JAMB UTME - Mathematics - 1998

Akọwa Nkọwa

$\begin{array}{ccc}\text{Class intervals}& F& \text{Class boundary}\\ 5-7& 17& 4.5-9.5\\ 10-14& 32& 9.5-14.5\\ 15-19& 25& 14.5-19.5\\ 20-24& 24& 19.5-24.5\end{array}$
mode = 9 + $\frac{D_1}{D_2+D_1}$ x C
= 9.5 + $\frac{5(32-17)}{2\left(32\right)-17-25}$
= 9.5 + $\frac{75}{27}$
= 12.27
$\approx$ 2.3

Akọwa Nkọwa

Number of red balls = 16,
Number of blue balls = 20
Let x represent the No of white balls to be added
∴ Total number of balls = 36 + x
2(36 + x) = 80
= 2x + 80 - 72
= 8
x = $\frac{8}{2}$
= 4

Akọwa Nkọwa

To solve this equation, we can start by simplifying the expression inside the square root symbol by taking the common denominator. √x/(x-2) - √(x-2)/(x-2) - 1 = 0 We can simplify this further by combining the two terms inside the square root, which have a common denominator. [√x - √(x-2)]/(x-2) - 1 = 0 Now we can take the common denominator of the two terms inside the parenthesis and simplify. [√x - √(x-2) - (x-2)]/(x-2) = 0 Simplifying the numerator further, [√x - √(x-2) - x + 2]/(x-2) = 0 [√x - x + 2 - √(x-2)]/(x-2) = 0 [√x - x + 2] = √(x-2) Squaring both sides of the equation, (√x - x + 2)² = x - 2 Expanding and simplifying, x² - 2x(√x + 1) + 3 = 0 We can now use the quadratic formula to solve for x: x = [2(√x + 1) ± √(4x - 8)]/2 x = (√x + 1) ± √(x - 2) However, we need to make sure that the solution we get satisfies the original equation. We can check by substituting the value of x back into the original equation. After testing each option, we find that the only solution that satisfies the original equation is x = 9/4.

Akọwa Nkọwa

10112 + x7 = 2510 = 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 2o
= 8 + 0 + 2 + 1
= 1110
x7 = 2510 - 1110
= 1410
$\begin{array}{cc}7& 14\\ 7& 2R0\\ & 0R2\end{array}$
X = 207

Akọwa Nkọwa

$\frac{150}{Z}$ = tan 60o,
Z = $\frac{150}{tan{60}^o}$
= $\frac{150}{3}$
= 50$\sqrt{3}$cm
$\frac{150}{XxZ}$ = tan45o = 1
X + Z = 150
X = 150 - Z
= 150 - 50$\sqrt{3}$
= 50( $\sqrt{3}$ - $\sqrt{3}$)m

Akọwa Nkọwa

The volume of the cylindrical jug is given by V = πr^2h, where r = 5cm (radius = diameter/2) and h = 18cm. Thus, V = π(5cm)^2(18cm) = 450π cm^3. The volume of each cylindrical tin is given by V = πr^2h, where r = 3cm (radius = diameter/2) and h = 10cm. Thus, V = π(3cm)^2(10cm) = 90π cm^3. Since the jug contains oil that can fill 450π/90π = 5 cylindrical tins, the market woman sells 5 tins of oil. Therefore, the amount she makes by selling all the oil is 5 × ₦15.00 = ₦75.00. Since she bought the jug for ₦50.00, her profit is ₦75.00 - ₦50.00 = ₦25.00. Therefore, the market woman made ₦25.00 by selling all the oil. Hence, the answer is option D: ₦25.00.

Akọwa Nkọwa

m = $\frac{dy}{dv}$ = 6x - 5
∫dy = ∫(6x - 5)dx
y = 3x2 - 5x + C
when x = 2, y = 5
∴ 5 = 3(2)2 - 5(2) +C
C = 3
∴ y = 3x2 - 5x + 3

Akọwa Nkọwa

$\frac{1}{3x}$ + $\frac{1}{2}$ > $\frac{1}{4x}$
= $\frac{2+3x}{6x}$ > $\frac{1}{4x}$
= 4(2 + 3x) > 6x = 12x2 - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < $\frac{1}{6}$ = x < $\frac{1}{6}$ (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > $\frac{1}{6}$
Combining solutions in cases(1) and (2)
= x > 0, x < $\frac{1}{6}$ = 0 < x < $\frac{1}{6}$

Akọwa Nkọwa

To factorize the expression r² - r(2p + q) + 2pq, we need to look for two numbers whose product is 2pq and whose sum is -r(2p + q). Let's try to break up -r(2p + q) into two parts such that their product is 2pq. We can write -r(2p + q) as -2rpq - rq². Then, we can rewrite the expression as: r² - 2rpq - rq² + 2pq Now, we can group the first two terms and the last two terms together and factor them separately: r(r - 2pq) - q(r - 2pq) We can see that r - 2pq is a common factor, so we can factor it out: (r - 2pq)(r - q) Therefore, the factorization of r² - r(2p + q) + 2pq is (r - 2pq)(r - q). So the correct option is (c) (r - q)(r - 2p).

Akọwa Nkọwa

[$\frac{1}{0.03}$ + $\frac{1}{0.024}$]
= [$\frac{1}{0.03\times 0.024}$]-1
= [$\frac{0.024}{0.003}$]-1
= $\frac{0.03}{0.024}$
= $\frac{30}{24}$ = 1.25

Akọwa Nkọwa

To find the variance of the numbers k, k+1, k+2, we can use the formula for variance which is the average of the squared differences from the mean. First, we need to find the mean of the three numbers. Mean = (k + k + 1 + k + 2) / 3 = (3k + 3) / 3 = k + 1 So, the mean is k + 1. Next, we find the squared differences from the mean for each number: For k, the difference from the mean is k - (k+1) = -1. The squared difference is (-1)^2 = 1. For k+1, the difference from the mean is (k+1) - (k+1) = 0. The squared difference is 0^2 = 0. For k+2, the difference from the mean is (k+2) - (k+1) = 1. The squared difference is 1^2 = 1. Now we can find the variance: Variance = [(1^2 + 0^2 + 1^2) / 3] = 2/3 = 0.67 (rounded to two decimal places) Therefore, the answer is option (A) 2/3 or as a percentage approximately 66.7%.

Akọwa Nkọwa

PQ = $\left(\begin{array}{cc}2& 3\\ 4& 5\end{array}\right)$$\left(\begin{array}{cc}u& 4+u\\ -2v& v\end{array}\right)$
= $\left(\begin{array}{cc}(2u-6v& 2(4+u)+3v)\\ 4u-10v& 4(4+u)+5v\end{array}\right)$
= $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
2u - 6v = 1.....(i)
4u - 10v = 0.......(ii)
2(4 + u) + 3v = 0......(iii)
4(4 + u) + 5v = 1......(iv)
2u - 6v = 1 .....(i) x 2
4u - 10v = 0......(ii) x 1
$\frac{\text{4u - 12v = 0}}{\text{-4u - 10v = 0}}$
-2v = 2 = v = -1
2u - 6(-1) = 1 = 2u = 5
u = -$\frac{5}{2}$
∴ (U, V) = (-$\frac{5}{2}$ - 1)

Akọwa Nkọwa

let y = $\frac{x}{cosx}$ = x sec x
y = u(x) v (x0
$\frac{dy}{dx}$ = U$\frac{dy}{dx}$ + V$\frac{du}{dx}$
dy x [secx tanx] + secx
x = x secx tanx + secx

Akọwa Nkọwa

$\left|\begin{array}{ccccc}& 1& 2& 3& 4\\ 1& 1& 2& 3& 4\\ 2& 2& 4& 6& 8\\ 3& 3& 6& 9& 12\\ 4& 4& 8& 12& 16\end{array}\right|$
P (product of x and y not greater than 6) = $\frac{10}{16}$
= $\frac{5}{8}$

Akọwa Nkọwa

$\frac{k}{\sqrt{3}+\sqrt{2}}$ = k$\sqrt{3-2}$
$\frac{k}{\sqrt{3}+\sqrt{2}}$ x $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
= k$\sqrt{3-2}$
= k($\sqrt{3}-\sqrt{2}$)
= k$\sqrt{3-2}$
= k$\sqrt{3}$ - k$\sqrt{2}$
= k$\sqrt{3-2}$
k2 = $\sqrt{2}$
k = $\frac{2}{\sqrt{2}}$
= $\sqrt{2}$

Akọwa Nkọwa

mean (x) = $\frac{1+x+1+2x+1}{3}$
= $\frac{3x+3}{3}$
= 1 + x
$\begin{array}{ccc}X& (X-X)& (X-X{)}^2\\ 1& -x& x^2\\ x+1& 0& 0\\ 2x+1& x& x^2\\ & & 2x^2\end{array}$
S.D = $\sqrt{\frac{\sum (x-7{)}^2}{\sum f}}$
= ${\sqrt{\left(6\right)}}^2$
= $\frac{2x^2}{3}$
= 2x2
= 18
x2 = 9
∴ x = $\pm$ $\sqrt{9}$
= $\pm$3

Akọwa Nkọwa

Akọwa Nkọwa

c$\frac{1}{3}$ = a$\frac{1}{2}$b
= a$\frac{1}{2}$b x a-2
= a-$\frac{3}{2}$
= (c$\frac{1}{3}$)3
= (a-$\frac{3}{2}$)$\frac{1}{3}$
c = a-$\frac{1}{2}$

Akọwa Nkọwa

pm2 + qm + 1 = (m - 1) Q(x) + 2
p(1)2 + q(1) + 1 = 2
p + q + 1 = 2
p + q = 1.....(i)
pm2 + qm + 1 = (m - 1)Q(x) + 4
p(-1)2 + q(-1) + 1 = 4
p - q + 1 = 4
p - q = 3....(ii)
p + q = 1, p - q = -3
2p = -2, p = -1
-1 + q = 1
q = 2

Akọwa Nkọwa

Akọwa Nkọwa

$\left|\begin{array}{ccc}x& 1& 0\\ 1-x& 2& 3\\ 1& 1+x& 4\end{array}\right|$ = x$\left|\begin{array}{cc}2& 3\\ 1+x& 4\end{array}\right|$ - $\left|\begin{array}{cc}1-x& 3\\ 1& 4\end{array}\right|$ = 0
= x[8 - 3(1 + x)] - [4(1 - x)-3] - 0 = x[5 - 3x] - [1 - 4x]
= 5x - 3x2 -1 + 4x
= -3x2 + 9X - 1

Akọwa Nkọwa

∫${}_2^{\pi }$(sec2 x - tan2x)dx
∫${}_2^{\pi }$ dx = [X]${}_2^{\pi }$
= $\pi$ - 2 + c
when c is an arbitrary constant of integration

Akọwa Nkọwa

A$△$ = $\frac{1}{2}$ x 8 x h = 20

= $\frac{1}{2}$ x 8 x h = 4h

h = $\frac{20}{4}$

= 5cm

A$△$(PQTS) = L x H

A$△$PQRT = A$△$QSR + A$△$PQTS

20 + 50 = 70cm2

ALTERNATIVE METHOD

A$△$PQRT = $\frac{1}{2}$ x 5 x 28

= 70cm2

Akọwa Nkọwa

To find the mean (m), you need to add up all the numbers in the set and then divide by the total number of numbers. In this case, the sum is 47 and the total number of numbers is 8, so the mean is 47/8 = 5.875. To find the median (n), you need to arrange the numbers in order from smallest to largest and then find the middle number. In this case, the numbers in order are: 2, 3, 5, 6, 7, 7, 8, 9. The middle number is 7, so the median is 7. To find m + 2n, you just need to substitute the values of m and n into the expression and solve. m + 2n = 5.875 + 2(7) = 19.875. To the nearest whole number, m + 2n is 20. Therefore, the answer is: 19.

Akọwa Nkọwa

2p - 10 = $\frac{p+1+1-4P^2}{2}$ (Arithmetic mean)
= 2(2p - 100 = p + 2 - 4P2)
= 4p - 20 = p + 2 - 4p2
= 4p2 + 3p - 22 = 0
= (p - 2)(4p + 11) = 0
∴ p = 2 or -$\frac{4}{11}$

Akọwa Nkọwa

x $\ast$ y = xy - y - x, x $\ast$ 3 = 3x - 3 - x = 2x - 3
2 $\ast$ x = 2x - x - 2 = x - 2
∴ 2x - 3 = x - 2
x = -2 + 3
= 1

Akọwa Nkọwa

To solve the problem, we need to use the distance formula between two points in a coordinate plane. The distance formula is given by: d = sqrt[(x2 - x1)^2 + (y2 - y1)^2] where d is the distance between the two points (x1, y1) and (x2, y2). Using the given points, we have: (x2, y2) = (-x, 2) (x1, y1) = (x, 3) Substituting these values into the distance formula, we get: d = sqrt[(-x - x)^2 + (2 - 3)^2] Simplifying the expression inside the square root: d = sqrt[4x^2 + 1] We are given that d = 5, so we can substitute that into the equation and solve for x: 5 = sqrt[4x^2 + 1] 25 = 4x^2 + 1 24 = 4x^2 6 = x^2 x = sqrt(6) or x = -sqrt(6) Since we are only interested in the positive value of x, the answer is x = sqrt(6). Therefore, the correct option is:

Akọwa Nkọwa

Since PQ and RS intersect inside the circle at right angles, then the line joining the point of intersection to the center of the circle will bisect both chords. Let O be the center of the circle, and let T be the point of intersection of the two chords. Then, angle QTR = 90 degrees and the angle subtended by chord PQ at the center O is twice angle QPR. Therefore, angle POQ = 2 * angle QPR = 70 degrees (since angle QPR = 35 degrees). Similarly, angle ROS = 70 degrees. Since PQ and RS are chords of a circle, then angle POQ = angle PTS and angle ROS = angle TQS. Thus, angle PTS + angle TQS = 140 degrees. Also, angle PTS + angle PTQ + angle QTS = 180 degrees (because they form a straight line). Therefore, angle TQS = 180 - 140 - 90 = 50 degrees. Since angle PQT = angle RQT (because they are opposite angles), then angle PQS = angle RQS = (180 - angle QTS)/2 = (180 - 50)/2 = 65 degrees. Therefore, the answer is 55 degrees.

Akọwa Nkọwa

Akọwa Nkọwa

$\frac{11+2}{6x^2-x-1}$ = $\frac{11+2}{3x+1}$
= $\frac{A}{3x+1}$ + $\frac{B}{2x-1}$
11x = 2 = A(2x - 1) + B(3x + 1)
put x = $\frac{1}{2}$
= -$\frac{-5}{3}$
= -$\frac{-5}{3}$A $\to$ A = 1
∴ $\frac{11x+2}{6x^2-x-1}$ = $\frac{1}{3x+1}$ + $\frac{3}{2x-1}$

Akọwa Nkọwa

The cost of normal work = 35r
The cost of overtime = q x 2r = 2qr
The man's total weekly earning = 35r + 2qr
= r(35 + 2q)

Akọwa Nkọwa

$\frac{x+a}{x?a}$ = m
x + a = mx - ma
a + ma = mx - x
a(m + 1) = x(m - 1)
$\frac{a}{x}$ = $\frac{m?1}{m+a}$

Akọwa Nkọwa

$\begin{array}{cccccccccccc}x& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ f& 1& 9& 4& 7& 10& 8& 7& 9& 8& 2& 1\end{array}$
no pupils who failed the test = 1 + 3 + 4 + 7 + 10

= 25

5 of pupils who fail = $\frac{25}{60}$ x 100%

= 41.70%

Akọwa Nkọwa

Let the G.P be a, ar, ar2, S3 = $\frac{1}{2}$S
a + ar + ar2 = $\frac{1}{2}$($\frac{a}{1-r}$)
2(1 + r + r)(r - 1) = 1
= 2r3 = 3
= r3 = $\frac{3}{2}$
r($\frac{3}{2}$)$\frac{1}{3}$ = $\sqrt{\frac{3}{2}}$

Akọwa Nkọwa

Akọwa Nkọwa

The maximum value of the function y = 6sin(2x-25) occurs when the argument of the sine function is equal to 90 degrees (or pi/2 radians), because the maximum value of the sine function is 1. So we need to solve the equation 2x-25 = 90. Adding 25 to both sides gives 2x = 115, and then dividing by 2 gives x = 57.5 degrees. Therefore, the answer is x = 57.5 degrees, which is the value of x where the function 6sin(2x-25) attains its maximum value in the given range.

Akọwa Nkọwa

To solve the problem, we need to use the formula for the volume of a cylinder, which is V = πr²h, where V is the volume, r is the radius, and h is the height of the cylinder. We are given the diameter of the drum, which is 56 cm. To find the radius, we need to divide the diameter by 2: radius (r) = diameter / 2 = 56 cm / 2 = 28 cm We are also given that the drum contains 123.2 litres of oil when full. To convert litres to cubic centimeters, we need to multiply by 1000: 123.2 litres * 1000 = 123200 cubic centimeters Now we can use the formula for the volume of a cylinder to find the height (h): V = πr²h h = V / (πr²) h = 123200 / (π28²) h ≈ 123200 / 2463.47 h ≈ 50.00 cm Therefore, the height of the drum is approximately 50.00 cm.

Akọwa Nkọwa

Akọwa Nkọwa

30 + x = 100

x = 100 - 30

= 70o

Akọwa Nkọwa

$\frac{10}{\sin {30}^o}=\frac{15}{\sin x}=\frac{10}{0.5}=\frac{15}{\sin x}$

$\frac{15}{20}=\sin x$

sin x = $\frac{15}{20}=\frac{3}{4}$

N.B x = < PRQ

Akọwa Nkọwa

m = $\frac{y_2-y_1}{x_2-x_2}=\frac{3-0}{0-2}=\frac{-3}{2}$

= $\frac{y-y_1}{x-x_1}$

m = $\frac{y-3}{x}$ $\ge$ $\frac{-3}{2}$

2(y - 3) $\ge$ - 3x = 2y - 6 $\ge$ - 3x

= 2y + 3x $\le$ 6 ; x $\le$ 0, y $\le$ 0

Akọwa Nkọwa

To find the midpoint of a line segment, we need to find the average of the endpoints. The x-intercept of the line y = 4x + 3 is found by setting y = 0 and solving for x: 0 = 4x + 3 x = -3/4 So the x-coordinate of the midpoint is the average of -3/4 and 0: x = (-3/4 + 0)/2 = -3/8 To find the y-coordinate of the midpoint, we plug in x = -3/8 to the equation of the line: y = 4(-3/8) + 3 = -3/2 + 3 = 3/2 So the midpoint is (-3/8, 3/2). Therefore, the answer is (-3/8, 3/2).

Akọwa Nkọwa

< TWR = < QTR = 40${}^{\circ }$ (alternate segment)

< TWR = < TXR = 40${}^{\circ }$(Angles in the same segments)

< YXR = 40${}^{\circ }$ + 32${}^{\circ }$ = 72${}^{\circ }$

< YXR + < YTR = 180${}^{\circ }$(Supplementary)

72${}^{\circ }$ + < YTR = 180${}^{\circ }$

< YTR = 180${}^{\circ }$ - 72${}^{\circ }$

= 108${}^{\circ }$

Akọwa Nkọwa

To find dy/dx, we need to differentiate y with respect to x. We can start by using the chain rule, which states that if we have a function of the form f(g(x)), the derivative of that function with respect to x is f'(g(x)) * g'(x). In this case, we have y = 243(4x + 5)-2, which can be written as y = 243/(4x + 5)^2. Using the chain rule, we have: dy/dx = d/dx(243/(4x + 5)^2) = -2 * 243 / (4x + 5)^3 * d/dx(4x + 5) = -2 * 243 / (4x + 5)^3 * 4 = -1944 / (4x + 5)^3 Now, to find the value of dy/dx when x = 1, we just need to substitute x = 1 into the expression we found above: dy/dx = -1944 / (4(1) + 5)^3 = -1944 / 729 = -8/3 Therefore, the answer is option A: -83. To summarize, we used the chain rule to differentiate y with respect to x, which gave us an expression for dy/dx in terms of x. We then substituted x = 1 to find the value of dy/dx at that point.

Akọwa Nkọwa

Based on the information given in the pie chart, the monthly expenditure on school fees can be represented by the fraction 1/10 (since it is one-tenth of the total expenditure) and the monthly expenditure on housing can be represented by the fraction 2/10 (since it is twice the amount of school fees). To find out how much the worker spends on housing, we need to first calculate the total amount of money he spends on all his monthly expenses. From the pie chart, we see that the total monthly expenditure is represented by the fraction 7/10. If we let x represent the amount of money the worker spends on housing each month, we can set up the following equation to solve for x: x + (1/10)*7200 = (2/10)*7200 Simplifying the equation, we get: x + 720 = 1440 Subtracting 720 from both sides, we get: x = 720 Therefore, the worker spends ₦7200/10 = ₦720 on school fees each month, and ₦720 * 2 = ₦1440 on housing each month. So the answer is ₦2000.

Akọwa Nkọwa

log4(y - 1) + log4($\frac{1}{2}$x) = 1
log4(y - 1)($\frac{1}{2}$x) $\to$ (y - 1)($\frac{1}{2}$x) = 4 ........(1)
log2(y + 1) + log2x = 2
log2(y + 1)x = 2 $\to$ (y + 1)x = 22 = 4.....(ii)
From equation (ii) x = $\frac{4}{y+1}$........(iii)
put equation (iii) in (i) = y (y - 1)[$\frac{1}{2}(\frac{4}{y-1}$)] = 4
= 2y - 2
= 4y + 4
2y = -6
y = -3
x = $\frac{4}{-3+1}$
= $\frac{4}{-2}$
X = 2
therefore x = -2, y = -3

Akọwa Nkọwa

p(x) = $\frac{2}{5}$ p(y) = $\frac{1}{3}$
p(x or y) = p(x ∪ y)
= p(x) + p(y)
= $\frac{2}{5}$ + $\frac{1}{3}$
= $\frac{11}{5}$

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