2010 - WAEC Mathematics Past Questions and Answers - page 3

sum of exterior angles = 360^o

Sum of interior angle = (n - 2) x 180

360 = \(\frac{1}{2}\) x(n - 2) x 180(90^o)

360 = \(\frac{1}{2}\) x(n - 2) x 90^o

\(\frac{360}{90}\) = a - 2

4 = n - 2

n = 4 + 2 = 6

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 = \(\frac{\sqrt{9y^2N^2 - 4m}}{2}\)

sequence: -2, 4, -8, 16........{GP}

a = -2; r = \(\frac{4}{-2}\) = -2

nth term T_n = ar^n-1

T_n = (-2)(-2)^^n-1

T_n = (-2)^{1 + n - 1}

T_n = (-2)ⁿ

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.

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%

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

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

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

5 1 6 2_seven
-2 6 4 4_seven
--------
2 2 1 5
--------

the missing number is 2

(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\(\frac{1}{2}\)