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.

1,206.

The diagram is a polygon. Find the largest of its interior angles

A.

300o

B.

100o

C.

120o

D.

150o

Correct answer is D

Sum of interior angles = 3602

i.e 3x + 2x + 2x + x + 4x = 360o

12x = 360

x = \(\frac{360}{12}\)

x = 30o

x + c = 180

30o + c = 180

c = 180 - 30

c = 150o

1,207.

The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?

A.

-\(\frac{1}{2} \leq x\) < 3

B.

-\(\frac{1}{2} < x \leq 3\)

C.

-\(\frac{1}{2} < x < 3\)

D.

-\(\frac{1}{2} \leq x \leq 3\)

Correct answer is C

2x2 - 5x - 3 = 0

2x2 - 6x + x - 3 = 0

2x(x - 3) + 1(x - 3) = 0

(2x + 1)(x - 3) = 0

2x + 1 = 0

2x = -1

x = -\(\frac{1}{2}\)

x - 3 = 0

-\(\frac{1}{2}\) < x < 3

1,208.

In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ

A.

65o

B.

45o

C.

25o

D.

19o

Correct answer is D

< RSQ = < RPQ = 48o (angle in the same segment)

< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))

< SQR = 480

< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))

i.e. < QRS + 48o + 48o = 180

< QRS = 180 - (48 + 48) = 180 - 96 = 84o

but < PRQ + < PRS = < QRS

< PRQ = < QRS - < PRS - 84 - 65

= 19o

1,209.

The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.

A.

2.32cm

B.

1.84cm

C.

0.62cm

D.

0.26cm

Correct answer is C

No explanation has been provided for this answer.

1,210.

The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS

A.

144o

B.

72o

C.

40o

D.

36o

Correct answer is A

Reflex < POS = 2x (angle at centre is twice that at circumference)

reflex < POS + < POS = 350o(angles at a point)

i.e. 2x + 8x = 360o

10x = 360o

x = \(\frac{360}{10}\)

= 36o

< PRS = \(\frac{1}{2}\)

< POS(< at centre twice that circumference)

= \(\frac{1}{2}\) x 8x = 4x

4 x 36o

< PRS = 144