Mathematics Questions & Answers - Free Online Practice

1,666.

In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY

8cm

6cm

4cm

3cm

View answer & explanation

Correct answer is B

No explanation has been provided for this answer.

1,667.

In the figure, PS = QS = RS and QSR - 100^o, find QPR

40^o

50^o

80^o

100^o

View answer & explanation

Correct answer is B

Since PS = QS = RS

S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius

QPR \(\pm\) \(\frac{100^o}{2}\) = 50^o

1,668.

In a class of 30 students, the marks scored in an examination are displayed in the histogram. What percentage of the student scored more than 40%?

14%

40%

46\(\frac{3}{4}\)%

53\(\frac{1}{3}\)%

View answer & explanation

Correct answer is C

This histogram is transferred into this frequency table
\(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\)

Students who scored more than 40 = 6 + 6 + 2 = 14

i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)%

1,669.

If PST is a straight line and PQ = QS = SR in the diagram, find y.

24^o

48^o

72^o

84^o

View answer & explanation

Correct answer is A

< PSQ = < SQR = < SRQ = 24^o

< QSR = 180^o - 48^o = 132^o

< PSQ + < QSR + y + 180 (angle on a straight lines)

24 + 132 + y = 180^o = 156^o + y = 180

y = 180^o - 156^o

= 24^o

1,670.

If PQR is a straight line with QS = QR, calculate TPQ, If QT\\SR and TQS = 3y^o

62^o

56\(\frac{3}{2}\)^o

20\(\frac{3}{2}\)^o

18^o

View answer & explanation

Correct answer is C

Since QS = QR

then, angle SQR = angle SRQ

2 SQR = 180 - 56, SQR = \(\frac{124}{2}\) = 62^o

QTP = 62^o

QTP = 62^o, corresponding angle

3y + 56 + 62 = 180 = 3y = 180 - 118

3y = 62 = 180

3y = 180 - 118

3y = 62

y = \(\frac{62}{3}\)

= 20\(\frac{3}{2}\)