JAMB UTME - Mathematics - 2000
Ajụjụ 1 Ripọtì
If P3446 - 23P26 = 2PP26, find the value of the digit P.
Akọwa Nkọwa
Covert everything to base 10 and collect like terms, such that:
210P - 42P = 434 + 406
168P = 840
P = 840/168 = 5
Ajụjụ 2 Ripọtì
In a youth club with 94 members, 60 like modern music, and 50 like traditional music. The number of members who like both traditional and modern music is three times those who do not like any type of music. How many members like only one type of music?
Akọwa Nkọwa
Use a venn diagram:
60-3x+3x+50-3x = 94-x.
110-3x+x = 94
-2x = 94-110
=>-2x = -16, this x = -8.
Members that like only one game:
= 60 - 3x + 50 - 3x
= 60 - 3x8 + 50 - 3x8
= 60 - 24 + 50 - 24
= 36 + 26 = 62
Ajụjụ 3 Ripọtì
A bowl is designed by revolving completely the area enclosed by y = x2 - 1, y = 3 and x ≥ 0 around the axis. What is the volume of this bowl?
Akọwa Nkọwa
∫30 π(y+1) dy = π[y2 + y30
= π(9/2 + 3) = 15π/2
Ajụjụ 5 Ripọtì
Find the inverse of p under the binary operation * defined by p*q = p + q - pq, where p and q are real numbers and zero is the identity
Akọwa Nkọwa
If P-1 is the inverse of P and O is the identity, Then P-1 * P = P * P-1 = 0
i.e. P-1 + P - P-1.P = 0
P-1 - P-1.P = -P
P-1(1 - P) = -P
P-1 = -P/(P-1)
= P/(P-1)
Ajụjụ 7 Ripọtì
Let P = {1, 2, u, v, w, x}; Q = {2, 3, u, v, w, 5, 6, y} and R = {2, 3, 4, v, x, y}.
Determine (P-Q) ? R
Akọwa Nkọwa
To determine (P-Q) ? R, we need to first find the set difference (P-Q), which is the set of elements that are in P but not in Q. So, we need to remove all the elements in Q from P. (P-Q) = {1, w, x} Then, we take the intersection of (P-Q) and R, which means we need to find all the elements that are common to both (P-Q) and R. (P-Q) ? R = {x} Therefore, the answer is {x}.
Ajụjụ 8 Ripọtì
Solve the inequality 2 - x > x2.
Ajụjụ 9 Ripọtì
The expression ax2 + bx + c equals 5 at x = 1. If its derivative is 2x + 1, what are the values of a, b, c respectively?
Akọwa Nkọwa
At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5;
f(1) => a + b + c = 5 .....(1)
Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2)
From (2),=> b = 1, and 2ax = 2x, => a = 1.
substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3
Thus a = 1, b = 1 and c = 3
Ajụjụ 10 Ripọtì
The 3rd term of an A.P is 4x - 2y and the 9th term is 10x - 8y. Find the common difference.
Akọwa Nkọwa
In an Arithmetic Progression (A.P), the difference between any two adjacent terms is constant. Let d be the common difference of the A.P, then: - The 3rd term is the first term plus two common differences, i.e. a + 2d = 4x - 2y - The 9th term is the first term plus eight common differences, i.e. a + 8d = 10x - 8y We can now solve for d by subtracting the first equation from the second: a + 8d - (a + 2d) = 10x - 8y - (4x - 2y) 6d = 6x - 6y d = (x - y) Therefore, the common difference is x - y, which is option (C).
Ajụjụ 11 Ripọtì
A man wishes to keep his money in a savings deposit at 25% compound interest so that after three years he can buy a car for ₦150,000. How much does he need to deposit?
Akọwa Nkọwa
Amount A = P(1+r)n;
A = ₦150,000, r = 25%, n = 3.
150,000 = P(1+0.25)3 = P(1.25)3
P = 150,000/1.253 =₦76,800.00
Ajụjụ 12 Ripọtì
In a regular polygon, each interior angle doubles its corresponding exterior angle. Find the number of sides of the polygon.
Akọwa Nkọwa
2x + x = 180°, => 3x = 180°, and thus x = 60°
Each exterior angle = 60° but size of ext. angle = 360°/n
Therefore 60° = 360°/n
n = 360°/60° = 6 sides
Ajụjụ 13 Ripọtì
In how many ways can a delegation of 3 be chosen from 5 men and 3 women. If at least 1 man and 1 woman must be included?
Ajụjụ 14 Ripọtì
In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH.
Akọwa Nkọwa
To solve this problem, we need to use the property of a circle that states that a diameter is twice the radius of the circle. Since FH is a diameter, then FH = 2OH, where OH is the radius of the circle. Since GE is a chord that meets FH at right angle at the point N, then N is the midpoint of FH. Therefore, NH = NF = 8 cm. Let's use the Pythagorean theorem to find OH. Since GE is a chord of the circle, we can draw radii from the center O to the endpoints of the chord so that we have a right triangle. Let x be the length of OE. Then, HG = 2x, and OH = x + 12 (since FH = 2OH = 2(x + 12) = 2x + 24). Using the Pythagorean theorem, we have: x^2 + 12^2 = (2x)^2 x^2 + 144 = 4x^2 3x^2 = 144 x^2 = 48 x = sqrt(48) = 4sqrt(3) Therefore, OH = x + 12 = 4sqrt(3) + 12. Finally, we can calculate FH: FH = 2OH = 2(4sqrt(3) + 12) = 8sqrt(3) + 24 ≈ 32.39 cm. Therefore, the answer is 32 cm.
Ajụjụ 15 Ripọtì
If α and β are the roots of the equation 3x2 + 5x - 2 = 0, find the value of 1/α + 1/β
Akọwa Nkọwa
We know that if α and β are the roots of the quadratic equation ax^2 + bx + c = 0, then α + β = -b/a and α × β = c/a. In this equation, 3x^2 + 5x - 2 = 0, we have a = 3, b = 5, and c = -2. So, α + β = -b/a = -5/3 And, α × β = c/a = -2/3 We need to find the value of 1/α + 1/β = (α + β)/(α × β) = (-5/3) / (-2/3) = 5/2. Therefore, the value of 1/α + 1/β is 5/2.
Ajụjụ 16 Ripọtì
Simplify 3(2n+1) - 4(2n-1) 2n+1 - 2n
Akọwa Nkọwa
Start by expanding 3(2n+1) - 4(2n-1) 2n+1 - 2n:
3 x 2n x 21 - 22 x 2n x 2-1 2n x 2 -2n
Solving the equation above gives;
2n x 21 x 2 2n = 2 x 2 = 4
Ajụjụ 17 Ripọtì
If the mean of the numbers 0, (x+2), (3x+6), and (4x+8) is 4, find their mean deviation.
Akọwa Nkọwa
The mean deviation of a set of numbers is the average of the absolute deviations from their mean. To find the mean deviation of the given set of numbers, we need to first find their mean. Mean = (0 + (x+2) + (3x+6) + (4x+8))/4 Mean = (8x + 16)/4 Mean = 2x + 4 We are given that the mean is 4. Therefore, 2x + 4 = 4 2x = 0 x = 0 Substituting x = 0 in the set of numbers, we get: 0, 2, 6, 8 Mean = (0+2+6+8)/4 = 4 To find the mean deviation, we need to find the deviation of each number from the mean and then take the average of their absolute values. Deviations from the mean: 0-4 = -4 2-4 = -2 6-4 = 2 8-4 = 4 Mean deviation = (|-4| + |-2| + |2| + |4|)/4 Mean deviation = (12/4) Mean deviation = 3 Therefore, the answer is 3.
Ajụjụ 18 Ripọtì
Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR
Akọwa Nkọwa
QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o
Ajụjụ 19 Ripọtì
An equilateral triangle of side √3cm is inscribed in a circle. Find the radius of the circle.
Akọwa Nkọwa
In an equilateral triangle, all the sides are of equal length and each angle is 60 degrees. Let's draw an equilateral triangle with side √3 cm inscribed in a circle. Since the triangle is equilateral, we know that the circumcenter (center of the circle that passes through all the vertices of the triangle) is also the centroid (point of intersection of the medians). The median of an equilateral triangle is the line segment from a vertex to the midpoint of the opposite side. Therefore, the circumcenter is also the midpoint of any side of the triangle. Let's choose one side of the triangle and label its midpoint as point M. We can draw a perpendicular line from M to the opposite vertex of the triangle, which will bisect the side and form a right triangle with one leg being half of the side of the triangle and the other leg being the radius of the circle. Using the Pythagorean theorem, we can solve for the radius: (radius)^2 = (√3/2)^2 + (1/2)^2 (radius)^2 = 3/4 + 1/4 (radius)^2 = 1 radius = 1 cm Therefore, the radius of the circle is 1 cm. Answer: 1 cm.
Ajụjụ 20 Ripọtì
Evaluate (1/2 - 1/4 - 1/8 - 1/16 + ...) - 1
Akọwa Nkọwa
The given expression is a geometric series with the first term as 1/2 and the common ratio as -1/2. Using the formula for the sum of an infinite geometric series with |r| < 1, we can find that the sum of the series is: S = a/(1-r) = (1/2)/(1-(-1/2)) = (1/2)/(3/2) = 1/3 Therefore, the expression (1/2 - 1/4 - 1/8 - 1/16 + ...) - 1 simplifies to: 1/3 - 1 = -2/3 So, the answer is -2/3.
Ajụjụ 21 Ripọtì
The variance of x, 2x, 3x, 4x and 5x is
Akọwa Nkọwa
Hint: prepare a three-columned table, one for (x), another for the (deviation), and the last for the (squared-deviation).
Sum of (x) = 15x, algebraic sum of (deviation) = zero (0)
Sum of (squared-deviation) = 10x2
Variance = ∑(squared-deviation)/n = 10x2/5 = 2x2
Ajụjụ 22 Ripọtì
if (x - 1), (x + 1) and (x - 2) are factors of the polynomial ax3 + bx2 + cx - 1, find a, b, c in that order.
Akọwa Nkọwa
This is a polynomial of the 3rd order, thus x should have three answers. Use the factors given to get values of x as 1, -1 and -2.
Form three equations, and carry out elimination and subsequent substitution to get a = -1/2, b = 1, and c = 1/2
Ajụjụ 23 Ripọtì
A binary operation * is defined by a*b = ab. If a*2 = 2-a, find the possible values of a.
Akọwa Nkọwa
a * b = ab
a * 2 = a2 = 2 - a
a2 + 2a - a - 2 = 0
a(a+2) - 1(a+2) = 0
(a-1)(a+2) = 0
a = 1, -2
Ajụjụ 24 Ripọtì
Evaluate 5-3log52 x 22log23
Akọwa Nkọwa
5-3log52 x 22log23i Let -3log52 = p => log52-3 = p
∴2-3 = 5p
∴5-3log52 = 5log52-3 = 5pii 22log23 = q => log532 = q
∴32 = 2q
∴222log23 = 2q
= 32 = 9
i x ii = 2-3 x 9 = 1/23 x 9
= 1/8 x 9
= 9/8 = 11/8
Ajụjụ 26 Ripọtì
Find the value of \( \displaystyle \int_{0}^{\pi}\frac{\cos^{2}\theta-1}{\sin^{2}\theta}\,d\theta \)
Ajụjụ 27 Ripọtì
A frustrum of pyramid with square base has its upper and lower sections as squares of sizes 2m and 5m respectively and the distance between them 6m. Find the height of the pyramid from which the frustrum was obtained.
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 10, 9, 7, 10, 6, 5
Akọwa Nkọwa
To find the range, we need to subtract the smallest number from the largest number in the set. The smallest number in the set is 4 and the largest number is 10. So, the range is 10 - 4 = 6. To find the mode, we need to determine the number that occurs the most frequently in the set. In this case, the number 10 occurs the most frequently, appearing 7 times. So, the mode is 10. To find the sum of the range and the mode, we simply add the two values together. Thus, the sum of the range and the mode is 6 + 10 = 16. Therefore, the answer is option (A) 16.
Ajụjụ 29 Ripọtì
P is a point on one side of the straight line UV and P moves in the same direction as UV. If the straight line ST is on the locus of P and angle VUS = 60°, find angle UST.
Akọwa Nkọwa
Make a good sketch of the question, and follow this steps.
Since ST is parallel to UV => angle USR = 50°(alternate to VUS.
Angle UST = 180° - 50° = 130° (angle on a straight line)
Ajụjụ 30 Ripọtì
If 31410 - 2567 = 340x, find x.
Akọwa Nkọwa
31410 - 2567 = 340x,
Convert 2567 and 340x to base 10, such that:
314 - 139 = 3x2 + 4x
=> 3x2 + 4x - 175 = 0 (quadratic)
Factorising, (x - 7) (3x + 25) = 0,
either x = 7 or x = -25/3 ( but x cannot be negative)
Therefore, x = 7.
Ajụjụ 31 Ripọtì
In how many ways can the word MATHEMATICS be arranged?
Akọwa Nkọwa
Number of letter in MATHEMATICS = 11
Number of letter M = 2
Number of letter E = 2
Number of letter A = 2
Arrangement = 11!/(2! 2! 2!) ways
Ajụjụ 32 Ripọtì
X and Y are two events. The probability of X or Y is 0.7 and that of X is 0.4. If X and Y are independent, find the probability of Y.
Akọwa Nkọwa
P (X or Y) = P(X) + P(Y), when they are independent as given.
0.7 = 0.4 + P(Y)
P(Y) = 0.7 - 0.4 = 0.30
Ajụjụ 33 Ripọtì
The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
If \(U = \{x : x\text{ is an integer and }1 \le x \le 20\}\)
\(E_1 = \{x : x\text{ is a multiple of }3\}\)
\(E_2 = \{x : x\text{ is a multiple of }4\}\)
and an integer is picked at random from \(U\), find the probability that it is not in \(E_2\)
Akọwa Nkọwa
U = {1, 2, 3, 4, 5,..., 20}
E1 = {3, 6, 9, 12, 15, 18}
E2 = {4, 8, 12, 16, 20}
P(E1) = 5/20
P(not E1) = 1 - (5/20) = 15/20 = 3/4
Ajụjụ 35 Ripọtì
A trader realizes 10x - x2 naira profit from the sale of x bags on corn. How many bags will give him the desired profit?
Akọwa Nkọwa
The trader's profit function can be represented as P(x) = 10x - x^2, where x is the number of bags sold and P(x) is the profit obtained from selling x bags of corn. To find the number of bags that will give him the desired profit, we need to maximize the profit function P(x). We can do this by finding the derivative of P(x) with respect to x, and then setting it to zero to find the critical point(s) of the function. P'(x) = 10 - 2x Setting P'(x) to zero gives: 10 - 2x = 0 Solving for x gives: x = 5 Therefore, the trader will make the maximum profit when he sells 5 bags of corn. This means that the answer is, 5.
Ajụjụ 36 Ripọtì
Simplify \( \frac{1}{1+\sqrt{5}} - \frac{1}{1-\sqrt{5}} \)
Akọwa Nkọwa
11+√5
- 11−√5
= 3−√5−3−√5(3+√5)(3−√5
= −2√59−5
= −2√54
= - 12√5
Ajụjụ 38 Ripọtì
A ship sails a distance of 50km in the direction S50°E and then sails a distance of 50km in the direction ₦40°E. FInd the bearing of the ship from the original position.
Ajụjụ 39 Ripọtì
If the population of a town was 240,000 in January 1998 and it increased by 2% each year, what would be the population of the town in January, 2000?
Akọwa Nkọwa
To solve this problem, we can use the formula for exponential growth, which is: A = P(1 + r)^t where A is the final amount, P is the initial amount, r is the annual growth rate as a decimal, and t is the time in years. In this problem, we are given P = 240,000, r = 0.02, and t = 2 (since we want to find the population after two years). We can plug these values into the formula and simplify: A = 240,000(1 + 0.02)^2 A = 240,000(1.02)^2 A = 240,000(1.0404) A = 249,696 Therefore, the population of the town in January 2000 would be 249,696. So the answer is 249,696.
Ajụjụ 40 Ripọtì
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
A predator moves in a circle of radius \( \sqrt{2} \) centre \((0,0)\), while a prey moves along the line \(y = x\). If \(0 \le x \le 2\), at which point(s) will they meet?
Akọwa Nkọwa
The predator moves in a circular path with a radius of √2 centered at the origin (0,0), while the prey moves along the line y = x. The prey's movement is restricted between x = 0 and x = 2. At some point, the predator will catch up to the prey, and they will meet. The predator will catch the prey when the distance between them is less than or equal to √2. We can find the equation of the circle centered at (0,0) with a radius of √2: x^2 + y^2 = (√2)^2 x^2 + y^2 = 2 We can substitute y = x into the equation to get: x^2 + x^2 = 2 2x^2 = 2 x^2 = 1 x = ±1 Therefore, the possible points of intersection are (1,1) and (-1,-1), but we know that the prey can only be at x values between 0 and 2, so the only possible point of intersection is (1,1). Therefore, the answer is (1,1) only.
Ajụjụ 42 Ripọtì
If y = 2x - sin2x, find dy/dx when x = π/4
Akọwa Nkọwa
y = 2x cos2x - sin2x
dy/dx = 2 cos2x +(-2x sin2x) - 2 cos2x
= 2 cos2x - 2x sin2x - cos2x
= -2x sin2x
= -2 x (π/4) sin2 x (π/4)
= -(π/2) x 1 = -(π/2)
Ajụjụ 43 Ripọtì
If the volume of a hemisphere is increasing at a steady rate of 18π m/s, at what rate is its radius changing when its is 6m?
Akọwa Nkọwa
We can solve this problem using the formula for the volume of a hemisphere, which is given by V = (2/3)πr^3. To find the rate of change of the radius, we need to differentiate this equation with respect to time t. Thus, we have: dV/dt = (2/3)π(3r^2)(dr/dt) where dV/dt is the rate of change of volume (18π m/s), and r = 6 m is the radius of the hemisphere. Solving for dr/dt, we get: dr/dt = (3dV/dt)/(4πr^2) = (3(18π))/(4π(6)^2) = 0.25 m/s Therefore, the rate of change of the radius when the volume of the hemisphere is increasing at a steady rate of 18π m/s and its radius is 6 m is 0.25 m/s. Thus, the correct option is: 0.25 m/s.
Ajụjụ 44 Ripọtì
Given that the various faces of a fair dice 1, 2, 3, 4, 5, 6 appeared 30, 43, 54, 40, 41, 32 times respectively in a single toss. Picture the figures as being represented in a simple table with number (X) against frequency (f).
If a pie chart is used to depict the data, the angle corresponding to 4 is?
Akọwa Nkọwa
To find the angle corresponding to 4 in the pie chart, we need to calculate the percentage of times 4 appeared in the toss. Total frequency = 30 + 43 + 54 + 40 + 41 + 32 = 240 Frequency of 4 = 40 Percentage of times 4 appeared = (Frequency of 4 / Total frequency) x 100% = (40/240) x 100% = 16.67% The pie chart represents a circle of 360°. Therefore, to find the angle corresponding to 4 in the pie chart, we need to calculate 16.67% of 360°. Angle corresponding to 4 = (16.67/100) x 360° = 60° Therefore, the angle corresponding to 4 in the pie chart is 60°. Option (d) is the correct answer.
Ajụjụ 45 Ripọtì
3y = 4x - 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K.
Akọwa Nkọwa
Grad of 3y = 4x - 1
y = 4x/3 - 1/3
Grad = 4/3
Grad of Ky = x + 3
y = x/k + 3/4
Grad = 1/k
Since two lines are perpendicular,
1/k = -3/4
-3k = 4
k = -4/3
Ajụjụ 46 Ripọtì
if P and Q are fixed points and X is a point which moves so that XP = XQ, the locus of X is
Akọwa Nkọwa
Given two fixed points P and Q, the locus of points X which are equidistant from P and Q is the perpendicular bisector of PQ. This is because any point on the perpendicular bisector of PQ will be equidistant from both P and Q by definition. Therefore, if X is a point which moves so that XP = XQ, it must lie on the perpendicular bisector of PQ. Therefore, the correct answer is "the perpendicular bisector of PQ".
Ajụjụ 47 Ripọtì
A function f(x) passes through the origin and its first derivative is 3x + 2. What is f(x)?
Akọwa Nkọwa
Hints:
1. Integrate the given first derivative of f(x) at the boundaries, (0,0)
Then solve accordingly to get f(x) = y = (3x2/)2 + 2x
