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.
On the curve, the points at which the gradient of the curve is equal to zero are
c, d, f. i, l
b, e, g, j, m
a, b, c, d, f, i, j, l
c, d, f, h, i, l
Correct answer is B
The gradient of any curve is equal to zero at the turning points. i.e. maximum or minimum points. The points in the above curve are b, e, g, j, m
36\(\pi\)cm 3
54\(\pi\)cm 2
18\(\pi\)cm 2
108\(\pi\)cm 2
Correct answer is A
The volume of the solid = vol. of cone + vol. of hemisphere
volume of cone = \(\frac{1}{3} \pi r^2 h\)
= \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\)
vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\)
= \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\)
vol. of solid = 18\(\pi\) + 18\(\pi\)
= 36\(\pi\)cm3
In the figure, XR and YQ are tangents to the circle YZXP if ZXR = 45° and YZX = 55°, Find ZYQ
135O
125O
100O
90O
Correct answer is A
< RXZ = < ZYX = 45O(Alternate segment)
< ZYQ = 90 + 45
= 135°
In the figure, PS = RS = QS and QRS = 50°. Find QPR
25O
40O
50O
65O
Correct answer is A
In the figure PS = RS = QS, they will have equal base QR = RP
In angle SQR, angle S = 50O
In angle QRP, 65 + 65 = 130O
Since RQP = angle RPQ = \(\frac{180 - 130}{2}\)
= \(\frac{50}{2} = 25^o\)
QPR = 25°
In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR
\(\frac{47}{5}\)
5
\(\frac{32}{5}\)
\(\frac{22}{5}\)
Correct answer is A
From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)
QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\)
= \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\)
= \(\frac{47}{5}\)