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**Question 1**
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A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of #22,800.00 was paid as import duty, calculate the cost of the engine.

**Question 2**
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In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t.

**Question 3**
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A fair die is tossed twice what is the probability of get a sum of at least 10.

**Question 4**
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In **△LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).**

**Find the area of △LMN **

**Question 5**
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consider the statements:

P = All students offering Literature(L) also offer History(H);

Q = Students offering History(H) do not offer Geography(G).

Which of the Venn diagram correctly illustrate the two statements?

**Question 6**
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What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)?

**Question 7**
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make u the subject in x =\(\frac{2u-3}{3u + 2}\)

**Question 9**
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The height of an equilateral triangle of side is 10 3√ cm. calculate its perimeter.

**Question 10**
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Simplify 2√7- 14/√7+7/21

**Question 11**
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The distance d between two villages east more than 18 KM but not more than 23KM.

which of these inequalities represents the statements?

**Question 12**
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In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t.

**Question 14**
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In the diagram, **?XYZ is produced to T. if |XY| = |ZY| and ?XYT = 40°, find ?XZT**

**Question 15**
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A box contains 40 identical balls of which 10 are red and 12 are blue. if a ball is selected at random from the box what is the probability that it is neither red nor blue?

**Question 16**
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\(\overline{XY}\) is a line segments with the coordinates X (- 8,- 12) and Y(p,q). if the midpoint of \(\overline{XY}\) is (-4,-2) find the coordinates of Y.

**Answer Details**

The midpoint of a line segment is the point that is exactly halfway between the two endpoints of the segment. To find the midpoint, we take the average of the x-coordinates and the y-coordinates of the endpoints. So, for line segment XY with endpoints X (-8,-12) and Y (p,q), the midpoint is (-4,-2). Therefore, the average of the x-coordinates of X and Y is -4: (-8 + p)/2 = -4 Solving for p, we find: p = -8 + 2 * -4 = 0 Similarly, the average of the y-coordinates of X and Y is -2: (-12 + q)/2 = -2 Solving for q, we find: q = -12 + 2 * -2 = 8 Therefore, the coordinates of Y are (0,8).

**Question 17**
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A solid brass cube is melted and recast as a solid cone of height h and base radius r. If the height of the cube is h, find r in terms of h.

**Question 19**
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The sum of the interior angles of a regular polygon with k sides is (3k-10) right angles. Find the size of the exterior angle?

**Question 20**
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A cyclist moved at a speed of Xkm/h for 2 hours. He then increased his speed by 2 km/h for the next 3 hours.

If the total distance covered is 36 km, calculate his initials speed.

**Question 21**
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For what value of x is \(\frac{4 - 2x}{x + 1}\) undefined.

**Answer Details**

A fraction is undefined when its denominator is equal to zero. Therefore, we need to find the value of x that makes the denominator x + 1 equal to zero.

x + 1 = 0

x = -1

Therefore, the value of x that makes the fraction undefined is x = -1

**Question 22**
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Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx)

**Question 23**
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If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.

**Question 24**
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The circumference of a circular track is 9km. A cyclist rides round it a number of times and stops after covering a distance of 302km. How far is the cyclist from the starting point?

**Question 25**
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500 tickets were sold for a concert tickets for adults and children were sold at $4.50 and $3.00 respectively if the total receipts for the concerts was $1987.50 how many tickets for adults were sold?

**Answer Details**

Let's assume that x is the number of tickets sold for adults, and y is the number of tickets sold for children. We know that the total number of tickets sold is 500, so we can write an equation: x + y = 500 We also know that the price of an adult ticket is $4.50 and the price of a child ticket is $3.00. So we can write another equation based on the total receipts: 4.5x + 3y = 1987.5 Now we have two equations with two unknowns, and we can solve for x, the number of adult tickets sold. One way to do this is to use the first equation to solve for y: y = 500 - x Then we can substitute this expression for y into the second equation: 4.5x + 3(500 - x) = 1987.5 Simplifying and solving for x: 4.5x + 1500 - 3x = 1987.5 1.5x = 487.5 x = 325 So the number of adult tickets sold is 325. Therefore, the correct answer is option (A): 325.

**Question 26**
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if tanθ = \(frac{3}{4}\), 180° < θ < 270°, find the value of cosθ.

**Question 28**
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From a point T, a man moves 12km due west and then moves 12km due south to another point Q. Calculate the bearing of T from Q.

**Question 30**
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A trapezium of parellel sides 10cm and 21cm and height 8cm is inscribed in a circle of radius 7cm. calculate the area of the region not covered by the trapezium.

*π =\(\frac{22}{7}\) *

**Question 31**
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Mensah is 5 years old and joyce is thrice as old as mensah. In how many years will joyce be twice as old as Mensah?

**Question 32**
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Which of the following is not an exterior angle of a regular polygon?

**Question 33**
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The equation of a line is given as 3 x - 5y = 7. Find its gradient (slope)

**Answer Details**

The equation of a line in the slope-intercept form is y = mx + c, where m is the gradient (slope) of the line and c is the y-intercept. However, the given equation of the line is not in the slope-intercept form, but in the standard form. To find the slope of the line, we need to rewrite the equation in the slope-intercept form. We can do this by solving the equation for y: 3x - 5y = 7 -5y = -3x + 7 y = (3/5)x - 7/5 Comparing this with the slope-intercept form, we can see that the gradient (slope) of the line is 3/5. Therefore, the answer is 3/5.

**Question 34**
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correct 0.007985 to three significant figures.

**Question 35**
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In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.

If angle ABC equals 158°, find ?ADE

**Question 36**
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The diagonal of a rhombus are 12cm and 5cm. calculate its perimeter

**Question 37**
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Find, correct to two decimal, the mean of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\).

**Question 38**
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**Question 39**
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Find The quadratic Equation Whose Roots Are -2q And 5q.

**Question 41**
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find the first quartile of 7,8,7,9,11,8,7,9,6 and 8.

**Question 42**
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If 4x+2y=16 and 6x-2y=4 , find the value of (y-x).

**Question 43**
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A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area

**Question 44**
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If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n.

**Question 45**
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A man will be (x+10)years old in 8years time. If 2years ago he was 63 years., find the value of x

**Answer Details**

A man will be (x+10) years old in 8years time.

As at today, he is x + 2 years of age.

The man was 63 years old 2 years ago, so he is 63+2=65 now.

8 years from now, he will be 65+8=73.

He will be (x+10) years old when he is 73. So

x+10=73

x=73-10=63

**Question 46**
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if p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q.

**Question 47**
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If \(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{p}{(x-3)(x -2)}\), find p.

**Question 48**
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The pie chart represents the distribution of fruits on display in the shop if there are 60 apples on display how many oranges are there?

**Question 49**
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Height(cm) | 160 | 161 | 162 | 163 | 164 | 165 |

No. of players | 4 | 6 | 3 | 7 | 8 | 9 |

the table shows the height of 37 players of a basketball team calculates correct to one decimal place the mean height of the players.

**Question 50**
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.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics.

If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information

(ii) how many students study both subjects

(iii) find the fraction of the class that study Biology but not Physics.

(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15m by 8 m which was to be carpeted.

The cost of carpeting was Gh¢ 890.00 per square metre. If a total of GH 216,120.00 was spent on painting and carpeting, how much was the cost of painting?

(a)

(i) Here is a Venn diagram to represent the information given:

B / \ / \ P/_____\ | | 80 ---

where **P** represents the students who study Physics and **B** represents those who study Biology. The region inside the circle **P** but outside the intersection represents students who only study Physics, while the region inside the circle **B** but outside the intersection represents students who only study Biology. The intersection represents students who study both subjects.

(ii) To find the number of students who study both subjects, we need to find the size of the intersection. Let **x** be the number of students who study both subjects. Then we can write two equations based on the given information:

- \(\frac{3}{4}\times80\) students study Biology.
- \(\frac{3}{5}\times80\) students study Physics.

Using these equations, we can solve for **x**:

- Biology students: \(\frac{3}{4}\times80 = 60\)
- Physics students: \(\frac{3}{5}\times80 = 48\)
- Total students: 80
- Students who study both subjects:
**x**

From the diagram, we can see that the total number of students is 80, so we have:

60 + **x** + 48 = 80

Simplifying this equation, we get:

**x** = 12

Therefore, 12 students study both subjects.

(iii) To find the fraction of the class that study Biology but not Physics, we need to find the size of the region inside the circle **B** but outside the intersection, and divide by the total number of students. Let **y** be the number of students who study Biology but not Physics. Then we can write another equation based on the given information:

- \(\frac{3}{4}\times80\) students study Biology.
**x**students study both subjects.

Using these equations, we can solve for **y**:

- Biology students: \(\frac{3}{4}\times80 = 60\)
- Students who study both subjects:
**x**= 12 - Students who study Biology but not Physics:
**y**

From the diagram, we can see that:

**y** + **x** = 60

Substituting **x** = 12, we get:

**y** + 12 = 60

Simplifying this equation, we get:

**y** = 48

Therefore, 48 students study Biology but not Physics. The fraction of the class that study Biology but not Physics is:

\(\frac{48}{80} = \frac{3}{5

**Answer Details**

(a)

(i) Here is a Venn diagram to represent the information given:

B / \ / \ P/_____\ | | 80 ---

where **P** represents the students who study Physics and **B** represents those who study Biology. The region inside the circle **P** but outside the intersection represents students who only study Physics, while the region inside the circle **B** but outside the intersection represents students who only study Biology. The intersection represents students who study both subjects.

(ii) To find the number of students who study both subjects, we need to find the size of the intersection. Let **x** be the number of students who study both subjects. Then we can write two equations based on the given information:

- \(\frac{3}{4}\times80\) students study Biology.
- \(\frac{3}{5}\times80\) students study Physics.

Using these equations, we can solve for **x**:

- Biology students: \(\frac{3}{4}\times80 = 60\)
- Physics students: \(\frac{3}{5}\times80 = 48\)
- Total students: 80
- Students who study both subjects:
**x**

From the diagram, we can see that the total number of students is 80, so we have:

60 + **x** + 48 = 80

Simplifying this equation, we get:

**x** = 12

Therefore, 12 students study both subjects.

(iii) To find the fraction of the class that study Biology but not Physics, we need to find the size of the region inside the circle **B** but outside the intersection, and divide by the total number of students. Let **y** be the number of students who study Biology but not Physics. Then we can write another equation based on the given information:

- \(\frac{3}{4}\times80\) students study Biology.
**x**students study both subjects.

Using these equations, we can solve for **y**:

- Biology students: \(\frac{3}{4}\times80 = 60\)
- Students who study both subjects:
**x**= 12 - Students who study Biology but not Physics:
**y**

From the diagram, we can see that:

**y** + **x** = 60

Substituting **x** = 12, we get:

**y** + 12 = 60

Simplifying this equation, we get:

**y** = 48

Therefore, 48 students study Biology but not Physics. The fraction of the class that study Biology but not Physics is:

\(\frac{48}{80} = \frac{3}{5

**Question 51**
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In the diagram, \(\overline{AD}\) is a diameter of a circle with Centre O. If ABD is a triangle in a semi-circle ∠OAB=34",

find: (a) ∠OAB (b) ∠OCB

**(a) ∠OAB = 34°:** Given in the problem.

**(b) ∠OCB:** We know that in a circle, opposite angles are equal. So, ∠OCB = ∠OAB = 34°.

**Explanation:** When a straight line cuts a circle at two points, it is called a chord of the circle. And when the chord is a diameter of the circle, it is called the diameter. The angle formed between the chord and the line drawn from the center of the circle to the midpoint of the chord is called the angle subtended by the chord at the center of the circle. And it is equal to half of the angle formed by the chord at the circumference of the circle.

So, in this case, ABD is a triangle inscribed in a semicircle. And ∠OAB is half of the central angle subtended by the chord AB. Hence, ∠OCB, which is opposite to ∠OAB is equal to ∠OAB.

**Answer Details**

**(a) ∠OAB = 34°:** Given in the problem.

**(b) ∠OCB:** We know that in a circle, opposite angles are equal. So, ∠OCB = ∠OAB = 34°.

**Explanation:** When a straight line cuts a circle at two points, it is called a chord of the circle. And when the chord is a diameter of the circle, it is called the diameter. The angle formed between the chord and the line drawn from the center of the circle to the midpoint of the chord is called the angle subtended by the chord at the center of the circle. And it is equal to half of the angle formed by the chord at the circumference of the circle.

So, in this case, ABD is a triangle inscribed in a semicircle. And ∠OAB is half of the central angle subtended by the chord AB. Hence, ∠OCB, which is opposite to ∠OAB is equal to ∠OAB.

**Question 52**
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