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Mathematics
UTME / JAMB 1997
Permutations, Combinations & Probability
Solution
In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges, if one fruit is chosen at random, what is the probability that the fruit is either a grape or a banana?
A
\(\frac{17}{30}\)
B
\(\frac{11}{30}\)
C
\(\frac{6}{30}\)
D
\(\frac{5}{30}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Pgrape or Pbanana = \(\frac{6}{30}\) + \(\frac{11}{30}\)
= \(\frac{17}{30}\)
Source: Myschool Past Questions Archive (Record #30867)
Mathematics
UTME / JAMB 1997
Permutations, Combinations & Probability
Solution
A number is selected at random between 10 and 20, both numbers inclusive. Find the probability that the number is an even number
A
\(\frac{5}{11}\)
B
\(\frac{1}{2}\)
C
\(\frac{6}{11}\)
D
\(\frac{7}{10}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
n(even numbers between 10 and 20 inclusive) = 6
n(numbers between 10 and 20) = 11
P(even) = \(\frac{6}{11}\)
Source: Myschool Past Questions Archive (Record #30868)
Mathematics
UTME / JAMB 1997
Indices, Logarithms & Surds
Solution
Find the standard derivation of the following data -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
A
2
B
3
C
\(\sqrt{10}\)
D
\(\sqrt{11}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
x = \(\frac{\sum x}{N}\)
= \(\frac{0}{11}\)
= 0
\(\begin{array}{c|c} x & (x - x) & (x - x)^2 \\\hline -5 & -5 & 25 \\ -4 & -4 & 16 \\-3 & -3 & 9 \\ -2 & -2 & 4 \\ -1 & -1 & 1\\ 0 & 0 & 0\\ 1 & 1 & 1\\ 2 & 2 & 4\\ 3 & 3 & 9\\ 4 & 4 & 16 \\5 & 5 & 25\\ \hline & & 110\end{array}\)
S.D = \(\sqrt{\frac{\sum(x - x)^2}{\sum f}}\)
= \(\sqrt{\frac{110}{11}}\)
= \(\sqrt{10}\)
Source: Myschool Past Questions Archive (Record #30870)
Mathematics
UTME / JAMB 1997
Indices, Logarithms & Surds
Diagram
The figure shows circles of radii 3cm and 2cm with centres at X and Y respectively. The circles have a transverse common tangent of length 25cm. Calculate XY
Attached Figure / Diagram
A
\(\sqrt{630}\)cm
B
\(\sqrt{626}\)cm
C
\(\sqrt{615}\)cm
D
\(\sqrt{600}\)cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #33069)
Mathematics
UTME / JAMB 1997
Mensuration (Areas, Perimeters & Volumes)
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
Trigonometry, Elevation & Bearings
Diagram
Solution
In the figure, XYZ is a triangle with XY = 5cm, XZ = 2cm and XZ is produced to E making the angle < YZE = 150o. If the angle XYZ = \(\theta\), calculate the value of sin\(\theta\)
Attached Figure / Diagram
A
\(\frac{3}{5}\)
B
\(\frac{1}{2}\)
C
\(\frac{2}{5}\)
D
\(\frac{1}{5}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{5}{\sin 30^o} = \frac{2}{\sin \theta} = \frac{5}{\frac{1}{2}} = \frac{2}{\sin \theta}\)
10 = \(\frac{2}{\sin \theta}\)
sin\(\theta\) = \(\frac{2}{10}\)
= \(\frac{1}{5}\)
Source: Myschool Past Questions Archive (Record #33071)
Mathematics
UTME / JAMB 1997
Statistics (Data Representation, Central Tendency & Dispersion)
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
Statistics (Data Representation, Central Tendency & Dispersion)
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
Indices, Logarithms & Surds
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
Sets, Logic & Venn Diagrams
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
Trigonometry, Elevation & Bearings
Solution
(a) Given that \(\sin x = \frac{5}{13}, 0^\circ \leq x \leq 90^\circ\), find \(\frac{\cos x - 2 \sin x }{2\tan x}\).
(b)
The diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30^\circ and 70^\circ respectively and PR = 500m, calculate, correct to 3 significant figures :
(i) |QP| ; (ii) the height of the mountain.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
\(\sin x = \frac{5}{13}\)
\(AB = \sqrt{13^{2} - 5^{2}} = 12\)
\(\cos x = \frac{12}{13}\)
\(\tan x = \frac{5}{12}\)
\(\frac{\cos x - 2 \sin x}{2 \tan x} = \frac{\frac{12}{15} - 2(\frac{5}{13})}{2(\frac{5}{12})}\)
= \(\frac{\frac{12}{13} - \frac{10}{13}}{\frac{5}{6}}\)
= \(\frac{2}{13} \times \frac{6}{5} = \frac{12}{65}\)
(b) (i)
< QRP = 180^\circ - 70^\circ = 110^\circ (angles on a straight line)
< RQP = 180^\circ - (110^\circ + 30^\circ) = 40^\circ
\(\frac{PR}{\sin 40^\circ} = \frac{PQ}{\sin 110^\circ}\)
\(\frac{500}{\sin 40^\circ} = \frac{PQ}{\sin 110^\circ}\)
\(PQ = \frac{500 \sin 110^\circ}{\sin 40^\circ} = 730.94m\)
\(\approxeq 731 m\)
(ii) \(\sin 30^\circ = \frac{QN}{PQ} = \frac{QN}{730.94}\)
\(QN = 730.94 \times 0.5 = 365.47 m\)
\(\approxeq 365 m\)
Source: Myschool Past Questions Archive (Record #54471)
Mathematics
WAEC / WASSCE 1997
Fractions, Decimals, Percentages & Approximations
Solution
The table below shows how a company's sales manager spent his 1995 annual salary.
Food
30%
Rent
18%
Car Maintenance
25%
Savings
12%
Taxes
5%
Others
10%
(a) Represent this information on a pie chart.
(b) Find his savings at the end of the year if his annual salary was N60,000.00.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Angle of sector:
\(Food = \frac{30}{100} \times 360^\circ = 108^\circ\)
\(Rent = \frac{18}{100} \times 360^\circ = 64.8^\circ\)
\(\text{Car Maintenance} = \frac{25}{100} \times 360^\circ = 90^\circ\)
\(Savings = \frac{12}{100} \times 360^\circ = 43.2^\circ\)
\(Taxes = \frac{5}{100} \times 360^\circ = 18^\circ\)
\(Others = \frac{10}{100} \times 360^\circ = 36^\circ\)
(a)
(b) Annual Salary = N60,000.00
\(\therefore Savings = N(\frac{12}{100} \times 60,000) = N7,200\)
Source: Myschool Past Questions Archive (Record #54472)
Mathematics
WAEC / WASSCE 1997
Fractions, Decimals, Percentages & Approximations
Solution
(a) Given that \(\frac{5y - x}{8y + 3x} = \frac{1}{5}\), find the value of \(\frac{x}{y}\) to two decimal places.
(b) If 3 is a root of the quadratic equation \(x^{2} + bx - 15 = 0\), determine the value of b. Find the other root.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{5y - x}{8y + 3x} = \frac{1}{5}\)
\(5(5y - x) = 8y + 3x \implies 25y - 5x = 8y + 3x\)
\(25y - 8y = 3x + 5x \implies 17y = 8x\)
\(\therefore \frac{x}{y} = \frac{17}{8} = 2.125 \approxeq 2.13\)
(b) \(x^{2} + bx - 15 = 0\)
Since x = 3 is a root of the equation, f(3) = 0.
\(3^{2} + 3b - 15 = 0 \implies 3b = 6\)
\(b = 2\)
\(\therefore\) The equation is \(x^{2} + 2x - 15 = 0\)
\(x^{2} - 3x + 5x - 15 = 0 \implies x(x - 3) + 5(x - 3) = 0\)
\((x - 3)(x + 5) = 0\)
The second root of the equation is x = -5.
Source: Myschool Past Questions Archive (Record #54476)
Mathematics
WAEC / WASSCE 1997
Indices, Logarithms & Surds
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
Plane Geometry, Polygons & Circle Theorems
Solution
(a) Using a ruler and a pair of compasses only, construst \(\Delta\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75^\circ. Measure |AC|.
(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
|AC| = 8.35 cm, < BAD = 70^\circ.
Source: Myschool Past Questions Archive (Record #54482)