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Mathematics
WAEC / WASSCE 1994
Inequalities & Linear Programming
Solution
If sin\( \theta \) = K find tan\(\theta\), 0^\circ \(\leq\) \(\theta\) \(\leq\) 90^\circ.
A
1-K
B
\( \frac{k}{k - 1} \)
C
\( \frac{k}{\sqrt{1 - k^2}} \)
D
\( \frac{k}{1 - k} \)
E
\( \frac{k}{\sqrt{ k^2 - 1}} \)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)
Source: Myschool Past Questions Archive (Record #22006)
Mathematics
UTME / JAMB 1994
Binary Operations, Matrices & Determinants
Solution
The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is
A
-67
B
-57
C
-1
D
3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
((1 x5x1) +( 3x6x2) + (3x4x0)) - ((3x5x2) + (1x6x0) + (3x4x1))
( 5 + 36) - (30 +12) = 41 - 42
= -1
Source: WAEC / JAMB / NECO Classroom Archive (Record #31019)
Mathematics
UTME / JAMB 1994
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize a2x - b2y - b2x + a2y
A
(a - b)(x + y)
B
(y - x)(a - b)(a + b)
C
(x - y)(a - b)(a + b)
D
(x + y)(a - b)(a + b)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
a2x - b2y - b2x + a2y = a2x - b2x - b2y + a2y Rearrange
= x(a2 - b2) + y(a2 - b2)
= (x + y)(a2 - b2)
= (x + y)(a + b)(a - b)
Source: WAEC / JAMB / NECO Classroom Archive (Record #31007)
Physics
UTME / JAMB 1994
Work, Energy, Power & Simple Machines
Diagram
Solution
In the fig above, XY is of length 1m, The value of R at balance point Z is
Attached Figure / Diagram
A
3.0Ω
B
13.3Ω
C
15.0Ω
D
30.0Ω
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
At null deflection, p.d across R = p.d across xz
p.d across 20\Omega = p.d across zy
\(\frac{R}{20}\) = \(\frac{60}{40}\)
R = 30\Omega
Source: WAEC / JAMB / NECO Classroom Archive (Record #41724)
Chemistry
UTME / JAMB 1994
Periodic Table & Periodicity
Solution
Which of the following is the correct order of decreasing activity of the metals Fe, Ca, Al and Na?
A
Fe> Ca> Al> Na
B
Na> Ca> Al> Fe
C
Al> Fe> Na> Ca
D
Ca> Na> Fe>, Al
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
The order of reactivity of metals is as follows, Potassium > Sodium > Lithium > Calcium > Magnesium > Aluminium > Zinc > Iron > Copper > Silver > Gold.
Source: WAEC / JAMB / NECO Classroom Archive (Record #13594)
Chemistry
UTME / JAMB 1994
Air & Environmental Chemistry
Solution
In an electrolytic set-up to protect iron from corrosion, the iron is?
A
made cathode
B
made anode
C
used with a metal of lower electropositive potential
D
initially coated with tin
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
In cathodic protection,iron object is made cathode by connecting it with reactive elements like magnesium,zinc.
Source: WAEC / JAMB / NECO Classroom Archive (Record #13591)
Chemistry
UTME / JAMB 1994
Electrolysis & Electrochemical Cells
Solution
What volume of Oxygen measured at s.t.p would be liberated on electrolysis by 9650 coulombs of electricity?
(Molar volume of gas = 22.4 dm\(^3\), 1F = 96,500 C/mol)
A
22.4 dm3
B
11.2 dm3
C
1.12 dm3
D
0.560 dm3
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
For a diatomic divalent element like oxygen, the following reaction takes place:
2H\(_2\)0 → O\(_2\) + 4H\(^+\) + 4e\(^-\)
⇒ 4e\(^-\) are transferred to liberate 1 mole of O\(_2\)
So, to calculate the volume of O\(_2\)
4 Faradays (4 x 96,500) of electricity to produce 22.4 dm\(^3\) of O\(_2\) at S.T.P
4 X 96,500 C = 22.4 dm\(^3\)
9650 C = x dm\(^3\)
x = \(\frac{9650 X 22.4}{ 4 X 96,500}\)
x = 0.56dm\(^3\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #13543)
Chemistry
UTME / JAMB 1994
Non-Metals & Their Compounds
Diagram
Solution
In the diagram above, R is a mixture of?
Attached Figure / Diagram
A
potassium tetraxochlorate (VII) and concentrated H2SO4
B
potassium trioxochlorate (V) and concentrated H2SO4
C
potassium tetratoxomanganate (VII) and concentrated HCI
D
manganese (IV) oxide and concentrated HCI
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
R is concentrated hydrochloric acid and potassium tetraoxomanganate(VII)
(See diagram above)
Source: WAEC / JAMB / NECO Classroom Archive (Record #31514)
Chemistry
UTME / JAMB 1994
Metals, Transition Elements & Extraction
Diagram
Solution
2HCL(ag) + CaCO3(ag) → CaCL2(ag) + H2O(1) + CO from the reaction above, which of the following curves represents the consumption of calcium trioxocarbonate (IV) as dilute HCL is added to it?
Attached Figure / Diagram
A
L
B
M
C
N
D
P
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
The concentration of the \(CaCO_3\) decreases as reaction time decreases. therefore option D is correct.
Source: WAEC / JAMB / NECO Classroom Archive (Record #31491)
Mathematics
WAEC / WASSCE 1994
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify: \(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)
A
0
B
1
C
2
D
3
E
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)
= \(\frac{7}{3} \div \frac{8}{3} \times \frac{8}{7}\)
= \(\frac{7}{3} \times \frac{3}{8} \times \frac{8}{7}\)
= 1
Source: Myschool Past Questions Archive (Record #23459)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
Which of the following is equal to \(\frac{72}{125}\)
A
\( \frac{2^3 \times 3^2}{5^3} \)
B
\( \frac{2^4 \times 3}{5^3} \)
C
\( \frac{2^3 \times 3}{5^3} \)
D
\( \frac{2^4 \times 3}{5^5} \)
E
\( \frac{2^2 \times 3^2 \times 4^2}{5^2} \)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}\)
= \(\frac{2^3 \times 3^2}{5^3}\)
Source: Myschool Past Questions Archive (Record #23460)
Mathematics
WAEC / WASSCE 1994
Fractions, Decimals, Percentages & Approximations
Solution
Express in \( \frac{8.75}{0.025} \)standard form
A
3.5 x 10-3
B
3.5 x 10-2
C
3.5 x 101
D
3.5 x 102
E
3.5 x 103
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{8.75}{0.025}\)
= \(\frac{8750}{25}\)
= \(350\)
= \(3.5 \times 10^2\)
Source: Myschool Past Questions Archive (Record #23461)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
Evaluate \( \frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
A
6
B
5
C
4
D
3
E
2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}\)
= \(\frac{3}{2^{-1}}\)
= 6
Source: Myschool Past Questions Archive (Record #23463)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
Simplify: 16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)
A
0
B
2
C
4
D
10
E
20
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
16\(^{\frac{5}{4}}\) x 2\(^{-3}\) x 3\(^0\)
= \((2^4)^{\frac{5}{4}} \times 2^{-3} \times 1\)
= \(2^5 \times 2^{-3} \times 1\)
= \(2^2\)
= 4.
Source: Myschool Past Questions Archive (Record #23466)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
simplify; 2lo\(g_3\) 6 + lo\(g_3\) 12 - lo\(g_3\) 16
A
2
B
3
C
2 - 2log32
D
3 - log32
E
4 - log32
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
2lo\(g_3\) 6 + lo\(g_3\) 12 - lo\(g_3\) 16
= \(\log_3 6^2 + \log_3 12 - \log_3 16\)
= \(\log_{3} (\frac{36 \times 12}{16})\)
= \(\log_{3} (27)\)
= \(\log_{3} 3^3\)
= \(3 \log_{3} 3\)
= 3
Source: Myschool Past Questions Archive (Record #23468)