 5**
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If dy/dx = x + cos x, find y

**Answer Details**

We need to integrate the given expression to find the value of y. Integrating both sides with respect to x, we get: ∫dy = ∫(x + cos x)dx y = 1/2 * x^2 + sin x + c where c is the constant of integration. Therefore, the correct answer is: - x^2/2 + sin x + c

**Question 6**
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Evaluate ∫0−4(1−2x)dx

**Answer Details**

∫0−4(1−2x)dx=[x−x2]0−4=(0−0+C)−(−4(−4)2+C)=C−(−4−16+C)=C−(−20+C)=C+20−C=20

**Question 7**
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Differentiate (x2 - 1/x)2 with respect to x

**Answer Details**

The given expression is (x^2 - 1)/x^2. To differentiate this expression with respect to x, we can use the quotient rule of differentiation. The quotient rule states that the derivative of (f(x)/g(x)) = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2. Applying the quotient rule, we get: [(x^2)*2x - (x^2 - 1)*2x]/(x^2)^2 Simplifying the expression further, we get: [2x^3 - 2x^3 + 2x]/x^4 = 2/x^3 Therefore, the answer is option C: 4x^2 - 2 - 2/x^3.

**Question 8**
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In the diagram P, Q, R, S are points on the circle RQS = 30o. PRS = 50o and PSQ = 20o. What is the value of xo + yo?

**Answer Details**

Draw a line from P to Q

< PQS = < PRS (angle in the sam segment)

< PQS = 50o

Also, < QSR = < QPR(angles in the segment)

< QPR = xo

x + y + 5= = 180(angles in a triangle)

x + y = 180 - 50

x + y = 130o

**Question 9**
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Convert 22324 to base six

**Answer Details**

1st convert to base 10

22324 = 2x43 + 2x42 + 3x41 + 2x40

= 2x64 + 2x16 + 3x4 + 2x1

= 128 + 32 + 12 + 2

= 174 convert to base 6

6/174

6/29 R 0

6/4 R 5

6/0 R 5

= 4506

**Question 10**
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The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English English and Health education as the their favourite subjects?

**Answer Details**

(52160=36160 $\frac{52}{160}=\frac{36}{160}$)% = 88160 $\frac{88}{160}$%

= 55%

**Question 11**
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The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?

**Answer Details**

To find the values of n and m, we need to first determine the mode and the number of students who scored 4 or more. The mode is the most frequently occurring score in the set. In this case, the score of 4 appears most frequently, with a total of 12 students achieving that score. Therefore, the mode is 4. To find the number of students who scored 4 or more, we need to add up the frequencies of all scores that are 4 or greater. Adding the frequencies of scores 4, 5, 6, and 7 gives a total of 33 students. Therefore, the values of (n, m) are (33, 4).

**Question 12**
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A binary operation θ defined on the set of real number is such that xθy = xy/6 for all x, y ∈ R. Find the inverse of 20 under this operation when the identity element is 6

**Answer Details**

X ⊕ y = xy/6 = xe/6 = x

Xe = 6x

e = 6

X ⊕ y = xx1/6 where x1 = 20

Xx20/6 = 6

X = 36/20

X = 9/5

**Question 13**
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If y = x2 - x - 12, find the range of values of x for which y ≥ $\ge $ 0

**Answer Details**

y = x2 - x - 12

= (x - 4)(x - 3)

∴ x = 4 or x = -3

-3 < x < 4

But y ≥
$\ge $ 0

∴ -3 < x ≤
$\le $ 4

**Question 14**
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What is the locus of points equidistant from the lines *ax + bc + c* = 0?

**Answer Details**

The given equation of a line can be rewritten in the standard form of a line, which is y = mx + b, where m is the slope and b is the y-intercept. To rewrite the equation, we can solve for y: ax + bc + c = 0 ax + bc = -c y = -(a/b)x - (c/b) The locus of points equidistant from the lines y = -(a/b)x - (c/b) is a line perpendicular to the given lines, with the distance from the origin equal to the absolute value of c/b. This is because the perpendicular bisector of a line segment is equidistant from the endpoints of the segment, and in this case the "endpoints" are the given lines. Since the slope of the given lines is -(a/b), the slope of the perpendicular bisector is b/a. The perpendicular bisector passes through the origin (0,0), so we can find its equation by using the point-slope form of a line: y - 0 = (b/a)x y = (b/a)x This is equivalent to the equation y = mx, where m = b/a, which is the equation of a line passing through the origin with slope b/a. Therefore, the answer is: a line ax - ay + q = 0.

**Question 15**
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If the locus of the points which are equidistant from point P and Q meets line PQ at point N, then PN equals

**Answer Details**

The answer is NQ. By definition, the locus of points that are equidistant from two points P and Q is the perpendicular bisector of the line segment PQ. Therefore, if this locus intersects line PQ at point N, then N must be the midpoint of the segment PQ. So, PN and NQ are equal, and since N is the midpoint of PQ, PN is half of PQ. Thus, PN is equal to NQ.

**Question 16**
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Calculate the logarithm to base 9 of 3-4 * 92 * (81)-1

**Answer Details**

3−4×92×81−1=log9(3−4×92×81−1)=log9(134×92×181)=log9(181×811×181)=log9181=log9192=log99−2=−2log99−2×1=−2

**Question 17**
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The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?

**Answer Details**

P(GTN) = 5/20

P(Qtel) = 3/20

∴P(GTN or Qtel) = (5/20) + (3/20)

= 8/20

= 2/5

**Question 18**
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If tan θ = 5/4, find sin2θ - cos2θ.

**Answer Details**

Tan θ = 5/4

x2 = 52 + 42

= 25 + 16

= 41

x = √41

sin2θ - cos2θ = (5/√41)2 - (4/√41)2

= 25/41 - 16/41

= 9/41

**Question 19**
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simplify 79?2313+2545

**Answer Details**

79−2313+2545=7−6913+2545=1913+(25×54)=1913+12\=192+36=1956=19×65=215

**Question 20**
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Find the value of k if the expression kx3 + x2 - 5x - 2 leaves a remainder 2 when it is divided by 2x + 1

**Answer Details**

F(x) = Q x D + R

Kx3 + x2 - 5x – 2 = Q(2x+1)+R

If 2x+1 = 0 implies x = -1/2

∴k(-1/2)3 + (-1/2)2 -5(-1/2) -2

= Q(2(-1/2) + 1) +2

K(-1/8) + 1/4 + 5/2 - 2 = Q(-1+1)+2

-k/8 + 1/4 + 5/2 - 2 = 0+2

(-k+2+20-16) / 8 = 2

(-k+6) / 8 = 2

-k+6 = 2*8

-k+6 = 16

-k = 16-6

-k = 10

∴k = -10

**Question 21**
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The solution set of the shaded area above is

**Answer Details**

Y = x implies y ≤ x

Y + x = 4 implies y = 4 – x

∴y = 4 – x

∴y ≤ x, y + x ≤ 4 and y ≥ 0

**Question 22**
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If the mean of five consecutive integers is 30, find the largest of the numbers

**Answer Details**

Let's call the smallest of the five consecutive integers "x". Since the numbers are consecutive, we know that the next four integers will be x + 1, x + 2, x + 3, and x + 4. To find the mean of these five numbers, we add them up and divide by 5: (x + (x+1) + (x+2) + (x+3) + (x+4)) / 5 = 30 Simplifying the left side: (5x + 10) / 5 = 30 Canceling the 5's on the left side: x + 2 = 30 Subtracting 2 from both sides: x = 28 Therefore, the five consecutive integers are 28, 29, 30, 31, and 32, and the largest of these numbers is 32. So the correct option is: **32**.

**Question 23**
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In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find

**Answer Details**

To solve this problem, we need to use the fact that the angle between a tangent and a chord of a circle is equal to the angle formed by the chord in the opposite segment. Let's call the point where the tangent MN touches the circle point T. Then, angle MOT is 90 degrees because OT is a radius of the circle and MT is a tangent, and these two lines are perpendicular at point T. Also, angle MTS is 55 degrees because MN is tangent to the circle and makes an angle of 55 degrees with PS. Therefore, angle MTN is 180 - 90 - 55 = 35 degrees because the angles in a triangle add up to 180 degrees. Now, let's consider the triangle OTN. We know that angle OTN is 90 degrees because OT is a radius of the circle and TN is a tangent, and these two lines are perpendicular at point T. We also know that angle MTN is 35 degrees. Therefore, angle OTM is 180 - 90 - 35 = 55 degrees because the angles in a triangle add up to 180 degrees. Finally, we can see that angle PTS is equal to angle OTM because they both intercept the same arc PS. Therefore, angle PTS is also 55 degrees. Since angle PTS is half of angle POS (which is equal to 110 degrees because it intercepts a semicircle), angle POS is 2 * 110 = 220 degrees. Finally, the angle between the tangent MN and the chord PS is equal to angle MTS, which is 55 degrees. Therefore, the answer is option (C) 35o.

**Question 24**
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If T = 2 π √1/g, make g the subject of the formula

**Answer Details**

T = 2 π √1/gT/2π = √1/g

(T/2π)2 = (√1/g)2T2/4π2 = 1/g

T2 x g = 4π2 x l

G = (4π2 x l) / T2

**Question 25**
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In the parallelogram PQRS given, find angle < SQR

**Answer Details**

< QRS = 30∘ $\circ $(corresponding < ), < QSR = 50∘ $\circ $

< SQR = 180∘ $\circ $ - (30 + 50)∘ $\circ $ (Sum of < S in a △ $\u25b3$)

180∘ $\circ $ - 80∘ $\circ $ = 100∘

**Question 26**
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If (*K*2)6 * 36 = 35(*K*4)5, what is the value of k?

**Answer Details**

(K2)6 * 36 = 35(K4)5

k*61 + 2*60 + 3*50 = 3*50 + K*51 + 4*50

6K + 2 = 5K + 4

6K - 5k = 4 - 2

K = 2

**Question 27**
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Find the value of x for which the function 3x3 - 9x is minimum

**Answer Details**

= 3x3 - 9x2

dy/dx = 9x2 - 18x

As dy/dx = 0

9x2 - 18x = 0

9x(x-2) = 0

9x = 0 which implies x = 0

x-2 = 0 implies x = 2

d2y/dx2 = 18x - 18

when x = 0

d2y/dx2< 0 ∴ x is is minimum

dy/dx = 9x2 - 18x

As dy/dx = 0

9x2 - 18x = 0

9x(x-2) = 0

9x = 0 which implies x = 0

x-2 = 0 implies x = 2

d2y/dx2 = 18x - 18

when x = 0

d2y/dx2< 0 ∴ x is is minimum

when x = 2d2y/dx2 = 18

∴ > 0 x is minimum

**Question 28**
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In triangle XYZ, ∠XYZ = 15∘
$\circ $, ∠XZY = 45∘
$\circ $ and lXYl = 7 cm. Find lYZl.

**Answer Details**

= 180 - 60

= 120∘ $\circ $

Using sine rule

xsinx=7sinzxsin120=7sin45x=7sin120sin45x=7sin(180−120)sin45x=7sin60sin45=(7(

= 120∘ $\circ $

Using sine rule

xsinx=7sinzxsin120=7sin45x=7sin120sin45x=7sin(180−120)sin45x=7sin60sin45=(7(