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Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2
A
75.00
B
15.00
C
8.66
D
3.87
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\)
= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\)
hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2 = (\(2\sqrt{15} - \sqrt{15}\))2
= (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\))
= 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\)
= 4 x 15 - 2 x 15 - 2 x 15 + 15
= 60 - 30 - 30 + 15
= 15
Source: Myschool Past Questions Archive (Record #46046)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Which of following is a valid conclusion from the premise. "Nigeria footballers are good footballers"?
A
Joseph plays football in Nigeria therefore he is a good footballer
B
Joseph is a good footballer therefore he is a Nigerian footballer
C
Joseph is a Nigerian footballer therefore he is a good footballer
D
Joseph plays good football therefore he is a Nigerian footballer
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
From the venn diagram, Nigeria footballers from a subset of good footballers.
Source: Myschool Past Questions Archive (Record #46076)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
On a map, 1cm represent 5km. Find the area on the map that represents 100km2.
A
2cm2
B
4cm2
C
8cm2
D
8cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
On a map, 1cm represents 5km. Then it follows that 1cm2 represents 25km2. Acm2 represents 100km2. By apparent cross-multiplication, 1cm2 x 100km2 = Acm2x 25km2
therefore A = \(\frac{100}{25}\) = 4cm2
Source: Myschool Past Questions Archive (Record #46078)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Simplify; \(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)
A
32n
B
9
C
3n
D
3n + 1
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)
= \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\)
= 3\(^{n - 1 + 3n + 3 - 4n}\)
= 3\(^{4n - 4n - 1 + 3}\)
= 32
= 9
Source: Myschool Past Questions Archive (Record #46096)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined
A
6, -7
B
3, -6
C
3, -7
D
-3, -7
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\)
Factorize the denominator;
Y2 + 7y - 3y - 21
= y(y + 7) -3 (y + 7)
= (y - 3)(y + 7)
Hence the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined
when y2 + 4y - 21 = 0
ie. y = 3 or -7
Source: Myschool Past Questions Archive (Record #46100)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?
A
1
B
3
C
7
D
17
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
2x + y = 7...(1)
3x - 2y = 3...(2)
From (1), y = 7 - 2x for y in (2)
3x - 2(7 - 2x) = 3
3x - 14 + 4x = 3
7x + 3 + 14 = 17
x = \(\frac{17}{7}\)
Hence, 7 x \(\frac{17}{7}\)
= 17 - 10
= 7
Source: Myschool Past Questions Archive (Record #46103)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\)
A
\(\frac{x + 1}{x(1 - x)}\)
B
\(\frac{3x - 1}{ x(1 - x)}\)
C
\(\frac{3x + 1}{ x(1 - x)}\)
D
\(\frac{x - 1}{ x(1 - x)}\)
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\)
= \(\frac{2x - 1 + x}{x(1 - x)}\)
= \(\frac{3x - 1}{x(1 - x)}\)
Source: Myschool Past Questions Archive (Record #46106)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)
A
s = \(\frac{mrp}{nr + m^2}\)
B
s = \(\frac{nr + m^2}{mrp}\)
C
s = \(\frac{nrp}{mr + m^2}\)
D
s = \(\frac{nrp}{nr + m^2}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
P = S + \(\frac{sm^2}{nr}\)
P = S(1 + \(\frac{m^2}{nr}\))
P = S(1 + \(\frac{nr + m^2}{nr}\))
nrp = S(nr + m2)
S = \(\frac{nrp}{nr + m^2}\)
Source: Myschool Past Questions Archive (Record #46107)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
Factorize; (2x + 3y)2 - (x - 4y)2
A
(3x - y)(x + 7y)
B
(3x + y)(2x - 7y)
C
(3x + y)(x - 7y)
D
(3x - y)(2x + 7y)
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
(2x + 3y)2 - (x - 4y)2
= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)
= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)
= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2
= 3x2 + 20xy - 7y2
= 3x2 + 21xy - xy - 7y2
= 3x(x + 7y) - y(x + 7y)
= (3x - y)(x + 7y)
Source: Myschool Past Questions Archive (Record #46109)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take \(\pi = \frac{22}{7}\)]
A
2.6cm
B
3.5cm
C
3.6cm
D
7.0cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Curved surface area of cylinder = 2\(\pi\)rh
110 = 2 x \(\frac{22}{7}\) x r x 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
Source: Myschool Past Questions Archive (Record #46112)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x
A
3
B
5
C
6
D
8
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Volume of pyramid = \(\frac{1}{3}\)lbh
90 = \(\frac{1}{3} \times x \times 6 \times 15\)
x = \(\frac{90 \times 33}{6 \times 15}\)
= 3
Source: Myschool Past Questions Archive (Record #46113)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
A
\(4\sqrt{11}\)
B
\(4\sqrt{10}\)
C
\(2\sqrt{17}\)
D
\(2\sqrt{13}\)
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
|PQ| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\)
= \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\)
= \(\sqrt{4^2 + 6^2}\)
= \(\sqrt{16 + 36}\)
= \(\sqrt{52}\)
= 2\(\sqrt{13}\)
Source: Myschool Past Questions Archive (Record #46119)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Diagram
Solution
In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50º. Find the value of < UXY.
Attached Figure / Diagram
A
70o
B
80o
C
105o
D
110o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the diagram above, \(x_1\) = 50º(angles in alternate segment)
\(x_1\) = \(x_2\)(base angles of isos. \(\Delta\))
< UXY + \(x_2\) + 50º = 180º(sum of angles on a straight line)
< UXY + 50º + 50º = 180º
< UXY + 100º = 180º
< UXY = 180º - 100º
= 80º
Source: Myschool Past Questions Archive (Record #46153)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Diagram
Solution
In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177o. Calculate < PST.
Attached Figure / Diagram
A
54o
B
44o
C
34o
D
27o
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
In the diagram above, x1 = 180o - 117o = 63o(opposite angles of a cyclic quad.)
x1 = x2 (base angles of isos. \(\Delta\))
x\(_1\) + x\(_2\) + \(\alpha\) = 180º (sum of angles of a \(\Delta\)
63º + 63º + \(\alpha\) = 180º
\(\alpha\) = 180º - (63 + 63)º
= 54º
< PST = 54º (aalternate angles)
Source: Myschool Past Questions Archive (Record #46173)
Mathematics
WAEC / WASSCE 2016
Number and Numeration
Solution
The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range
A
16
B
14
C
13
D
11
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
First, arrange the marks in order of magnitude; 3, 4, 5, 7, 10, 13, 14, 16
Hence range = 16 - 3 = 13
Source: Myschool Past Questions Archive (Record #46193)