JAMB UTME - Mathematics - 1994
Question 1 Report
The grades A1, A2, A3, C4 and F earned by students in a particular course are shown in the pie chart. What percentage of the students obtained a C4 grade?
Answer Details
Question 2 Report
Evaluate \( \int_{-1}^{1} (2x + 1)^2 \, dx \)
Answer Details
= ∫1−1 (4x2 + 4x + 1)dx
= ∫1−1 [4x33 + 2x2 + c]
= [43 + 3 + c] - [4 + 13 + c]
= 83 + 3 + -1 - C
= 83 + 2
= 143
= 423
Question 3 Report
Simplify \( \sqrt{48} - \frac{9}{\sqrt{3}} + \sqrt{75} \)
Answer Details
Rearrange = √48 + √75 - 9√3
= (√16 x √3) + (√25 x √3) - 9√3
=4√3 + 5√3 - 9√3
Rationalize → 9√3 = 9√3 x √3√3
= 9√3√9 - 9√3√3
= 3√3
Question 4 Report
The equation of the line in the graph is
Answer Details
y2 = 0, y1 = 4
x2 = 3 and x1 = 0
y2−y1x2−x1=0−43−0=−43
Equation of straight line = y = mx + c
where m = gradient and c = y
intercept = 4
y = 4x + 43 , multiple through by 3
3y = 4x + 12
Question 5 Report
In the diagram. Find h
Answer Details
S = a+b+c2 = 5+6+72
182=9
A△ √9×4×3×2
√216=6√6cm3
A△ = 12×6×h
6√6=12×7×h
h = 12h√6
Question 6 Report
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Question 7 Report
Given that \( \sqrt{2} = 1.1414 \), find without using tables, the value of \( \frac{1}{\sqrt{2}} \)
Question 8 Report
In the diagram, O is the centre of the circle. If SOQ is a diameter and < PRS is \(38^\circ\), what is the value of < PSQ
Answer Details
Question 9 Report
Solve for r in the following equation \( \frac{1}{r-1} + \frac{2}{r+1} = \frac{3}{r} \)
Answer Details
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
Question 10 Report
Evaluate \( \frac{\log_{5}(0.04)}{\log_{3}18-\log_{3}2} \)
Answer Details
Question 11 Report
If \(y = 3t^3 + 2t^2 - 7t + 3\), find \(\frac{dy}{dt}\) at \(t = -1\)
Answer Details
dydt = 9t2 + 4t - 7
When t = -1
dydt = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
Question 12 Report
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| \(f\) | 2 | 1 | 2 | 1 | 2 |
Find the variance of the frequency distribution above
Answer Details
x = ?fx?f
= 248
= 3
Variance (62) = ?f(¯x?x)2?f
= 188
= 94
Question 13 Report
A binary operation \( \oplus \) is defines on the set of all positive integers by \( a \oplus b = ab \) for all positive integers a, b. Which of the following properties does NOT hold?
Answer Details
The set of all national rules Q, is closed under the operations, additions, subtraction, multiplication and division. Since a ⊕ b = ab; b ⊕ a = ba = ab
The number 1 is the identity element under multiplication
Question 14 Report
Evaluate \( \frac{1}{3} \div \left[\frac{5}{7}\left(\frac{9}{10}-1+\frac{3}{4}\right)\right] \)
Answer Details
13÷[57(910?1010+34)]
= 13÷[57(?110+34)]
= 13÷[57(?2+1520)]
= 13÷[57×1320]
13+[1328] = 13×2813
= 2839
Question 15 Report
Answer Details
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
Question 16 Report
he determination of the matrix \(\begin{pmatrix}1 & 3 & 3 \\ 4 & 5 & 6 \\ 2 & 0 & 1\end{pmatrix}\) is
Answer Details
2 x 5 x 3 = 30
0 x 6 = 022 downward arrows = -ve
-1(1 x 5x - 1) = 5 - (2 x 6 x 2) = -24
= -(4 x 0 x 1)
= 0−19
therefore 22 - 19 = 3
Question 17 Report
Find the value of \( \log_{10} r + \log_{10} r^2 + \log_{10} r^4 + \log_{10} r^8 + \log_{10} r^{16} + \log_{10} r^{32} = 63 \)
Answer Details
log10r63 = 63
63 = 1063
∴ r = 10
Question 18 Report
Find the inequality which represents the shaded portion in the diagram
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Question 19 Report
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= 8 + 8 + PQ
length PQ = θ360×2πr
= 45360 x 2 x π x 8
= 2π
Perimeter of sector 2r + L
Where l = length of arc and r = radius
∴ P = 2(8) + 2π
= 16 + 2π
Question 20 Report
Answer Details
< STP = < RQV = 30o
But SQR + 30o + 50o = 180o
SQR = 180 - 80
= 100o
Question 21 Report
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \(f\) | \(y+2\) | \(y-2\) | \(2y-3\) | \(y+4\) | \(3y-4\) |
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y
Answer Details
Question 22 Report
In the frustum of the cone, the top diagram is twice the bottom diameter. If the height of the frustum is h centimeters, find he height of the cone.
Answer Details
2 x r = r (x + h)
Total height of cone = x + h
but x = h
total height = 2h
Question 23 Report
Simplify \( \frac{(2m-u)^2-(m-2u)^2}{5m^2-5u^2} \)
Answer Details
= 2m?u+m?2u)(2m?u?m+2u)5(m+u)(m?u)
= 3(m?u)(m+u)5(m+u)(m?u)
= 35
Question 24 Report
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Question 25 Report
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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
Question 26 Report
| Age in years | 10 | 11 | 12 |
| Number of pupils | 6 | 27 | 7 |
The table above shows the number of pupils in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
Answer Details
Question 27 Report
Answer Details
∴ (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. 27p+9q=−27−18p=−18
∴ p = 1
Question 28 Report
Solve for x and y \(\begin{pmatrix}1 & 1 \\ 3 & y\end{pmatrix}\begin{pmatrix}x \\ 1\end{pmatrix} = \begin{pmatrix}4 \\ 1\end{pmatrix}\)
Answer Details
x = 4 - 1
= 3
3x + y =1
3(3) = y = 1
= 9 + y = 1
y = 1 - 9
= -8
Question 29 Report
Answer Details
∴ (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
Question 31 Report
Answer Details
Question 32 Report
| Class Interval | 1 − 5 | 6 − 10 | 11 − 15 | 16 − 20 | 21 − 25 |
| Frequency | 6 | 15 | 20 | 7 | 2 |
Estimate the median of the frequency distribution above
Answer Details
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 + [502−2120 ]5
= 11 + (25−2120 )5
= 11 + ((4)20 )
11 + 1 = 12
Question 33 Report
Answer Details
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
Question 34 Report
Answer Details
Let the sum of the 12 numbers be x and the 13th number be y.
x12=3⟹x=36
36+y13=5⟹36+y=65
y=65−36=29
Question 35 Report
Without using table, solve the equation \(8x^{-2} = \frac{2}{25}\)
Answer Details
= 200x-2 = 2
= 100x-2 = 1
x-2 = 1100
x-2 = 10-2
x = 10
Question 36 Report
Evaluate \( \frac{0.36 \times 5.4 \times 0.63}{4.2 \times 9.0 \times 2.4} \)
Answer Details
= 36420×5490×63240
= 670×1830×2180
= 272000
= 0.0135
≈ = 0.013
Question 37 Report
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40 = 55 - x
x = 55 - 40
= 15
∴ P(both) 1540
= 38
Question 38 Report
Answer Details
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.
Question 39 Report
Answer Details
Question 40 Report
If a = 1, b = 3, solve for x in the equation \( \frac{a}{a-x} = \frac{b}{x-b} \)
Answer Details
11?x = 3x?3
? 3(1 - x) = x - 3
3 - 3x = x - 3
Rearrange 6 = 4x; x = 64
= 32
Question 41 Report
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Question 42 Report
The equation of the graph is
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when y = 0, x = 3 (3, 0)
Question 43 Report
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Question 44 Report
Find the range of values of x for which \( \frac{1}{x} > 2 \) is true
Answer Details
1x
> 2 = xx2
> 2
x > 2x2
= 2x2 < x
= 2x2 - x < 0
= x(2x - 10 < 0
Case 1(+, -) = x > 0, 2x - 1 < 0
x > 0, x < 12
(solution)
Case 2(-, 4) = x < 0, 2x - 1 > 0
x < 0, x , 12
= 0
Question 45 Report
Find P if \( \frac{x-3}{(1-x)(x+2)} = \frac{p}{1-x} + \frac{Q}{x+2} \)
Answer Details
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 = −23
Question 46 Report
In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS
Answer Details
RQT = 180? - (50 + 60)
= 180? - 110?
= 70?
Since RQT = RTS = 70?
Question 47 Report
| \( \oplus \mathit{mod}\ 10 \) | 2 | 4 | 6 | 8 |
| 2 | 4 | 8 | 2 | 6 |
| 4 | 8 | 6 | 4 | 2 |
| 4 | 8 | 6 | 4 | 2 |
| 6 | 2 | 4 | 6 | 8 |
| 8 | 6 | 2 | 8 | 4 |
The multiplication table above has modulo 10 on the set S = (2, 4, 6, 8). Find the inverse of 2
Answer Details
12 = 2 which is also the value required
Question 48 Report
Given that for sets A and B, in a universal set E, \(A \subseteq B\) then \(A \cap (A \cap B)^{1}\) is
Answer Details
A(A ∩ B)1 = A ∩ A1
The intersection of complement of a set P and P1 has no element
i.e. n(A ∩ A1) = ϕ
