JAMB UTME - Mathematics - 2004

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Number of times 5 was obtained = 20
P(obtaining 5) = 20/100
= 1/5

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To find the ratio of the number of grains sowed to the number harvested, we need to divide the number of grains sowed by the number of grains harvested. Number of grains sowed = 5000 Number of grains harvested = 5000 cobs × 500 grains per cob = 2,500,000 So the ratio of the number of grains sowed to the number harvested is: 5000 : 2,500,000 Simplifying the ratio by dividing both sides by 5000, we get: 1 : 500 Therefore, the answer is: 1 : 500.

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$\begin{array}{ccc}\text{class-marks(x)}& \text{(freq. (f)}& \text{cum. freq.}\\ 9& & \mathrm{3.....3}\\ 14& & \mathrm{6....3}+6=9\\ 19& & \mathrm{9.....9}+9=18\\ 24& & \mathrm{3.....18}+3=21\\ 29& & \mathrm{2.....21}+2=23\\ 34& & \mathrm{2.....23}+2=25\\ & & 25\end{array}$
% of students scoring more than 20 marks

= $\frac{7}{25}$ x 100% = 28%

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Coordinates of the line are (x1 y1)(0,2) respectively and (x2 y2)(2,0) respectively
Gradient (m) of the line = (y2 - y1) / (x2 - x1) = (0-6) / (2-0) = -6/2
= -3
Equation of the line y-y1 = m(x-x1)
y - 6 = -3(x - 0)
y - 6 = -3x
y + 3x = 6
inequality; from the y axis, the shaded region is below the line
∴ y + 3x ∠ 6 and line is a broken line

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The range of a distribution is the difference between the maximum and minimum values. In this case, the maximum weight is 18 kg and the minimum weight is 15 kg, so the range is 18 kg - 15 kg = 3 kg. Therefore, the answer is option (B) 3.

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The locus of a point which is 5cn from the line LM is a pair of lines on opposite sides of LM and parallel to it, each distance 5cm from LM

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The height of the pole is 5√3m and its shadow on the ground is 5m. Let x be the angle of elevation of the sun. From the geometry of the situation, we can see that tan(x) = opposite/adjacent = (5√3)/5 = √3. Taking the inverse tangent of both sides, we get: x = tan^(-1)(√3) ≈ 60° Therefore, the angle of elevation of the sun is 60°. So the answer is (D) 60°.

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xo = 35 + 29 (Exterior angle = sum of two interior opposite angles)
x = 55o (∠ at the center twice ∠ at circumference) y = 110o
∠PSO = ∠SPO (base ∠S of 1sc Δ b/c PO = SO)
∴ ∠PSO = (180 - 110)/2
= 35o

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y ∝ 1/x
y = K/x
K = xy
K = (1/2) * 4 ( when y = 4 and x = 1/2)
K = 2
∴y = 2/x
10 = 2/x (when y = 10)
10x = 2
x = 2/10
x = 1/5

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The nth term of the sequence Qn is Qn = 3*2^(n-2), and the mth term of the sequence Um is Um = 3*2^(2m-3). We are asked to find the product of Q2 and U2, which means we need to find Q2 and U2 and then multiply them together. Substituting n=2 in the formula for Qn, we get: Q2 = 3*2^(2-2) = 3*2^0 = 3*1 = 3 Substituting m=2 in the formula for Um, we get: U2 = 3*2^(2*2-3) = 3*2^1 = 3*2 = 6 So the product of Q2 and U2 is: Q2 * U2 = 3 * 6 = 18 Therefore, the answer is (A) 18.

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tan θ - 7/3.5 = 2
tan θ= 2.0000
θ = 63o

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To factorize the expression, we can group the terms as follows: ac - 2bc - a + 4b^2 = a(c - 1) - 2b(c - 2b) Now we can factor out the common factors: a(c - 1) - 2b(c - 2b) = (c - 1)(a - 2b) So the completely factored form of the expression is: ac - 2bc - a + 4b^2 = (c - 1)(a - 2b) Therefore, the correct answer is option A: (a - 2b)(c - a - 2b).

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Let's assume the packet of chalk has a total of 15 units (the lowest common multiple of 5 and 15). The first teacher received 2/5 of this, which is 6 units of chalk. The remainder is 9 units of chalk (15 - 6). The second teacher received 2/15 of this remainder, which is (2/15) x 9 = 2/5 units of chalk. So the third teacher received the remaining chalk, which is 9 - 2/5 = 35/5 - 2/5 = 33/5 units of chalk. Therefore, the fraction of chalk that the third teacher received is 33/5, which can be simplified to 6 3/5 or 13/5. So the answer is option D: 13/25.

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x = -3
substitute x = -3 in 3x3 + 5x2 - 11x + 4
3(-3)3 + 5(-3)2 - 11(-3) + 4
-81 + 45 + 33 + 4
-81 + 82
= 1

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P×Q=√PQ4×(8×32)=4×√8×32=4×√256=4×16=√4×16=√64=8

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To find the derivative of the given function, we need to apply the product rule of differentiation, which is: d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x) Here, f(x) = (2 + 3x) and g(x) = (1 - x) Taking the derivatives of the functions separately, we have: f'(x) = 3 g'(x) = -1 Substituting the values in the product rule formula, we get: d/dx [(2 + 3x)(1 - x)] = (2 + 3x)(-1) + (1 - x)(3) Simplifying the expression, we get: d/dx [(2 + 3x)(1 - x)] = -2 - 3x + 3 - 3x d/dx [(2 + 3x)(1 - x)] = -6x + 1 Therefore, the correct answer is: -6x + 1.

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In a circle, the ratio of the arc length to the circumference of the circle is equal to the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution (360 degrees). Using this property, we can find the value of the angle θ in the following way: The circumference of the circle is given by: C = 2πr = 2 x (22/7) x 15 = 94.29 cm The given arc length is 22 cm. Therefore, the ratio of the arc length to the circumference of the circle is: 22/94.29 = 0.2332 Let the angle of one full revolution be 360 degrees. Then, the ratio of the angle subtended by the arc at the center of the circle to the angle of one full revolution is also 0.2332. Thus, we can write: θ/360 = 0.2332 Multiplying both sides by 360, we get: θ = 360 x 0.2332 = 83.952 degrees (approximately) Therefore, the value of θ is approximately 84 degrees. The correct option is: -  84o

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To find the area of the trapezium PQRU, we need to know the length of its height or perpendicular distance between the parallel sides PQ and RU. We can find this height by using the area of the parallelogram PQTU, which has the same base PQ and height as the trapezium, and a known area of 32 cm². Area of parallelogram PQTU = base × height 32 cm² = 4 cm × height Height = 8 cm Now we can use the formula for the area of a trapezium: Area of trapezium PQRU = (sum of parallel sides × height) ÷ 2 = (PQ + RU) × height ÷ 2 = (4 cm + 10 cm) × 8 cm ÷ 2 = 56 cm² Therefore, the area of trapezium PQRU is 56 cm². Answer option (B) is correct.

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The midpoint of a line segment is the point that is exactly halfway between the endpoints of the line segment. The formula for finding the midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is: Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2) In this case, we are given the endpoints of the line segment PQ as P(-3, 5) and Q(5, -3). Using the formula for the midpoint, we can find the coordinates of the midpoint as: Midpoint = ((-3 + 5) / 2, (5 - 3) / 2) = (2 / 2, 2 / 2) = (1, 1) Therefore, the midpoint of the line joining P(-3, 5) and Q(5, -3) is (1, 1). Hence, the correct answer is option (A) (1, 1).

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6logx2 - 3logx3 = 3log50.2
= logx26 - 3logx33 = log5(0.2)3
= logx(64/27) = log5(1/5)3
logx(64/27) = log5(1/125)
let logx(64/27) = y
∴xy = 64/27
and log5(1/125) = y
∴ 5y = 1/125
5y = 125-1
5y = 5-3
∴ y = -3
substitute y = -3 in xy = 64/27
implies x-3 = 64/27
1/x3 = 64/27
64x3 = 27
x3 = 27/64
x3 = 3√27/64
x = 3/4

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The diagonal QS bisects the angle formed by PQ and QR
∴ [A]

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Let's say the number of white balls introduced is x. The total number of balls in the basket before introducing the white balls was 12 + 16 = 28. After introducing the x white balls, the total number of balls becomes 12 + 16 + x. Now, the probability of picking a white ball is given as 3/7. We can express this probability as: Number of white balls / Total number of balls = 3/7 Substituting the values, we get: x / (12 + 16 + x) = 3/7 Solving for x: 7x = 3(28 + x) 7x = 84 + 3x 4x = 84 x = 21 Therefore, the number of white balls introduced is 21. So, option (B) is the correct answer.

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There are a total of 30 + 22 + 18 = 70 medals in the container. The probability of selecting a gold medal is 30/70. The probability of not selecting a gold medal is equal to the probability of selecting either a silver or a bronze medal. Therefore, the probability of not selecting a gold medal is (22 + 18)/70 = 40/70. This fraction can be simplified to 4/7, which is the answer.

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Explanation:

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Sectorial angle of beans = 360 - (150 + 90 + 60)
= 360 -300
= 60${}^{\circ }$
if 150${}^{\circ }$ represents 3000 tonnes
1${}^{\circ }$ 3000/150
60${}^{\circ }$ will represents (3000/150) * (60/1)
= 1200 tonnes

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The inequality given is: -1 < 3 - 2x ≤ 5. We need to find the integer values of x that satisfy this inequality. First, let's solve for the left side of the inequality: -1 < 3 - 2x. Add 2x to both sides: 2x - 1 < 3 Add 1 to both sides: 2x < 4 Divide both sides by 2: x < 2 Now, let's solve for the right side of the inequality: 3 - 2x ≤ 5. Subtract 3 from both sides: -2x ≤ 2 Divide both sides by -2, remembering to flip the inequality: x ≥ -1 So, the integer values of x that satisfy the inequality are -1, 0, 1, and 2. Therefore, the correct option is: - -1, 0, 1,

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To find the sum of the first three terms of a G.P, we need to determine the common ratio (r) of the progression. Given that the first and fourth terms are 6 and 162 respectively, we can use the formula for the nth term of a G.P to obtain: a1 = 6, a4 = 162 a4 = a1 * r^3 162 = 6 * r^3 r^3 = 27 r = 3 Now, we can find the second and third terms of the progression using the common ratio: a2 = a1 * r = 6 * 3 = 18 a3 = a2 * r = 18 * 3 = 54 The sum of the first three terms of the G.P is: a1 + a2 + a3 = 6 + 18 + 54 = 78 Therefore, the correct answer is option (C) 78.

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To find the rate of change of the volume of a hemisphere with respect to its radius, we need to differentiate the formula for the volume of a hemisphere with respect to its radius. The formula for the volume of a hemisphere is V = (2/3)πr^3. Taking the derivative of this formula with respect to r gives us dV/dr = 2πr^2. Therefore, the rate of change of the volume of the hemisphere with respect to its radius is 2πr^2. When r = 2, the rate of change is 2π(2^2) = 2π(4) = 8π. Thus, the answer is 8π.

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A histogram is a graphical representation of a frequency distribution of a dataset, where the data is divided into intervals or bins and the frequency of each interval is represented by the area of a rectangle or bar. To construct a histogram, we need to follow certain rules, which are described by the statements: I. Rectangular bars of equal width II. The height of each rectangular bar is proportional to the frequency of the corresponding class interval. III. Rectangular bars have common sides with no gaps in between. Therefore, a histogram is described completely by statements I, II, and III. Statement I ensures that each bar is of equal width, which is necessary to represent the intervals or bins in a consistent way. Statement II ensures that the height of each bar accurately represents the frequency of the corresponding class interval. This is important because the area of the bar represents the frequency of the interval, which is the product of the width and height of the bar. Statement III ensures that the bars are adjacent to each other without any gaps, which is necessary to accurately represent the continuity of the data. Hence, the correct answer is option (A) I, II, and III.

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y = 2x2(2x - 1)
y = 4x3 - 2x2
dy/dx = 12x2 - 4x
at x = -1
dy/dx = 12(-1)2 - 4(-1)
= 12 + 4
= 16

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y = 3cos(x/3)
dy/dx = 3x(1/3)x - sin (x/3)
= - sin x/3
But x = 3π/2 ∴ -sin(x/3) = -sin(3π/6)
= -sin (3 * 180)/6
= - sin 90
= -1

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3√42x = 16
this implies that (3√42x)3 = (16)3
42x = 42*3
42x = 46
∴ 2x = 6
x = 3

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Let's start by defining some variables: - n: the number of students in the group - sum_age_students: the sum of the ages of the students We know that the mean age of the group of students is 15, so we can write: mean_age_students = sum_age_students / n = 15 From the problem, we also know that when the age of the teacher is added to the age of the students, the mean of their ages becomes 18. So we can write another equation: (mean_age_students + 45) = (sum_age_students + 45) / (n + 1) = 18 Now we have two equations with two variables (n and sum_age_students), and we can solve for them simultaneously. Let's start by simplifying the second equation: sum_age_students + 45 = 18(n + 1) sum_age_students = 18n + 27 Now we can substitute this expression for sum_age_students in the first equation: sum_age_students / n = 15 (18n + 27) / n = 15 18n + 27 = 15n 3n = 27 n = 9 Therefore, there are 9 students in the group.

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PQ=√(β−1)2+(1−α)23=√(β2−2β2+1+1−2α+α2)3=√(α2+β2−2α+2β+2)3=√(α2+β2−2(α+β)+2)3=√(α2+β2−2∗2+2)3=√(α2+β2−2)9=(α2+β2−2)α2+β2=9+2α2+β2=11

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A pentagon is a five-sided polygon, and the sum of its interior angles can be found using the formula: Sum of interior angles = (n-2) * 180 degrees where n is the number of sides in the polygon. For a pentagon, n = 5, so the formula becomes: Sum of interior angles = (5-2) * 180 degrees = 3 * 180 degrees = 540 degrees Now, we are given that the sum of the interior angles of a pentagon is 6x + 6y. Therefore, we can equate this expression with 540 degrees to get: 6x + 6y = 540 Dividing both sides by 6, we get: x + y = 90 Subtracting x from both sides, we get: y = 90 - x Therefore, the value of y in terms of x is y = 90 - x. Hence, the correct answer is option (A) y = 90 - x.

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The given series is a geometric series with the first term a = 1/2 and the common ratio r = 1/3. To find the sum to infinity of the series, we can use the formula: S = a/(1 - r) where S is the sum to infinity. Substituting the values of a and r, we get: S = (1/2)/(1 - 1/3) = (1/2)/(2/3) = 1/2 * 3/2 = 3/4 Therefore, the sum to infinity of the series 1/2, 1/6, 1/18, ... is 3/4. So the answer is (C) 3/4.

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The problem describes a direct variation between the length of a simple pendulum and the square of its period, which means that we can write the relationship as: L = kT^2 where L is the length of the pendulum, T is its period, and k is a constant of proportionality. To find the value of k, we can use the information given in the problem that a pendulum with a period of 4 seconds has a length of 64 cm. Substituting these values into the equation above, we get: 64 = k(4^2) 64 = 16k k = 4 Now that we know the value of k, we can use the same equation to find the length of a pendulum with a period of 9 seconds: L = 4(9^2) L = 324 Therefore, the length of the pendulum whose period is 9 seconds is 324 cm.