WAEC SSCE - General Mathematics - 1996

Find, correct to 1 decimal place, the volume of cylinder of height 8cm and base radius 3cm. [Take π = 3.142]

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The formula for the volume of a cylinder is V = πr²h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Substituting the given values in the formula, we have: V = π(3cm)²(8cm) V = 72π cm³ V ≈ 226.1cm³ (rounded to 1 decimal place, using π ≈ 3.142) Therefore, the correct option is (D) 226.2cm³.

Simplify: \(\frac{5}{x - y} - \frac{4}{y - x}\)

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To simplify \(\frac{5}{x-y}-\frac{4}{y-x}\), we first notice that \(y-x=-(x-y)\). Thus, we can rewrite the expression as \(\frac{5}{x-y}+\frac{4}{x-y}\). Now, we can combine the fractions by finding a common denominator, which is \(x-y\). Thus, we have \[\frac{5}{x-y}+\frac{4}{x-y}=\frac{5+4}{x-y}=\frac{9}{x-y}.\] Therefore, the simplified expression is \(\boxed{\frac{9}{x-y}}\).

One side of a rectangle is 8cm and the diagonal is 10cm. What is the area of the rectangle?

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We know that the diagonal of a rectangle divides it into two right triangles with the diagonal as the hypotenuse, and the sides of the rectangle as the legs of the right triangles. Let's call the other side of the rectangle "x". Using the Pythagorean theorem, we have: 10² = 8² + x² Simplifying and solving for x, we get: x = √(10² - 8²) = √36 = 6 Therefore, the area of the rectangle is: 8 x 6 = 48 cm² Hence, the answer is 48cm².

If x is a real number which of the following is more illustrated on the number line?

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Each interior angle of a regular nonagon(nine sided polygon) is equal to

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A nonagon has nine sides and nine interior angles. In a regular nonagon, all the sides and angles are equal. To find the measure of each interior angle of a regular nonagon, we can use the formula: Interior angle = (n - 2) x 180° / n where n is the number of sides of the polygon. Substituting n = 9 in the formula, we get: Interior angle = (9 - 2) x 180° / 9 = 7 x 20° = 140° Therefore, each interior angle of a regular nonagon is equal to 140°. So the correct answer is option (C) 140^o.

A boy measured the length and breath of a rectangular lawn as 59.6m and 40.3m respectively instead of 60m and 40m. What is the percentage error in his calculation of the perimeter of the lawn?

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The correct perimeter of the rectangular lawn can be calculated by adding twice the length and twice the breadth, i.e., 2 × length + 2 × breadth = 2 × 60m + 2 × 40m = 120m + 80m = 200m The perimeter calculated by the boy would be: 2 × 59.6m + 2 × 40.3m = 119.2m + 80.6m = 199.8m The difference between the correct perimeter and the calculated perimeter is: 200m - 199.8m = 0.2m To find the percentage error, we divide the difference by the correct perimeter and multiply by 100: (0.2m / 200m) × 100% = 0.1% Therefore, the percentage error in the boy's calculation of the perimeter of the lawn is 0.1%. The correct option is: 0.1%.

What percentage of observation lie outside interquartile range of any distribution?

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The interquartile range (IQR) is the range between the first quartile (25th percentile) and the third quartile (75th percentile) of a distribution. Half of the observations lie within this range, which means that the other half lie outside of it. Therefore, the percentage of observations that lie outside the IQR is 100% - 50% = 50%. So, the answer is 50%.

The trapezium PQRS parallel to SR,?PQS = 34^o and ?SPQ = 2 ?SRQ. Find the size of, SQR

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The angle of elevation of the top X of a vertical pole from a point P on a level ground is 60^o. The distance from P to the foot of the pole is 55m. Without using tables, find the height of the pole.

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We can use trigonometry to solve this problem. Let's draw a diagram to help us visualize the situation:

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We know that the angle of elevation of X from P is 60 degrees, which means that the angle XPX' is also 60 degrees, where X' is the foot of the pole. We also know that the distance from P to X' is 55m.

Let H be the height of the pole. Then we have:

tan 60 = H / PX'

We can simplify this expression using trigonometric ratios:

√3 = H / 55

Solving for H, we get:

H = 55√3

Therefore, the height of the pole is 55√3 meters. So the correct option is (c).

If the hypotenuse of a right-angle isosceles triangle is 2, what is the length of each of the other side?

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In a right-angle isosceles triangle, the two legs are congruent to each other. Let x be the length of each leg. By the Pythagorean theorem, we know that: x² + x² = 2² Simplifying the equation gives: 2x² = 4 Dividing both sides by 2, we have: x² = 2 Taking the square root of both sides gives: x = √2 Therefore, the length of each leg is √2, which is approximately 1.41. So, the correct answer is not in the options given, but it is approximately equal to, √2.

M varies directly as n and inversely as the square of p. If M = 3, when n = 2 and p = 1, find M in terms of n and p.

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We are given that "M varies directly as n and inversely as the square of p." This means that M is directly proportional to n and inversely proportional to the square of p. We can represent this relationship mathematically as: M ∝ n/p^2 where the symbol ∝ means "is proportional to". We are also given that M = 3 when n = 2 and p = 1. We can use this information to find the constant of proportionality k: M ∝ n/p^2 3 ∝ 2/1^2 3 ∝ 2 To find k, we can write: M = k(n/p^2) Substituting the values we know: 3 = k(2/1^2) k = 3/2 Now we can use k to find M in terms of n and p: M = (3/2)(n/p^2) Simplifying, we get: M = (3n)/(2p^2) Therefore, the answer is option D: M = 3n/2p^2.

The angle of a sector of a circle of radius 35cm is 288^o. Find the perimeter of the sector. [Take π = 22/7]

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The number of goals scored by a football team in 20 matches is shown in the table above

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The modal score is the score that occurs most frequently. Looking at the table, we can see that the score of 2 occurs the most frequently, appearing 6 times. Therefore, the modal goal scored is 2.

The table gives the distribution of outcomes obtained when a die was rolled 100 times.

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The number of goals scored by a football team in 20 matches is shown in the table above

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In the diagram above, ?SPQ = 79^o. Find ?SRQ

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In the diagram, ?PMN = ?PRQ and ?PNM = ?PQR. If /Pm/ = 3cm, /MQ/ = 7cm and /PN/ = 5cm, find /NR/

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The roots of a quadratic equation are -¹/₄ and 3. The quadratic equation is

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We know that if a quadratic equation has roots α and β, then the equation can be written as: (x - α)(x - β) = 0 Expanding the above expression, we get: x² - (α + β)x + αβ = 0 Here, the roots of the quadratic equation are -¹/₄ and 3. Therefore, α = -¹/₄ and β = 3. Substituting these values in the above equation, we get: x² - (α + β)x + αβ = 0 x² - (-¹/₄ + 3)x + (-¹/₄ × 3) = 0 x² - ¹¹/₄ x - ³/₄ = 0 Hence, the quadratic equation is 4x² - 11x - 3 = 0. Therefore, the correct option is: 4x² - 11x - 3 = 0

Simplify: \((2\frac{1}{6} - 1\frac{2}{3}) \div 2\frac{2}{3}\).

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In the diagram above, O is the center of the circle of radius 3.5cm, ∠POQ = 60°. What is the area of the minor sector POQ?

[Take π = 22/7].

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If tan x = \(2\frac{2}{5}\), find the value of sin x; 0 \(\leq\) x \(\leq\) 90^o

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If sin\(\theta\) cos\(\theta\), for 0^o \(\leq\) θ \(\leq\) 360^o, find the value of \(\theta\)

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The 6th term of a G.P is -2 and its first term is 18. What is the common ratio?

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Simplify 56x\(^{-4}\) \(\div\) 14x\(^{-8}\)

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To simplify the expression, we can use the rule of dividing powers with the same base. We divide the coefficients and subtract the exponents: 56x^-4 ÷ 14x^-8 = (56/14)x^(-4)-(-8) = 4x^-4+8 = 4x⁴ Therefore, the simplified expression is 4x⁴.

The cumulative frequency curve may be used to find the

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The cumulative frequency curve can be used to find the median. A cumulative frequency curve represents the cumulative frequencies of a dataset in ascending order, with the class boundaries on the x-axis and the cumulative frequency on the y-axis. To find the median from a cumulative frequency curve, we need to locate the point on the curve where the cumulative frequency is half of the total frequency. This point will correspond to the median class, and we can then use the formula for finding the median of grouped data to calculate the exact value of the median. Therefore, the answer is median.

How many sides has a polygon if the sum of its interior angle is 1440^o?

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To find the number of sides of a polygon given the sum of its interior angles, we can use the formula: Sum of interior angles = (n - 2) × 180° where "n" is the number of sides of the polygon. We are given that the sum of the interior angles of the polygon is 1440°. Substituting this value into the formula, we get: 1440° = (n - 2) × 180° Simplifying this equation, we can divide both sides by 180: 8 = n - 2 Adding 2 to both sides, we get: n = 10 Therefore, the polygon has 10 sides.

The marks obtained by pupils of class are grouped as shown below; 0 - 4, 5 - 9,10 -14, 15 -19. Which of the following is/are not true? l. The mid values of the grouped marks are 2, 7, 12 and 17. II The class interval is 4. III The class boundaries are 0.5, 4.5, 9.5, 14.5 and 19.5

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The mid values of the grouped marks can be found by taking the average of the upper and lower limits of each class interval. For example, the mid value of the first class interval (0 - 4) is (0+4)/2 = 2. Similarly, the mid values of the other class intervals can be calculated as 7, 12, and 17. The class interval is the difference between the upper limit of a class interval and the lower limit of the previous class interval. For example, the class interval between 0-4 and 5-9 is 5-4 = 1. The class interval in this case is not 4, but rather it is 5-0 = 5. The class boundaries are the values that separate one class interval from another. They are found by adding and subtracting half of the class interval from the upper and lower limits of each class interval. For example, the lower class boundary of the first class interval (0 - 4) is 0 - 0.5 = -0.5, and the upper class boundary of the second class interval (5 - 9) is 9 + 0.5 = 9.5. Therefore, the class boundaries for the given grouped marks are -0.5, 4.5, 9.5, 14.5, and 19.5. Therefore, the statement that is not true is II only, since the class interval is not 4. The correct class interval is 5, as explained above. So the answer is (B) II only.

The graph of 2y = 5x² - 3x² - 2 cuts the y axis at the point

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To find where the graph of the equation 2y = 5x² - 3x² - 2 intersects the y-axis, we substitute x = 0 and solve for y: 2y = 5(0)^2 - 3(0)^2 - 2 2y = -2 y = -1 Therefore, the graph of the equation intersects the y-axis at the point (0, -1). Hence, the correct option is (c) (0, -1).

In the diagram above, PQ is parallel to RST. /RS/ = /SQ/ = /TQ/ and ?PQR = 35^o.Calculate ?SQT

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