WAEC SSCE - General Mathematics - 1990

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The formula for the area of a sector is: A = (θ/360)πr² where A is the area of the sector, θ is the angle of the sector in degrees, r is the radius of the circle, and π is the mathematical constant pi. We are given that the radius of the circle is 7cm and the area of the sector is 44cm². Substituting these values into the formula above, we get: 44 = (θ/360) × (22/7) × 7² Simplifying this equation, we get: 44 = (θ/360) × 22 × 7 44 = (θ/360) × 154 Multiplying both sides by 360, we get: θ = (44 × 360) / 154 θ = 102.597 Rounding to the nearest degree, we get: θ ≈ 103^o Therefore, the angle of the sector correct to the nearest degree is 103^o. Answer is correct.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

We can solve for x by using the laws of exponents and taking the logarithm of both sides. First, we can rewrite 3²x as (3²)ⁿ, where n = 2x. So, we have: (3²)ⁿ = 27 3ⁿ² = 27 Now, we can take the logarithm of both sides of the equation. Let's use the natural logarithm, denoted as ln, which is the logarithm to the base e: ln(3ⁿ²) = ln(27) Using the power rule of logarithms, we can simplify the left-hand side: n² ln(3) = ln(27) Now, we can solve for n: n² = ln(27) / ln(3) n² = 3 Taking the square root of both sides, we get: n = ± √3 But we know that n = 2x, so we can substitute back: 2x = ± √3 Solving for x, we get: x = ± (1/2) √3 Since the question is asking for a real value of x, we can take the positive square root: x = (1/2) √3 ≈ 0.866 Therefore, x is approximately 0.866, which is option (B) 1.5 rounded to one decimal place.

Bayanin Amsa

Bayanin Amsa

To solve the equation 7y\(^2\) = 3y, we can start by rearranging it to get 7y\(^2\) - 3y = 0. We can then factor out y to get y(7y - 3) = 0. From here, we have two solutions: y = 0 or 7y - 3 = 0, which gives us y = 3/7. Therefore, the solutions to the equation 7y\(^2\) = 3y are y = 0 or y = 3/7. Therefore, the answer is: y = 0 or 3/7.

Bayanin Amsa

To find the probability that a student chosen at random is against the proposal, we need to add up the number of students from Group A and Group B who are against the proposal, and divide that by the total number of students. The number of students against the proposal in Group A is 32, and in Group B it is 48. So the total number of students against the proposal is: 32 + 48 = 80 The total number of students in both groups is: 128 + 32 + 96 + 48 = 304 So the probability that a student chosen at random is against the proposal is: 80 / 304 = 5/19 Therefore, the answer is option (C) 5/19.

Bayanin Amsa

To find the volume of a cylindrical container, we use the formula V = πr^2h, where V is the volume, r is the radius, and h is the height of the cylinder. In this problem, the radius of the container is given as 7cm, and the height is given as 5cm. Substituting these values into the formula, we get: V = πr^2h V = (22/7)(7^2)(5) V = (22/7)(49)(5) V = 22(5)(7) V = 770 Therefore, the volume of the container is 770 cm^3. So, the correct answer is option (E) 770cm^3.

Bayanin Amsa

Bayanin Amsa

In an arithmetic progression (A.P.), the terms increase or decrease by a constant difference called the common difference. In this problem, the first term is 2, and the common difference is 0.5. Therefore, the second term would be 2 + 0.5 = 2.5, the third term would be 2.5 + 0.5 = 3, and the fourth term would be 3 + 0.5 = 3.5. Therefore, the answer is option (C) 3.5.

Bayanin Amsa

If the bearing of point X from point Y is 074°, then the bearing of Y from X is the opposite direction, which is 180° away from 074°. To find the bearing of Y from X, we can add 180° to 74°, which gives us 254°. Therefore, the correct answer is 254°.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The total surface area of the cylindrical container is the sum of the areas of its top, bottom, and lateral surface. The top and bottom of the container are both circles, each with an area of πr², where r is the radius of the container. So the total area of the top and bottom is 2πr². The lateral surface of the container is a rectangle that has been rolled into a cylinder. The length of the rectangle is the circumference of the circle, which is 2πr. The height of the rectangle is the height of the container, which is given as 5cm. So the area of the lateral surface is 2πrh. Therefore, the total surface area of the container is: 2πr² + 2πrh Substituting the given values for r and h, we get: 2 × (22/7) × 7² + 2 × (22/7) × 7 × 5 = 2 × (22/7) × 49 + 2 × (22/7) × 35 = (22/7) × 2 × (49 + 35) = (22/7) × 168 = 528 cm² Therefore, the total surface area of the container is 528 cm². So, the correct option is (D) 528cm².

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The set H consists of all positive integers x such that x² is less than 3 and x is not equal to zero. The only positive integer that satisfies this condition is 1, because 1²=1 which is less than 3. Therefore, the set H contains only one element which is 1, so the answer is H = {1}. Option (a) is correct, and options (b), (c), (d), and (e) are incorrect. It is important to note that the symbol ∈ means "is an element of", and the symbol ≠ means "is not equal to". The symbol J ≤ H is not a valid statement because J is not a subset of H.

Bayanin Amsa

Bayanin Amsa

To find the value of θ, we can use the formula: θ = (arc length / radius) In this case, the arc length is given as 22cm, and the radius is given as 15cm. So we have: θ = (22 / 15) θ = 1.47 (approx) However, the answer options are in degrees, so we need to convert radians to degrees. We can use the formula: degrees = (radians × 180) / π Substituting θ = 1.47 and π = 22/7, we get: degrees = (1.47 × 180) / (22/7) degrees = 89.14 (approx) Therefore, the answer is closest to option (B) 84^o.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

In a cyclic quadrilateral, the opposite angles add up to 180 degrees. Therefore, we can find the value of angle PQR as follows: angle PSR + angle PQR = 180 (since PQRS is a cyclic quadrilateral) 86 + angle PQR = 180 angle PQR = 180 - 86 = 94 degrees We are given angle QPR as 38 degrees, and since angles in a triangle add up to 180 degrees, we can find angle PRQ as: angle PRQ = 180 - angle PQR - angle QPR angle PRQ = 180 - 94 - 38 angle PRQ = 48 degrees Therefore, the answer is option (C) 48^o.

Bayanin Amsa

To find out how many students studied both economics and geography, we need to use the principle of inclusion-exclusion. First, we add the number of students who studied economics and geography, as some students could have studied both subjects. Let's call this number "x." Then, we subtract x from the total number of students (80), which gives us the number of students who studied only economics or only geography. We know that 65 students studied economics, so the number of students who studied only economics is 65 - x. Similarly, the number of students who studied only geography is 50 - x. Since every student had to study at least one subject, the total number of students who studied only economics or only geography is equal to the sum of the students who studied only economics and the students who studied only geography: (65 - x) + (50 - x) Simplifying this expression, we get: 115 - 2x But we know that this number is equal to the total number of students who studied only economics or only geography, which is 80 minus the number of students who studied both subjects: 80 - x Therefore, we can set up an equation: 115 - 2x = 80 - x Solving for x, we get: x = 35 So 35 students studied both economics and geography.

Bayanin Amsa

Bayanin Amsa

We can simplify the expression using the laws of logarithms. First, we can simplify the numerator by using the fact that the square root of 8 is equal to 8 raised to the power of 1/2: log√8 = log(8^1/2) = 1/2 log 8 Now, we can substitute this into the original expression: \(\frac{\log \sqrt{8}}{\log 8} = \frac{1/2 \log 8}{\log 8}\) We can simplify this by canceling out the factor of log 8 in the numerator and denominator: \(\frac{1/2 \log 8}{\log 8} = \frac{1}{2}\) Therefore, the simplified expression is 1/2, which corresponds to option (B).

Bayanin Amsa

Bayanin Amsa

To evaluate (cos40° - sin30°), we first need to find the values of cos40° and sin30°. Using a mathematical table (such as a trigonometric table), we can look up the values of cos40° and sin30°: - cos40° = 0.7660 - sin30° = 0.5000 Now we can substitute these values into the expression: (cos40° - sin30°) = (0.7660 - 0.5000) = 0.2660 Therefore, the answer is option (D) 0.2660.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To convert a decimal number to standard form, we need to express it in the form of a x 10ⁿ, where a is a number between 1 and 10 and n is an integer. Starting with 0.00562, we need to move the decimal point to the right until we obtain a number between 1 and 10. This means moving the decimal point four places to the right: 0.00562 = 5.62 x 10^-3 Therefore, the answer is option A: 5.62 x 10^-3.

Bayanin Amsa

To solve the equation 2a\(^2\) - 3a - 27 = 0, we can use the quadratic formula which is: a = (-b ± sqrt(b\(^2\) - 4ac)) / 2a In this case, we have a = 2, b = -3, and c = -27. Substituting these values into the formula, we get: a = (-(-3) ± sqrt((-3)\(^2\) - 4(2)(-27))) / 2(2) Simplifying the expression under the square root, we get: a = (-(-3) ± sqrt(225)) / 4 which gives us: a = (3 ± 15) / 4 Therefore, we have two solutions: a = 3 and a = -6/2 = -3/2 Hence, the correct option is -3, 9/2.

Bayanin Amsa

To factorize 2e\(^2\) - 3e + 1, we can use the quadratic formula: e = (-b ± √(b² - 4ac)) / 2a where a = 2, b = -3, and c = 1. Plugging in the values, we get: e = (3 ± √(9 - 8)) / 4 e = (3 ± 1) / 4 So the roots are e = 1 and e = 1/2. Therefore, we can factorize 2e\(^2\) - 3e + 1 as: 2e\(^2\) - 3e + 1 = 2(e - 1)(e - 1/2) Simplifying this expression, we get: 2(e - 1)(2e - 1) Therefore, the factorization of 2e\(^2\) - 3e + 1 is (2e-1) (e-1), which corresponds to option (A).

Bayanin Amsa

The surface area of a sphere can be calculated using the formula: Surface area = 4πr² where r is the radius of the sphere. Substituting r=7cm and π = 22/7, we get: Surface area = 4 x (22/7) x 7² cm² Surface area = 4 x (22/7) x 49 cm² Surface area = 616 cm² Therefore, the surface area of the sphere is 616cm². So the answer is 616cm².

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

We can solve for P by isolating it on one side of the equation. Starting with: 1/3log₁₀ P = 1 We can multiply both sides by 3 to get rid of the 1/3: log₁₀ P = 3 Now, we can rewrite this equation in exponential form: 10³ = P Simplifying: P = 1000 Therefore, the value of P is 1000.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To find the median of a distribution, we first need to arrange the values in order from smallest to largest. In this case, the data is already presented in order, so we can simply identify the middle value of the data set. Since there are 5 values, the middle value will be the 3rd value. So, the median of the distribution is N12.00, which is the value in the middle of the data set. Therefore, the answer is (c) N12.00.

Bayanin Amsa

Bayanin Amsa

(a) Multiply out the numerator and divide:

In standard form:

(b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).

Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).

The eight poles, in order, are:

Bayanin Amsa

(a) Table of values for \(y = 2x^{2} - 5x + 1\)

(b) Plot the points
\[ (-3,34),\,(-2,19),\,(-1,8),\,(0,1),\,(1,-2),\,(2,-1),\,(3,4),\,(4,13),\,(5,26) \] using the given scales, and join them with a smooth curve.

(c) Table of values for \(y=x+6\)

Plot the points \((-2,4)\), \((0,6)\) and \((2,8)\), then draw a straight line through them.

(d)

(i) Least value of \(y = -2.1\), occurring at \(x = 1.3\), correct to one decimal place.

(ii) The solutions of \(2x^{2}-5x+1=x+6\) are:

\[ x=-0.7 \quad \text{and} \quad x=3.7 \]

Bayanin Amsa

(a) Construction

1. Draw \(AB=8.0\text{ cm}\).
2. At \(B\), construct \(\angle ABC=60^\circ\). On this ray, mark \(C\) such that \(BC=5.5\text{ cm}\).
3. With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc.
4. Through \(C\), construct a line parallel to \(AB\). Let it meet the arc at \(D\), choosing the intersection to the left of \(C\). Join \(AD\) and \(BD\) to complete trapezium \(ABCD\).
5. Construct perpendiculars from \(D\) and \(C\) to the straight line \(AB\), meeting it at \(P\) and \(Q\) respectively. Join \(P\) to \(Q\). Thus \(PQCD\) is a rectangle.

(b) Measurements from the construction

\[|AC|=7.0\text{ cm},\qquad |QB|=2.9\text{ cm}.\]

Bayanin Amsa

Take \(R = 6400\) km and \(\pi = 3.142\).

(a) Distance P to Q along the parallel of latitude \(40^\circ\)N. The longitude difference is \(38^\circ\text{E} + 22^\circ\text{W} = 60^\circ\). The radius of the parallel is \(R\cos 40^\circ\), so

(b) Distance Q to T along the meridian \(22^\circ\)W. The latitude difference is \(40^\circ - 28^\circ = 12^\circ\), and along a meridian the radius is \(R\):

(c) Average speed from P to T via Q over 12 hours. Total distance:

Bayanin Amsa

None

Bayanin Amsa

Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.

Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with

The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:

(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]

(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]

(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where

Bayanin Amsa

(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:

(b) Since every student speaks at least one of the languages,

\[n(H\cup I\cup Y)=40.\]

Using the inclusion-exclusion principle,

\[40=25+16+21-8-11-6+x\]

\[40=37+x\]

\[x=3.\]

Therefore, 3 students speak all three languages.

Hence the numerical entries in the Venn diagram are:

\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]

\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]