2016 - WAEC Mathematics Past Questions & Answers - page 1

1
If 23x + 101x = 130x, find the value of x
A
7
B
6
C
5
D
4
CORRECT OPTION: d
23x + 101x = 130x

2 x X1 + 3 x Xo + 1 x X2 + 0 x X1 + 1 x Xo

= 1 x Xo = 1 x X2 + 3 x X1 + 0 x Xo

= X2 + 3x + 0

2x + 3 = x2 + 0 + 1 + x2 + 3x

2x - 3x + x2 - x2 = -3 - 1

- x = -4

x = 4
2
Simplify: (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)
A
\(\frac{1}{60}\)
B
\(\frac{5}{72}\)
C
\(\frac{1}{10}\)
D
1\(\frac{7}{10}\)
CORRECT OPTION: c
(\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

= (\(\frac{9 - 8}{12} \times \frac{6}{5}\))

= \(\frac{1}{12} \times \frac{6}{5}\)

= \(\frac{1}{10}\)
3
Simplify:(\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))2
A
75.00
B
15.00
C
8.66
D
3.87
CORRECT OPTION: b
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
4
The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If the stone falls 45cm in 3 seconds, how far will it fall in 6 seconds?
A
90cm
B
135cm
C
180cm
D
225cm
CORRECT OPTION: c
d \(\alpha\) t2

d = t2 k

where k is a constant. d = 45cm, when t = 3s; thus 45 = 32 x t

k = \(\frac{45}{9}\) = 5

thus equation connecting d and t is d = 5t2

when t = 6s, d = 5 x 62

= 5 x 36

= 180cm
5
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
CORRECT OPTION: c
From the venn diagram, Nigeria footballers from a subset of good footballers.
6
On a map, 1cm represent 5km. Find the area on the map that represents 100km2.
A
2cm2
B
4cm2
C
8cm2
D
8cm2
CORRECT OPTION: b
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
7
Simplify; \(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)
A
32n
B
9
C
3n
D
3n + 1
CORRECT OPTION: b
\(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)

= \(\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}}\)

= 3\(^{n - 1 + 3n + 3 - 4n}\)

= 3\(^{4n - 4n - 1 + 3}\)

= 32

= 9
8
What sum of money will amount to D10,400 in 5 years at 6% simple interest?
A
D8,000.00
B
D10,000.00
C
D12,000.00
D
D16,000.00
CORRECT OPTION: a
A = P + 1

I = A - P

= 10,400 - P

Now using I = \(\frac{P \times T \times R}{100}\)

i.e. 10,400 - P = \(\frac{P \times 5 \times 6}{100}\)

= 100(10,400 - P) = 30P

10(10,400 - P) = 3P

104,000 - 10P = 3P

104,000 - 10P = 3P

104,000 = 3P + 10P

= 104,000 = 13P

P = \(\frac{104,000}{100}\)

P = D8,000
9
The roots of a quadratic equation are \(\frac{4}{3}\) and -\(\frac{3}{7}\). Find the equation
A
21x2 - 19x - 12 = 0
B
21x2 + 37x - 12 = 0
C
21x2 - x + 12 = 0
D
21x2 + 7x - 4 = 0
CORRECT OPTION: a
Let x = \(\frac{4}{3}\), x = -\(\frac{3}{7}\)

Then 3x = 4, 7x = -3

3x - 4 = 0, 7x + 3 = 0

(3x - 4)(7x + 3) = 0

21x2 + 9x - 28x - 12 = 0

21x2 - 19x - 12 = 0
10
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
CORRECT OPTION: c
\(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\)

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
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