Express 0.0000407 to 2 significant figures
The number 0.0000407 is 4.07 × 10^{-5}. To two significant figures, look at the first two digits 4 and 0. The next digit is 7, which is greater than or equal to 5, so round up the 0 to 1, making it 4.1 × 10^{-5} or 0.000041.
If x varies inversely as y and y varies directly as z, what is the relationship between x and z?
Since x varies inversely as y, x = k / y for some constant k. y varies directly as z, so y = m z for some constant m. Substituting, x = k / (m z) = (k/m) / z, so x ∝ 1/z.
Evaluate (3 1/4 × 1 3/5) / (11 1/3 - 5 1/3)
3 1/4 = 13/4, 1 3/5 = 8/5, so numerator (13/4) × (8/5) = 104/20 = 26/5. Denominator 11 1/3 = 34/3, 5 1/3 = 16/3, difference (34/3 - 16/3) = 18/3 = 6. Then (26/5) / 6 = 26/5 × 1/6 = 26/30 = 13/15.
Using the addition and multiplication tables in modulo 5, solve the equation (n ⊕ 4) ⊕ 3 = 0 (mod 5)
Assuming ⊕ denotes addition modulo 5, (n + 4 mod 5) + 3 mod 5 ≡ 0 mod 5. This simplifies to n + 7 ≡ 0 mod 5, or n + 2 ≡ 0 mod 5, so n ≡ -2 ≡ 3 mod 5. Thus, n = 3.
The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age?
Let Tunde's age be k, then Ola's age is 2k. Ola : Musa = 4:5, so 2k / m = 4/5, where m is Musa's age. Then 2k = (4/5) m, m = (5/2) (2k) = 5k/2. Ratio T:M = k : 5k/2 = 2:5.
If M = {x: 3 ≤ x < 8} and N = {x: 8 < x ≤ 12}, which of the following is true? I. 8 ∈ M ∩ N II. 8 ∈ M ∪ N III. M ∩ N = ∅
I. 8 is not in M (x < 8) or N (x > 8), so not in intersection. II. 8 is not in M or N, so not in union. III. No common elements, so intersection is empty. Only III is true.
Given that a = log 7 and b = log 2, express log 14 in terms of a and b
log 14 = log (7 × 2) = log 7 + log 2 = a + b.
If x = 2/3 and y = -6, evaluate xy - y/x
xy = (2/3) × (-6) = -4. y/x = (-6) / (2/3) = -6 × (3/2) = -9. Then xy - y/x = -4 - (-9) = -4 + 9 = 5.
Solve the equation: 1/(5x) + 1/x = 3
Combine terms: (1 + 5)/(5x) = 3, so 6/(5x) = 3. Then 5x = 6/3 = 2, x = 2/5.
A sum of ₦18,100 was shared among 5 boys and 4 girls with each boy taking ₦20.00 more than each girl. Find a boy's share
Let girl's share be g, boy's g + 20. 5(g + 20) + 4g = 18100, 5g + 100 + 4g = 18100, 9g = 18000, g = 2000, boy = 2020.
One factor of 7x² + 33x - 10 is
To factor 7x² + 33x - 10, find two numbers that multiply to 7 × (-10) = -70 and add to 33: 35 and -2. Rewrite as 7x² + 35x - 2x - 10 = 7x(x + 5) - 2(x + 5) = (7x - 2)(x + 5). Thus, one factor is 7x - 2.
Solve: -1/4 < (3/4)(3x - 2) < 1/4
Divide by 3/4 (positive): -1/4 ÷ 3/4 = -1/3 < 3x - 2 < 1/4 ÷ 3/4 = 1/3. Add 2: 5/3 < 3x < 7/3. Divide by 3: 5/9 < x < 7/9.
Simplify: 3x - (p - x) - (r - p)
3x - p + x - r + p = 4x - r.
An arc of a circle of radius 7.5cm is 7.5cm long. Find, correct to the nearest degree, the angle which the arc subtends at the centre of the circle. Take π = 22/7
Arc length = r θ (radians), 7.5 = 7.5 θ, θ = 1 radian. To degrees: 1 × (180/π) = 180 × 7/22 ≈ 57.27°, nearest 57°.
Water flows out of a pipe at a rate of 40π cm³ per second into an empty cylindrical container of base radius 4cm. Find the height of water in the container after 4 seconds
Volume in 4 s = 40π × 4 = 160π cm³. Base area = π (4)^2 = 16π cm². Height h = volume / area = 160π / 16π = 10 cm.
The dimensions of a water tank are 13cm, 10cm, and 70cm. If it is half-filled with water, calculate the volume of water in litres
Volume = (13 × 10 × 70) / 2 = 9100 / 2 = 4550 cm³ = 4.55 litres (since 1000 cm³ = 1 litre).
If the curved surface area of a solid hemisphere is equal to its volume, find the radius
Curved surface area = 2π r², volume = (2/3) π r³. Set equal: 2π r² = (2/3) π r³. Divide both sides by 2π r² (r ≠ 0): 1 = (1/3) r, r = 3 cm.
Which of the following is true about a parallelogram?
In a parallelogram, adjacent angles sum to 180° (supplementary), and opposite angles are equal. The statement for A is true.
In a right-angled triangle with angles 30° and 60°, the side opposite 30° is 5 cm. Find the hypotenuse.
In 30-60-90 triangle, sides 1 : √3 : 2. Opposite 30° is half hypotenuse, so hypotenuse = 2 × 5 = 10 cm.
The exterior angle of a regular hexagon is
Sum of exterior angles = 360°, 6 equal exterior angles, each 360°/6 = 60°.
The exterior angle of a regular quadrilateral is
A regular quadrilateral is a square. Sum of exterior angles = 360°, 4 equal exterior angles, each 360°/4 = 90°.
Two angles of a triangle are 40° and 37°. Find the measure of the third angle.
Sum of angles in triangle = 180°, so third angle = 180° - 40° - 37° = 103°.
Calculate the gradient (slope) of the line joining points (-1, 1) and (2, -2)
Slope m = (y2 - y1)/(x2 - x1) = (-2 - 1)/(2 - (-1)) = (-3)/3 = -1.
If P (2, 3) and Q (2, 5) are points on a graph, calculate the length PQ
Distance = √[(2-2)^2 + (5-3)^2] = √[0 + 4] = 2 units.
A bearing of 320° expressed as a compass bearing is
320° is 40° west of north, so N40°W.
Given that cos 30° = sin 60° = √3/2 and sin 30° = cos 60° = 1/2, evaluate (tan 60° - 1)/(1 - tan 30°)
tan 60° = √3, tan 30° = 1/√3. Numerator √3 - 1, denominator 1 - 1/√3 = (√3 - 1)/√3. So (√3 - 1) / [(√3 - 1)/√3] = √3.
A stationary boat is observed from a height of 100m. If the horizontal distance between the observer and the boat is 80m, calculate, correct to two decimal places, the angle of depression of the boat from the point of observation
tan θ = opposite / adjacent = 100 / 80 = 1.25, θ = arctan(1.25) ≈ 51.34°.
The average age of a group of 25 girls is 10 years. If one girl, aged 12 years and 4 months, joins the group, find, correct to one decimal place, the new average age of the group
Total age = 25 × 10 = 250 years. New girl = 12 + 4/12 = 12 1/3 ≈ 12.333 years. New total = 250 + 12.333 = 262.333, average = 262.333 / 26 ≈ 10.09 ≈ 10.1 years.
The probability that a candidate passes a mathematics test is 5/6. What is the ratio of the probability that he passes the test to the probability that he fails the test?
Probability of failure = 1 - 5/6 = 1/6. Ratio passes:fails = (5/6):(1/6) = 5:1.
Find the median of a distribution with frequencies 7, 4, 18, 12, 8, 11 for marks 0, 1, 2, 3, 4, 5 respectively
Total n = 60, median position (60/2) = 30th value. Cumulative frequencies: 0:7, 1:11, 2:29, 3:41. The 30th falls in mark 3, so median = 3.
Find the first quartile of a distribution with frequencies 7, 4, 18, 12, 8, 11 for marks 0, 1, 2, 3, 4, 5 respectively
First quartile position = (60 + 1)/4 = 15.25th value. Cumulative: up to 1:11, up to 2:29. The 15th is in mark 2, so Q1 = 2.0.
In a class of 45 students, 28 offer Chemistry and 25 offer Biology. If each student offers at least one of the two subjects, calculate the probability that a student selected at random from the class offers Chemistry only
Number offering both = 28 + 25 - 45 = 8. Chemistry only = 28 - 8 = 20. Probability = 20/45 = 4/9.
In what number base was the addition 1 + nn = 100, where n > 0, done?
In base b, nn_b = n b + n = n(b + 1), 100_b = b². So 1 + n(b + 1) = b², n(b + 1) = b² - 1 = (b - 1)(b + 1), n = b - 1, b = n + 1.
Simplify √2(√6 + 2√2) - 2√3
√2 × √6 = √12 = 2√3, √2 × 2√2 = 2 × 2 = 4, so 2√3 + 4 - 2√3 = 4.
Three exterior angles of a polygon are 30°, 40°, and 60°. If the remaining exterior angles are 46° each, name the polygon
Sum of exterior angles = 360°. Given sum 130°, remaining sum 230°, number of remaining angles = 230 / 46 = 5, total angles = 3 + 5 = 8, octagon.
The sum of three interior angles of a quadrilateral is 230°. Find the measure of the fourth angle
The sum of interior angles of a quadrilateral is 360°, so fourth angle = 360° - 230° = 130°.
Simplify the expression (a + b)(a - b)
(a + b)(a - b) = a² - ab + ab - b² = a² - b².
The 6th term of the sequence: 2/3, 7/15, 4/15, ...
This is an arithmetic progression with first term a = 2/3 and common difference d = -1/5. The 6th term is a + 5d = 2/3 + 5(-1/5) = 2/3 - 1 = -1/3.
The diagonal of a square is 60cm. Calculate its perimeter
Side s, diagonal s√2 = 60, s = 60/√2 = 30√2. Perimeter = 4s = 120√2 cm.
The roots of a quadratic equation are -1/2 and 2/3. Find the equation
Sum of roots = -1/2 + 2/3 = 1/6 = -b/a. Product = (-1/2)(2/3) = -1/3 = c/a. For a=6, b=-1, c=-2, equation 6x² - x - 2 = 0.
Now practice in exam mode
You've studied the answers — now test yourself under real exam conditions with the timer running.
Start WAEC Mathematics 2017 Quiz