Correct Answer: B. 2.0

Explanation

mean = \(\bar{x}\) = \(\frac{\sum x}{n}\)

= \(\frac{5 + 3 + 0 + 7 + 2 + 1}{6}\)

\(\frac{18}{6}\) = 3

\(\begin{array}{c|c}

x & x - \bar{x} & |x - \bar{x}|\\ \hline 5 & -2 & 2 \\ \hline 3 & 0 & 0 \\ \hline 0 & -3 & 3 \\ \hline 7 & 4 & 4 \\ \hline 2 & -1 & 1 \\ \hline 1 & -2 & 2 \\ \hline & & & \sum|x - \bar{x}| \\ \hline & & & 12

\end{array}\)

hence, MD = \(\frac{\sum|x - \bar{x}}{n}|\)

\(\frac{12}{6}\)

= 2

Correct Answer: C. \(72 cm^2\)

Explanation

Let each side be l, then area = \(l^2\)

Using Pythagoras theorem

\(l^2 + l^2 = 12^2\)

\(2l^2 = 144\)

divide both sides by 2

\(l^2 = 72\)

Therefore, the Area of the square is \(72 cm^2\)

Correct Answer: D. 2x²-3x-2=0

Explanation

If x = \(-\frac{1}{2}\) and 2; then

\(x + \frac{1}{2} = 0\) and \(x - 2 = 0\)

\(\implies (x + \frac{1}{2})(x - 2) = 0\)

\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)

\(x^2 - \frac{3}{2}x - 1 = 0\)

\(2x^2 - 3x - 2 = 0\)

Correct Answer: A. 8m

Explanation

Given, Length of minor arc = 5.5m

Angle subtended by minor arc = 360° - 210° = 150°

\(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \)

\(\frac{55r}{21} = 5.5\)

\(r = \frac{5.5 \times 21}{55}\)

r = 2.1m

Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\)

= \(7.7m \approxeq 8m\) (to the nearest metre)

Correct Answer: C. \(\frac{-3}{2}\)

Explanation

\(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0

using 6x as lcm

→ \(\frac{6+8-5+6x}{6x}\)

→ \(\frac{9+6x}{6x}\) = 0

9+6x = 0

6x = -9

x = \(\frac{-9}{6}\) or \(\frac{-3}{2}\)

Correct Answer: D. 5x + 3y - 1 = 0

Explanation

Midpoint between P(4, 5) and Q(-6, -1) = \(\frac{4 - 6}{2}, \frac{5 - 1}{2} = (-1, 2)\)

Gradient of PQ = \(\frac{5 - (-1)}{4 - (-6)} = \frac{3}{5}\)

Gradient of line perpendicular to PQ = \(\frac{-1}{\frac{3}{5}} = -\frac{5}{3}\)

\(line = \frac{y - 2}{x + 1} = \frac{-5}{3}\)

\(3y - 6 = -5x - 5\)

\(5x + 3y - 6 + 5 = 0 \implies 5x + 3y - 1 = 0\)

Correct Answer: B. Isosceles triangle

Explanation

Let A = (1, -3), B = (6, 2) and C = (0,4)

\(d =\sqrt{ (y_2 - y_1)^2 + (x_2 - x_1)^2}\)

|AB| = \(\sqrt{(2 - (-3))^2 + (6 - 1)^2}\)

= \(\sqrt{(2 + 3)^2 + (6 - 1)^2}\)

= \(\sqrt{5^2 + 5^2}\)

= \(\sqrt{25 + 25}\)

= \(\sqrt50\)

= 5\(\sqrt2\) units

|BC| = \(\sqrt{(4 - 2)^2 + (0 - 6)^2}\)

= \(\sqrt{2^2 + (-6)^2}\)

= \(\sqrt{4 + 36}\)

= \(\sqrt40\)

= 2\(\sqrt10\) units

|AC| = \(\sqrt{(4 - (-3)^2 + (0 - 1)^2}\)

= \(\sqrt{(4 + 3)^2 + (0 - 1)^2}\)

= \(\sqrt{7^2 + (-1)^2}\)

= \(\sqrt{49 + 1}\)

= \(\sqrt50\)

= 5\(\sqrt{2}\) units

Since two sides are equal (|AB| = |AC|), then the triangle is Isosceles.

Correct Answer: C. \(\frac{10!}{4!2!2!}\)

Correct Answer: D. 20 700πcm³

Correct Answer: C. x = \(\sqrt[3] \frac{3yz + b}{a}\)

Explanation

\(y = \frac{ax^3 - b}{3z}\)

cross multiply

\(ax^3 - b\) = 3yz

\(ax^3\) = 3yz + b

divide both sides by a

\(x^3 = \frac{3yz + b}{a}\)

take cube root of both sides

therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)