JAMB UTME - Mathematics - 1994

Akọwa Nkọwa

1x $\frac{1}{x}$ > 2 = xx2 $\frac{x}{x^2}$ > 2

x > 2x2

= 2x2 < x

= 2x2 - x < 0

= x(2x - 10 < 0

Case 1(+, -) = x > 0, 2x - 1 < 0

x > 0, x < 12 $\frac{1}{2}$ (solution)

Case 2(-, 4) = x < 0, 2x - 1 > 0

x < 0, x , 12 $\frac{1}{2}$ = 0

Akọwa Nkọwa

The sum of angles of a quadrilateral is 360o
∴ (3y - x - z)o + 3xo + (2z - 2y - x)o + po = 360o
Where P is the fourth angle
3y - x - z + 3x + 2z - 2y - x + p = 360o
p = 360 - (x + y + z)
∴ p = (360 - x - y - z)o

Akọwa Nkọwa

Since (x - 1), is a factor, when the polynomial is divided by (x - 1), the remainder = zero
∴ (x - 1) = 0
x = 1
Substitute in the polynomial the value x = 1
= p(1)3 + q(1)2 + 11(1) - 6 = 0
p + q + 5 = 0 .....(i)
Also since x - 3 is a factor, ∴ x - 3 = 0
x = 3
Substitute p(3)3 + q(3)2 + 11(3) - 6 = 0
27p + 9q = -27 ......(2)
Combine eqns. (i) and (ii)
Multiply equation (i) by 9 to eliminate q
9p + 9q = -45
Subt. $\frac{27p+9q=-27}{-18p=-18}$
∴ p = 1

Akọwa Nkọwa

$\frac{(n+1)(n+2)}{2}$
If n = 1, the expression becomes 3
n = 2, the expression becomes 6
n = 4, the expression becomes 15
n = 5, the expression becomes 21

Akọwa Nkọwa

Since SOQ is a diameter of the circle, angle SPQ = 90 degrees (angles subtended by a diameter are always 90 degrees). Therefore, angle PSQ is the difference between angle PRS and angle SPQ. Angle PSQ = angle PRS - angle SPQ = 38 - 90 = -52 degrees. However, since angles in a circle add up to 360 degrees, we can add 360 degrees to -52 to get the equivalent angle in the circle. Angle PSQ = -52 + 360 = 308 degrees. Therefore, the value of angle PSQ is 308 degrees. So the correct option is: 52∘.

Akọwa Nkọwa

Internal dimension are 50cm, 36cm and 20cm
internal volume = 50 x 36 x 20cm3
1000 x 36cm3
= 36000cm3
External dimension are 54cm x 40cm x 22cm
= 2160cm2 x 22cm = 47520cm3
Volume of wood = Ext. volume - Int. volume
= 47,520cm3 - 36,000cm3
= 11,520cm3

Akọwa Nkọwa

Akọwa Nkọwa

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30o

But SQR + 30o + 50o = 180o

SQR = 180 - 80

= 100o

Akọwa Nkọwa

8x-2 = $\frac{2}{25}$
= 200x-2 = 2
= 100x-2 = 1
x-2 = $\frac{1}{100}$
x-2 = 10-2
x = 10

Akọwa Nkọwa

$\frac{1}{3}÷\left[\frac{5}{7}\right(\frac{9}{10}?1+\frac{3}{4}\left)\right]$
$\frac{1}{3}÷\left[\frac{5}{7}\right(\frac{9}{10}?\frac{10}{10}+\frac{3}{4}\left)\right]$
= $\frac{1}{3}÷\left[\frac{5}{7}\right(\frac{?1}{10}+\frac{3}{4}\left)\right]$
= $\frac{1}{3}÷\left[\frac{5}{7}\right(\frac{?2+15}{20}\left)\right]$
= $\frac{1}{3}÷[\frac{5}{7}\times \frac{13}{20}]$
$\frac{1}{3}+\left[\frac{13}{28}\right]$ = $\frac{1}{3}\times \frac{28}{13}$
= $\frac{28}{39}$

Akọwa Nkọwa

y = x3 - 27, y = -27 $\to$ (0, -27)

when y = 0, x = 3 (3, 0)

Akọwa Nkọwa

We can use the identity a^2 - b^2 = (a+b)(a-b) to factorize the expression. Let's rewrite the expression using this identity: a^2x - b^2y - b^2x + a^2y = (a^2x + a^2y) - (b^2x + b^2y) Now we can see that we have a common factor of (a^2 - b^2) in both terms. (a^2 - b^2)(x + y) = (a + b)(a - b)(x + y) So the correct answer is: (a + b)(a - b)(x + y)

Akọwa Nkọwa

We know that the gradient of a curve is given by its derivative. Therefore, we need to find the derivative of the given curve and equate it to 2 to find the point where the gradient is 2. y = 2x^2 - 2x + 3 dy/dx = 4x - 2 Equating dy/dx to 2, we get: 4x - 2 = 2 4x = 4 x = 1 Substituting x = 1 in the original equation, we get: y = 2(1)^2 - 2(1) + 3 y = 3 Therefore, the point where the curve has a gradient of 2 is (1, 3). So, the correct option is: (1, 3).

Akọwa Nkọwa

40 = 20 - x + x + 35 - x
40 = 55 - x
x = 55 - 40
= 15
∴ P(both) $\frac{15}{40}$
= $\frac{3}{8}$

Akọwa Nkọwa

The inverse of 2 is 6 since 2 x 6 = 12; under mod 10
12 = 2 which is also the value required

Akọwa Nkọwa

$\frac{0.36\times 5.4\times 0.63}{4.2\times 9.0\times 2.4}$
= $\frac{36}{420}\times \frac{54}{90}\times \frac{63}{240}$
= $\frac{6}{70}\times \frac{18}{30}\times \frac{21}{80}$
= $\frac{27}{2000}$
= 0.0135
$\approx$ = 0.013

Akọwa Nkọwa

A $\subset$ B means A is contained in B i.e. A is a subset of B(A $\cap$ B)1 = A1
A(A $\cap$ B)1 = A $\cap$ A1
The intersection of complement of a set P and P1 has no element
i.e. n(A $\cap$ A1) = $\varphi$

Akọwa Nkọwa

$\frac{a}{a?x}$ = $\frac{b}{x?b}$
$\frac{1}{1?x}$ = $\frac{3}{x?3}$
? 3(1 - x) = x - 3
3 - 3x = x - 3
Rearrange 6 = 4x; x = $\frac{6}{4}$
= $\frac{3}{2}$

Akọwa Nkọwa

y = 3t3 + 2t2 - 7t + 3
$\frac{dy}{dt}$ = 9t2 + 4t - 7
When t = -1
$\frac{dy}{dt}$ = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2

Akọwa Nkọwa

$\sqrt{48}$ - $\frac{9}{\sqrt{3}}$ + $\sqrt{75}$
Rearrange = $\sqrt{48}$ + $\sqrt{75}$ - $\frac{9}{\sqrt{3}}$
= (√16 x √3) + (√25 x √3) - $\frac{9}{\sqrt{3}}$
=4√3 + 5√3 - $\frac{9}{\sqrt{3}}$
Rationalize $\to$ 9√3 = $\frac{9}{\sqrt{3}}$ x $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{9\sqrt{3}}{\sqrt{9}}$ - $\frac{9\sqrt{3}}{\sqrt{3}}$
= 3√3

Akọwa Nkọwa

To find the common difference of the arithmetic progression, we can use the formula: nth term = a + (n - 1)d where a is the first term, d is the common difference, and n is the term number. In this case, we know that the first term is 1, and the sixth term is 11. So we can plug these values into the formula: 11 = 1 + (6 - 1)d Simplifying the right-hand side, we get: 11 = 1 + 5d Subtracting 1 from both sides, we get: 10 = 5d Dividing both sides by 5, we get: d = 2 Therefore, the common difference of the arithmetic progression is 2.

Akọwa Nkọwa

$\frac{(2m?u{)}^2?(m?2u{)}^2}{5m^2?5u^2}$
= $\frac{2m?u+m?2u\left)\right(2m?u?m+2u)}{5(m+u)(m?u)}$
= $\frac{3(m?u)(m+u)}{5(m+u)(m?u)}$
= $\frac{3}{5}$

Akọwa Nkọwa

To find the probability that a pupil chosen at random is at least 11 years old, we need to add up the number of pupils in the 11-year-old and 12-year-old age groups, since they are the ones who are at least 11 years old. From the table, we can see that there are 27 pupils who are 11 years old and 7 pupils who are 12 years old. Therefore, the total number of pupils who are at least 11 years old is 27 + 7 = 34. To find the probability, we divide the number of pupils who are at least 11 years old by the total number of pupils in the class. From the table, we can see that the total number of pupils in the class is 6 + 27 + 7 = 40. Therefore, the probability that a pupil chosen at random is at least 11 years old is: 34/40 = 17/20 So the answer is: 17/20.

Akọwa Nkọwa

Akọwa Nkọwa

a $\oplus$ b = ab
The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a $\oplus$ b = ab; b $\oplus$ a = ba = ab
The number 1 is the identity element under multiplication

Akọwa Nkọwa

The shaded area in the diagram represents the region below the line passing through the points (1, 0) and (0, 2). To find the equation of the line, we first need to find its slope: slope = (change in y) / (change in x) slope = (0 - 2) / (1 - 0) slope = -2 Next, we use the point-slope form of the equation of a line to find the equation of the line: y - 0 = -2(x - 1) y = -2x + 2 Now we can test each inequality option to see which one represents the shaded region. We can do this by picking a point in the shaded region, plugging in its coordinates into the inequality, and checking if the inequality is true. For example, the point (0, 0) is in the shaded region, so we plug in x=0 and y=0 into each inequality: - 2(0) - 0 - 2 ≥ 0 is false - 2(0) - 0 - 2 ≤ 0 is true - 2(0) - 0 - 2 < 0 is true - 2(0) - 0 - 2 > 0 is false Therefore, the inequality that represents the shaded portion in the diagram is 2x - y - 2 ≤ 0.

Akọwa Nkọwa

${\int }_{-1}^1$(2x + 1)2dx
= ${\int }_{-1}^1$(4x2 + 4x + 1)dx
= ${\int }_{-1}^1$[$\frac{4x^3}{3}$ + 2x2 + c]
= [$\frac{4}{3}$ + 3 + c] - [4 + $\frac{1}{3}$ + c]
= $\frac{8}{3}$ + 3 + -1 - C
= $\frac{8}{3}$ + 2
= $\frac{14}{3}$
= $\frac{42}{3}$

Akọwa Nkọwa

To evaluate the expression, we need to use the properties of logarithms. First, we can simplify the expression inside the parentheses of the logarithm: log5(0.04) = log5(4/100) = log5(4) - log5(100) = log5(2^2) - log5(10^2) = 2log5(2) - 2 Next, we can simplify the second part of the expression: log318 - log32 = log3(18/32) = log3(9/16) = log3(3^2) - log3(2^4) = 2log3(3) - 4 Substituting these simplified expressions back into the original expression, we get: 2log5(2) - 2 - (2log3(3) - 4) Simplifying further, we get: 2log5(2) - 2 - 2log3(3) + 4 Combining like terms, we get: 2log5(2) - 2log3(3) + 2 Now, we can plug in the values of log5(2) and log3(3) using a calculator: log5(2) ≈ 0.4307 and log3(3) = 1 Substituting these values, we get: 2(0.4307) - 2(1) + 2 ≈ 0.8614 - 2 + 2 ≈ -0.1386 Therefore, the answer is -1, since it is the only option that is negative.

Akọwa Nkọwa

log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63
log10r63 = 63
63 = 1063
∴ r = 10

Akọwa Nkọwa

To find the mean deviation of a set of numbers, we first need to find the mean or average of those numbers. The mean of the numbers 4, 5, and 9 is: (mean) = (4 + 5 + 9) / 3 = 6 Next, we need to find the deviation of each number from the mean. To do this, we subtract the mean from each number: 4 - 6 = -2 5 - 6 = -1 9 - 6 = 3 To avoid positive and negative deviations cancelling out, we take the absolute value of each deviation: | -2 | = 2 | -1 | = 1 | 3 | = 3 Then, we find the mean of these absolute deviations by summing them up and dividing by the number of numbers: (2 + 1 + 3) / 3 = 2 Therefore, the mean deviation of the set of numbers 4, 5, 9 is 2. So, the correct option is 2.

Akọwa Nkọwa

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180${}^{?}$ - (50 + 60)

= 180${}^{?}$ - 110${}^{?}$

= 70${}^{?}$

Since RQT = RTS = 70${}^{?}$

Akọwa Nkọwa

Let the sum of the 12 numbers be x and the 13th number be y.

x12=3⟹x=36 $\frac{x}{12}=3⟹x=36$

36+y13=5⟹36+y=65 $\frac{36+y}{13}=5⟹36+y=65$

y=65−36=29

Akọwa Nkọwa

To find the values of p and q, we can use the midpoint formula, which states that the midpoint of a line segment between two points (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2). Here, we are given that M(4,q) is the midpoint of the line joining L(p,-2) and N(q,p). So we know that: - The x-coordinate of M is the average of the x-coordinates of L and N: (p+q)/2 = 4 - The y-coordinate of M is the average of the y-coordinates of L and N: (-2+p)/2 = q Simplifying the first equation, we get: (p+q)/2 = 4 p+q = 8 q = 8-p Substituting this into the second equation, we get: (-2+p)/2 = q (-2+p)/2 = 8-p -2+p = 16-2p 3p = 18 p = 6 Therefore, the values of p and q are: - p = 6 - q = 8-p = 2 So the correct option is p = 6, q = 2.

Akọwa Nkọwa

Median = L + [$\frac{\frac{N}{2}-f}{fm}$]h
N = Sum of frequencies
L = lower class boundary of median class
f = sum of all frequencies below L
fm = frequency of modal class and
h = class width of median class
Median = 11 + [$\frac{\frac{50}{2}-21}{20}$]5
= 11 + ($\frac{25-21}{20}$)5
= 11 + ($\frac{\left(4\right)}{20}$)
11 + 1 = 12

Akọwa Nkọwa

Perimeter = OP + OQ + PQ
= 8 + 8 + PQ
length PQ = $\frac{\theta }{360\times 2\pi r}$
= $\frac{45}{360}$ x 2 x $\pi$ x 8
= 2$\pi$
Perimeter of sector 2r + L
Where l = length of arc and r = radius
∴ P = 2(8) + 2$\pi$
= 16 + 2$\pi$

Akọwa Nkọwa

$\frac{1}{r-1}$ + $\frac{2}{r+1}$ = $\frac{3}{r}$
Multiply through by r(r -1) which is the LCM
= (r)(r + 1) + 2(r)(r - 1)
= 3(r - 1)(r + 1)
= r2 + r + 2r2 - 2r
3r2 - 3 = 3r2
r = 3r2 - 3
-r = -3
∴ r = 3

Akọwa Nkọwa

$\frac{x-3}{(1-x)(x+2)}$ = $\frac{p}{1-x}$ + $\frac{Q}{x+2}$
Multiply both sides by LCM i.e. (1 - x(x + 2))
∴ x - 3 = p(x + 2) + Q(1 - x)
When x = +1
(+1) - 3 = p(+1 + 2) + Q(1 - 1)
-2 = 3p + 0(Q)
3p = -2
∴ p = $\frac{-2}{3}$

Akọwa Nkọwa

$\left(\begin{array}{cc}1& 1\\ 3& y\end{array}\right)$$\left(\begin{array}{c}x\\ 1\end{array}\right)$ = $\left(\begin{array}{c}4\\ 1\end{array}\right)$ = x + 1 = 4
x = 4 - 1
= 3
3x + y =1
3(3) = y = 1
= 9 + y = 1
y = 1 - 9
= -8

Akọwa Nkọwa

$\begin{array}{cccccc}x& f& fx& \overline{x}?x& (\overline{x}?x{)}^2& f(\overline{x}?x{)}^2\\ 1& 2& 2& ?2& 4& 8\\ 2& 1& 2& ?1& 1& 1\\ 3& 2& 6& 0& 0& 0\\ 4& 1& 4& 1& 1& 1\\ 2& 2& 10& 2& 4& 8\\ & 8& 24& & & 18\end{array}$
x = $\frac{?fx}{?f}$
= $\frac{24}{8}$
= 3
Variance (62) = $\frac{?f(\overline{x}?x{)}^2}{?f}$
= $\frac{18}{8}$
= $\frac{9}{4}$

Akọwa Nkọwa

$\frac{x}{r}$ = $\frac{x+h}{2r}$

2 x r = r (x + h)

Total height of cone = x + h

but x = h

total height = 2h

Akọwa Nkọwa

Gradient of line = $\frac{\text{Change in y}}{\text{Change in x}}=\frac{y_2-y_1}{x_2-x_1}$

y2 = 0, y1 = 4

x2 = 3 and x1 = 0

$\frac{y_2-y_1}{x_2-x_1}=\frac{0-4}{3-0}=\frac{-4}{3}$

Equation of straight line = y = mx + c

where m = gradient and c = y

intercept = 4

y = 4x + $\frac{4}{3}$, multiple through by 3

3y = 4x + 12

Akọwa Nkọwa

The given scenario can be visualized as follows:

          A (top of cliff)
          /|
         / |
        /  |
       /   | 10m
      /    |
     /θ    |   
    /      |
   /       |
  B--------C (boat on water surface)

Here, the angle of depression of point B from point A is given as 30 degrees. We are required to find the distance between point B and point C, denoted by BC.

We know that the tangent of an angle is the ratio of the opposite side to the adjacent side. In this case, the opposite side is AB and the adjacent side is BC.

Thus, we have:

tan 30° = AB / BC

AB is the height of the cliff, which is given as 10m.

Hence, we have:

1/√3 = 10 / BC

Solving for BC, we get:

BC = 10√3 meters

Therefore, the boat is 10√3 meters away from the foot of the cliff.

Hence, the answer is 10√3m.

Akọwa Nkọwa

The sine function is a periodic function, which means that its values repeat after a certain interval. Specifically, the sine function has a period of 360 degrees (or 2π radians), which means that sin(x) = sin(x ± 360) for any angle x. In this case, we want to find the value of sin(-690). To do so, we can add or subtract multiples of 360 degrees to get an angle in the range between -360 and 360 degrees. Adding 2*360 degrees to -690 gives us an equivalent angle of 30 degrees, since -690 + 2*360 = -690 + 720 = 30. Now we can use the unit circle or a calculator to find the sine of 30 degrees. The exact value of sin(30) is 1/2, which means that sin(-690) is also equal to 1/2. Therefore, the answer is: - 1/2 Note that none of the given options match this answer exactly, so it's possible that there is a mistake in the question or answer choices.

Akọwa Nkọwa

A$△$ = $\sqrt{S(S-a)(S-b)(S-c)}$ (Hero's Formula)

S = $\frac{a+b+c}{2}$ = $\frac{5+6+7}{2}$

$\frac{18}{2}=9$

A$△$ $\sqrt{9}\times 4\times 3\times 2$

$\sqrt{216}=6\sqrt{6}c{m}^3$

A$△$ = $\frac{1}{2}\times 6\times h$

6$\sqrt{6}=\frac{1}{2}\times 7\times h$

h = $\frac{12}{h}\sqrt{6}$

Akọwa Nkọwa

Since XY is a fixed line and
XPY = 90o P is on one side of XY
P1P2P3......Pn are all possible cases where
XPY = 90o the only possible tendency is a semicircle because angles in semicircle equals 90o

Akọwa Nkọwa

4 x 2 = -8 upward arrows = +ve
2 x 5 x 3 = 30
0 x 6 = $\frac{0}{22}$ downward arrows = -ve
-1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24
= -(4 x 0 x 1)
= $\frac{0}{-19}$
therefore 22 - 19 = 3

Akọwa Nkọwa

Akọwa Nkọwa

$\frac{1}{\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$ x $\frac{\sqrt{2}}{\sqrt{2}}$
= $\frac{\sqrt{2}}{2}$
= $\frac{1.414}{2}$
= 0.707