2017 - WAEC Mathematics Past Questions and Answers - page 4
If the simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% annum is N500.00, calculate the total amount due after 6 years at the same rate
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
Calculate the variance of 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9
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
A circular pond of radius 4m has a path of width 2.5m round it. Find, correct to two decimal places, the area of the path. [Take\(\frac{22}{7}\)]
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}\)
Fig. 1 and Fig. 2 are the addition and multiplication tables respectively in modulo 5. Use these tables to solve the equation (n \(\oplus 4\))
(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)
The diagram shows a circle centre O. if <STR = 29 and <RST = 450, calculate the value of <STO
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\(^o\)
In the diagram, XY is a straight line. <POX = <POQ and <ROY = <QOR. Find the value of <POQ + <ROY.
<POX = <POQ; <ROY = QOR
2 <POQ + 2 <ROY = 180
2(<POQ = <ROY) = 180
<POQ + <ROY = 90
The diagram shows a circle O. If < ZYW = 33\(^o\) , find < ZWX
In ZY = 90\(^o\) < subtends In a semi O
ZWY = 180 - (90\(^o\) + 33)
= 57
ZWX = 57 + 33 = 90\(^o\)
In the diagram, PQ and PS are tangents to the circle O. If PSQ = m, <SPQ = n and <SQR = 33\(^o\), find the value of (m + n)
< SQP = 180 - (90 + 33) < on a ----
= 180 - (123)
= 57\(^o\)
Therefore, (m + n) = 123\(^o\)
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 7 | 4 | 18 | 12 | 8 | 11 |
The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, Find the median of the distribution
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 7 | 4 | 18 | 12 | 8 | 11 |
The table gives the distribution of marks obtained by a number of pupils in a class test. Using this information, find the first quartile
First quartile = \(\frac{n}{4} = \frac{60}{4}\) = 15
The 15th value is 2