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.

Find the mean deviation of a set of numbers: 14, 15, 16, 17, 18, and 19.

A.

2.5

B.

1.7

C.

1.5

D.

3.5

Correct answer is **C**

x̄ = \(\frac{ 14 + 15 + 16 +17 + 18 + 19}{6} = \frac{99}{6}\) = 16.5

M . D = \(\frac{ (14 - 16.5) + (15 - 16.5) + (16 - 16.5) + (17 - 16.5) + (18 - 16. 5) + (19 - 16.5) }{ 6}\)

M.D = \(\frac{(-2.5) + (-1.5) + (-0.5) + (0.5) + (1.5) + (2.5)}{6}\)

Taking the absolute value of the deviations

M.D = \(\frac{2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5}{6}\)

M.D = \(\frac{9}{6}\) = 1.5

A.

288

B.

400

C.

300

D.

225

Correct answer is **D**

\(M ∝ n^2\sqrt{q}\)

\(M = Kn^2\sqrt{q}\)

K = \(\frac{M}{n^2\sqrt{q}}\)

K = \(\frac{24}{2^2\sqrt4}\)

k = \(\frac{24}{8} = 3\)

Now, let's find M when n = 5 and q = 9

M = \(Kn^2\sqrt{q}\)

M = \( 3\times5^2\sqrt9\)

\(M = 3\times25\times3\)

Therefore, M = 225.

The diagonals of a rhombus are 16 cm and 12 cm find the length of the side.

A.

20cm

B.

8cm

C.

14cm

D.

10cm

Correct answer is **D**

In a rhombus, the diagonals are perpendicular bisectors of each other, and they bisect the angles of the rhombus. This means that a rhombus is essentially made up of four congruent right-angled triangles.

We can use the Pythagorean theorem to find the length of one side of the rhombus (s)

\(s^2 = 8^2 + 6^2\)

\(s^2 = 64 + 36\)

\(s^2 = 100\)

s = \(\sqrt{100}\)

s =10 cm

So, the length of each side of the rhombus is 10 cm.

If 2x - 3y = -11 and 3x + 2y = 3, evaluate \( (y - x)^2\)

A.

16

B.

25

C.

9

D.

4

Correct answer is **A**

2x - 3y = -11 --- (i)

3x + 2y = 3 --- (ii)

Multiply equation (i) by 3 and equation (ii) by 2

6x - 9y = -33 --- (iii)

6x + 4y = 6 --- (iv)

Subtract equation (iii) from (iv)

13y = 39

y = \(\frac{39}{13}\) = 3

substitute (3) for y in equation (ii)

3x + 2(3) = 3

3x + 6 = 3

3x = 3 - 6

3x = -3

x = \(\frac{-3}{3}\) = - 1

Now,

\((y - x)^2 = (3 - (-1))^2\)

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

= \(4^2\)

= 16