How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
The diagram shows a circle O. If < ZYW = 33\(^o\) , find < ZWX
33\(^o\)
57\(^o\)
90\(^o\)
100\(^o\)
Correct answer is C
In ZY = 90\(^o\) < subtends In a semi O
ZWY = 180 - (90\(^o\) + 33)
= 57
ZWX = 57 + 33 = 90\(^o\)
In the diagram, XY is a straight line.
60\(^o\)
90\(^o\)
100\(^o\)
120\(^o\)
Correct answer is B
<POX = <POQ; <ROY = QOR
2 <POQ + 2 <ROY = 180
2(<POQ = <ROY) = 180
<POQ + <ROY = 90
The diagram shows a circle centre O. if <STR = 29 and <RST = 45, calculate the value of <STO
12\(^o\)
15\(^o\)
29\(^o\)
34\(^o\)
Correct answer is A
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\(^o\)
1
2
3
4
Correct answer is C
(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)
7.83\(m^2\)
32.29\(m^2\)
50.29\(m^2\)
82.50\(m^2\)
Correct answer is D
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}\)