Mathematics Questions & Answers - Free Online Practice

1,716.

PQRS isa cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50o with the tangent as shown in the figure. What is the size of QRS?

50^o

40^o

110^o

80^o

100^o

View answer & explanation

Correct answer is E

< SPQ = 80o

< SPQ + < SRQ = 180(Supplementary)

80 +
< QRS = 180o - 80o

= 100o

1,717.

If O is the centre of the circle in the figure, find the value of x

260

100

130

View answer & explanation

Correct answer is C

From the diagram; The value of x = 360^o - 2(130^o)

= 360 - 260

= 100^o

1,718.

PQRS is a desk of dimensions 2m x 0.8 which is inclined at 30o to the horizontal. Find the inclination of the diagonal PR to the horizontal

30^o 35'

30^o 32'

15^o 36'

10^o

10^o 42'

View answer & explanation

Correct answer is E

tan\(\theta\) = \(\frac{0.8}{2}\) = (0.4)

\(\theta\) = tan^-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10^o 42'

1,719.

The figure FGHK is a rhombus. What is the value of angle X?

30^o

90^o

150^o

120^o

60^o

View answer & explanation

Correct answer is D

< HKF = 60^o, < KFG = 120^o

< KFG = < KHG = x(opposite angles)

x = 120^o

1,720.

Which of the following equations represents the graph?

y = 1 + 2x + 3x²

y = 1 - 2x + 3x²

y = 1 + 2x - 3x²

y = 1 - 2x - 3x²

y = 3x² + 2x - 1

View answer & explanation

Correct answer is E

The roots of the function are 1 and \(\frac{1}{3}\)

sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\)

product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\)

x² - (sum of roots)x + (product of roots) = 0

x² + (-\(\frac{2}{3}\))x - (-\(\frac{1}{3}\)) = 0

x² + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0

3x² + 2x - 1 = 0