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**Question 1**
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The diagram is a circle with centre O. Find the area of the shaded portion.

**Answer Details**

To find the area of the shaded portion, we need to subtract the area of the triangle from the area of the sector. First, we need to find the radius of the circle. Since the diameter is given as 6cm, the radius is half of it which is 3cm. Next, we need to find the angle of the sector. We can do this by using the properties of the isosceles triangle. The angle at the centre of the circle is twice the angle at the circumference of the circle. So, the angle at the centre of the circle is: 2 x (180 - 90) = 180 degrees This means that the angle of the sector is 180 degrees. Now we can calculate the area of the sector: Area of sector = (angle/360) x πr^2 = (180/360) x π(3)^2 = 4.5π cm^2 Next, we need to find the area of the triangle. Since the triangle is isosceles, the base angles are equal, and we can use trigonometry to find the height. The base of the triangle is equal to the diameter of the circle which is 6cm. Using trigonometry, we know that: tan(45) = height/base height = base x tan(45) height = 6 x 1 height = 6cm Now we can calculate the area of the triangle: Area of triangle = (1/2) x base x height = (1/2) x 6 x 6 = 18 cm^2 Finally, we can calculate the area of the shaded portion by subtracting the area of the triangle from the area of the sector: Area of shaded portion = Area of sector - Area of triangle = 4.5π - 18 = (9/2)(π - 2) cm^2 Therefore, the answer is 9(π−2)cm2.

**Question 2**
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What is the n-th term of the sequence 2, 6, 12, 20...?

**Answer Details**

The given sequence is formed by adding consecutive even numbers starting from 2. The first term is 2, the second term is 2 + 4, the third term is 2 + 4 + 6, and so on. Thus, we can observe that the n-th term is obtained by adding the first n even numbers. The sum of the first n even numbers is n(n+1), which can be proved using mathematical induction. Therefore, the n-th term of the sequence is n(n+1). Hence, the correct option is: n2 + n.

**Question 3**
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Solve the equation y2−11y+24=0

**Answer Details**

To solve the equation: y^2 - 11y + 24 = 0 We can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by: x = (-b ± sqrt(b^2 - 4ac)) / (2a) In this case, we have a = 1, b = -11, and c = 24. Substituting these values into the quadratic formula, we get: y = (-(-11) ± sqrt((-11)^2 - 4(1)(24))) / (2(1)) Simplifying this expression, we get: y = (11 ± sqrt(121 - 96)) / 2 y = (11 ± sqrt(25)) / 2 y = (11 ± 5) / 2 So the solutions to the equation are: y = 8 or y = 3 Therefore, the equation y^2 - 11y + 24 = 0 has two solutions: y = 8 and y = 3.

**Question 4**
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p = ∣∣ ∣∣x302y3424∣∣ ∣∣ $\left|\begin{array}{ccc}x& 3& 0\\ 2& y& 3\\ 4& 2& 4\end{array}\right|$

q = ∣∣ ∣∣x2z3y203z∣∣ ∣∣ $\left|\begin{array}{ccc}x& 2& z\\ 3& y& 2\\ 0& 3& z\end{array}\right|$

pq is equivalent to

**Answer Details**

= ∣∣ ∣∣030213422∣∣ ∣∣
$\left|\begin{array}{ccc}0& 3& 0\\ 2& 1& 3\\ 4& 2& 2\end{array}\right|$

Q = ∣∣ ∣∣024312032∣∣ ∣∣
$\left|\begin{array}{ccc}0& 2& 4\\ 3& 1& 2\\ 0& 3& 2\end{array}\right|$ = pT

pq = ppT

**Question 5**
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The sketch is the curve of y = ax2 + bx + c. Find a, b and c respectively

**Answer Details**

Given the graph and the curve y = ax2 + bx + c the roots are x - 2 and 2 while its equation (x + 2)(x - 2) = y

y = x2 - 4 i.e. y = x2 + 0x - 4

a = 1, b = 0 and c = -4

**Question 6**
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For an arithmetical sequence, the first term is 2 and the common difference is 3. Find the sum of the first 11 terms

**Answer Details**

An arithmetic sequence is a sequence of numbers where the difference between any two consecutive terms is constant. In this case, the first term is 2 and the common difference is 3, so the sequence is: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32 To find the sum of the first 11 terms of this sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an) where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, n = 11, a1 = 2, and an = 32 (the 11th term can be found by adding the common difference 3, 10 times to the first term 2). So, substituting the values in the formula: Sn = 11/2 * (2 + 32) Sn = 11/2 * 34 Sn = 187 Therefore, the sum of the first 11 terms is 187. So the correct answer is.

**Question 7**
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The probability of an event P is 34 $\frac{3}{4}$ while that of another event Q is 16 $\frac{1}{6}$. If the probability of both P and Q is 12 $\frac{1}{2}$. What is the probability of either P or Q.

**Question 8**
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The radius of a circle is given as 5cm subject to an error of 0.1cm. What is the percentage error in the area of the circle?

**Answer Details**

Area = πr2=π×5×5=25π
$\pi {r}^{2}=\pi \times 5\times 5=25\pi $

Area error = π×0.12=0.01π
$\pi \times {0.1}^{2}=0.01\pi $

∴
$\therefore $ % error = 0.01π25π×10
$\frac{{\textstyle 0.01\pi}}{{\textstyle 25\pi}}\times 10$

=0.0125×100=125

**Question 9**
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If x = (all prime factors of 44) and y = (all prime factors of 60), the elements of X ∪ Y and X ∩ Y respectively are

**Answer Details**

The prime factors of 44 are 2, 2, 11. The prime factors of 60 are 2, 2, 3, 5. To find X ∪ Y (the union of X and Y), we need to list all the distinct prime factors from both 44 and 60. These are 2, 3, 5, 11. Therefore, X ∪ Y is (2, 3, 5, 11). To find X ∩ Y (the intersection of X and Y), we need to list all the prime factors that are common to both 44 and 60. These are 2. Therefore, X ∩ Y is (2). Therefore, the answer is (2, 3, 5, 11) and (2), which corresponds to option D.

**Question 10**
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Simplify (1.25×10−4)×(2.0×10−1)(6.25×105)

**Answer Details**

(1.25×10−4)×(2.0×10−1)(6.25×105)
$\frac{(1.25\times {10}^{-4})\times (2.0\times {10}^{-1})}{(6.25\times {10}^{5})}$ = 1.25×26.25
$\frac{1.25\times 2}{6.25}$ x 104 - 1 - 5

2.506.25
$\frac{2.50}{6.25}$ x 10-2 = 250625
$\frac{250}{625}$ x 10-2

0.4 x 10-2 = 4.0 x 10-3

**Question 11**
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ind then equation line through (5, 7) parallel to the line 7x + 5y = 12

**Answer Details**

Equation (5, 7) parallel to the line 7x + 5y = 12

5Y = -7x + 12

y = −7x5
$\frac{-7x}{5}$ + 125
$\frac{12}{5}$

Gradient = −75
$\frac{-7}{5}$

∴ Required equation = y−7x−5
$\frac{y-7}{x-5}$ = −75
$\frac{-7}{5}$ i.e. 5y - 35 = -7x + 35

5y + 7x = 70

**Question 12**
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Find n if 34n = 100112

**Answer Details**

To find n if 34n = 100112, convert both sides to base 10

= 3n + 4 = (1 x 24) + (0 x23) + (0 x 22) + (1 x 21) + 1 x 2o

= 3n + 4 = 16 + 0 + 0 + 2 + 1

3n + 4 = 19

3n = 15

n = 5

**Question 13**
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The angle between latitudes 30oS and 13oN is

**Answer Details**

The distance between two latitudes is always the same, and this distance is equal to 1/360th of the circumference of the earth. The circumference of the earth is approximately 40,000 km, so 1/360th of the circumference is approximately 111 km. To find the angle between latitudes 30°S and 13°N, we need to add the distance between these latitudes to the angle at the center of the earth. The distance between 30°S and the equator (0°) is 30°, and the distance between the equator and 13°N is 13°, for a total distance of 43°. Therefore, the angle between latitudes 30°S and 13°N is 43°. So, the answer is

43o

**Question 14**
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If y = x sin x, Find d2yd2x

**Answer Details**

To find the second derivative of y = x sin x, we need to differentiate the function twice with respect to x. First, let's find the first derivative: y' = (x cos x) + (sin x) Using the product rule and the derivative of sin x. Next, we can find the second derivative: y'' = [(x cos x) + (sin x)]' = (cos x - x sin x) + cos x Using the product rule and the derivative of cos x. Therefore, the second derivative of y = x sin x is y'' = 2 cos x - x sin x.

**Question 15**
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A cliff on the bank of a river is 300 meter high. if the angle of depression of a point on the opposite side of the river is 60∘ $\circ $, find the width of the river.

**Answer Details**

The problem involves finding the width of a river given the height of a cliff and the angle of depression of a point on the opposite side of the river. The angle of depression is the angle formed between the horizontal line and the line of sight from the point on the opposite side of the river to the top of the cliff. We can use trigonometry to solve the problem. Let x be the width of the river. Then we have a right triangle with the height of the cliff as the opposite side, x as the adjacent side, and the angle of depression as 60 degrees. Using the tangent function, we have: tan(60) = opposite/adjacent sqrt(3) = 300/x x = 300/sqrt(3) x = 100sqrt(3) Therefore, the width of the river is 100sqrt(3) meters. So, the correct option is: - 100 - 75√3 m - 100√3m - 200√3m (100√3m) is the correct answer.

**Question 16**
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If the scores of 3 students in a test are 5, 6 and 7, find the standard deviation of their scores

**Answer Details**

To find the standard deviation of the scores, we need to first calculate the mean (average) of the scores: mean = (5 + 6 + 7) / 3 = 6 Next, we calculate the deviation of each score from the mean: deviation of 5 = 5 - 6 = -1 deviation of 6 = 6 - 6 = 0 deviation of 7 = 7 - 6 = 1 To calculate the standard deviation, we take the square root of the average of the squared deviations: standard deviation = √[(1^2 + 0^2 + (-1)^2) / 3] = √(2/3) ≈ 0.82 Therefore, the answer is option (D), √(2/3).

**Question 17**
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Obtain a maximum value of the function f(x) = x3 $3$ - 12x + 11.

**Answer Details**

To obtain the maximum value of the given function f(x) = x^3 - 12x + 11, we need to find its critical points. The critical points of a function are the points where the derivative of the function is either zero or undefined. So, let's find the derivative of the given function: f'(x) = 3x^2 - 12 Now, we'll set f'(x) = 0 and solve for x: 3x^2 - 12 = 0 x^2 - 4 = 0 (x - 2)(x + 2) = 0 So, the critical points of the function are x = 2 and x = -2. To determine whether these are maximum or minimum points, we'll take the second derivative of the function: f''(x) = 6x Now, we'll substitute the critical points into the second derivative: f''(2) = 12 f''(-2) = -12 Since f''(2) is positive, the critical point x = 2 is a minimum point. Since f''(-2) is negative, the critical point x = -2 is a maximum point. Therefore, the maximum value of the given function is obtained when x = -2. Now, we'll substitute x = -2 into the original function: f(-2) = (-2)^3 - 12(-2) + 11 f(-2) = -8 + 24 + 11 f(-2) = 27 So, the maximum value of the given function is 27, and the correct option is (D).

**Question 18**
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Ice forms on a refrigerator ice box at rate of (4 - 0.61)g per minute after 1 minute. If initially, there were 2g of ice, find the mass of ice formed in 5 minutes.

**Question 19**
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Resolve 3x2+x−2 $\frac{3}{{x}^{2}+x-2}$ into partial fractions

**Answer Details**

3x2+x−2
$\frac{3}{{x}^{2}+x-2}$ = 3(x−1)(x+1)
$\frac{3}{(x-1)(x+1)}$

Ax−1
$\frac{A}{x-1}$ + Bx+2
$\frac{B}{x+2}$

A(x + 2) + B(x - 1) = 3

when x = 1, 3A = 3 →
$\to $ a = 1

when x = -2, -3B = 3 →
$\to $ B = -1

3 = 1 - 1

= 1x−1+1x+2

**Question 20**
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A man invested a sum of ₦280.00 partly at 5% and partly at 4%, if the total interest is ₦12.80 per annum, find the amount invested at 5%.

**Answer Details**

Let x be the amount invested at 5% and y be the amount invested at 4%. Then we have the following system of equations: x + y = 280 --- (1) (since the man invested a total of ₦280.00) 0.05x + 0.04y = 12.80 --- (2) (since the total interest earned is ₦12.80) To solve for x, we can use the first equation to get y = 280 - x, and substitute into the second equation to get: 0.05x + 0.04(280 - x) = 12.80 Simplifying and solving for x, we get: 0.05x + 11.20 - 0.04x = 12.80 0.01x = 1.60 x = 160 Therefore, the amount invested at 5% is ₦160.00.

**Question 21**
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a student blows a balloon and its volume increases at a rate of π
$\pi $(20 - t2)cm3S-1 after t seconds. If the initial volume is 0 cm3, find the volume of the balloon after 2 seconds

**Answer Details**

The volume of the balloon increases at a rate of π(20-t^2) cm^3S^-1 after t seconds. If we want to find the volume of the balloon after 2 seconds, we need to integrate the given rate with respect to time from 0 to 2, since we want to know the change in volume from the initial volume of 0 cm^3 after 2 seconds. Integrating the given rate with respect to time, we get: ∫[0,2] π(20-t^2) dt = π[20t - (t^3/3)] from 0 to 2 Plugging in the values, we get: π[20(2) - (2^3/3)] - π[20(0) - (0^3/3)] = π[40 - 8/3] = π[120/3 - 8/3] = π[112/3] = 37.33π Therefore, the volume of the balloon after 2 seconds is approximately 37.33π cubic centimeters.

**Question 22**
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Simplify 5√18 $18$ - 3√72 $72$ + 4√50

**Answer Details**

To simplify these expressions, we need to simplify the numbers under the radical sign first by factoring out their perfect square factors. For the first expression 5√18, we can factor out the perfect square factor of 9, which leaves us with 5√2√2√2 or 10√2. For the second expression, we can simplify √72 to √(36*2), and since 36 is a perfect square, we can factor it out, which leaves us with 3√2*6, or 3√2*2√3, which simplifies to 6√6. For the third expression, we can simplify √50 to √(25*2), and since 25 is a perfect square, we can factor it out, which leaves us with 2√2*5, or 2√2*√5, which simplifies to 2√10. Putting it all together, we have: 5√18 = 10√2 - 3√72 = 6√6 - 4√50 = 2√10 Now we can substitute these simplified expressions back into the original expressions: 10√2 - 6√6 + 2√10 To simplify this expression, we can group like terms. The coefficients of √6 are -6 and 0, since there are no other terms with √6. The coefficients of √10 are 2 and 0, since there are no other terms with √10. The coefficient of √2 is 10. So our simplified expression is: 10√2 - 6√6 + 2√10 = 10√2 - 6√6 + 2√10 = 10√2 - 6√(2*3) + 2√(2*5) = 10√2 - 6√2√3 + 2√2√5 = (10-6√3+2√5)√2 Therefore, the answer is 17√2.

**Question 23**
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A student has 5 courses to take from Mathematics and physics. There are 4 courses in Mathematics and 3 in Physics which he can choose his courses so that he takes exactly two courses in Physics?

**Answer Details**

To solve this problem, we need to use the combination formula, which is nCr = n! / r!(n-r)!. Here, we have 4 Mathematics courses and 3 Physics courses. The student needs to select 2 Physics courses out of 3, which can be done in 3C2 ways (i.e., 3 different ways). For each of the 3 ways the student selects 2 Physics courses, he must also select 3 Mathematics courses out of 4, which can be done in 4C3 ways (i.e., 4 different ways). Therefore, the total number of ways the student can select exactly two courses in Physics is 3C2 * 4C3 = 3 * 4 = 12. Hence, the correct answer is 12.

**Question 24**
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If x = 3 - √3 $3$, find x2 + 36x2

**Answer Details**

x = 3 - √3
$3$

x2 = (3 - √3
$3$)2

= 9 + 3 - 6√34
$34$

= 12 - 6√3
$3$

= 6(2 - √3
$3$)

∴ x2 + 36x2
$\frac{36}{{x}^{2}}$ = 6(2 - √3
$3$) + 366(2−√3)
$\frac{36}{6(2-\sqrt{3})}$

6(2 - √3
$3$) + 62−√3
$\frac{6}{2-\sqrt{3}}$ = 6(- √3
$3$) + 6(2+√3)(2−√3)(2+√3)
$\frac{6(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$

= 6(2 - √3
$3$) + 6(2+√3)4−3
$\frac{6(2+\sqrt{3})}{4-3}$

6(2 - √3
$3$) + 6(2 + √3
$3$) = 12 + 12

= 24

**Question 25**
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Given that θ $\theta $ is an acute angle and sin θ $\theta $ = mn $\frac{m}{n}$, find cos θ

**Question 26**
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In the figure, PQr is a semicircle while PQ and Qr are chords. QS is the perpendicular from Q to the diameter PR. What is the expression for QS?

**Question 27**
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Five people are to be arranged in a row for a group photograph. How many arrangements are the if a married couple in the group insist on sitting next to each other?

**Question 28**
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Find all values of x satisfying the inequality -11 ≤ $\le $ 4 - 3x ≤ $\le $ 28

**Answer Details**

To solve the inequality -11 ≤ 4 - 3x ≤ 28, we need to isolate x in the middle of the inequality. First, we subtract 4 from all parts of the inequality: -15 ≤ -3x ≤ 24. Then, we divide by -3, remembering to flip the direction of the inequality signs when dividing by a negative number: 5 ≥ x ≥ -8. Therefore, the correct option is -8 ≤ x ≤ 5.

**Question 29**
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In the figure, if XZ is 10cm, calculate RY in cm.

**Question 30**
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p = ∣∣ ∣∣x302y3424∣∣ ∣∣ $\left|\begin{array}{ccc}x& 3& 0\\ 2& y& 3\\ 4& 2& 4\end{array}\right|$

q = ∣∣ ∣∣x2z3y203z∣∣ ∣∣ $\left|\begin{array}{ccc}x& 2& z\\ 3& y& 2\\ 0& 3& z\end{array}\right|$ Where pT is the transpose p calculate /pT/ when x = 0, y = 1 and z = 2

**Answer Details**

= ∣∣ ∣∣030213422∣∣ ∣∣
$\left|\begin{array}{ccc}0& 3& 0\\ 2& 1& 3\\ 4& 2& 2\end{array}\right|$

PT = ∣∣ ∣∣024213032∣∣ ∣∣
$\left|\begin{array}{ccc}0& 2& 4\\ 2& 1& 3\\ 0& 3& 2\end{array}\right|$

/pT/ = ∣∣ ∣∣024313032∣∣ ∣∣
$\left|\begin{array}{ccc}0& 2& 4\\ 3& 1& 3\\ 0& 3& 2\end{array}\right|$

= 0[2 - 6] - 2[6 - 0] + 4[9 - 0]

= 0 - 12 + 36 = 24

**Question 31**
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Find the sum to infinity to the following series 3 + 2 + 43 $\frac{4}{3}$ + 89 $\frac{8}{9}$ + 1617 $\frac{16}{17}$ + .....

**Answer Details**

3 + 2 + 43
$\frac{4}{3}$ + 89
$\frac{8}{9}$ + 1617
$\frac{16}{17}$ + .....

a = 3

r = 23
$\frac{2}{3}$

s α
$\alpha $ = a1−r
$\frac{a}{1-r}$ = 31−23
$\frac{3}{1-\frac{2}{3}}$

= 313
$\frac{3}{\frac{1}{3}}$

= 3 x 3

= 9

**Question 32**
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Make t the subject of formula S = ut + 12at2