2010 - WAEC Mathematics Past Questions and Answers - page 3
21
The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n
A
6
B
8
C
9
D
12
correct option: a
sum of exterior angles = 360o
Sum of interior angle = (n - 2) x 180
360 = \(\frac{1}{2}\) x(n - 2) x 180(90o)
360 = \(\frac{1}{2}\) x(n - 2) x 90o
\(\frac{360}{90}\) = a - 2
4 = n - 2
n = 4 + 2 = 6
Users' Answers & CommentsSum of interior angle = (n - 2) x 180
360 = \(\frac{1}{2}\) x(n - 2) x 180(90o)
360 = \(\frac{1}{2}\) x(n - 2) x 90o
\(\frac{360}{90}\) = a - 2
4 = n - 2
n = 4 + 2 = 6
22
If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular
A
\(\frac{\sqrt{9y^2 N^2 - 2m}}{3}\)
B
\(\frac{\sqrt{9y^2 N^2 - 4m}}{2}\)
C
\(\frac{\sqrt{9y^2 N^2 - 3m}}{2}\)
D
\(\frac{\sqrt{9y^2 N - 3m}}{2}\)
correct option: b
y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\)
3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
Users' Answers & Comments3yN = 2(\(\sqrt{x^2 + m})\)
\(\frac{3yN}{2} = \sqrt{x^2 + m}\)
(\(\frac{3yN}{2})^2 = ( \sqrt{x^2 + m})\)
\(\sqrt{\frac{9y^2N^2}{4} - \frac{m}{1}}\)
x = \(\frac{\sqrt{9Y^2N^2 - 4m}}{4}\)
x = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)
23
The nth term of the sequence -2, 4, -8, 16.... is given by
A
Tn = 2n
B
Tn = (-2)n
C
Tn = (-2n)
D
Tn = n
correct option: b
sequence: -2, 4, -8, 16........{GP}
a = -2; r = \(\frac{4}{-2}\) = -2
nth term Tn = arn-1
Tn = (-2)(-2)^n-1
Tn = (-2)1 + n - 1
Tn = (-2)n
Users' Answers & Commentsa = -2; r = \(\frac{4}{-2}\) = -2
nth term Tn = arn-1
Tn = (-2)(-2)^n-1
Tn = (-2)1 + n - 1
Tn = (-2)n
24
How many times, correct to the nearest whole number, will a man run round circular track of diameter 100m to cover a distance of 1000m?
A
3
B
4
C
5
D
6
correct option: a
No. of times = \(\frac{\text{Total distance}}{\text{Circumference of circle}}\)
= \(\frac{\text{Total distance}}{\pi d}\)
= \(\frac{1000m}{\frac{22}{7} \times 100m}\)
= \(\frac{1000 \times 7}{2200} = 3.187\)
= 3(approx.) nearest whole no.
Users' Answers & Comments= \(\frac{\text{Total distance}}{\pi d}\)
= \(\frac{1000m}{\frac{22}{7} \times 100m}\)
= \(\frac{1000 \times 7}{2200} = 3.187\)
= 3(approx.) nearest whole no.
25
Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a
A
profit of 13%
B
loss of 12%
C
loss of 10%
D
loss of 5%
correct option: c
s.p = N6900
%profit = 15%
%profit = \(\frac{s.p - c.p}{c.p}\) x 100%
15% = \(\frac{6900 - c.p}{c.p}\) x 100%
\(\frac{15}{100}\)c.p = N6900 - c.p
0.15 c.p = N6900 - c.p
1.15c.p + c.p = N6900
c.p = \(\frac{6900}{1.15}\)
= 6000.00
Now new S.P = N6600
profit = s.p - c.p = 6000 - 6600
= 600
%profit = \(\frac{600}{6600}\) x 100%
= 10%
Users' Answers & Comments%profit = 15%
%profit = \(\frac{s.p - c.p}{c.p}\) x 100%
15% = \(\frac{6900 - c.p}{c.p}\) x 100%
\(\frac{15}{100}\)c.p = N6900 - c.p
0.15 c.p = N6900 - c.p
1.15c.p + c.p = N6900
c.p = \(\frac{6900}{1.15}\)
= 6000.00
Now new S.P = N6600
profit = s.p - c.p = 6000 - 6600
= 600
%profit = \(\frac{600}{6600}\) x 100%
= 10%
26
The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R
A
18
B
20
C
22
D
28
correct option: b
mean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\)
50 = \9\frac{sum}{R}\)
sum = 50R.....(1)
Sum of ages of the men that left = 55 + 63 = 188
remaining sum = 50R - 118
remaining no. of men = R - 2
now mean age = 50 - 1 = 49 years
49 = \(\frac{50R - 118}{R - 2}\)
49(R - 2) = 50R - 118
49R - 50R = -188 - 98
-R = -20
R = 20
Users' Answers & Comments50 = \9\frac{sum}{R}\)
sum = 50R.....(1)
Sum of ages of the men that left = 55 + 63 = 188
remaining sum = 50R - 118
remaining no. of men = R - 2
now mean age = 50 - 1 = 49 years
49 = \(\frac{50R - 118}{R - 2}\)
49(R - 2) = 50R - 118
49R - 50R = -188 - 98
-R = -20
R = 20
27
\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\ \hline y & 1 & 2 & 3 & 4\end{array}\).
The table is for the relation y = mx + c where m and c are constants. What is the equation of the line described in the tablet?
The table is for the relation y = mx + c where m and c are constants. What is the equation of the line described in the tablet?
A
y = 2x
B
y = x + 1
C
y = x
D
y = \(\frac{1}{2}x + 1\)
correct option: d
y = mx + c; when x = 0; y = 1
1 = m(0) + c; 1 = 0 + c; c = 1
when x = 2; y = 2
2 = m(2) + c; 2 = 2m + c; but c = 1
2 = 2m + 1
2 - 1 = 2m
2m = 1
m = \(\frac{1}{2}\)
y = \(\frac{1}{2}\)x + 1
Users' Answers & Comments1 = m(0) + c; 1 = 0 + c; c = 1
when x = 2; y = 2
2 = m(2) + c; 2 = 2m + c; but c = 1
2 = 2m + 1
2 - 1 = 2m
2m = 1
m = \(\frac{1}{2}\)
y = \(\frac{1}{2}\)x + 1
28
What is the value of x when y = 5?
A
8
B
9
C
10
D
11
correct option: a
when y = 5; x = ?; y = \(\frac{1}{2}\)x + 1
5 = \(\frac{1}{2}\)x + 1
5 - 1 = \(\frac{1}{2}\)x
4 = \(\frac{1}{2}\)x
x = 4 x 2
x = 8
Users' Answers & Comments5 = \(\frac{1}{2}\)x + 1
5 - 1 = \(\frac{1}{2}\)x
4 = \(\frac{1}{2}\)x
x = 4 x 2
x = 8
29
The subtraction below is in base seven. Find the missing number.
5 1 6 2seven
-2 6 4 4seven
--------
2 * 1 5
--------
5 1 6 2seven
-2 6 4 4seven
--------
2 * 1 5
--------
A
2
B
3
C
4
D
5
correct option: a
5 1 6 2seven
-2 6 4 4seven
--------
2 2 1 5
--------
the missing number is 2
Users' Answers & Comments-2 6 4 4seven
--------
2 2 1 5
--------
the missing number is 2
30
If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x
A
1\(\frac{1}{2}\)
B
\(\frac{1}{2}\)
C
-\(\frac{3}{2}\)
D
-1\(\frac{1}{2}\)
correct option: a
(x - p)(2x + 1) = 0
2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)
Users' Answers & Comments2x2 + x - 2px - p = 0
2x2 + x (1 - 2p) - p = 0
2x2 - (2p - 1)x - p = 0
divide through by 2
x2 - \(\frac{(2p - 1)}{2}\)x - \(\frac{p}{2}\) = 0
compare to x2 - (sum of roots)x + product of roots = 0
sum of roots = \(\frac{2p - 1}{2}\)
But sum of roots = 1
Given; \(\frac{2p - 1}{2}\) = 1
2p - 1 = 2 x 1
2p - 1 = 2
2p = 2 + 1 = 3
p = \(\frac{3}{2}\)
p = 1\(\frac{1}{2}\)