Mathematics questions and answers

Mathematics Questions and Answers

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.

1,596.

In the figure PQR a straight line segment, PQ = QT. Triangle PQT is an isosceles triangle, < SRQ is 75o and < QPT IS 25o. Calculate the value of < RST

A.

50o

B.

25o

C.

55o

D.

45o

Correct answer is C

< T = \(\frac{x}{1}\) = 25o (PQ = QT)

< SQR = 2(25o) = 50o (sum of interior angle)

< Q + < R + < S = 180o

50o + 75o + < S = 180o = 125o + < S = 180o

< S = 180o - 125o = 55o

1,597.

Triangle SPT is the solution of the linear inequalities

A.

2y - x - 2 \(\leq\) 0, y + 2x + 2 \(\geq\) 0, x \(\geq\) 0

B.

2y - x - 2 \(\geq\) 0, y + 2x + 2 \(\leq\) 0, -2 \(\geq\) x \(\geq\) -1

C.

-2 \(\geq\) x \(\geq\) 2, y \(\leq\) 0, y + 2x + 2 \(\geq\) 0, x \(\geq\) 0

D.

2y - x - 2 \(\geq\) 0, y + 2x + 2 \(\geq\) 0, y \(\leq\) 0, x \(\geq\) 0

Correct answer is C

No explanation has been provided for this answer.

1,598.

The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs?

A.

17.5 - 20.5

B.

16.5 - 19.5

C.

15.5 - 19.5

D.

15.5 - 18.5

Correct answer is A

No explanation has been provided for this answer.

1,599.

If the diagram is the graph of y = x2, the shaded area is

A.

64 square units

B.

\(\frac{126}{4}\) square units

C.

32 square units

D.

\(\frac{64}{3}\) square units

Correct answer is D

A = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)

= \(\frac{64}{3}\) square units

1,600.

In the diagram, EFGH is a circle centre O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH

A.

32cm

B.

26cm

C.

20cm

D.

16cm

Correct answer is B

\(\bigtriangleup\)OEN = \(\bigtriangleup\)OGN

OE = OG = r(radii)

(r - 8)2 + 122 = r2

r2 - 16r + 64 + 144 = r2

16r = 64 + 144 = r2

16r = 64 + 144 = 208

\(\frac{208}{16}\) = 13cm

FH = 2FO = 2OH = 2R

= 2 x 13cm

= 26cm