JAMB UTME - Mathematics - 2005

Akọwa Nkọwa

To find the equation of the perpendicular at the point (4,3) to the line y + 2x = 5, we need to follow these steps: 1. Determine the slope of the given line: To do this, we can rearrange the equation y + 2x = 5 into slope-intercept form, y = -2x + 5. Thus, the slope of this line is -2. 2. Determine the slope of the perpendicular line: The slopes of perpendicular lines are negative reciprocals of each other. Therefore, the slope of the perpendicular line is 1/2. 3. Use the point-slope form of a line to write the equation of the perpendicular line: The point-slope form of a line is y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line. Since we know the slope of the perpendicular line is 1/2 and it passes through the point (4,3), we can substitute these values into the point-slope form to get y - 3 = 1/2(x - 4). 4. Simplify the equation: We can simplify the equation by multiplying both sides by 2 to eliminate the fraction, giving us 2y - 6 = x - 4. Then, we can rearrange it into standard form, which is Ax + By = C, by adding 4 to both sides and switching the positions of x and y to get 2y + x = 10. Therefore, the equation of the perpendicular at the point (4,3) to the line y + 2x = 5 is 2y + x = 10, which is option (D).

Akọwa Nkọwa

We are given that the radius of a circular disc is increasing at a rate of 0.5cm/sec. We need to find the rate at which the area of the disc is increasing when the radius is 6cm. The formula for the area of a circle is A = πr^2, where A is the area and r is the radius. We can use the chain rule of differentiation to find the rate of change of the area with respect to time. dA/dt = dA/dr * dr/dt We know that dr/dt = 0.5 cm/sec (given) and we need to find dA/dt when r = 6 cm. dA/dr = 2πr (differentiating A = πr^2 with respect to r) So, dA/dt = (2πr) * (0.5) = πr * 1 = 6π cm^2/sec (substituting r = 6cm) Therefore, the rate at which the area of the disc is increasing when its radius is 6cm is 6π cm^2/sec. is the correct answer.

Akọwa Nkọwa

There are a total of 10 balls in the box, and we are selecting 2 balls at random with replacement, which means that after each ball is selected, it is returned to the box before the next ball is selected. The probability of selecting a blue ball on the first draw is 5/10 = 1/2. Since we are replacing the ball, the probability of selecting another blue ball on the second draw is also 1/2. Therefore, the probability of selecting two blue balls is (1/2) x (1/2) = 1/4. Similarly, the probability of selecting a red ball on the first draw is 3/10. Since we are replacing the ball, the probability of selecting another red ball on the second draw is also 3/10. Therefore, the probability of selecting two red balls is (3/10) x (3/10) = 9/100. Finally, to obtain the probability of obtaining two blue or two red balls, we add the probabilities of these two mutually exclusive events. Therefore, the probability of obtaining two blue or two red balls is 1/4 + 9/100 = 17/50. So the correct option is  17/50.

Akọwa Nkọwa

Akọwa Nkọwa

Let x = the number of oranges
The 1st received 1/3 of x = 1/3x
∴Remainder = x - 1/3x = 2x/3
The 2nd received 2/3 of 2x/3 = 2/3 * 2x/3 = 4x/3
The 3rd received 12 oranges
∴1/3x + 4x/9 + 12 = x
(3x + 4x + 108)/9 = x
3x + 4x + 108 = 9x
7x + 108 = 9x
9x - 7x = 108
2x = 108
x = 54 oranges

Akọwa Nkọwa

y = sin (2x3 + 3x - 4)
let u = 2x3 + 3x - 4
∴du/dx = 6x2
y = sin U
dy/du = cos U
dy/dx = du/dx * dy/dy
∴dy/dx = (6x2 + 3) cos U
= (6x2 + 3)cos(2x3 + 3x - 4)

Akọwa Nkọwa

Explanation:

Akọwa Nkọwa

Solution

Akọwa Nkọwa

31/2 - (21/3 * 11/4) + 3/5
= 7/2 - (7/3 * 5/4) + 3/5
= 7/2 - 35/12 + 3/5
= L.C.M = 60
= (210 - 175 + 36)/60
= 71/60
= 111/60

Akọwa Nkọwa

Akọwa Nkọwa

QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o

Akọwa Nkọwa

To find the value of m+2r, we first need to find the value of m and r. m is the model height, which is simply the average of the given heights: m = (1.35 + 1.25 + 1.35 + 1.40 + 1.35 + 1.50 + 1.35 + 1.50 + 1.20) / 9 m = 1.35 r is the range of heights, which is the difference between the highest and lowest height: r = 1.50 - 1.20 r = 0.30 Now we can calculate m+2r: m+2r = 1.35 + 2(0.30) m+2r = 1.95 Therefore, the value of m+2r is 1.95.

Akọwa Nkọwa

Angle of Excellent
= 360 - (120+80+90)
= 360 - 290
= 70${}^{\circ }$
If 360${}^{\circ }$ represents 36 students
1${}^{\circ }$ will represent 36/360
50${}^{\circ }$ will represent 36/360 * 70/1
= 7

Akọwa Nkọwa

Log102 = 0.3010 and Log103 = 0.4771
Log104.5 = Log1041/2
= Log109/2
= Log109 - Log102
= log1032 - Log102
= 2Log103 - Log102
= 2(0.4771) - 0.3010
= 0.9542 - 0.3010
= 0.6532

Akọwa Nkọwa

We are told that the weight W of a metal bar varies jointly as its length L and the square of its diameter d, which can be expressed as:

W = kLd2

where k is a constant of variation. We need to find an expression for d in terms of W and L, given that W = 140 when d = 4/3 and L = 54.

To find the value of k, we can substitute the given values into the equation and solve for k as follows:

140 = k × 54 × (4/3)^2

140 = k × 54 × 16/9

k = 140 × 9 / (54 × 16)

k = 35 / 64

Now we can substitute the value of k into the original equation and solve for d in terms of W and L:

W = kLd2

d2 = W / (kL)

d = √(W / (kL))

Substituting the value of k, we get:

d = √((64/35) × W / L)

Simplifying, we get:

d = √((64W) / (35L))

Therefore, the expression for d in terms of W and L is:

d = √((64W) / (35L))

So the answer is option (A) √42W/5L.

Akọwa Nkọwa

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{{280}^o}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

Akọwa Nkọwa

In a regular polygon, all interior angles are congruent (i.e., have the same measure). Let's call the measure of one interior angle of the polygon "x". Since the polygon is regular, we know that it has n sides, where n is a positive integer. We can use the formula for the sum of interior angles of a polygon, which is: Sum of interior angles = (n-2) * 180 degrees Since each interior angle in this polygon measures 140 degrees, we can set up an equation: x = 140 We can then solve for the number of sides by using the fact that the sum of the interior angles is also equal to: (n) * (x) = n * 140 And since the sum of the interior angles is also equal to: (n-2) * 180 We can set these two expressions equal to each other and solve for n: n * 140 = (n-2) * 180 140n = 180n - 360 40n = 360 n = 9 Therefore, the polygon has 9 sides, and the sum of the interior angles is: (n-2) * 180 = (9-2) * 180 = 1260 degrees So the correct answer is 1260 degrees.

Akọwa Nkọwa

m = 3, p = -3, q = 7 and r = 5/2
m(p+q+r) = 3(-3 + 7 + 5/2)
= 3(4 + 5/2)
= 3(4 + 21/2)
= 3 * 61/2
= 3 * 13/2
= 39/2
= 19.50

Akọwa Nkọwa

The curved surface area of a cone is given by the formula: πrℓ where r is the radius of the circular base of the cone and ℓ is the slant height. In this case, the diameter of the circular base is 10 cm, so the radius is 5 cm. The height of the cone is 12 cm, and to find the slant height, we can use the Pythagorean theorem: ℓ² = r² + h² Substituting the values, we get: ℓ² = 5² + 12² ℓ² = 169 ℓ = 13 cm Now, we can use the formula for the curved surface area of the cone: Curved surface area = πrℓ Curved surface area = π × 5 × 13 Curved surface area = 65π cm² Therefore, the curved surface area of the given cone with circular base diameter 10 cm and height 12 cm is 65π cm². is the correct answer.

Akọwa Nkọwa

To find the number of possible ways to seat seven people P,Q,R,S,T,U, and V at a circular table, we can use the formula (n-1)!, where n is the number of people. Since the people are seated at a circular table, we need to use circular permutations. A circular permutation is where the order of the arrangement matters, but rotations of the same arrangement are considered the same. In other words, if we rotate a circular arrangement, it is still considered the same arrangement. Using the formula for circular permutations, the number of ways to seat seven people at a circular table is (7-1)! = 6!, which is equal to 720. Therefore, there are 720 possible ways to seat seven people P,Q,R,S,T,U, and V at a circular table. "720," is the correct answer.

Akọwa Nkọwa

Akọwa Nkọwa

To find a polynomial with the given zeros, we can use the fact that the product of the factors (x+2), (x+1), and (x-3) will give us the desired polynomial. Multiplying out these factors, we get: (x+2)(x+1)(x-3) = (x^2 + 3x + 2)(x-3) = x^3 - 9x^2 - x + 18 Therefore, the polynomial that has zeros of -2, -1, and 3 is x^3 - 7x - 6.

Akọwa Nkọwa

f(x) = 2 + x - x2dy/dx = 1-2x
As dy/dx = 0
1-2x = 0
2x = 1
x = 1/2
At x = 1/2
f(x) = 2 + x - x2
= 2 + 1/2 -(1/2)2
= 2 + 1/2 - 1/4
= (8+2-1) / 4 = 9/4

Akọwa Nkọwa

To find the area bounded by the given curves, we need to find the points of intersection of the two curves. Setting the equations of the curves equal to each other gives: x^2 - x + 3 = 3 Simplifying, we get: x^2 - x = 0 x(x-1) = 0 So, x = 0 or x = 1. The curves intersect at the points (0, 3) and (1, 3). The area between the curves is given by: ∫(y = x^2-x+3)dy from y=3 to y=9 = [y^2/2 - xy + 3y] from y=3 to y=9 = [(81-9)/2 - 9 + 27] - [(9-9)/2 - 0 + 9] = 17/6 units (sq) Therefore, the answer is (a) 17/6 units (sq).

Akọwa Nkọwa

To solve this problem, we can use the relationship between the length of a chord and the angle it subtends in a circle. Specifically, if a chord of length $c$ subtends an angle of $\theta$ degrees at the center of a circle of radius $r$, then: $$c = 2r\sin\left(\frac{\theta}{2}\right)$$ In this problem, we are given that the angle subtended by the chord is $60^\circ$ and the radius of the circle is $14$ cm. Thus, we can plug in these values to get: $$c = 2\cdot 14\cdot \sin\left(\frac{60}{2}\right) = 2\cdot 14\cdot \sin(30)$$ Recall that $\sin(30) = \frac{1}{2}$, so we have: $$c = 2\cdot 14\cdot \frac{1}{2} = 14$$ Therefore, the length of the chord is $\boxed{14}$ cm. To summarize, the length of a chord in a circle can be found using the formula $c = 2r\sin\left(\frac{\theta}{2}\right)$, where $r$ is the radius of the circle and $\theta$ is the angle subtended by the chord at the center of the circle. Applying this formula to the given problem, we find that the length of the chord is 14 cm.

Akọwa Nkọwa

To find the sum of the first 20 terms of the series 8, 12, 16, ....., 96, we need to find the common difference between the terms first. We can do this by subtracting any two consecutive terms, such as: 12 - 8 = 4 16 - 12 = 4 ... 96 - 92 = 4 So, the common difference between the terms is 4. Next, we can use the formula for the sum of an arithmetic series to find the sum of the first 20 terms: S = (n/2)(a1 + an) Where S is the sum of the series, n is the number of terms, a1 is the first term, and an is the nth term. Using this formula, we have: S = (20/2)(8 + 96) S = 10(104) S = 1040 Therefore, the sum of the first 20 terms of the series 8, 12, 16, ....., 96 is 1040.

Akọwa Nkọwa

${13}_m+{24}_m={41}_m$
$1\times m+3\times m^0+2\times m+4\times m^0=4\times m+1\times m^0$
$m+3+2m+4=4m+1$
$3m+7=4m+1\Rightarrow m=6$

Akọwa Nkọwa

Akọwa Nkọwa

The locus of points equidistant from two intersecting lines PQ and ST is the perpendicular bisector of the line segment joining PQ and ST. To understand why, consider a point P on line PQ and a point S on line ST. Let M be the midpoint of the line segment PS. Any point Q on line PQ that is equidistant from lines PQ and ST must be located on the perpendicular bisector of PS. Similarly, any point T on line ST that is equidistant from lines PQ and ST must also be located on the perpendicular bisector of PS. Therefore, the locus of points equidistant from lines PQ and ST lies on the perpendicular bisector of PS. Since the perpendicular bisector of PS is also the perpendicular bisector of line segment PQ and ST, option A (perpendicular bisector of PQ) is correct.

Akọwa Nkọwa

The bearing of R from S = 180 + 44 = 224${}^{\circ }$

Akọwa Nkọwa

t = time taken and N = number of men
t ∝ 1/N
t = K/N
K = Nt
K = 30 * 6
K = 180
∴t = 180/N
4 = 180/N
4N = 180
N = 180/4
45 men

Akọwa Nkọwa

πr2 = 360o

Akọwa Nkọwa

To find the identity elements of the operation *, we need to find values of x such that a * x = x * a = a for any value of a. Let's start by considering a * x: a * x = ax + 2(a + x + 1) Now let's set this equal to a and solve for x: ax + 2(a + x + 1) = a ax + 2x + 2a + 2 = a ax + 2x + a = -2 x(a + 2) = -2 - a x = (-2 - a)/(a + 2) So any value of x that satisfies this equation is an identity element of the operation *. Now let's consider x * a: x * a = xa + 2(x + a + 1) We can substitute the value of x that we just found into this equation to get: x * a = (-2 - a)a/(a + 2) + 2((-2 - a)/(a + 2) + a + 1) Simplifying this expression, we get: x * a = (-2a - a^2 - 4)/(a + 2) So any value of x that satisfies both equations (a * x = a and x * a = a) is an identity element of the operation *. To find the values of a that satisfy these equations, we can substitute x = (-2 - a)/(a + 2) into the equation a * x = a: a * ((-2 - a)/(a + 2)) = a -2 - a + 2 = a(a + 2) a^2 + 3a + 2 = 0 (a + 1)(a + 2) = 0 So the values of a that satisfy the equation are -1 and -2. Substituting these values into the equation for x that we found earlier, we get: x = (-2 - (-1))/(-1 + 2) = -1 x = (-2 - (-2))/(-2 + 2) is undefined Therefore, the identity element for this operation is -1.

Akọwa Nkọwa

To solve the inequality 7x - 3 > 25 + 3x, we need to isolate the variable 'x' on one side of the inequality. First, we can simplify by subtracting 3x from both sides, giving us: 4x - 3 > 25 Next, we can add 3 to both sides to get: 4x > 28 Finally, we can solve for 'x' by dividing both sides by 4: x > 7 Therefore, the range of values of 'x' that satisfy the inequality is x > 7, meaning any value of 'x' that is greater than 7 will make the inequality true.

Akọwa Nkọwa

If we have 6 different coloured chalks, we can arrange them in 6! = 720 ways. However, in this case, we have 2 chalks that are the same colour, which means we are double counting some arrangements. To see how many arrangements we are double counting, imagine we label the two same-coloured chalks as A and A'. We can arrange the chalks in 6! ways, but for each of these arrangements, we can swap A and A' and get the same arrangement. So each arrangement is counted twice, and we need to divide by 2 to get the correct number of arrangements. Therefore, the number of arrangements is 6!/2 = 360. So the answer is (D) 360.

Akọwa Nkọwa

Akọwa Nkọwa

To find the derivative of y with respect to x, we need to apply the chain rule and power rule of differentiation. First, we take the derivative of the outer function, which is squaring the quantity (1-2x). This gives us: 2(1-2x) Then, we multiply this result by the derivative of the inner function (1-2x). This gives us: 2(1-2x)(-2) Combining these results, we have: dy/dx = 2(1-2x)(-2) To find the value of dy/dx at x=-1, we substitute -1 for x in the expression above: dy/dx = 2(1-2(-1))(-2) = -54 Therefore, the answer is -54.

Akọwa Nkọwa

Two games played only = 5 + 7 + 3
= 15

Akọwa Nkọwa

We need to find the probability that an integer between 1 and 20 chosen at random is divisible by both 2 and 3. The integers that are divisible by 2 and 3 are: 6, 12, and 18. Therefore, there are three integers that satisfy this condition. Since there are 20 integers in total, the probability of selecting one of these three integers at random is 3/20. Therefore, the answer is option (C) 3/20.

Akọwa Nkọwa

To find the size of ∠EOG, we need to use the fact that the sum of the angles around point O is 360 degrees. So, we can start by finding the size of ∠FOG and ∠EOH, which are vertical angles to ∠KOF and ∠LOH respectively: ∠FOG = ∠KOF = 52∘ ∠EOH = ∠LOH = 85∘ Now, we can use the fact that opposite angles in a parallelogram are equal to find the size of ∠EOG. Since EF is parallel to HG, we have: ∠EOG = 180 - ∠FOG - ∠EOH Substituting the values we found earlier, we get: ∠EOG = 180 - 52 - 85 ∠EOG = 43 degrees Therefore, the size of ∠EOG is 43 degrees.

Akọwa Nkọwa

Akọwa Nkọwa

Let's denote the first term of the AP by 'a', and the common difference by 'd'. We're told that the sum of the first four terms of the AP is 42, so we can write an equation: a + (a+d) + (a+2d) + (a+3d) = 42 Simplifying this equation gives: 4a + 6d = 42 2a + 3d = 21 --- equation (1) We're also told that the 7th term of the AP is twice the third term, which we can write as: a + 6d = 2(a + 2d) Simplifying this equation gives: a = 4d --- equation (2) Now we can substitute equation (2) into equation (1) to get an equation in terms of 'd' only: 2(4d) + 3d = 21 Solving for 'd', we get: d = 3 Therefore, the common difference of the AP is 3.

Akọwa Nkọwa

Akọwa Nkọwa

${\int }_0^{\frac{\pi }{2}}$sin 2x dx = [-1/2cos 2x + C]${}_0^{\frac{\pi }{2}}$
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1

Akọwa Nkọwa

Akọwa Nkọwa

The sum of the interior angles of a polygon with n sides is given by the formula (n-2) x 180 degrees. Since the polygon in this question is regular, all its interior angles are equal, and therefore each interior angle of this polygon measures (n-2) x 180 degrees / n. To find the size of one exterior angle, we can use the fact that the sum of the exterior angles of any polygon is always 360 degrees. Therefore, each exterior angle of this polygon measures 360 degrees / n. Using the fact that the sum of the interior angles of the polygon is 1800 degrees, we can write: (n-2) x 180 degrees = 1800 degrees Solving for n, we get n=12. Therefore, the polygon has 12 sides. Using the formula for the exterior angle of a regular polygon, we can calculate the size of one exterior angle: 360 degrees / 12 = 30 degrees. Therefore, the size of one exterior angle of the polygon is 30 degrees.

Akọwa Nkọwa

p + 40o = 100o (exterior ∠ = sum of two interior opp ∠s)
p = 100o - 40o
P = 60o
But q + p = 180o (∠s on a straight line)
q + 60o
q = 180o - 60o
q = 120o
x = q (corresponding ∠)
∴x = 120o

Akọwa Nkọwa

Akọwa Nkọwa

$mean=\frac{60+5x}{18+x}3.5=\frac{60+5x}{18+x}\frac{7}{2}=\frac{60+5x}{18+x}$
7(18+x) = 2(60+5x)
126 + 7x = 120 + 10x
10x - 7x = 126 - 120
3x = 6
x = 2

Akọwa Nkọwa

P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

Q = {-1, 3, 5, 7, 11, 13, 17, 19, 23}

P ∩ $\cap$ Q = {3, 5, 7, 11, 13, 17, 19}