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.

1,751.

If \(3x - \frac{1}{4})^{\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is

A.

x > \(\frac{1}{3}\)

B.

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

C.

x < \(\frac{1}{4}\)

D.

x < \(\frac{9}{16}\)

E.

x > \(\frac{9}{16}\)

Correct answer is E

\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)

= \(3x - 2 > \frac{1}{4} - x\)

= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)

\(x > \frac{9}{16}\)

1,752.

The minimum point on the curve y = x2 - 6x + 5 is at

A.

(1, 5)

B.

(2, 3)

C.

(-3, -4)

D.

(3, -4)

Correct answer is D

Given the curve \(y = x^{2} - 6x + 5\)

At minimum or maximum point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(\frac{\mathrm d y}{\mathrm d x} = 2x - 6\)

\(2x - 6 = 0 \implies x = 3\)

Since 3 > 0, it is a minimum point.

When x = 3, \(y = 3^{2} - 6(3) + 5 = -4\)

Hence, the turning point has coordinates (3, -4).

1,753.

A force of 5 units acts on a particle in the direction to the east and another force of 4 units acts on the particle in the direction north-east. The resultants of the two forces is

A.

\(\sqrt{3}\) units

B.

3 units

C.

\(\sqrt{41 + 20 \sqrt{2}}\) units

D.

\(\sqrt{41 - 20 \sqrt{2}}\) units

Correct answer is C

Force to the east = 5 units

force to the North - east = 4 units.

Resultant of the two forces is the square root

52 + 42 = 41 and plus the sum of its resistance

5 x 4\(\sqrt{2}\) = 20\(\sqrt{2}\)

= \(\sqrt{41 + 20 \sqrt{2}}\) units

1,754.

Simplify \(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\)

A.

\(\frac{4ab}{a - b}\)

B.

\(\frac{-4ab}{a^2 - b^2}\)

C.

\(\frac{-4ab}{a^{-2} - b}\)

D.

\(\frac{4ab}{a^{-2} - b^{-2}}\)

Correct answer is B

\(\frac{a - b}{a + b}\) - \(\frac{a + b}{a - b}\) = \(\frac{(a - b)^2}{(a + b)}\) - \(\frac{(a + b)^2}{(a - b0}\)

applying the principle of difference of two sqrt. Numerator = (a - b) + (a + b) (a - b) - (a + b)

= (a = b + a = b)(a = b - a = b)

2a(-2b) = -4ab

= \(\frac{-4ab}{(a + b)(a - b)}\)

= \(\frac{-4ab}{a^2 - b^2}\)

1,755.

If the four interior angles of a quadrilateral are (p + 10)°, (p - 30)°, (2p - 45)°, and (p + 15)°, then p is

A.

125o

B.

82o

C.

135o

D.

105o

E.

60o

Correct answer is B

Sum of interior angles of a polygon = (2n - 4) x 90

where n is the no. of sides

sum of interior angles of the quad. = ({2 x 4} -4) x 90

(8 - 4) x 90 = 4 x 90o

= 360o

(p + 10)o + (p - 30)o + (2p - 45)o + (p + 15)o = 360o

p + 10o + p - 30o + 2p - 45o + p + 15 = 360o

5p = 360o + 50o

= 82o