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Mathematics
UTME / JAMB 1997
Number and Numeration
Solution
A cone with a sector angle of 45^\circ is cut out of a circle of radius r cm. Find the base radius of the cone.
A
\(\frac{r}{16}\) cm
B
\(\frac{r}{6}\) cm
C
\(\frac{r}{8}\) cm
D
\(\frac{r}{2}\) cm
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
The formula for the base radius of a cone formed from the sector of a circle = \(\frac{r \theta}{360^\circ}\)
= \(\frac{r \times 45^\circ}{360^\circ}\)
= \(\frac{r}{8} cm\)
Source: Myschool Past Questions Archive (Record #30825)
Mathematics
UTME / JAMB 1997
Number and Numeration
Diagram
Solution
Find the volume of the prism
Attached Figure / Diagram
A
990cm3
B
880cm3
C
550cm3
D
495cm3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Volume of prism = base area x height
Base area = area of a trapezium = \(\frac{1}{2}(a + b)\times h = \frac{1}{2}(10 + 8)\times 5 = 9 \times 5 = 45cm^2\)
Height = 11cm
Therefore, volume of prism = 45 x 11 = \(495cm^3\)
Source: Myschool Past Questions Archive (Record #33070)
Mathematics
UTME / JAMB 1997
Number and Numeration
Diagram
Solution
The pie chart shows the income of a civil servant in a month. If his monthly income is N6000, find his monthly basic salary
Attached Figure / Diagram
A
N2050
B
N2600
C
N3100
D
N3450
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
360º - (60º + 60º + 67º + 50º = 237º)
360º - 237º = 123º
B. Salary = \(\frac{123}{360}\times N6000\)
= N2,050
Source: Myschool Past Questions Archive (Record #33072)
Mathematics
UTME / JAMB 1997
Number and Numeration
Diagram
Solution
In an examination, the result of a certain school is as shown in the histogram above. How many candidates did the school present?
Attached Figure / Diagram
A
12
B
16
C
18
D
19
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
3 + 5 + 8 + 1 + 2 = 19
Source: Myschool Past Questions Archive (Record #33073)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
(a) Copy and complete the binary multiplication table:
x
10
11
100
101
10
100
1000
11
110
1100
100
10000
10100
(b) Convert \(11.011_{two}\) to a number in base ten.
(c) Simplify \(\frac{9.6 \times 10^{18}}{0.24 \times 10^{5}}\) and express your answer in the form \(P \times 10^{m}\) where 1 < P < 10 and m is an integer.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
x
10
11
100
101
10
100
110
1000
1010
11
110
1001
1100
1111
100
1000
1100
10000
10100
(b) \(11.011_{two} = 1 \times 2^{1} + 1 \times 2^{0} + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3}\)
= \(2 + 1 + 0 + \frac{1}{4} + \frac{1}{8}\)
= \(3\frac{3}{8}\)
= \(3.375\)
(c) \(\frac{9.6 \times 10^{18}}{0.24 \times 10^{5}} \equiv \frac{9.6 \times 10^{18}}{2.4 \times 10^{4}}\)
= \(\frac{9.6}{2.4} \times 10^{18 - 4}\)
= \(4 \times 10^{14}\)
Source: Myschool Past Questions Archive (Record #54467)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.
(b)
The Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \(P \cap Q' \cap R\) ; (ii) \(P' \cap Q \cap R'\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P)
\(T_{6} = a + 5d = 35 ... (1)\)
\(T_{13} = a + 12d = 77 ... (2)\)
(2) - (1) : \(7d = 42 \implies d = 6\)
\(a + 5(6) = 35 \implies a + 30 = 35\)
\(a = 5\)
\(\therefore T_{20} = a + 19d = 5 + 19(6)\)
= \(5 + 114 = 119\)
(b)
Source: Myschool Past Questions Archive (Record #54470)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\).
(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :
(i) both were boys ; (ii) the two were of different sexes.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\)
No
Log
15.05
\(1.1775 = 1.1775\)
\(\sqrt{0.00695}\)
\(\bar{3}.8420 \div 2 = \bar{2}.9210\)
= 0.0985
\(6.95 \times 10^{2}\)
- 2.8420
Antilog = 0.001805
= \(\bar{3}.2565\)
\(\therefore \frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}} \approxeq 0.00181\) (3 sig. figs)
(b) No of boys = 2, No of girls = 3
\(\therefore\) Total students = 5
(i) P(both are boys) = \(\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)
(ii) P(both are of different sexes) = P(first a boy, then a girl) or P(first a girl, then a boy)
= \(\frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} \)
= \(\frac{6}{20} + \frac{6}{20}\)
= \(\frac{12}{20} = \frac{3}{5}\)
Source: Myschool Past Questions Archive (Record #54481)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
The solid is a cylinder surmounted by a hemispherical bowl. Calculate its
(a) total surface area ;
(b) volume (Take \(\pi = \frac{22}{7}\))
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) Area of curved surface of cylinder = \(2\pi r h\)
= \(2 \times \frac{22}{7} \times 7 \times 10cm^{2} = 440 cm^{2}\)
Area of the base of cylinder = \(\pi r^{2}\)
= \(\frac{22}{7} \times 7 \times 7 = 154 cm^{2}\)
Surface area of hemisphere = \(\frac{4\pi r^{2}}{2} = 2 \pi r^{2}\)
= \(2 \times \frac{22}{7} \times 7 \times 7 = 308 cm^{2}\)
\(\therefore\) Total surface area = 440 + 154 + 308 = 902cm\(^{2}\)
(b) Volume of cylinder = \(\pi r^{2} h\)
= \(\frac{22}{7} \times 7 \times 7 \times 10 = 1540 cm^{3}\)
Volume of hemisphere = \(\frac{1}{2}(\frac{4\pi r^{3}}{3})\)
= \(\frac{4 \times 22 \times 7 \times 7 \times 7}{2 \times 7 \times 3}\)
= \(718.67 cm^{3}\)
Total volume = 1540 + 718.67 = 2258.67 cm\(^{3}\).
Source: Myschool Past Questions Archive (Record #54491)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
(a) PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56^\circ. Find (i) SQT (ii) PQT.
(b) In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) From the figure < QRS = 180^\circ - 56^\circ = 124^\circ (opposite angles of a parallelogram)
\(\therefore\) < RQS = < PQS = \(\frac{180^\circ - 124^\circ}{2} = 28^\circ\) (base angles of an isosceles triangle)
\(\therefore\) < SQT = < RSQ = 28^\circ (alternate angles PS // QT)
(ii) < PQT = 90^\circ - 28^\circ = 62^\circ (< PQS is in a semi-circle)
(b)
\(\tan < BDC = \frac{30}{40} = 0.75\)
\(< BDC = \tan^{-1} (0.75) = 36.87^\circ\)
\(\therefore < BDF = 90^\circ - 36.87^\circ = 53.13^\circ \approxeq 53^\circ\)
\(\tan < ADF = \frac{14}{30} = 0.4667\)
\(< ADF = \tan^{-1} (0.4667) = 25.02^\circ \approxeq 25^\circ\)
Source: Myschool Past Questions Archive (Record #54493)
Mathematics
WAEC / WASSCE 1997
Number and Numeration
Solution
The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.
Marks (in %)
Freq
44 - 46
2
47 - 49
5
50 - 52
11
53 - 55
20
56 - 61
42
62 - 64
46
65 - 67
36
68 - 70
9
71 - 73
3
(a) Make a cumulative frequency for the distribution
(b) Draw the cumulative frequency curve.
(c) From your graph, estimate the median mark.
(d) The cut-off mark was 63%. What percentage of the candidates was selected?
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
Marks (in %)
Freq
Class boundaries
Cum Freq
44 - 46
2
43.5 - 46.5
2
47 - 49
5
46.5 - 49.5
7
50 - 52
11
49.5 - 52.5
18
53 - 55
20
52.5 - 55.5
38
56 - 61
42
55.5 - 61.5
80
62 - 64
46
61.5 - 64.5
126
65 - 67
36
64.5 - 67.5
162
68 - 70
9
67.5 - 70.5
171
71 - 73
3
70.5 - 73.5
174
Source: Myschool Past Questions Archive (Record #54494)
Mathematics
WAEC / WASSCE 1996
Number and Numeration
Diagram
Solution
In the diagram, ∠PMN = ∠PRQ and ∠PNM = ∠PQR. If /Pm/ = 3cm, /MQ/ = 7cm and /PN/ = 5cm, find /NR/
Attached Figure / Diagram
A
1cm
B
3cm
C
31/2
D
5cm
E
6cm
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
△PQR is similar to △PNM
\(\frac{PQ}{PN}\) = \(\frac{PR}{PM}\)
\(\frac{7 + 3 }{5}\) \(\frac{5 + NR}{3}\)
cross multiply
10 x 3 = (5 x 5) + 5 x NR
5NR = 30 - 25 = 5
NR = \(\frac{5}{5}\) = 1 cm.
Source: WAEC / JAMB / NECO Classroom Archive (Record #23822)
Mathematics
WAEC / WASSCE 1996
Number and Numeration
Solution
The 6th term of a G.P is \(\frac{-2}{27}\) and its first term is 18. What is the common ratio?
A
-1/2
B
-1/3
C
1/4
D
2
E
3
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
a, ar, ar\(^2\), ar\(^3\), ar\(^4\), ar\(^5\)
6th term = ar\(^5\) = \(\frac{-2}{27}\)
1st term = a = 18
=> ar\(^5\)
= 18r\(^5\) =\(\frac{-2}{27}\)
r\(^5\) = \(\frac{-2}{27 * 18}\)
→ \(\frac{-1}{27 * 9}\)
r\(^5\) = \(\frac{-1}{243}\)
r = \(^5\)\sqrt[\(\frac{-1}{243}\)]
r = \(\frac{-1}{3}\)
Source: Myschool Past Questions Archive (Record #23791)
Mathematics
WAEC / WASSCE 1996
Number and Numeration
Solution
Simplify: \((2\frac{1}{6} - 1\frac{2}{3}) \div 2\frac{2}{3}\).
A
3/16
B
7/16
C
1 13/24
D
2 11/24
E
10 2/5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\((2\frac{1}{6} - 1\frac{2}{3}) \div 2\frac{2}{3}\)
= \((\frac{13}{6} - \frac{5}{3}) \div \frac{8}{3}\)
= \(\frac{3}{6} \times \frac{3}{8}\)
= \(\frac{3}{16}\)
Source: Myschool Past Questions Archive (Record #23793)
Mathematics
WAEC / WASSCE 1996
Number and Numeration
Solution
Simplify 56x\(^{-4}\) \(\div\) 14x\(^{-8}\)
A
2x-12
B
3x-3
C
4x-4
D
4x-3
E
4x4
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(56x^{-4} \div 14x^{-8}\)
= \(4x^{- 4 - (-8)}\)
= \(4x^{4}\)
Source: Myschool Past Questions Archive (Record #23794)
Mathematics
WAEC / WASSCE 1996
Number and Numeration
Solution
Simplify Log\(_{10}\)4 + Log\(_{10}\)25
A
1
B
2
C
3
D
4
E
5
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Log\(_{10}\)4 + Log\(_{10}\)25 = Log\(_{10}\)(4 + 25)
= Log\(_{10}\)100 = Log\(_{10}\)10\(^2\)
= 2Log\(_{10}\)10 = 2 x 1 = 2
Source: Myschool Past Questions Archive (Record #23795)