Explanation

(a) Perimeter of a circle = \(2\pi r\).

Length of a side of square = 2r in terms of r.

Sum of perimeter of circle and square = \((2 \pi r) + (4 \times 2r)\)

\(2r (\pi + 4) = 100\)

\(r ( \frac{22}{7} + 4 ) = 50\)

\(r = 50 \div \frac{50}{7}\)

\(r = 7 cm\)

(b) (i) L = \(4 \times breadth\)

B = breadth.

60 cm = 2(4B + B)

2(5B) = 60 cm

\(B = \frac{60}{10} = 6 cm\)

\(L = 4 \times 6 = 24 cm\).

(ii)

Let the diagonal be d.

\(d^{2} = 6^{2} + 24^{2}\)

\(d^{2} = 36 + 576\)

\(d^{2} = 612\)

\(d = \sqrt{612}\)

= \(24.74 cm \approxeq 24.7 cm\).

Correct Answer: A. 56.3°

Explanation

Using the dot product:

a.b = |a||b|cos θ

a.b = 5(-2) + 12(3) = -10 + 36 = 26

|a| = √(52 + 122) = √(25 + 144)

|a| = √169 = 13

|b| = \(√((-2)^2 + 3^2)\) = √(4 + 9)

|b| = √13

= 26 = 13√13 x cos θ

= \(\frac{26}{13\sqrt13}\) = cos θ

=\(\frac{2}{\sqrt13}\) = cos θ

= θ = \(cos^{-1} (\frac{2}{\sqrt13})\)

∴ θ = 56.3º

Correct Answer: c. N750

Correct Answer: D. 7.55

Correct Answer: D. 10000

Explanation

log\(_{10}\) a = 4

a = 10\(^4\)

= 10000

Correct Answer: B. \(6(-\sqrt{3} i - j)\)

Explanation

\(F = F \cos \theta i + F \sin \theta j\)

\(r = (12, 210°) = 12 \cos 210 i + 12 \sin 210 j\)

= \(- 6\sqrt{3} i - 6j\)

= \(6(-\sqrt{3} i - j)\)

Correct Answer: C. 7

Explanation

2x + y = 7...(1)

3x - 2y = 3...(2)

From (1), y = 7 - 2x for y in (2)

3x - 2(7 - 2x) = 3

3x - 14 + 4x = 3

7x + 3 + 14 = 17

x = \(\frac{17}{7}\)

Hence, 7 x \(\frac{17}{7}\)

= 17 - 10

= 7

Explanation

(a) u = 10ms\(^{-1}\), a = 0.2m\(^{-2}\), s = 100m, t = ?

using S = ut + \(\frac{1}{2} at^2\)

100 = 10t + \(\frac{1}{2}(0.2)t^2\)

100 = 10t + 0.1t\(^2\)

0.1t\(^2\) + 10t - 100 = 0

t = \(\frac{-10 \pm \sqrt{10^2 - 4(0.1)(-100)}}{2(0.1)}\)

t = \(\frac{-10\pm \sqrt{100 + 40}}{0.2}\)

t = \(\frac{-10 \pm \sqrt{140}}{0.2}\)

t = \(\frac{-10 \pm 11.83}{0.2}\)

t = \(\frac{-10 - 11.82}{2}\)

t = \(\frac{1.83}{0.2}\)

t = \(\frac{21.83}{2}\)

t = 9.15 seconds

(b)(i) S = t\(^2\) - 5 + 6

v = \(\frac{ds}{dt}\)

= 2t - 5

at t = 0, y = 2(0) - 5

= - 5ms\(^{-1}\)

The initial velocity is 5 in the negative direction

(ii) When t = 0, S = t\(^2\) - 5t + 6

S = 0\(^2\) - 5(0) + 6

S = 6

The distance is 6m

Correct Answer: A. (7, -2)

Explanation

Let (x1, y1) be the image of the point (x, y) under the given transformations.

x1 = 3x - y

y1 = x + 4y

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix} \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}\)

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix} \begin{vmatrix} 7 \\ -2 \end{vmatrix}\)

Correct Answer: B. \(\frac{4}{7}\)

Explanation

\(P(X?Y)=\frac{5}{8}\)

\(P(X?Y)=P(X)\times P(Y)\)

Since X and Y are independent events, the probability of their union (X ? Y) can be calculated as:

\(P(X?Y)=P(X)+P(Y)-P(X?Y)\)

\(=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{1}{2}=P(Y)(1-\frac{1}{8})\)

\(=\frac{1}{2}=P(Y)(\frac{7}{8})\)

\(=P(Y)=\frac{1}{2}÷\frac{7}{8}\)

\(∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}\)