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Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Simplify T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\)
A
\(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
B
\(\frac{R_1 R_2 R_3}{R_2R_3 + R_1R_2 + 4R_1 R_2}\)
C
\(\frac{16R_1 R_2 R_3}{R_2R_3 + R_1R_2 + R_1 R_2}\)
D
\(\frac{4R_1 R_2 R_3}{4R_2R_3 + R_1R_2 + 4R_1 R_2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
T = \(\frac{4R_2}{R_1^{-1} + R_2^{-1} + 4R_3^{-1}}\) = \(\frac{4R_2}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{4}{R_3}}\)
= \(\frac{4R_2}{\frac{R_2R_3 + R_1R_3 + 4R_1R_2}{R_1R_2R_3}}\)
= \(\frac{4R_2 \times R_1 R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
= \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1R_2}\)
T = \(\frac{4R_1 \times R_2 R_3}{R_2R_3 + R_1R_3 + 4R_1 R_2}\)
Source: Myschool Past Questions Archive (Record #32596)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Solve the given equation \((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)
A
27
B
9
C
\(\frac{1}{27}\)
D
18
E
81
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\((\log_{3} x)^{2} - 6(\log_{3} x) + 9 = 0\)
Let \(\log_{3} x = a\).
\(a^{2} - 6a + 9 = 0\)
\(a^{2} - 3a - 3a + 9 = 0\)
\(a(a - 3) - 3(a - 3) = 0\)
\((a - 3)(a - 3) = 0\)
\(\implies a = 3 (twice)\)
\(\log_{3} x = 3 \implies x = 3^{3} = 27\)
Source: Myschool Past Questions Archive (Record #32603)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Diagram
Solution
In the figure, find the area of XYZW
Attached Figure / Diagram
A
60cm2
B
54cm2
C
27cm2
D
54\(\sqrt{2}\)cm2
E
27\(\sqrt{2}\)cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of trapezium XYZW = \(\frac{1}{2}\)(a + b)h
h = 6 sin 30º = 3
area = \(\frac{1}{2}\)(10 + 8) x 3
\(\frac{1}{2}\)(18) x 4
= 27cm\(^2\)
Source: Myschool Past Questions Archive (Record #32754)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Evaluate \((2^{0} + 4^{-\frac{1}{2}})^{2}\)
A
\(\frac{1}{4}\)
B
\(\frac{5}{4}\)
C
\(\frac{9}{4}\)
D
4
E
9
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\((2^{0} + 4^{-\frac{1}{2}})^{2}\)
= \((1 + (\frac{1}{4})^{\frac{1}{2}})^{2}\)
= \((1 + \frac{1}{2})^{2}\)
= \((\frac{3}{2})^{2} = \frac{9}{4}\)
Source: Myschool Past Questions Archive (Record #32607)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Which of the following is NOT a factor of 12\(^{4}\) - 5\(^{4}\)?
A
7
B
13
C
17
D
49
E
169
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
124 - 54 = (122)2 - (52)2
= (122 + 52) x (122 - 52)
(144 + 25)(144 - 25)
= (169)(119)
Factors of above 169, 17, 13, 7
Source: Myschool Past Questions Archive (Record #32608)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Rationalize the denominator of the given expression \(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} + \sqrt{a}}\)
A
1 + 2a - 2\(\sqrt{a(1 + a)}\)
B
\(\sqrt{1(1 + a)}\)
C
2a - 2\(\sqrt{a(1 + a)}\)
D
1 + 2a - 2\(\sqrt{a + b}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} + \sqrt{a}}\) = \(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} + \sqrt{a}}\) x \(\frac{\sqrt{1 + a} - \sqrt{a}}{\sqrt{1 + a} - \sqrt{a}}\)
= \(\frac{\sqrt{1 + a + a}}{1 + a - a}\)
= 2a + a(1 + a)
= 1 + 2a - 2\(\sqrt{a(1 + a)}\)
Source: Myschool Past Questions Archive (Record #32609)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Simplify \(\frac{log_{10}8}{log_{10}4}\)
A
log102
B
log104
C
\(\frac{3}{2}\)
D
2
E
None of the above
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{log_{10}8}{log_{10}4}\) = \(\frac{log_{10}2^3}{log_{10}2^2}\)
= \(\frac{3 log_{10}2}{2log_{10}2}\)
= \(\frac{3}{2}\)
Source: Myschool Past Questions Archive (Record #32618)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
If it is given that \(5^{x + 1} + 5^x = 150\), then the value of x is equal to
A
3
B
4
C
1
D
2
E
\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(5^{x + 1} + 5^x = 150\)
according to the laws of indices
\(5^1(5^x) + 5^x = 150\)
let \(5^x\) = y
5(y) + y = 150
6y = 150
y = \(\frac{150}{6}\)
= 25
but y = \(5^x\) = 25
\(5^x\) = 25 = \(5^2\)
equating powers since bases are the same
x = 2
Source: Myschool Past Questions Archive (Record #32631)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Solve the system of equation 2\(^{x + y}\) = 32, \(3^{3y - x}\) = 27
A
(3, 2)
B
(-3, 2)
C
(3, -2)
D
(-3, -2)
E
(2, 2)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(2^{x + y} = 32, 3^{3y - x}\) = 27
\(2^{x + y} = 2^5\)
\(3^{3y - x} = 3^3\)
equating powers, we have
x + y = 5 ------------1
3y - x = 3 ------------2
solving simultaneously(any method)
x = 3 and y = 2
Source: Myschool Past Questions Archive (Record #32632)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
Given log 2 = 0.69, log3 = 1, 10 and log7 = 1.90, all to a fixed base, find log 10.5 to the same base without using tables.
A
1.03
B
2.31
C
3.69
D
10.5
E
25
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
log 10.5 = log \(\frac{21}{2}\)
= log 21 - log 2
= log(3 x 7) - log 2
= log 3 + log 7 - log 2
= 1.10 + 1.90 - 0.69
= 3 - 0.69
= 2.31
Source: Myschool Past Questions Archive (Record #32636)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\) and \(\theta\) is less than 90º, calculate \(\frac{tan(90 - \theta)}{cos^2\theta}\)
A
\(\frac{1}{2}\)
B
\(\frac{4}{\sqrt{3}}\)
C
\(\sqrt{3}\)
D
4\(\sqrt{3}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
If Sin \(\theta\) = \(\frac{\sqrt{3}}{2}\), then \(\theta\) = 60º
∴ \(\frac{tan(90 - \theta)}{cos^2\theta}\) = \(\frac{tan(90 - 60)}{cos^2 60}\) = \(\frac{tan30}{cos^2 60}\)
= \(\frac{1}{\sqrt3} \times (\frac{2}{1})^2 = \frac{4}{\sqrt{3}}\)
Source: Myschool Past Questions Archive (Record #32751)
Mathematics
UTME / JAMB 1980
Indices, Logarithms & Surds
Diagram
Solution
Find the area of the shades segments in the figure
Attached Figure / Diagram
A
\(\sqrt{3}
B
4 \(\pi - \sqrt{3}\)
C
-\(\frac{2}{3} \pi\)
D
\(\frac{2\pi}{3}\) -3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of section = \(\frac{60^o}{360^o}\) x 11r2
= \(\frac{60}{360} \times \pi \times 2^2\)
= \(\frac{1}{6}\) x 4
= \(\frac{4\pi}{6}\)
= \(\frac{2\pi}{3}\)
Area of triangle = \(\frac{1}{2x}\)
= 2 x 28.......60
Segment Area = Area of section - Area of triangle
= \(\frac{2\pi}{3}\) -3
Source: Myschool Past Questions Archive (Record #32752)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
Find the value of (4\(^{\frac{1}{2}}\))\(^6\)
A
1
B
2
C
4
D
6
E
64
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
The powers will multiply when the bracket is expanded
The multiplied powers: \(\frac{1}{2}\) * 6 → 3
: 4\(^3\) = 4 * 4* 4
→ 64
Source: WAEC / JAMB / NECO Classroom Archive (Record #32670)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
Find the value of log\(_{10}\)\(\frac{1}{40}\), given that log10\(_4\) = 0.6021
A
1.3979
B
2.3979
C
-1.6021
D
2.6021
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
log\(_{10}\)\(\frac{1}{40}\) = log\(_{10}\)1 - (log\(_{10}\)4 + log\(_{10}\)10)
= 0 - 1.6021
= -1.6021
Source: WAEC / JAMB / NECO Classroom Archive (Record #32671)
Mathematics
UTME / JAMB 1979
Indices, Logarithms & Surds
Solution
\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\) equals
A
2 x 10-5
B
2 x 10-1
C
2 x 101
D
2 x 10-4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\((1.28 \times 10^{4}) \div (6.4 \times 10^{2})\)
= \(\frac{1.28 \times 10^{4}}{6.4 \times 10^{2}}\)
\(\equiv \frac{12.8 \times 10^{3}}{6.4 \times 10^{2}}\)
= \(2 \times 10^{1}\)
Source: Myschool Past Questions Archive (Record #32645)