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.
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 = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\)
= \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\)
= \(\frac{47}{5}\)
56cm2
112cm2
120cm2
176cm2
Correct answer is C
From the figure, PS = QR = YT = 7cm
Area of parallelogram PQRS = 56cm
56 = base x height, where base = 7
7 x h = 56cm,
h = \(\frac{56}{7}\)
= 8cm
Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm
Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8
\(\frac{1}{2}\) x 30 x 8 = 120cmsq
What is the equation of the quadratic function represented by the graph?
y = x2 + x - 2
y = -x2 - x + 2
y = x2 - x - 2
y = -x2 - x + 2
Correct answer is C
The required equation is y = x2 - x - 2
i.e. B where the graph touches the graph touches the x-axis y = 0
x2 - x - 2 = 0 = (x + 1)(x - 2) = 0
Hence roots of the equation are -1 and 2 as shown in the graph
In the figure, PQ is a parallel to ST and QRS = 40o. Find the value of x
55o
60o
65o
75o
Correct answer is A
From the figure, 3x + x - 40o = 180o
4x = 180o + 40o
4x = 220o
x = \(\frac{220}{4}\)
= 55o