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Mathematics
UTME / JAMB 1989
Indices, Logarithms & Surds
Diagram
Solution
OXYZW is a pyramid with a square base such that OX = OY= OZ = OW = 5cm and XY = XW = YZ = WZ = 6cm. Find the height OT
Attached Figure / Diagram
A
2\(\sqrt{5}\)
B
3
C
4
D
4\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
xz\(^2\) = 6\(^2\) + 6\(^2\)
36 + 36 = 72
xz = \(\sqrt{72}\)
6\(\sqrt{2}\) = XT
\(\frac{6\sqrt{2}}{2}\) = \(\frac{3}{\sqrt{2}}\)
OT\(^2\) = 5\(^2\) + (3\(\sqrt{2}\))2 = 25 + 18
OT = 4\(\sqrt{3}\)
Source: Myschool Past Questions Archive (Record #31615)
Mathematics
WAEC / WASSCE 1988
Indices, Logarithms & Surds
Solution
Simplify 36\(^\frac{1}{2}\) x 64\(^{-\frac{1}{3}}\) x 5\(^0\)
A
0
B
1\24
C
2/3
D
11/2
E
71/2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
36\(^\frac{1}{2}\) x 64\(^{-\frac{1}{3}}\) x 5\(^0\)
= \(6 \times \frac{1}{4} \times 1\)
= \(1\frac{1}{2}\)
Source: Myschool Past Questions Archive (Record #22012)
Mathematics
WAEC / WASSCE 1988
Indices, Logarithms & Surds
Solution
Points P and Q respectively 24m north and 7m east point R. Calculate |PQ| in meters
A
20
B
24
C
25
D
31
E
84
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
PQ\(^2\) = 24\(^2\) + 7\(^2\)
PQ\(^2\) = 625
PQ = 25
Source: Myschool Past Questions Archive (Record #22043)
Mathematics
WAEC / WASSCE 1988
Indices, Logarithms & Surds
Solution
Evaluate log106 + log1045 - log1027 without using logarithm tables
A
o
B
1
C
1.1738
D
1.3802
E
10
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
log106 + log1045 - log1027
log10 \(\frac{6 \times 45}{27}\)
= log102 x 5
log1010 = 1
Source: Myschool Past Questions Archive (Record #22092)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
If \(1P03{_4} = 115_{10}\) find P
A
o
B
1
C
2
D
3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(1 x 4^3 + P x 4^2 + 0 x 4 + 3 = 115_{10}\)
16p + 67 = 115 p = \(\frac{48}{16}\)
= 3
Source: Myschool Past Questions Archive (Record #30761)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
What is the solution of the equation x2 - x - 1 + 0?
A
x = 1.6 and x = -0.6
B
x = -1.6 and x = 0.6
C
x = 1.6 and x = 0.6
D
x = -1.6 and x = -0.6
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(x^{2} - x - 1 = 0\)
Using the quadratic formula,
\(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\)
a = 1, b = -1, c = -1.
\(x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}\)
\(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\)
\(x = \frac{1 + 2.24}{2} ; x = \frac{1 - 2.24}{2}\)
\(x = \frac{3.24}{2}; x = \frac{-1.24}{2}\)
\(x = 1.62 ; x = -0.61 \)
\(x \approxeq 1.6; -0.6\)
Source: Myschool Past Questions Archive (Record #31637)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
Evaluate \(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\)
A
\(\frac{2}{5}\)
B
\(\frac{5}{3}\)
C
3\(\sqrt{5}\)
D
3\(\sqrt{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{8^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{10^{\frac{2}{3}}}\) = \(\frac{(2^3)^{\frac{1}{3}} \times 5^{\frac{2}{3}}}{(2 \times 5)^{\frac{2}{3}}}\)
= \(\frac{2 \times 5^{\frac{2}{3}}}{2^{\frac{2}{3}} \times 5^{\frac{2}{3}}}\)
= \(2^{1 - \frac{2}{3}}\) x \(5^{\frac{2}{3} - \frac{2}{3}}\)
= \(2^{\frac{1}{3}}\) x \(5^0\)
= \(\sqrt[3]{2}\)
Source: Myschool Past Questions Archive (Record #31698)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\), evaluate; without using logarithm tables, \(log_{10} 4.5\)
A
0.3010
B
0.4771
C
0.6532
D
0.9542
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
If \(log_{10} 2 = 0.3010\) and \(log_{10} 3 = 0.4771\),
\(\log_{10} 4.5 = \log_{10} (\frac{3 \times 3}{2})\)
\(log_{10} 3 + log_{10} 3 - log_{10} 2 = 0.4771 + 0.4771 - 0.3010\)
= 0.6532
Source: Myschool Past Questions Archive (Record #31704)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
Find m such that (m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\)
A
1
B
2
C
3
D
4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
(m + \(\sqrt{3}\))(1 - \(\sqrt{3}\))2 = 6 - 2\(\sqrt{2}\)
(m + \(\sqrt{3}\))(4 - 2\(\sqrt{3}\)) = 6 - 2\(\sqrt{2}\)
= 6 - 2\(\sqrt{3}\)
4m - 6 + 4 - 2m\(\sqrt{3}\) = 6 - 2\(\sqrt{3}\)
comparing co-efficients,
4m - 6 = 6.......(i)
4 - 2m = -2.......(ii)
in both equations, m = 3
Source: Myschool Past Questions Archive (Record #31706)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle
A
\(\sqrt{3}\)cm
B
\(\sqrt{7}\)cm
C
3cm
D
7cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -\(\frac{1}{2}\)
= 5 + 2 = 7
PR = \(\sqrt{7}\)cm
Source: Myschool Past Questions Archive (Record #31739)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
Evaluate without using tables (0.008) -\(\frac{1}{3}\) x (0.16) - \(\frac{3}{2}\)
A
\(\frac{625}{8}\)
B
\(\frac{8}{625}\)
C
\(\frac{1}{8}\)
D
8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(0.008) -\(\frac{1}{3}\) x (0.16) - \(\frac{3}{2}\) = (8 x 10-3)10-3 x (16 x 10-2)-\(\frac{3}{2}\)
= \(\frac{(2^3)}{10^3}\) - 3 x \(\frac{(2^4)}{10}\) - \(\frac{-3}{2}\)
= \(\frac{625}{8}\)
Source: Myschool Past Questions Archive (Record #31859)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
Simplify without using tables \(\frac{Log_26 - Log_23}{Log_28 - 2Log_2\frac{1}{2}}\)
A
\(\frac{1}{5}\)
B
\(\frac{1}{2}\)
C
\(\frac{-1}{2}\)
D
\(\frac{log_23}{log_27}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
log\(_2\)6 - log\(_2\)3 = log\(_2\) (\(\frac{6}{3}\))
= log\(_2\)2... A
log\(_2\)8 - 2log\(_2\)\(\frac{1}{2}\)
log\(_2\)8 - log\(_2\)\(\frac{1}{2}\)\(^2\)
log\(_2\)8 - log\(_2\)\(\frac{1}{4}\)
= log\(_2\) 32... B
\(\frac{A}{B}\) = \(\frac{log_22}{log_232}\)
N.B. log\(_2\)2 = 1
= \(\frac{1}{5}\)
Source: Myschool Past Questions Archive (Record #31862)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
Simplify without using tables \(\frac{2\sqrt{14} \times 3\sqrt{21}}{7\sqrt{24} \times 2\sqrt{98}}\)
A
\(\frac{3\sqrt{14}}{4}\)
B
\(\frac{3\sqrt{2}}{4}\)
C
\(\frac{3\sqrt{14}}{28}\)
D
\(\frac{3\sqrt{2}}{28}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{2\sqrt{14} \times 3\sqrt{21}}{7 \sqrt{24} \times 2\sqrt{98}}\)
= \(\frac{6\sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}\)
= \(\frac{3\sqrt{3}}{14\sqrt{6}}\)
= \(\frac{3\sqrt{3}}{14\sqrt{2} \times \sqrt{3}}\)
= \(\frac{3\sqrt{2}}{28}\)
Source: Myschool Past Questions Archive (Record #31880)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
The sine, cosine and tangent of 210o are respectively
A
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
B
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{3}\)
C
\(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
D
\(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
210o = 180o - 210o = - 30o
From ratio of sides, sin -30o = -\(\frac{1}{2}\)
Cos 210o = 180o - 210o = -30o
= cos -30o = \(\frac{-3}{2}\)
But tan 30o = \(\frac{1}{\sqrt{3}}\), rationalizing this
= \(\frac{1}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{3}\)
∴ = \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)
Source: Myschool Past Questions Archive (Record #31933)
Mathematics
UTME / JAMB 1987
Indices, Logarithms & Surds
Solution
The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4\(\sqrt3\)cm
A
6cm
B
5cm
C
4cm
D
3cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Base of pyramid of a square of side 8cm vertex directly above the centre edge = \(4\sqrt{3}\)cm
From the diagram, the diagonal of one base is AC\(^2\) = 8\(^2\) + 8\(^2\)
AC\(^2\) = 64 + 64 = 128
AC = \(8\sqrt{2}\)
but OC = \(\frac{1}{2}\)AC = \(\frac{1}{2}\) x 8\(\sqrt{2}\) = \(4\sqrt{2}\)cm
OE = h = height
h\(^2\) = EC\(^2\) - OC\(^2\)
h\(^2\) = (\(4\sqrt{3}\))\(^2\) - (\(4\sqrt{2}\))\(^2\)
h\(^2\) = 16 x 3 - 16 x 2
48 - 32 = 16
h = \(\sqrt{16}\)
= 4cm
Source: Myschool Past Questions Archive (Record #31975)