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Mathematics
UTME / JAMB 1981
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Marks scored by some children in an arithmetic test are:5, 3, 6, 9, 4, 7, 8, 6, 2, 7, 8, 4, 5, 2, 1, 0, 6, 9, 0, 8.
The arithmetic mean of the marks is
A
6
B
5
C
8
D
7
E
10
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Add all the No.s together to give 100
\(\sum\)xf = 100
N = 20
x = \(\frac{\sum xf}{N}\)
= \(\frac{100}{20}\)
= 5
Source: Myschool Past Questions Archive (Record #32556)
Mathematics
UTME / JAMB 1981
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The weights of 30 new-born babies are given as follow: 6, 9, 5, 7, 6, 7, 5, 8, 9, 5, 7, 5, 8, 7, 8, 7, 5, 6, 5, 7, 6, 9, 9, 7, 8, 8, 7, 8, 9, 8. The mode is
A
6
B
5
C
8
D
7
E
10
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
x
5
6
7
8
9
f
6
4
8
7
5
Mode is the No having the highest frequency = 7
Source: Myschool Past Questions Archive (Record #32557)
Mathematics
UTME / JAMB 1981
Trigonometry, Elevation & Bearings
Solution
If \(\sin x^\circ = \frac{a}{b}\), what is \(\sin (90 - x)^\circ\)?
A
\(\frac{\sqrt{b^2 - a^2}}{b}\)
B
1\(\frac{-a}{b}\)
C
\(\frac{b^2 - a^2}{b}\)
D
\(\frac{a^2 - b^2}{b}\)
E
\(\sqrt{b^2 - a^3}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\sin x = \frac{a}{b}\)
\(\sin^{2} x + \cos^{2} x = 1\)
\(\sin^{2} x = \frac{a^{2}}{b^{2}}\)
\(\cos^{2} x = 1 - \frac{a^{2}}{b^{2}} = \frac{b^{2} - a^{2}}{b^{2}}\)
\(\therefore \cos x = \frac{\sqrt{b^{2} - a^{2}}}{b}\)
\(\sin (90 - x) = \sin 90 \cos x - \cos 90 \sin x\)
= \((1 \times \frac{\sqrt{b^{2} - a^{2}}}{b}) - (0 \times \frac{a}{b})\)
= \(\frac{\sqrt{b^{2} - a^{2}}}{b}\)
Source: Myschool Past Questions Archive (Record #32564)
Mathematics
UTME / JAMB 1981
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
7 pupils of average age 12 years leave a class of 25 pupils of average age 14 years. If 6 new pupils of average age 11years join the class, what is the average age of the pupils now in the class?
A
13years
B
12 years 7\(\frac{1}{2}\) months
C
13 years 5 months
D
13 years 10 months
E
11 years
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Total age of the 7 pupils = 7 x 12 = 84
Total age of the 25 pupils = 25 x 14 = 350
Total age of the 6 pupils = 6 x 11 = 66
7 pupils leaving a class of 25 pupils given 18 pupils
Total age of the 18 pupils = 350 - 84 = 266
6 pupils joining 18 pupils = 24
Total age of 24 pupils = 266 + 66 = 332
Average age of 24 pupils now in class \(\frac{332}{24}\)
= 13yrs 10 months
Source: Myschool Past Questions Archive (Record #32566)
Mathematics
UTME / JAMB 1981
Fractions, Decimals, Percentages & Approximations
Solution
A sum of money invested at 5% per annum simple interest amounts to $285.20 after 3 years. How long will it take the same sum to amount to $434.00 at 7\(\frac{1}{2}\)% per annum simple interest?
A
7\(\frac{1}{2}\) years
B
10 years
C
5 years
D
12 years
E
14 years
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(A = P(1 + \frac{R}{100})^{T}\)
\(285.20 = P(1 + \frac{5}{100})^{3}\)
\(285.20 = P(1.05)^{3} \implies 1.16P = 285.20\)
\(P = \frac{285.20}{1.16} = $245.86\)
Given Amount = $434.00
Interest = Amount - Principal
Interest = $434.00 - $245.86 = $188.14
\(T = \frac{100I}{PR}\)
\(T = \frac{100 \times 188.14}{245.86 \times 7.5}\)
\(T = \frac{18814}{1843.95} = 10.2\)
Approximately 10 years.
Source: Myschool Past Questions Archive (Record #32567)
Mathematics
UTME / JAMB 1981
Fractions, Decimals, Percentages & Approximations
Solution
By selling an article for N45.00 a man makes a profit of 8%. For how much should he have sold it in order to make a profit of 32%?
A
N180.00
B
N59.00
C
N63.00
D
N58.00
E
N55.00
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
When C.P = N45.00 at 8% profit
C.P = \(\frac{100}{108}\) x 45
at profit of 32%, S.P = \(\frac{132}{100}\) x c.p
= \(\frac{132}{100}\) x \(\frac{100 \times 45}{108}\)
= \(\frac{5940}{108}\)
= N55.00
Source: Myschool Past Questions Archive (Record #32568)
Mathematics
UTME / JAMB 1981
Trigonometry, Elevation & Bearings
Solution
An isosceles triangle of sides 13cm, 13cm, 10cm is inscribed in a circle. What is the radius of the circle?
A
7cm
B
12cm
C
8cm
D
7cm
E
69cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In \(\Delta DAC, \stackrel\frown{DAC} = \theta\)
\(\sin \theta = \frac{5}{13}\)
\(\theta = 22.6^\circ\)
\(< DOC = 22.6^\circ \times 2 = 45.2^\circ\)
\(\sin 45.2 = \frac{5}{r} \implies r = \frac{5}{\sin 45.2}\)
\(r = 7.046cm\)
= \(7\frac{1}{24} cm\)
Source: Myschool Past Questions Archive (Record #32570)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
A micrometer is defined as one millionth of a millimeter. A length of 12,000 micrometres may be represented as
A
0.00012m
B
0.0000012m
C
0.000012m
D
0.00000012m
E
0.000000012m
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
1 UM = 10\(^{-6}\)mm
= 10\(^{-9}\)m
1.2 x 10\(^4\) x 10\(^{-9}\)m
= 1.2 x 10\(^{-5}\)m = (0.000012)m
Source: Myschool Past Questions Archive (Record #32572)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
In one and a half hours, the minute hand of a clock rotates through an angle of
A
90o
B
180o
C
640o
D
450o
E
540o
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
1 hr, = 60 mins, 60 mins = 360o
30 mins = \(\frac{360^o}{1}\) c \(\frac{30}{60}\)
= 180o
90 mins = 360 + 180o
= 540o
Source: Myschool Past Questions Archive (Record #32575)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Simplify \(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\)
A
2\(\sqrt{2} - \sqrt{3}\)
B
3(\(\sqrt{6}\) - 1)
C
\(\sqrt{6}\) - 3
D
-\(\frac{1}{2}\)
E
\(\frac{-\sqrt{3}}{\sqrt{2} - \sqrt{2}}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) - \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\)
\(\frac{\sqrt{2}}{\sqrt{3} - \sqrt{2}}\) = \(\frac{\sqrt{2}}{\sqrt{3}}\) - \(\frac{\sqrt{3}}{\sqrt{2}}\)
\(\frac{\sqrt{3} + \sqrt{2}}{3 + \sqrt{2}}\) = \(\sqrt{6}\) + 2
\(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) = \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) x \(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
= 5 - 2\(\sqrt{6}\)
\(\sqrt{6}\) + 2 - (5 - 2 \(\sqrt{6}\)) = \(\sqrt{6}\) + 2 - 5 + 2\(\sqrt{6}\)
= 3\(\sqrt{6}\) - 3
= 3(\(\sqrt{6}\) - 1)
Source: Myschool Past Questions Archive (Record #32582)
Mathematics
UTME / JAMB 1981
Linear, Simultaneous & Quadratic Equations
Solution
Solve \(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
A
x = -1 or 3
B
x = 1 or 3
C
x = 1 or -5
D
x = -1 or 5
E
x = -1 or -3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{1}{x + 1}\) - \(\frac{1}{x + 3}\) = \(\frac{1}{4}\)
\(\frac{x + 3 - x - 1}{(x + 1)(x + 3)}\) = \(\frac{1}{4}\)
\(\frac{2}{x^2 + 4x + 3}\) = \(\frac{1}{4}\)
= x2 + 4x + 3 = 8
x2 + 4x - 5 = 0
= (x - 1)(x + 5) = 0
x = 1 or -5
Source: Myschool Past Questions Archive (Record #32583)
Mathematics
UTME / JAMB 1981
Plane Geometry, Polygons & Circle Theorems
Solution
In \(\bigtriangleup\)XYZ, XY = 3cm, XZ = 5cm and YZ = 7cm. If the bisector of XYZ meets XZ at W, what is the length of XW?
A
1.5cm
B
2.5cm
C
3cm
D
4cm
E
None of the above
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Let XW = a, a :7
\(\frac{a}{5 - a}\) = \(\frac{3}{7}\)
7a = 15 - 3a
10a = 15
a = \(\frac{15}{10}\)
= \(\frac{3}{2}\)
= 1.5cm
Source: Myschool Past Questions Archive (Record #32584)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
If \(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\), find y when x = 4.
A
8
B
\(\sqrt{65}\)
C
4\(\sqrt{2}\)
D
3
E
1
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(\log_{2} y = 3 - \log_{2} x^{\frac{3}{2}}\)
When x = 4,
\(\log_{2} y = 3 - \log_{2} 4^{\frac{3}{2}}\)
\(\log_{2} y = 3 - \log_{2} 2^{3}\)
\(\log_{2} y = 3 - 3 \log_{2} 2 = 3 - 3 = 0\)
\(\log_{2} y = 0 \implies y = 2^{0} = 1\)
Source: Myschool Past Questions Archive (Record #32585)
Mathematics
UTME / JAMB 1981
Indices, Logarithms & Surds
Solution
Given that 10\(^x\) = 0.2 and log\(_{10}\)2 = 0.3010, find x
A
-1.3010
B
-0.6990
C
0.6990
D
1.3010
E
0.02
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Given that log\(_{10}\)2 = 0.3010, If 10\(^x\) = 0.2
log\(_{10}\)10\(^x\) = log\(_{10}\) 0.2 = x = log\(_{10}\)0.2
But 0.2 = 2 x 10\(^{-1}\), so log\(_{10}\)0.2 = log\(_{10}(2 \times 10^{-1})\), x = log\(_{10}\)2 + log\(_{10}\)10\(^{-1}\)
x = 0.3010 - 1
x = - 0.6990
Source: Myschool Past Questions Archive (Record #32586)
Mathematics
UTME / JAMB 1981
Linear, Simultaneous & Quadratic Equations
Solution
Two cars X and Y start at the same point and travel towards a point P which is 150km away. If the average speed of Y is 60km per hour and x arrives at P 25 minutes earlier than Y. What is the average speed of X?
A
51\(\frac{3}{7}\)km per hour
B
72km per hour
C
37\(\frac{1}{2}\) km per hour
D
66km per hour
E
75km per hour
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Distance travelled by both x and y = 150 km.
Speed of y = 60km per hour, Time = \(\frac{dist.}{Speed}\)
Time spent by y = \(\frac{150}{60}\)
= 2\(\frac{3}{6}\)hrs
= \(\frac{1}{2}\) hrs
if x arrived at p 25 minutes earlier than y, then time spent by x = 2 hour 5 mins
= 125 mins
x average speed = \(\frac{150}{125}\) x \(\frac{60}{1}\)
= \(\frac{9000}{125}\)
= 72 km\hr
Source: Myschool Past Questions Archive (Record #32587)