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**Question 1**
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In the diagram, a cylinder is surmounted by a hemispherical bowl. Calculate the volume of the solid

**Answer Details**

V = πr2h+23πr3 $\pi {r}^{2}h+\frac{2}{3}\pi {r}^{3}$

N.B h = r

V = πx(3)2 $\pi x(3{)}^{2}$ x 20 = 23×π×(3)3 $\frac{2}{3}\times \pi \times (3{)}^{3}$

180π+18π=198cm3

**Question 2**
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The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have?

**Answer Details**

The sum of the interior angles of a polygon is given by the formula (n-2) x 180 degrees, where n is the number of sides of the polygon. In this problem, the sum of the interior angles is given as 20 right angles. One right angle is equal to 90 degrees, so 20 right angles would be 20 x 90 = 1800 degrees. Therefore, we can set up the equation (n-2) x 180 = 1800 and solve for n: (n-2) x 180 = 1800 n-2 = 10 n = 12 So the polygon has 12 sides. To summarize, the sum of the interior angles of a polygon can be calculated using (n-2) x 180 degrees, where n is the number of sides. Using this formula, we can solve for n when the sum of the interior angles is given. In this case, the polygon has 12 sides.

**Question 3**
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The probability of a student passing any examination is 2/3. If the students takes three examination, what is the probability that he will not pass any of them

**Answer Details**

The probability of passing any one of the three examinations is 2/3. Therefore, the probability of not passing any one of the exams is 1 - 2/3 = 1/3. Since the student takes three examinations, the probability of not passing any of them is the probability of not passing the first exam multiplied by the probability of not passing the second exam multiplied by the probability of not passing the third exam. So, the probability of not passing any of the three exams is (1/3) x (1/3) x (1/3) = 1/27. Therefore, the answer is option (D), which is 1/27.

**Question 4**
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In the diagram PQ is parallel to RS. What is the value of α+β+γ

**Question 7**
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If x varies directly as √n and x = 9 when n = 9, find x when n = (17/9)

**Answer Details**

If x varies directly as the square root of n, we can write the equation as: x = k√n where k is a constant of proportionality. To find k, we can use the given values of x and n: 9 = k√9 Solving for k, we get: k = 3 So the equation becomes: x = 3√n To find x when n = 17/9, we substitute n into the equation: x = 3√(17/9) = 3(√17/√9) = 3(√17/3) = √17 Therefore, x = √17.

**Question 8**
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A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45 degrees. if the distance between the hunter and the tree is 10.4 m, find the height of the tree.

**Question 9**
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Make r subject of the formula given that xr+a=ar

**Question 11**
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The time taken to do a piece of work is inversely proportional to the number of men employed. if it takes 45 men to do a piece of work in 5 days, how long will it take 25 men?

**Answer Details**

The time taken to do a piece of work is inversely proportional to the number of men employed, which means that the more men there are, the less time it will take to complete the work. In this case, we are told that 45 men can do the work in 5 days. If we let x be the number of days it takes for 25 men to do the work, then we can set up a proportion: 45 men x 5 days = 25 men x (number of days) Solving for x, we get: x = (45 x 5) / 25 = 9 days Therefore, it would take 25 men 9 days to do the same amount of work that 45 men can do in 5 days. The answer is 9 days.

**Question 12**
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Find the equation of the set of points which are equidistant from the parallel lines x = 1 and x = 7

**Answer Details**

To find the set of points which are equidistant from the parallel lines x=1 and x=7, we can begin by finding the midpoint of the segment connecting the two parallel lines. The midpoint of the segment joining the two parallel lines is ((1+7)/2, 0) = (4,0). Now, let (x,y) be any point that is equidistant from the two parallel lines. Then, the distance from (x,y) to the line x=1 is |x-1|, and the distance from (x,y) to the line x=7 is |x-7|. Since the point (x,y) is equidistant from the two lines, we have: |x-1| = |x-7| Solving for x, we get: x-1 = -(x-7) or x-1 = x-7 Solving each equation for x, we get: x = 4 or x = -6 Since the distance from (x,y) to x=1 is the same as the distance from (x,y) to x=7, it follows that the set of points that are equidistant from the two parallel lines is the vertical line passing through the midpoint of the segment joining the two parallel lines, namely, the line x=4. Therefore, the equation of the set of points which are equidistant from the parallel lines x=1 and x=7 is x = 4. Thus, the correct option is x = 4.

**Question 13**
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Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8

**Answer Details**

To find the x-intercept of the line 2y = 4x - 8, we set y = 0 and solve for x. So, we have: 2(0) = 4x - 8 4x = 8 x = 2 This means the x-intercept is (2, 0). To find the y-intercept of the line, we set x = 0 and solve for y. So, we have: 2y = 4(0) - 8 2y = -8 y = -4 This means the y-intercept is (0, -4). The midpoint of two points (x1, y1) and (x2, y2) is: ((x1 + x2)/2, (y1 + y2)/2) Using this formula, we can find the midpoint of the x and y intercepts: ((2 + 0)/2, (0 + (-4))/2) = (1, -2) Therefore, the coordinates of the midpoint of the x and y intercepts of the line 2y = 4x - 8 is (1, -2). Answer option number 2.

**Question 14**
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Solve for x in the equation x3 - 5x2 - x + 5 = 0

**Answer Details**

To solve this equation, we need to use algebraic methods to isolate the variable x on one side of the equation. First, we can combine like terms on the left-hand side of the equation: x^3 - 5x^2 - x + 5 = 0 Next, we can try to factor the equation, looking for factors that will make the equation equal to zero. By trial and error, we can see that (x-1) is a factor of the equation: (x-1)(x^2 - 4x - 5) = 0 Now we can use the zero product property, which tells us that if the product of two factors is zero, then at least one of the factors must be zero. So we can set each factor equal to zero and solve for x: x-1 = 0 or x^2 - 4x - 5 = 0 Solving for x-1, we get x = 1. Solving for x in the quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 1, b = -4, and c = -5. Plugging these values into the quadratic formula, we get: x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / 2(1) Simplifying, we get: x = (4 ± sqrt(36)) / 2 x = (4 ± 6) / 2 x = 5 or x = -1 Therefore, the solutions for x are 1, 5, and -1.

**Question 15**
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How many three-digit numbers can be formed from 32564 without repeating any of the digits?

**Question 16**
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A trader bought goats for N4000 each. He sold them for N180,000 at a loss of 25%. How many goats did he buy?

**Answer Details**

Let's start by using a bit of math to understand the problem. We know that the trader bought each goat for N4000, and sold them for N180,000. We also know that he sold them at a loss of 25%. If the trader sold the goats at a loss of 25%, it means that he sold them for only 75% of their original value. We can calculate the original value of the goats by dividing the selling price by 75% or multiplying it by 4/3. So, the original value of the goats = 180,000 * (4/3) = 240,000. Since the trader bought each goat for N4000, we can find out how many goats he bought by dividing the total value of the goats (N240,000) by the price he paid for each goat (N4000). Number of goats = 240,000 / 4000 = 60. Therefore, the trader bought 60 goats. To summarize, the trader bought 60 goats and sold them for N180,000 at a loss of 25%. We found the answer by calculating the original value of the goats and dividing it by the price the trader paid for each goat.

**Question 17**
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Without using tables, evaluate (343)1/3 x (

0.14)-1 x (25)-1/2

**Answer Details**

We can simplify the expression using the laws of exponents and basic arithmetic operations. First, let's evaluate each exponent: - (343)^(1/3) = 7, since 7^3 = 343 - (0.14)^(-1) = 1/0.14 ≈ 7.14 - 25^(-1/2) = 1/√25 = 1/5 Next, let's substitute these values back into the original expression: (343)^(1/3) x (0.14)^(-1) x 25^(-1/2) = 7 x 7.14 x 1/5 Multiplying these three numbers gives us: 7 x 7.14 x 1/5 ≈ 10 Therefore, the answer is 10.

**Question 18**
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Simplify (√0.7 + √70)2

**Answer Details**

To simplify (√0.7 + √70)², we can use the formula (a + b)² = a² + 2ab + b², where a = √0.7 and b = √70. So, we have: (√0.7 + √70)² = (√0.7)² + 2(√0.7)(√70) + (√70)² Simplifying, we get: 0.7 + 2√(0.7 × 70) + 70 = 0.7 + 2√49 + 70 = 0.7 + 2(7) + 70 = 84.7 Therefore, the answer is 84.7.

**Question 19**
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In the diagram, XZ is the diameter of the circle XYZW, with centre O and radius 152
$\frac{15}{2}$cm. If XY = 12cm, find the area of the triangle XYZ

**Question 20**
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Simplify 35÷(27×43÷49)

**Answer Details**

35÷(27×43÷49)=35÷(27×43×94)=35÷67=35×76=710

**Question 21**
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The range of the data: k+2, k-3, k+4, k-2, k, k-5, k+3, k-1, and k+6 is

**Answer Details**

To find the range of a data set, we need to subtract the smallest value from the largest value. In this case, the smallest value is k-5 and the largest value is k+6. Therefore, the range is (k+6) - (k-5) = k+6 - k+5 = 11. So the answer is 11.

**Question 22**
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Calculate the mean deviation of the set of numbers 7, 3, 14, 9, 7, and 8.

**Answer Details**

To find the mean deviation of a set of numbers, we first need to find the mean (average) of the set. Mean = (7+3+14+9+7+8) / 6 = 48/6 = 8 Next, we need to find the deviation of each number from the mean. To do this, we subtract the mean from each number: (7-8), (3-8), (14-8), (9-8), (7-8), (8-8) -1, -5, 6, 1, -1, 0 Notice that some of the deviations are negative and some are positive. To avoid canceling out the deviations that are opposite in sign, we take the absolute value of each deviation: 1, 5, 6, 1, 1, 0 Finally, we find the mean of these absolute deviations: Mean Deviation = (1+5+6+1+1+0) / 6 = 14/6 = 7/3 Therefore, the mean deviation of the given set of numbers is 7/3.

**Question 23**
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In the diagram above, PST is a straight line, PQ = QS = RS. If ∠RST = 72∘
$\circ $, find x

**Answer Details**

In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)

In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)

∴ ∠RQS = X + X

= 2X

Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,

72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)

∴ 72 = x + 2x

72 = 3x

x = 72/3

x = 24∘

**Question 24**
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In a school, 220 students offer Biology or Mathematics or both, 125 offer Biology and 110 Mathematics. How many offer Biology but not Mathematics?

**Question 25**
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The slope of the tangent to the curve y=2x2−2x+5 $y=2{x}^{2}-2x+5$ at the point (1,6) $(1,6)$ is

**Question 26**
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A bucket is 12 cm in diameter at the top, 8 cm in diameterat the bottom and 4 cm deep. Calculate its volume.

**Answer Details**

To calculate the volume of the bucket, we need to use the formula for the volume of a frustum of a cone which is given by: V = (1/3) × π × h × (r₁² + r₂² + r₁ × r₂) where h is the height of the frustum, r₁ and r₂ are the radii of the top and bottom faces, respectively. In this case, the height of the frustum is 4 cm, the top radius is 6 cm (half of the diameter), and the bottom radius is 4 cm. Substituting these values into the formula, we get: V = (1/3) × π × 4 × (6² + 4² + 6 × 4) V = (1/3) × π × 4 × (36 + 16 + 24) V = (1/3) × π × 4 × 76 V = 304π/3 cubic centimeters Therefore, the volume of the bucket is 304π/3 cubic centimeters. Answer (1) is correct.

**Question 27**
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If tan θ = 4/3, calculate sin2θ - cos2θ.

**Question 28**
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The venn diagram above shows the numbers of students offering music and history in a class of 80 students. If a student is picked at random from the class, what is the probability that he offers Music only?

**Answer Details**

30 - x + x + 40 - x = 80 - 20

70 - x = 60

- x = 60 - 70

- x = - 10

∴ x = 10

Music only = 30 - x

= 30 - 10

20

P(music only) = 20/80

= 1/4

= 0.25

**Question 29**
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The triangle POR is

**Answer Details**

< PRQ = 180∘ $\circ $ - 128∘ $\circ $ = 52∘ $\circ $ = < R

< P + < R + < Q = 180∘ $\circ $

76∘ $\circ $ + 52∘ $\circ $ + < Q = 180∘ $\circ $

128∘ $\circ $ + < Q = 180∘ $\circ $

< Q = 180∘ $\circ $ - 128∘ $\circ $

= 52∘ $\circ $

△ $\u25b3$PQR is an isosceles

**Question 30**
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The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is

**Answer Details**

This is a geometric series with first term 1 and common ratio 1/3. The formula for the sum of an infinite geometric series is S = a/(1-r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 1/(1 - 1/3) = 3/2. Therefore, the answer is 3/2.

**Question 31**
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No. of days | 1 | 2 | 3 | 4 | 5 | 6 |

No. of students | 20 | x | 50 | 40 | 2x | 60 |

The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term

**Answer Details**

20 + X + 50 + 40 + 2X + 60 = 260

3X + 170 = 260

3X = 260 - 170

3x = 90

x = 30

Absent for at least 4 days

4 | 5 | 6 |

40 | 2x | 60 |

i.e 40 + 2x + 60 = 40 + (2 x 30) + 60

= 40 + 60 + 60

= 160

**Question 32**
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The acres for rice, pineapple, cassava, cocoa, and palm oil in a certain district are given respectively as 2, 5, 3, 11, and 9. What is the angle sector for cassava in a pie chart?

**Answer Details**

To find the angle sector for cassava in a pie chart, we need to calculate what percentage of the total area is occupied by cassava. First, we need to find the total area of the pie chart, which represents the total acres of all the crops in the district. The total acres can be found by adding up the acres for each crop: 2 + 5 + 3 + 11 + 9 = 30 So the total area of the pie chart is 30 acres. Next, we need to find what percentage of the total area is occupied by cassava. We can do this by dividing the acres of cassava by the total acres and then multiplying by 100: (3 / 30) x 100 = 10% Therefore, the angle sector for cassava in the pie chart is 10% of the total area of the pie chart. To find the angle in degrees, we need to multiply the percentage by 360 (the total number of degrees in a circle): 10% x 360 = 36° So the angle sector for cassava in the pie chart is 36°.

**Question 33**
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(0.21 x 0.072 x 0.00054) (0.006 x 1.68 x 0.063), correct to 4.s.f

**Answer Details**

To find the product of the two numbers, we can simply multiply them together: (0.21 x 0.072 x 0.00054) x (0.006 x 1.68 x 0.063) = 0.000007728 x 0.000634176 Next, we can use a calculator to evaluate this expression and round to 4 significant figures: 0.000007728 x 0.000634176 = 0.00000489479045568 Rounding to 4 significant figures, we look at the fifth digit (9) and round up because it is greater than 5. Therefore, the final answer is: 0.01286 So the answer is 0.01286, correct to 4 significant figures.

**Question 34**
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If -2 is the solution of the equation 2x + 1 - 3c = 2c + 3x - 7, find the value of c.

**Answer Details**

To find the value of c, we need to substitute -2 for x in the given equation and then solve for c. Substituting -2 for x, we get: 2(-2) + 1 - 3c = 2c + 3(-2) - 7 Simplifying the equation, we get: -4 + 1 - 3c = 2c - 6 - 7 Combining like terms, we get: -3c - 3 = 2c - 13 Moving all the c terms to one side and all the constant terms to the other side, we get: -3c - 2c = -13 + 3 Simplifying, we get: -5c = -10 Dividing both sides by -5, we get: c = 2 Therefore, the value of c is 2.

**Question 35**
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A chord of a circle subtends an angle of 12o degrees at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector.

**Question 36**
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Find the mean of the data: 7, -3, 4, -2, 5, -9, 4, 8, -6, 12

**Question 37**
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In the diagram are two concentric circles of radii r and R respectively with centre O. If r = 25
$\frac{2}{5}$R, express the area of the shaded portion in terms of π
$\pi $ and R

**Answer Details**

Ashaded portion $\text{shaded portion}$ = πR2−πr2 $\pi {R}^{2}-\pi {r}^{2}$ r = 25 $\frac{2}{5}$R

Ashaded portion $\text{shaded portion}$ = πR2−π(23R)2=πR2−4πR225 $\pi {R}^{2}-\pi (\frac{2}{3}R{)}^{2}=\pi {R}^{2}-\frac{4\pi {R}^{2}}{25}$

= πR2(1−23)=2125πR2

**Question 38**
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In the diagram above , XZ is the diameter of the circle XZW, with center O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ

**Question 39**
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If the 9th term of an A.P is five times the 5th term, find the relationship between a and d.

**Answer Details**

Let the fifth term of the AP be 'a + 4d'. Then, the ninth term will be 'a + 8d'. The problem states that the ninth term is five times the fifth term, which can be represented mathematically as: a + 8d = 5(a + 4d) Simplifying the equation gives: a + 8d = 5a + 20d Subtracting 'a' and 20d from both sides gives: -12d = -4a Dividing both sides by -4 gives: 3d = a So the relationship between 'a' and 'd' is a = 3d. Thus, the answer is a + 3d = 0.

**Question 40**
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The mean of a set of six numbers is 60. If the mean of the first five is 50, find the sixth number in the set.

**Answer Details**

The mean of a set of six numbers is 60. Therefore, the sum of the six numbers is 6 x 60 = 360. The mean of the first five is 50. Therefore, the sum of the first five numbers is 5 x 50 = 250. To find the sixth number, we can subtract the sum of the first five numbers from the sum of all six numbers: 6th number = sum of all six numbers - sum of the first five numbers 6th number = 360 - 250 6th number = 110 Therefore, the sixth number in the set is 110.

**Question 41**
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If dy/dx = 2x - 3 and y = 3 when x = 0, find y in terms of x.

**Answer Details**

To solve this problem, we need to integrate both sides of the equation with respect to x. dy/dx = 2x - 3 Integrating both sides with respect to x gives: y = x^2 - 3x + c where c is the constant of integration. To find the value of c, we use the fact that y = 3 when x = 0: 3 = 0^2 - 3(0) + c c = 3 Therefore, the equation of y in terms of x is: y = x^2 - 3x + 3

**Question 42**
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Using the graph find the values of p and q if px + qy ≥ $\ge $ 4

**Answer Details**

m = y2−y1x2−x1=2−00−(4)=24