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.
TQ is tangent to circle XYTR, < YXT = 32o, RTQ = 40o. find < YTR
108o
121o
140o
148o
Correct answer is A
< TWR = < QTR = 40o (alternate segment)
< TWR = < TXR = 40o(Angles in the same segments)
< YXR = 40o + 32o = 72o
< YXR + < YTR = 180o(Supplementary)
72o + < YTR = 180o
< YTR = 180o - 72o
= 108o
35cm2
65cm2
70cm2
140cm2
Correct answer is C
A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20
= \(\frac{1}{2}\) x 8 x h = 4h
h = \(\frac{20}{4}\)
= 5cm
A\(\bigtriangleup\)(PQTS) = L x H
A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS
20 + 50 = 70cm\(^2\)
ALTERNATIVE METHOD
A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28
= 70cm\(^2\)
In diagram, PQ || ST and < PQR = 120o, < RST = 130o, find the angle marked x
50
65
70
80
Correct answer is C
x + 60o + 50o = 180o
x = 110o = 180o
x = 180o - 110o
= 70o
In the diagram, PR is a diameter of the circle PQRS. PST and QRT are straight lines. Find QRS
20o
25o
30o
35o
Correct answer is B
< PSR = \(\frac{1}{2}\)(180o) = 90o (angle substended by a semi-circle)
∴ < TSR = 90o
< SRT = 90o - 30o = 60o
< PRS = 90o - 35o = 55o
< PRQ = 180 - (60o + 55o) = 180 - 115o = 65o
< SQR = 35o(angles in the same segment)
< QSR + < SRQ + < SQR = 180o
< QSR + 120o + 35o = 180o
< QSR + 155 = 180o
< QSR = 180o - 155o
= 25o
x \(\leq\) 0, y \(\leq\) 0, 2y + 3x \(\leq\) 6
x \(\geq\) 0, y \(\geq\) 3, 3x + 2y \(\geq\) 6
x \(\geq\) 2, y \(\geq\) 0, 3x + 2y \(\leq\) 6
x \(\geq\) 0, y \(\geq\) 0, 3x + 2y \(\geq\) 6
Correct answer is A
m = \(\frac{y_2 - y_1}{x_2 - x_2} = \frac{3 - 0}{0 - 2} = \frac{-3}{2}\)
= \(\frac{y - y_1}{x- x_1}\)
m = \(\frac{y - 3}{x}\) \(\geq\) \(\frac{-3}{2}\)
2(y - 3) \(\geq\) - 3x = 2y - 6 \(\geq\) - 3x
= 2y + 3x \(\leq\) 6 ; x \(\leq\) 0, y \(\leq\) 0