2017 - WAEC Mathematics Past Questions and Answers - page 4

P = \(\frac{100l}{RT} = \frac{100 \times 500}{5 \times 2.5} = \frac{50,000}{12.5}\)

= 4,000

Annual interest is \(\frac{500}{5}\) = 100

for 6 year = 4,000 + 100 + 100 + 100 + 100 + 100 + 100

= N 4,600

x = \(\frac{2 + 3 + 3 + 4 + 5 + 5+ 5+ 7 + 7 + 9}{10}\)

=\(\frac{50}{10}\)

= 5

Variance = \(\frac{\sum{(x - x})^2}{N}\)

\(\frac{9 + 4+ 4+ 1 + 4 + 4 + 16}{10}\)

= \(\frac{42}{10}\)

= 4.2

Area of path = Area of (pond+path) - Area of pond

The area of the pond with the path: The radius = (4 + 2.5)m = 6.5m

Area = \(\pi \times r^{2}\) = \(\frac{22}{7} \times 6.5^{2} \approxeq 132.79m^{2}\)

Area of the pond = \(\frac{22}{7} \times 4^{2} \approxeq 50.29m^{2}\)

 Area of the path  = (132.79 - 50.29)m^{2} = 82.50m^{2}\)

(n \(\oplus\) 4) \(\oplus\) 3 = 0 (mod 5)

(3 \(\oplus\) 4) \(\oplus\) 3

12 \(\oplus\) 3 = 15 (mod 5)

(5 x 3 + 0) = 0 (mod 5)

SRT = 180 - (46 + 29) sum of < s in a

= 180 - 75

= 105

SOT = 2 x 46 < at the centre is twice all the circle = 92

RTO = 180 - (96 + 43)

= 41

STO = 41 - 29

= 12\(^o\)

<POX = <POQ; <ROY = QOR

2 <POQ + 2 <ROY = 180

2(<POQ = <ROY) = 180

<POQ + <ROY = 90

In ZY = 90\(^o\)  < subtends In a semi O

ZWY = 180 - (90\(^o\)  + 33)

= 57

ZWX = 57 + 33 = 90\(^o\) 

< SQP = 180 - (90 + 33)  < on a ----

= 180 - (123)

= 57\(^o\)

Therefore, (m + n) = 123\(^o\)

Median is \(\frac{n}{2} = \frac{6}{2}\)

= 3

Median = 3

First quartile = \(\frac{n}{4} = \frac{60}{4}\) = 15

The 15th value is 2