2016 - WAEC Mathematics Past Questions and Answers - page 4

First, arrange the marks in order of magnitude; 3, 4, 5, 7, 10, 13, 14, 16

Hence range = 16 - 3 = 13

Log₂(3x - 1) = 5

Log₂(3x - 1) = Log₂2⁵

Log₂(3x - 1) = Log₂32

3x - 1 = 32

3x = 32 + 1 = 33

x = \(\frac{33}{3}\)

= 11

Volume of sphere = Volume of cylinder

i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\)

\(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\)

r³ = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\)

r = 3\(\sqrt{27}\)

= 3

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm²

Area of \(\Delta\) QMN = Area of \(\Delta\) QNP

= Area of \(\Delta\) PNO (triangles between the same parallels)

Hence, area of the trapezium

3 x area of \(\Delta\) QNP

= 3 x 27

= 81cm²

Perimeter of a sector

= 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3

64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3

22 = \(\frac{33\theta}{90}\)

\(\theta = \frac{22 \times 30}{11}\)

= 60^o

From the venn diagram given,

M = (a, b, c), N = (c, f, g)

U = (a, b, c, d, e, f, g)

Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g)

= (f, g)

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9)

If 2(mod 9) y(mod 6), then y = 2 by comparism

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

= \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\

= \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\)

= \(\frac{p^2 - 2pr}{2p(p - 2r)}\)

= \(\frac{p(p - 2r)}{2p(p - 2r)}\)

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

In the diagram, < RPQ = 80^o(angles in same segment)

< SPR = 100^o - < RPQ

= 100 - 80

= 20^o

< SQR = < SPR = 20^o (same reason as above)

< SQR = 20^o