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Mathematics
WAEC / WASSCE 1995
Indices, Logarithms & Surds
Solution
(a) Using mathematical tables, find ; (i) \(2 \sin 63.35^\circ\) ; (ii) \(\log \cos 44.74^\circ\);
(b) Find the value of K given that \(\log K - \log (K - 2) = \log 5\);
(c) Use logarithm tables to evaluate \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i) \(\sin 63.35^\circ = 0.8938\)
\(2 \sin 63.35^\circ = 2 \times 0.8938 = 1.7876\)
(ii) \(\log \cos 44.74^\circ = \bar{1}.8514\)
(b) \(\log K - \log (K - 2) = \log 5\)
\(\log (\frac{K}{K - 2}) = \log 5\)
\(\therefore \frac{K}{K - 2} = 5 \implies K = 5(K - 2)\)
\(K = 5K - 10 \implies 10 = 5K - K = 4K\)
\(K = 2\frac{1}{2}\)
(c) \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)
No
Log
\((3.68)^{2}\)
\(0.5658 \times 2 = 1.316\)
\(6.705\)
\(0.8264 = 0.8264\)
= \(1.9580\)
\(\sqrt{0.3581}\)
\(\bar{1}.5540 \div 2 = \bar{1}.7770\)
\(151.7\)
\(\gets 2.1810\)
\(\therefore \frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}} = 151.7\)
Source: Myschool Past Questions Archive (Record #54526)
Mathematics
WAEC / WASSCE 1995
Linear, Simultaneous & Quadratic Equations
Solution
(a) Given that \(p = x + ym^{3}\), find m in terms of p, x and y.
(b) Using the method of completing the square, find the roots of the equation \(x^{2} - 6x + 7 = 0\), correct to 1 decimal place.
(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(p = x + ym^{3}\)
\(ym^{3} = p - x\)
\(m^{3} = \frac{p - x}{y}\)
\(m = \sqrt[3]{\frac{p - x}{y}}\)
(b) \(x^{2} - 6x + 7 = 0\)
\(\implies x^{2} - 6x = -7\)
\(x^{2} - 6x + (-3)^{2} = -7 + (-3)^{2}\)
\(x^{2} - 6x + 9 = 2\)
\((x - 3)^{2} = 2 \implies x - 3 = \pm \sqrt{2}\)
\(x = 3 \pm \sqrt{2} = 3 \pm 1.414\)
\(x = 4.414 ; 1.586\)
\(x \approxeq 4.4 ; 1.6\)
(c) Let the first odd number be z.
The next positive odd number = z + 2
\(\therefore z(z + 2) = 195\)
\(z^{2} + 2z = 195\)
\(z^{2} + 2z - 195 = 0 \implies z^{2} - 13z + 15z - 195 = 0\)
\(z(z - 13) + 15(z - 13) = 0 \implies \text{z = 13 or -15}\)
Since we are told the odd numbers are positive, then z = 13 is the suitable answer.
Hence the second odd number = 13 + 2 = 15.
The two odd numbers are 13 and 15.
Source: Myschool Past Questions Archive (Record #54527)
Mathematics
WAEC / WASSCE 1995
Trigonometry, Elevation & Bearings
Solution
(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = 2\cos 2x - 1\)
1.0
0
1.0
(b) Using a scale of 2cm = 30^\circ on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0^\circ \leq x \leq 180^\circ\).
(c) On the same axis, draw the graph of \(y = \frac{1}{180} (x - 360)\)
(d) Use your graphs to find the : (i) values of x for which \(2 \cos 2x + \frac{1}{2} = 0\); (ii) roots of the equation \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = 2\cos 2x - 1\)
1.0
0
-2
-3
-2
0
1.0
(b)
(c)
x
0^\circ
30^\circ
60^\circ
90^\circ
120^\circ
150^\circ
180^\circ
\(y = \frac{1}{180} (x - 360)\)
-2.0
-1.83
-1.67
-1.5
-1.33
-1.17
-1.0
(d)(i) \(2 \cos 2x + \frac{1}{2} = 0\)
\(2 \cos 2x + \frac{1}{2} - 1\frac{1}{2} = -1\frac{1}{2}\)
\(2 \cos 2x - 1 = -1\frac{1}{2}\)
The values of x at \(2 \cos 2x + \frac{1}{2} = 0\) are x = 51^\circ and 129^\circ.
(ii) \(2 \cos 2x - \frac{x}{180} + 1 = 0\)
\(2 \cos 2x - 1 = \frac{1}{180} (x - 360)\)
The values of x is where the two graphs intersect; which are at x = 54^\circ and 132^\circ.
Source: Myschool Past Questions Archive (Record #54528)
Mathematics
WAEC / WASSCE 1995
Plane Geometry, Polygons & Circle Theorems
Solution
(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105^\circ.
(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.
(c) Through D, construct a line I perpendicular to BC.
(d) If the line I meets AC at P, measure /BP/.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a - d)
Source: Myschool Past Questions Archive (Record #54529)
Mathematics
WAEC / WASSCE 1995
Trigonometry, Elevation & Bearings
Solution
(a) A man travels from a village X on a bearing of 060^\circ to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195^\circ. If Z is directly east of X, calculate, correct to three significant figures, the distance of :
(i) Y from Z ; (ii) Z from X .
(b) An aircraft flies due South from an airfield on latitude 36^\circN, longitude 138^\circE to an airfield on latitude 36^\circS, longitude 138^\circE.
(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.
[Take \(\pi = \frac{22}{7}\), R = 6400km].
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) < XZY = 180^\circ - (30^\circ + 45^\circ) = 105^\circ
\(\frac{20}{\sin 105^\circ} = \frac{YZ}{\sin 30^\circ}\)
\(YZ = \frac{20 \sin 30^\circ}{\sin 105^\circ} = 10.35km\)
\(\approxeq 10.4km \) (to 3 sig. figs)
(ii) \(\frac{20}{\sin 75} = \frac{XZ}{\sin 45}\)
\(XZ = \frac{20 \sin 45}{\sin 75} = 14.64km\)
\(\approxeq 14.6km\) ( to 3 sig. figs)
(b) Let P and Q be two airfields.
Difference in latitude = 36^\circ + 36^\circ = 72^\circ
(i) Distance between P and Q = \(\frac{72}{360} \times 2 \times \frac{22}{7} \times 6400\)
= \(\frac{22}{7} \times 2560\)
= \(8,045.71km \approxeq 8,050 km\) (3 sig. figs)
(ii) Time taken = \(\frac{8,045.71}{800} = 10.06 hours\)
\(\approxeq 10hours\) (to the nearest hour)
Source: Myschool Past Questions Archive (Record #54530)
Mathematics
WAEC / WASSCE 1995
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.
% Mark
11-20
21-30
31-40
41-50
51-60
61-70
71-80
81-90
No of
pupils
68
184
294
402
480
310
164
98
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a)(i)
Marks (x)
No of pupils (f)
Cum Freq
11 - 20
68
68
21 - 30
184
252
31 - 40
294
546
41 - 50
402
948
51 - 60
480
1428
61 - 70
310
1738
71 - 80
164
1902
81 - 90
98
2000
(ii)
(b)(i) Cut off mark = 68 marks for reading at 1700.
(ii) Candidates scoring more than 45% = 2000 - 720 = 1280
\(\therefore\) P(scoring at least 45%) = \(\frac{1280}{2000} = 0.64\)
Source: Myschool Past Questions Archive (Record #54531)
Mathematics
WAEC / WASSCE 1995
Permutations, Combinations & Probability
Solution
A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining
(a) two red balls ;
(b) two blue balls or two black balls ;
(c) one black and one red ball in any order.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
Number of blue balls = 5
Number of black balls = 3
Number of red balls = 2
= 10 balls.
(a) P(two red balls) = \(\frac{2}{10} \times \frac{2}{10} = \frac{1}{25}\)
(b) P(2 blue or 2 black balls) = P(2 blue) or P(2 black)
= \((\frac{5}{10} \times \frac{5}{10}) + (\frac{3}{10} \times \frac{3}{10})\)
= \(\frac{25}{100} + \frac{9}{100}\)
= \(\frac{34}{100} = \frac{17}{50}\)
(c) P( one red and one black in any order) = P(one red, one black) or P(one black, one red)
= \((\frac{2}{10} \times \frac{3}{10}) + (\frac{3}{10} \times \frac{2}{10})\)
= \(\frac{12}{100} = \frac{3}{25}\)
Source: Myschool Past Questions Archive (Record #54532)
Mathematics
WAEC / WASSCE 1994
Sequences & Series (AP & GP)
Solution
If the 2nd and 5th terms of a G.P are 6 and 48 respectively, find the sum of the first four for term
A
-45
B
-15
C
15
D
33
E
45
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
\(T_2\) = ar = 6
\(T_5\) = ar\(^{4}\) = 48
\(\frac{T_5}{T_2}\) = \(\frac{ar^{4}}{ar}\) = \(\frac{48}{6}\)
= r\(^{3}\) = 8
=> r = 2
\(S_n\) = \(\frac{a((r^n) - 1)}{r - 1}\)
\(S_4\) = \(\frac{a((r^4) - 1)}{r - 1}\)
\(S_4\) = \(\frac{3((2^4) - 1)}{2 - 1}\)
= 3(16 -1)
= 45
Source: Myschool Past Questions Archive (Record #21978)
Mathematics
WAEC / WASSCE 1994
Inequalities & Linear Programming
Solution
If sin\( \theta \) = K find tan\(\theta\), 0^\circ \(\leq\) \(\theta\) \(\leq\) 90^\circ.
A
1-K
B
\( \frac{k}{k - 1} \)
C
\( \frac{k}{\sqrt{1 - k^2}} \)
D
\( \frac{k}{1 - k} \)
E
\( \frac{k}{\sqrt{ k^2 - 1}} \)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\sin \theta = \frac{k}{1}\)
\(\implies 1^2 = k^2 + adj^2\)
\(adj = \sqrt{1 - k^2}\)
\(\therefore \tan \theta = \frac{k}{\sqrt{1 - k^2}}\)
Source: Myschool Past Questions Archive (Record #22006)
Mathematics
UTME / JAMB 1994
Binary Operations, Matrices & Determinants
Solution
The determination of the matrix \(\begin{pmatrix} 1 & 3 & 3 \\ 4 & 5 & 6\\ 2 & 0 & 1 \end{pmatrix}\) is
A
-67
B
-57
C
-1
D
3
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
((1 x5x1) +( 3x6x2) + (3x4x0)) - ((3x5x2) + (1x6x0) + (3x4x1))
( 5 + 36) - (30 +12) = 41 - 42
= -1
Source: WAEC / JAMB / NECO Classroom Archive (Record #31019)
Mathematics
UTME / JAMB 1994
Algebraic Expressions, Polynomials & Factorisation
Solution
Factorize a2x - b2y - b2x + a2y
A
(a - b)(x + y)
B
(y - x)(a - b)(a + b)
C
(x - y)(a - b)(a + b)
D
(x + y)(a - b)(a + b)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
a2x - b2y - b2x + a2y = a2x - b2x - b2y + a2y Rearrange
= x(a2 - b2) + y(a2 - b2)
= (x + y)(a2 - b2)
= (x + y)(a + b)(a - b)
Source: WAEC / JAMB / NECO Classroom Archive (Record #31007)
Mathematics
WAEC / WASSCE 1994
Plane Geometry, Polygons & Circle Theorems
Solution
Simplify: \(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)
A
0
B
1
C
2
D
3
E
4
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(2\frac{1}{3} \div 2\frac{2}{3} \times 1\frac{1}{7}\)
= \(\frac{7}{3} \div \frac{8}{3} \times \frac{8}{7}\)
= \(\frac{7}{3} \times \frac{3}{8} \times \frac{8}{7}\)
= 1
Source: Myschool Past Questions Archive (Record #23459)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
Which of the following is equal to \(\frac{72}{125}\)
A
\( \frac{2^3 \times 3^2}{5^3} \)
B
\( \frac{2^4 \times 3}{5^3} \)
C
\( \frac{2^3 \times 3}{5^3} \)
D
\( \frac{2^4 \times 3}{5^5} \)
E
\( \frac{2^2 \times 3^2 \times 4^2}{5^2} \)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{72}{125} = \frac{8 \times 9}{5 \times 5 \times 5}\)
= \(\frac{2^3 \times 3^2}{5^3}\)
Source: Myschool Past Questions Archive (Record #23460)
Mathematics
WAEC / WASSCE 1994
Fractions, Decimals, Percentages & Approximations
Solution
Express in \( \frac{8.75}{0.025} \)standard form
A
3.5 x 10-3
B
3.5 x 10-2
C
3.5 x 101
D
3.5 x 102
E
3.5 x 103
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{8.75}{0.025}\)
= \(\frac{8750}{25}\)
= \(350\)
= \(3.5 \times 10^2\)
Source: Myschool Past Questions Archive (Record #23461)
Mathematics
WAEC / WASSCE 1994
Indices, Logarithms & Surds
Solution
Evaluate \( \frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
A
6
B
5
C
4
D
3
E
2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\frac{27^{\frac{1}{3}}}{16^{-\frac{1}{4}}} \)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^4)^{-\frac{1}{4}}}\)
= \(\frac{3}{2^{-1}}\)
= 6
Source: Myschool Past Questions Archive (Record #23463)