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Mathematics
UTME / JAMB 1982
Number and Numeration
Solution
Two distinct sectors i the same circle substend 100o and 30o respectively at the centre of the circle. Their corresponding arcs are in ratio
A
10:3
B
1:100
C
3:1
D
5:2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The distinct sectors i the same circle substend 100o and 30o respectively at the centre of the circle. Their corresponding arcs are in ratio 100:30 = 10:3
Source: Myschool Past Questions Archive (Record #32544)
Mathematics
UTME / JAMB 1982
Number and Numeration
Diagram
Solution
TP and TQ are tangents to a circle centre 0, and r is a point on the circumference of the circle as shown in the figure. If angle PTQ = 45º, what is the magnitude of the angle PRO?
Attached Figure / Diagram
A
45\(^0\)
B
135\(^0\)
C
40\(^0\)
D
67\(\frac{1}{2}\)\(^0\)
E
52\(\frac{1}{2}\)\(^0\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\overline{PT}\) = \(\overline{PQ}\) ⇒ tangents from an external point are equal
\(T\hat{P}Q\) = \(T\hat{Q}P\) ⇒ base
Source: Myschool Past Questions Archive (Record #32764)
Mathematics
UTME / JAMB 1981
Number and Numeration
Solution
The ratio of the price of a loaf of bread to the price of a packet of sugar in 1975 was r : t. In 1980, the price of a loaf went up by 25% and that of a packet of sugar went up by 10%. Their new ratio is now
A
40r:50t
B
44r : 50t
C
50r : 44t
D
44r:55t
E
55r:44t
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Ratio of bread to sugar = r:t
25% increase in bread = \(\frac{125r}{100}\)
10% increase in sugar = \(\frac{100t}{100}\)
New ratio = \(\frac{125r}{100}\):\(\frac{110t}{100}\)
= 25r:22t
= 50r:44t
Source: Myschool Past Questions Archive (Record #32548)
Mathematics
UTME / JAMB 1981
Number and Numeration
Solution
Find the value of x satisfying \(\frac{x}{2}\) - \(\frac{1}{3}\) < \(\frac{2x}{5}\) + \(\frac{1}{6}\)
A
x < 5
B
x < 7\(\frac{1}{2}\)
C
x > 5
D
x > 7\(\frac{1}{2}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{x}{2} - \frac{1}{3} < \frac{2x}{5} + \frac{1}{6}\)
\(\frac{x}{2} - \frac{2x}{5} < \frac{1}{6} + \frac{1}{3}\)
\(\frac{x}{10} < \frac{1}{2}\)
\(2x < 10 \implies x < 5\)
Source: Myschool Past Questions Archive (Record #32550)
Mathematics
UTME / JAMB 1981
Number and Numeration
Solution
Which of the following fractions is less than one-third?
A
\(\frac{22}{63}\)
B
\(\frac{4}{11}\)
C
\(\frac{15}{46}\)
D
\(\frac{33}{98}\)
E
\(\frac{122}{303}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
All others are greater than 0.333 when converted to their fractions except \(\frac{15}{46}\)
Source: Myschool Past Questions Archive (Record #32552)
Mathematics
UTME / JAMB 1981
Number and Numeration
Solution
A cuboid has a diagonal of length 9cm and a square base of side 4cm. What is its height?
A
9cm
B
\(\sqrt{65}\)cm
C
\(4\sqrt{2}\)cm
D
7cm
E
6.5cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
The formula for calculating the diagonal of a cuboid is given by : \(\sqrt{l^2 + w^2 + h^2}\)
where w = width, l = length and h = height
9 = \(\sqrt{l^2 + w^2 + h^2}\)
81 = 4\(^2\) + 4\(^2\) + h\(^2\)
81 - 16 - 16 = h\(^2\)
h\(^2\) = 49
h = \(\sqrt{49}\) = 7cm
NOTE:[l = w = 4 (square base)]
Source: Myschool Past Questions Archive (Record #32553)
Mathematics
UTME / JAMB 1981
Number and Numeration
Solution
Evaluate correct to 4 decimal places 827.51 x 0.015
A
8.8415
B
12.4127
C
124.1265
D
12.4120
E
114.1265
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
827.51 x 0.015
By normal multiplication or use of four figure table,
827.51 x 0.015
= 12.4127 (to 4 decimal places).
Source: Myschool Past Questions Archive (Record #32554)
Mathematics
UTME / JAMB 1981
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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
Number and Numeration
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)