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Mathematics
WAEC / WASSCE 2007
Trigonometry, Elevation & Bearings
Diagram
Solution
(a) Simplify : \(\frac{x^{2} - y^{2}}{3x + 3y}\)
(b)
In the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(\frac{x^{2} - y^{2}}{3x + 3y}\)
\(\frac{(x + y)(x - y)}{3(x + y)}\) (Using difference of two squares)
= \(\frac{x - y}{3}\)
(b)(i)
\(\sin 37 = \frac{/PS/}{15}\)
\(/PS/ = 15 \times 0.6018\)
= \(9.03 cm\)
(ii) \(\cos 37 = \frac{/SK/}{15}\)
\(/SK/ = 15 \times 0.7986\)
= \(11.98 cm \)
\(\approxeq 12.0 cm\)
(iii) Area of the shaded portion = Area of rectangle PQRS - Area of triangle PSK.
/SR/ = 2(/SK/) = 2(11.98)
= 23.96 cm
Area of rectangle PQRS = \(23.96 \times 9.03 \)
= \(216.3588 cm^{2}\)
Area of triangle PKS = \(\frac{1}{2} \times 11.98 \times 9.03\)
= \(54.088 cm^{2}\)
Area of shaded portion : \((216.3588 - 54.088)cm^{2}\)
\(162.2708 cm^{2}\)
\(\approxeq 162 cm^{2}\)
Source: Myschool Past Questions Archive (Record #52996)
Mathematics
WAEC / WASSCE 2007
Trigonometry, Elevation & Bearings
Solution
(a)
In the diagram, AOB is a straight line. < AOC = 3(x + y)^\circ, < COB = 45^\circ, < AOD = (5x + y)^\circ and < DOB = y^\circ. Find the values of x and y.
(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53^\circ and 67^\circ. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(3(x + y) + 45 = 180^\circ\) ( sum of angles on a straight line)
\(3x + 3y + 45 = 180 \implies 3x + 3y = 135\)
\(3x = 135 - 3y \implies x = 45 - y\)
Also,
\(5x + y + y = 180\)
\(5x + 2y = 180..... (*)\)
Put x = 45 - y into (*) above,
\(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\)
\(225 - 3y = 180 \implies 3y = 45\)
\(y = \frac{45}{3} = 15^\circ\)
\(x = 45 - y\)
\(x = 45 - 15 = 30^\circ\)
(b)
Let the distance BD = x and CD = y.
In \(\Delta ABD\),
\(\tan 53 = \frac{33}{x}\)
\(x = \frac{33}{\tan 53}\)
= \( 24.868 cm\)
In \(\Delta ACD\),
\(\tan 67 = \frac{33}{y}\)
\(y = \frac{33}{\tan 67}\)
\(y = 14.01 cm\)
\(\therefore BC = x + y \)
\(24.868 + 14.01 = 38.878 cm\)
\(\approxeq 38.9 cm\)
Source: Myschool Past Questions Archive (Record #53004)
Mathematics
WAEC / WASSCE 2007
Sequences & Series (AP & GP)
Solution
(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.
(b)
In the diagram \(\overline{PQ} || \overline{YZ}\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \(\Delta\) XPQ = 24\(cm^{2}\).Calculate the area of the trapezium PQZY.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P)
3rd term = -9 ; 8th term = 26
\(\implies T_{3} = a + 2d = -9 ... (1)\)
\(\implies T_{8} = a + 7d = 26 ..... (2)\)
(i) Solving (2) - (1), we have
\(7d - 2d = 26 - (-9)\)
\(5d = 35 \implies d = 7\)
(ii) Putting d = 7 in (1) above, we have
\(a + 2(7) = -9\)
\(a + 14 = -9\)
\(a = - 9 - 14 = -23\)
(b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles)
\(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\)
\(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\)
\(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)
Source: Myschool Past Questions Archive (Record #53006)
Mathematics
WAEC / WASSCE 2007
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table
Number of absentees
0-4
5-9
10-14
15-19
20-24
Number of Days
1
5
10
9
5
Calculate the : (a) Mean
(b) Standard deviation , correct to two decimal places.
medium Difficulty
Verified Correct Answer: Option
Step-by-Step Mathematical Solution
\(x\)
mid-value
\(f\)
\(fx\)
\((x - \bar{x})\)
\((x - \bar{x})^{2}\)
\(f(x - \bar{x})^{2}\)
0 - 4
2
1
2
-12
144
144
5 - 9
7
5
35
-7
49
245
10 - 14
12
10
120
-2
4
40
15 - 19
17
9
153
3
9
81
20 - 24
22
5
110
8
64
320
\(\sum \)
30
420
830
(a) Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{420}{30} = 14\)
(b) Standard deviation = \(\sqrt{\frac{\sum f(x - \bar{x})^{2}}{\sum f}\)
= \(\sqrt{\frac{830}{30}}\)
= \(\sqrt{27.66}\)
= 5.26.
Source: Myschool Past Questions Archive (Record #53009)
Mathematics
UTME / JAMB 2007
Inequalities & Linear Programming
Diagram
Solution
The cumulative frequency for 5\(\leq\) x \(\leq\) 12 from the distribution above is
Attached Figure / Diagram
A
22
B
40
C
46
D
72
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
The cumulative frequency 5\(\leq\) x \(\leq\) 12 = 10 + 12 + 10 + 8 + 6 + 8 + 10 + 8 = 72
Source: Myschool Past Questions Archive (Record #72042)
Mathematics
UTME / JAMB 2007
Plane Geometry, Polygons & Circle Theorems
Solution
Evaluate \(\begin{pmatrix}3 & -2 \\-7 & 5 \end{pmatrix}\) + 2\(\begin{pmatrix}-2 & 4\\3 & -1 \end{pmatrix}\)
A
\(\begin{pmatrix}-1 & 6 \\-1 & 3 \end{pmatrix}\)
B
\(\begin{pmatrix}3 & 4 \\-2 & 6 \end{pmatrix}\)
C
\(\begin{pmatrix}-1 & 6 \\1 & 3 \end{pmatrix}\)
D
\(\begin{pmatrix}3 & 4 \\2 & 6 \end{pmatrix}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + 2 \begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix}\)
\(2 \begin{pmatrix}-2 & 4 \\ 3 & -1 \end{pmatrix} = \begin{pmatrix}2 \cdot -2 & 2 \cdot 4 \\ 2 \cdot 3 & 2 \cdot -1\end{pmatrix} = \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix}\)
\(\begin{pmatrix}3 & -2 \\ -7 & 5 \end{pmatrix} + \begin{pmatrix}-4 & 8 \\ 6 & -2\end{pmatrix} = \begin{pmatrix}3 + (-4) & -2 + 8 \\ -7 + 6 & 5 + (-2)\end{pmatrix}\)
\(= \begin{pmatrix}-1 & 6 \\ -1 & 3\end{pmatrix}\)
Thus, the final result is:
\(\begin{pmatrix}-1 & 6 \\ -1 & 3\end{pmatrix}\)
Source: Myschool Past Questions Archive (Record #72043)
Mathematics
UTME / JAMB 2007
Binary Operations, Matrices & Determinants
Solution
If f(x) = 3x - 2I, P = \(\begin{pmatrix}2 & 1 \\-1 & 0 \end{pmatrix}\) and I is 2 x 2 identity matrix. Evaluate f(p)
A
\(\begin{pmatrix}6 & -3 \\3 & 0 \end{pmatrix}\)
B
\(\begin{pmatrix}2 & 0 \\0 & 2 \end{pmatrix}\)
C
\(\begin{pmatrix}8 & 3 \\-3 & 2 \end{pmatrix}\)
D
\(\begin{pmatrix}4 & 3 \\-3 & -2 \end{pmatrix}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
I = \(\begin{pmatrix}1 & 0 \\0 & 1 \end{pmatrix}\)
Then, 3x - 2I becomes 3p - 2I (since we are finding f(p))
3\(\begin{pmatrix}2 & 1 \\-1 & 0 \end{pmatrix}\) - 2\(\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}\)
= \(\begin{pmatrix}6 & 3 \\-3 & 0 \end{pmatrix}\) - \(\begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}\)
= \(\begin{pmatrix}4 & 3 \\-3 & -2 \end{pmatrix}\)
Source: Myschool Past Questions Archive (Record #72044)
Mathematics
UTME / JAMB 2006
Plane Geometry, Polygons & Circle Theorems
Diagram
Solution
The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?
Attached Figure / Diagram
A
(33, 4)
B
(22, 4)
C
(33, 12)
D
(12, 4)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The mode (m) = 4.
No of students who scored 4 and above = 12 + 10 + 6 + 4 + 1 = 33.
\(\therefore\) (n, m) = (33, 4)
Source: Myschool Past Questions Archive (Record #8622)
Mathematics
UTME / JAMB 2006
Permutations, Combinations & Probability
Solution
A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?
A
15
B
20
C
30
D
45
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The question will be answered in
\(^6C_4 = \frac{6!}{(6-4)!4!}\)
=\(\frac{6!}{2!4!}\)
=\(\frac{6\times5\times4!}{2\times1\times4!}\) = 15ways.
Source: Myschool Past Questions Archive (Record #8623)
Mathematics
UTME / JAMB 2006
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
If the mean of five consecutive integers is 30, find the largest of the numbers
A
28
B
30
C
32
D
34
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Let the consecutive numbers be a, a+1, a+2, a+3, a+4
Mean = \(\frac{(a+a+1+a+2+a+3+a+4)}{5}\)
30 = \(\frac{5a+10}{5}\)
30\( \times 5 = 5a + 10\)
5a = 150 - 10
5a = 140
a = 28
∴ a + 4 = 28 + 4
= 32
Source: Myschool Past Questions Archive (Record #8639)
Mathematics
UTME / JAMB 2006
Permutations, Combinations & Probability
Solution
A bag contains 5 black, 4 white and x red marbles. If the probability of picking a red marble is 2/5, find the value of x
A
8
B
10
C
4
D
6
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Black = 5
White = 4
Red = x
Total = 9+x
P(red) = 2/5
x / (9+x) = 2/5
5x = 2(9+x)
5x = 18+2x
5x-2 = 18
3x = 18
x = 6
Source: Myschool Past Questions Archive (Record #8656)
Mathematics
UTME / JAMB 2006
Statistics (Data Representation, Central Tendency & Dispersion)
Solution
Find the variance of 2x, 2x-1 and 2x+1
A
2/3
B
2
C
√(2/3)
D
1
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(\sum x = 6x\)
\(\sum(x-\bar{x})^2\) = 2
\(\bar{x} = \frac{\sum x}{n}\)
= \(\frac{6x}{3}\)
= 2x
Variance = \(\frac{\sum(x-\bar{x})^2}{n}\)
= \(\frac{2}{3}\)
Source: Myschool Past Questions Archive (Record #8663)
Mathematics
UTME / JAMB 2006
Permutations, Combinations & Probability
Diagram
Solution
The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?
Attached Figure / Diagram
A
3/20
B
1/4
C
2/5
D
3/4
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5
Source: Myschool Past Questions Archive (Record #8671)
Mathematics
UTME / JAMB 2006
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
Solution
The pie chart above shows the expenditure of a family whose income is N30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?
Attached Figure / Diagram
A
N 28 000
B
N 25 500
C
N15 000
D
N 12 500
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90o
Expenditure on transport = 45o
x + 2x + 90 + 45 = 360o
3x + 135 = 360o
3x = 360 - 135
3x = 225
x = 225/3 = 75o
∴∠ for food = 2x = 2 * 75
= 150
360o = N30,000
1o = ?
1o = 30,000/360
150o = (30000/360) * (150/1)
= N12500
Source: Myschool Past Questions Archive (Record #8682)
Mathematics
UTME / JAMB 2006
Plane Geometry, Polygons & Circle Theorems
Solution
For what value of n is \(^{n+1}C_3\) = 4(\(^nC_3\))?
A
6
B
5
C
4
D
3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(^{n+1}C_3 = 4(^nC_3)\\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\\=\frac{n+1}{n-2}=\frac{4}{1}\\n+1 = 4(n-2)\\n+1 = 4n-8\\-3n = -9\\\frac{-9}{-3}\\n=3\)
Source: Myschool Past Questions Archive (Record #8692)