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Mathematics
WAEC / WASSCE 1988
Statistics (Data Representation, Central Tendency & Dispersion)
Diagram
Solution
The histogram below shows the number of candidates, in thousands, obtaining given ranges of marks in a State examination. Find the total number of candidates that sat for the examination
Attached Figure / Diagram
A
80,000
B
120,000
C
250,000
D
260,000
E
270,000
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
No of candidates = { 5 + 10 + 60 + 80 + 50 + 30 + 5 + 5 + 5}thousand = 270 x 1000 = 270000
Source: WAEC / JAMB / NECO Classroom Archive (Record #22228)
Mathematics
WAEC / WASSCE 1988
Plane Geometry, Polygons & Circle Theorems
Diagram
Which of the regions P, Q, R, S and T satisfies the following 0 < y < 1, y < x + 2, x < 0
Attached Figure / Diagram
A
P
B
Q
C
R
D
S
E
T
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #22185)
Mathematics
WAEC / WASSCE 1988
Mensuration (Areas, Perimeters & Volumes)
Solution
A cone is made from a sector of circle of radius 14cm and angle of 90o. What is the area of the curved surface of the cone? (Take x = 22/7)
A
22cm2
B
38.5cm2
C
77cm2
D
154cm2
E
308cm2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Length of arc = θ/360 * 2πr
=> r = 90/360 * 14/4 = 3.5
slant height t = 14cm
curved surface area of cone = πrl
=> 3.142 x 3.5 x 14 = 153.95
Source: Myschool Past Questions Archive (Record #22192)
Mathematics
WAEC / WASSCE 1988
Mensuration (Areas, Perimeters & Volumes)
Solution
A rope of length 18m is used to form a sector of a circle of radius 3.5m on a school playing field. What is the size of the angle of the sector, correct to the nearest degree?
A
33o
B
40o
C
90o
D
180o
E
270o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Circumference = 2\pir
2 x 22/7 x 3.5 ≅ 22m
22m = 360o
11m = 360/22 x 11 = 180o
Source: Myschool Past Questions Archive (Record #22199)
Mathematics
WAEC / WASSCE 1988
Trigonometry, Elevation & Bearings
Diagram
Solution
In the diagram above, PR is perpendicular from P to QS, PQ = 2cm, QR = 1cm and PR = RS. what is the size of the angle QPS?
Attached Figure / Diagram
A
135o
B
105o
C
90o
D
75o
E
60o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\Delta RPS\) = Isosceles triangle
\(\therefore < RPS = \frac{180^\circ - 90^\circ}{2}\) = 45^\circ
In \(\Delta QPR\),\(\sin < QPR = \frac{1}{2}\) = 30^\circ
\(\therefore < QPS = 30^\circ + 45^\circ = 75^\circ\)
Source: Myschool Past Questions Archive (Record #22208)
Mathematics
WAEC / WASSCE 1988
Mensuration (Areas, Perimeters & Volumes)
Solution
Cotonou and Niamey are on the same line of longitude and Niamey is 7o north of Cotonou. If the radius of the earth is 6400km, how far is Niamey north of Cotonou along the line of longitude, correct to the nearest kilometer? [Take π = 22/7]
A
391km
B
503km
C
782km
D
1006km
E
2012km
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
θ/360 * 2πR
= 7/360 * 2 * 22/7 * 6400 = 782km
Source: Myschool Past Questions Archive (Record #22226)
Mathematics
WAEC / WASSCE 1988
Sets, Logic & Venn Diagrams
Solution
If Q = (all perfect squares less than 30) and P = (all odd numbers from 1 to 10). Find Q ∩ P.
A
(1, 4, 9, 16, 25)
B
(1, 3, 4, 5, 7, 9, 16, 25)
C
(1, 3, 5, 7, 9)
D
(1, 9)
E
Φ
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Q = {1, 4, 9, 16, 25} P = { 1, 3, 5, 7, 9}
Q ∩ P = { 1, 9}
Source: Myschool Past Questions Archive (Record #22227)
Mathematics
WAEC / WASSCE 1988
Fractions, Decimals, Percentages & Approximations
Diagram
Solution
The histogram below shows the number of candidates, in thousands, obtaining given ranges of marks in a State examination. How many candidates scored at most 30%?
Attached Figure / Diagram
A
15,000
B
20,000
C
25,000
D
35,000
E
60,000
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Those who scored at most 30% = ( 5 + 10 + 20)thousand = 35000
Source: Myschool Past Questions Archive (Record #22229)
Mathematics
WAEC / WASSCE 1988
Permutations, Combinations & Probability
Solution
The surnames of 40 children in a class arranged in alphabetical order. 16 of the surnames begins with O and 9 of the surname begins with A, 14, of the letters of the alphabet do not appear as the first letter of a surname
What is the probability that the surname of a child picked at random from the class begins with either O or A?
A
5/8
B
7/8
C
9/16
D
14/25
E
39/40
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
No. of possible outcome = 40
No. of expected outcome = 16 + 9 = 25
Probability = 25/40 = 5/8
Source: Myschool Past Questions Archive (Record #22230)
Mathematics
WAEC / WASSCE 1988
Permutations, Combinations & Probability
Solution
The surnames of 40 children in a class arranged in alphabetical order. 16 of the surnames begins with O and 9 of the surname begins with A, 14, of the letters of the alphabet do not appear as the first letter of a surname
If more than one surname begins with a letter besides A and O, how may surnames begin with that letter
A
2
B
3
C
4
D
5
E
6
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
14 of the letters of the alphabet do not appear as the first letter of surname
∴ the no. of letters that appeared = 26-14 = 12 alphabets
15 surnames begin with 10 letters beside O and A
∴ 6 surnames begin with a letter
Source: Myschool Past Questions Archive (Record #22231)
Mathematics
UTME / JAMB 1988
Indices, Logarithms & Surds
Solution
If \(1P03{_4} = 115_{10}\) find P
A
o
B
1
C
2
D
3
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(1 x 4^3 + P x 4^2 + 0 x 4 + 3 = 115_{10}\)
16p + 67 = 115 p = \(\frac{48}{16}\)
= 3
Source: Myschool Past Questions Archive (Record #30761)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
If \(\frac{1}{p}\) = \(\frac{a^2 + 2ab + b^2}{a - b}\) and \(\frac{1}{q}\) = \(\frac{a + b}{a^2 - 2ab + b^2}\) Find \(\frac{p}{q}\)
A
\(\frac{a + b}{a - b}\)
B
\(\frac{1}{a^2 - b^2}\)
C
\(\frac{a - b}{a + b}\)
D
a2 - b2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{1}{p} = \frac{a^{2} + 2ab + b^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{a^{2} - 2ab + b^{2}}\)
\(\frac{1}{p} = \frac{(a + b)^{2}}{a - b}\)
\(\frac{1}{q} = \frac{a + b}{(a - b)^{2}}\)
\(\therefore p = \frac{a - b}{(a + b)^{2}}\)
\(\frac{p}{q} = p \times \frac{1}{q} = \frac{a - b}{(a + b)^{2}} \times \frac{a + b}{(a - b)^{2}}\)
= \(\frac{1}{(a + b)(a - b)}\)
= \(\frac{1}{a^{2} - b^{2}}\)
Source: Myschool Past Questions Archive (Record #31627)
Mathematics
UTME / JAMB 1988
Variations
Solution
If x varies inversely as the cube root of y and x = 1 when y = 8, find y when x = 3
A
\(\frac{1}{3}\)
B
\(\frac{2}{3}\)
C
\(\frac{8}{27}\)
D
\(\frac{4}{9}\)
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(x \propto \frac{1}{\sqrt[3]{y}} \implies x = \frac{k}{\sqrt[3]{y}}\)
When y = 8, x = 1
\(1 = \frac{k}{\sqrt[3]{8}} \implies 1 = \frac{k}{2}\)
\(k = 2\)
\(\therefore x = \frac{2}{\sqrt[3]{y}}\)
When x = 3,
\(3 = \frac{2}{\sqrt[3]{y}} \implies \sqrt[3]{y} = \frac{2}{3}\)
\(y = (\frac{2}{3})^{3} = \frac{8}{27}\)
Source: Myschool Past Questions Archive (Record #31629)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
if a = -3, b = 2, c = 4, evaluate \(\frac{a^3 - b^3 - c^{\frac{1}{2}}}{b - a - c}\)
A
37
B
\(\frac{-37}{5}\)
C
\(\frac{37}{5}\)
D
-37
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
\(\frac{a^{3} - b^{3} - c^{\frac{1}{2}}}{b - a - c} = \frac{(-3)^{3} - (2)^{3} - 4^{\frac{1}{2}}}{2 - (-3) - 4}\)
= \(\frac{-37}{1} = -37\)
Source: Myschool Past Questions Archive (Record #31630)
Mathematics
UTME / JAMB 1988
Plane Geometry, Polygons & Circle Theorems
Solution
If (g(y)) = \(\frac{y - 3}{11}\) + \(\frac{11}{y^2 - 9}\). what is g(y + 3)?
A
\(\frac{y}{11} + \frac{11}{y(y + 6)}\)
B
\(\frac{y}{11} + \frac{11}{y(y + 3)}\)
C
\(\frac{y + 30}{11} + \frac{11}{y(y + 3)}\)
D
\(\frac{y + 3}{11} + \frac{11}{y(y - 6)}\)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
\(g(y) = \frac{y - 3}{11} + \frac{11}{y^{2} - 9}\)
\(\therefore g(y + 3) = \frac{(y + 3) - 3}{11} + \frac{11}{(y + 3)^{2} - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y + 9 - 9}\)
\(g(y + 3) = \frac{y}{11} + \frac{11}{y^{2} + 6y}\)
= \(\frac{y}{11} + \frac{11}{y(y + 6)}\)
Source: Myschool Past Questions Archive (Record #31631)