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.

Simplify \(\frac{(a^2 - \frac{1}{a}) (a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)

A.

a^{\(\frac{2}{3}\)}

B.

a^{-\(\frac{1}{3}\)}

C.

\(a^{2}\) + 1

D.

a

E.

a^{\(\frac{1}{3}\)}

Correct answer is **E**

\(\frac{(a^2 - \frac{1}{a})(a^{\frac{4}{3}} + a^{\frac{2}{3}})}{a^2 - \frac{1}{a}^2}\)

= \(\frac{(\frac{a^2 - 1}{a})(\frac{a^2 + 1}{a^{\frac{2}{3}}})}{a^{4} - \frac{1}{a^2}}\)

= \(\frac{a^4 - 1}{a^{\frac{5}{3}}}\) x \(\frac{a^2}{a^4 - 1}\)

= a\(\frac{1}{3}\)

Without using tables, simplify \(\frac{1n \sqrt{216} - 1n \sqrt{125} - 1n\sqrt{8}}{2(1n3 - 1n5)}\)

A.

-3

B.

3

C.

\(\frac{3}{5}\)

D.

\(\frac{-3}{2}\)

E.

1n 6 - 2 1n 5

Correct answer is **C**

No explanation has been provided for this answer.

The locus of all points having a distance of 1 unit from each of the two fixed points a and b is

A.

A line parallel to the line ab

B.

A line perpendicular to the line ab through the mid-point of ab

C.

A circle through a and b with centre at the mid-point of ab

D.

A circle with centre at a and passes through b

E.

A circle in a plane perpendicular to ab and centre at the mid-point of the line ab

Correct answer is **B**

The locus of all points having a distance of 1 unit from each of the two fixed points. a and b is: a perpendicular to the (ab) through the mid-points ab

If (25)^{x - 1} = 64(\(\frac{5}{2}\))^{6}, then x has the value

A.

7

B.

4

C.

32

D.

64

E.

5

Correct answer is **B**

(25)^{x - 1} = 64(\(\frac{5}{2}\))^{6}

(\(\frac{5}{2}\))^{6} = (\(\frac{5^6}{64}\))

25^{x - 1} = 64 x (\(\frac{5^6}{64}\))

5^{2x - 2} = 5^{6}

2x - 2 = 6x

= \(\frac{8}{2}\)

= 4

Two triangles have the same areas if

A.

Two sides in one triangle are equal to two sides in the other

B.

Three sides in one triangle are equal to three sides in the other

C.

Two angles in the triangle are equal to two angles in the other

D.

Three angles in one triangle are equal to three angles in the other

E.

One side and the opposite angle in one triangle are equal to one side and the opposite angle in the other

Correct answer is **B**

Two triangles have the same area if three sides in one triangle equal to three in the other.

The quantity (x + y) is a factor of

A.

x^{2} + y^{2}

B.

x^{3} - y^{3}

C.

2x^{2} - 3xy + y^{2} - x + 1

D.

2x^{3} + 2x^{2}y - xy + 3x - y^{2} + 3y

E.

x^{5} - y^{5}

Correct answer is **D**

(x + y) is a factor 2x^{3} + 2x^{2}y - xy + 3x - y^2

A.

The volume of the sphere is greater than the volume of the cone

B.

The volume of the cone is less than the volume of the cylinder

C.

The total surface area of the cone is greater than that of the sphere

D.

The total surface area of the cylinder is less than that of the sphere

E.

The total sufeac area of the cone is equal to that of the cylinder

Correct answer is **C**

No explanation has been provided for this answer.

The set of value of x and y which satisfies the equations x^{2} - y - 1 = 0 and y - 2x + 2 = 0 is

A.

1, 0

B.

1, 1

C.

2, 2

D.

0, 2

E.

1, 2

Correct answer is **A**

x^{2} - y - 1 = 0.......(i)

y - 2x + 2 = 0......(ii)

By re-arranging eqn. (ii)

y = 2x - 2........(iii)

Subst. eqn. (iii) in eqn (i)

x^{2} - (2x - 2) - 1 = 0

x^{2} - 2x + 1 = 0

= (x - 1) = 0

When x - 1 = 0

x = 1

Sub. for x = 1 in eqn. (iii)

y = 2 - 2 = 0

x = 1, y = 0

A.

1cm

B.

\(\sqrt{\frac{3\pi}{24}}\)

C.

\(\frac{\pi}{24\sqrt{3}}\)

D.

24\(\sqrt{3}\)

Correct answer is **C**

The rise of water is equivalent to the volume of the sphere of radius \(\frac{1}{2}\)cm immersed x \(\frac{1}{\text{No. of sides sq. root 3}}\)

Vol. of sphere of radius = 4\(\pi\) x \(\frac{1}{8}\) x \(\frac{1}{3}\) - (\(\frac{1}{2}\))^{3}

= \(\frac{1}{8}\)

= \(\frac{\pi}{6}\) x \(\frac{1}{4\sqrt{3}}\)

= \(\frac{\pi}{24\sqrt{3}}\)

If sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3, then the angle \(\theta\) is equal to

A.

30^{o}

B.

45^{o}

C.

60^{o}

D.

90^{o}

E.

105^{o}

Correct answer is **B**

sec\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

Where 1 + tan\(^2\) \(\theta\) = sec\(^2\) \(\theta\)

1 + tan\(^2\) \(\theta\) + tan\(^2\) \(\theta\) = 3

2 tan\(^2\) \(\theta\) = 2

tan\(^2\) \(\theta\) = 1

tan\(\theta\) = √1

where √1 = 1

tan\(\theta\) = 1

And tan 45° = 1

∴ \(\theta\) = 45°

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