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.

1,326.

In the diagram, < WOX = 60o, < YOE = 50o and < OXY = 30o. What is the bearing of x from y?

A.

300o

B.

240o

C.

190o

D.

150o

Correct answer is A

The bearing of x from y = 270o + \(\theta\)

where \(\theta\) + 50o = y

in \(\bigtriangleup\) OXY

O + X + Y = 180o

Where O = 40o + 30o = 70o

70o + 30o + y = 180o

y + 100o = 180o

y = 180o - 100o = 30o

\(\theta\) + 50o = 80o

80o - 50o = 30o

The bearing of x from y = 270o + 30o = 300o

1,327.

In the diagram, 0 is the centre of the circle. Find the value x

A.

34

B.

29

C.

17

D.

14

Correct answer is D

POQ in a straight line

Hence, < POQ + < QOR = 180o

56o + < QOR = 180o

< QOR = 180o - 56o

= 124o

Now, in \(\bigtriangleup\) QOR OR = OQ = Radius

< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))

2x + 124 + 2x = 180o

4x + 124 = 180

4x = 180 - 124

4x = 56

x = \(\frac{56}{4}\)

x = 14o

1,328.

The diagram shows a rectangular cardboard from which a semi-circle is cut off. Calculate the area of the remaining part

A.

44cm2

B.

99cm2

C.

154cm2

D.

165cm2

Correct answer is B

Area of remaining = Area of rectangle = Area of semi-circle

22 x 8 - \(\frac{1}{2}\)xar2

ehere r - \(\frac{14}{2}\)cm = 7cm

Area of remaining = 176 - \(\frac{1}{2}\) x \(\frac{22}{4}\) x 7 x 7

= 176 - 77

= 99cm2

1,329.

The diagram is a net right rectangular pyramid. Calculate the total surface area

A.

208cm2

B.

112cm2

C.

92cm2

D.

76cm2

Correct answer is C

Total surface area = sum of the area of the \(\bigtriangleup\) S + Area of the rectangle

= 2 x Area of \(\bigtriangleup\) PTQ + 2 x Area of \(\bigtriangleup\) QUR + Area of rectangle PQRS

2 x \(\frac{1}{2}\)(8 x 4) + 2 x \(\frac{1}{2}\)(5 x 4) + 8 x 5

= 32 + 20 + 40

= 92cm2

1,330.

In the diagram, GI is a tangent to the circle at H. If EF//GI, calculate the size of < EHF

A.

126o

B.

72o

C.

64cmo

D.

32cmo

Correct answer is B

F = 54o (Alternate triangle angles)

< GHE = F = 54o(Angle between a chord and a tangent = angles in the alternate segment)

Now, < GHE + < EHF + < IHF = 180o(Angles on a straight line)

54o + < EHF + 54o = 180o

< EHF = 180o - 108o

= 72o