Correct Answer: C. 21

Explanation

Let's think of this list of numbers: 2, 4, 6, ... Where does it end? The first 5 even numbers goes to 10. That means that the last number in the first 20 will be the number 40. So the question is, Where is the middle? Well, this is an even number of values, so there is no actual middle. What we have to do, then is find the 10th and the 11th numbers and take their average. The 10th number is easy, based on what we just said. If the 5th is 10, then the 10th is 20. The 11th will just be two more than that, namely 22. To calculate the median, we just have to find the average of those two numbers: 20+22/2=21 If you prefer to write out the full list: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40

Correct Answer: C. 2/5

Explanation

The given set {1, 2, 3, 4 ...., 9, 10}

(x ? 7 ) = 7 7, 8, 9, 10 = 4

Total number of possible outcome = 10

4/10 = 2/5

Correct Answer: C. 36

Correct Answer: C. \(\frac{3}{8}\)

Correct Answer: D. \(-\frac{5}{8}\)

Explanation

\((\frac{1}{9})^{2x-1} = (\frac{1}{81})^{2-3x}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{2(2-3x)}\)

\((\frac{1}{9})^{2x-1} = (\frac{1}{9})^{4-6x}\)

Since the bases are equal, powers can be equated

= 2x - 1 = 4 - 6x

= 2x + 6x = 4 + 1

= 8x = 5

\(\therefore x = \frac{5}{8}\)

Correct Answer: D. x = \(\frac {-(9c + 5dy)}{4y - 3}\)

Explanation

y = \(\frac {3x - 9c}{4x + 5d}\)

\(\frac {y}{1} = \frac {3x - 9c}{4x + 5d}\)

= y(4x + 5d) = 3x - 9c

= 4xy + 5dy = 3x - 9c

= 4xy - 3x = -9c - 5dy

= (4y - 3)x = -9c - 5dy

= x = \(\frac {-9c - 5dy}{4y - 3}\)

x = \(\frac {-(9c + 5dy)}{4y - 3}\)

Explanation

(a) Area of trapezium PQRS = \(\frac{1}{2}(PQ + RS) \times 40\)

= \(\frac{1}{2}(20 + 60) \times 40\)

= \(40 \times 40\)

= \(1600 cm^{2}\)

(b) In \(\Delta\) SPT,

\(|SP|^{2} = |ST|^{2} + |TP|^{2}\) (Pythagoras theorem)

\(|ST|^{2} = |SP|^{2} - |TP|^{2}\)

= \(50^{2} - 40^{2}\)

= \(2500 - 1600\)

\(|ST|^{2} = 900\)

\(\therefore |ST| = 30 cm\)

\(|UR| = 60 cm - (30 + 20) cm\)

= 10 cm

Let < QRS = \(\alpha\),

\(\tan \alpha = \frac{40}{10} = 4\)

\(\alpha = \tan^{-1} (4)\)

= \(75.96° \approxeq 76°\)

Correct Answer: B. \(2,\frac{3}{2}\)

Explanation

\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}\)

\(p = \begin{bmatrix} x&4\\3&7\end{bmatrix}=7x-12, Q =\begin{bmatrix} x&3\\1&2x\end{bmatrix}=2x^2-3\)

|Q| = |P| + 3 (Given)

\(= 2\times2 - 3 = 7x - 12 + 3\)

\(= 2\times2 - 3 - 7x + 12 - 3 = 0\)

\(= 2\times2 - 7x + 6 = 0\)

\(= 2\times2 - 4x - 3x + 6 = 0\)

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

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

\(\therefore x=2,\frac{3}{2}\)

Correct Answer: C. x\(^2\) + y\(^2\) - 10x + 8y + 25 =0

Explanation

(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)

x\(^2\) - 2hx + y\(^2\) - 2ky + h\(^2\) + k\(^2\) = r\(^2\)

x\(^2\) - 2(3)x + y\(^2\) - 2(-4) y + 5\(^2\) + (-4)\(^2\) = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 25 + 16 = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 41 = r\(^2\)

at point (5,0)

5\(^2\) - 10(5) + 0\(^2\) + 8(0) + 41 = r\(^2\)

25 - 50 + 41 = r\(^2\)

16 = r\(^2\)

r = \(\sqrt{16}\)

= 4

x\(^2\) + y\(^2\) - 10x + 8y + 25 = 0

Explanation

They were expected to resolve the forces vertically and horizontally to have; (\(^{5 \cos 60^o}_{5 \sin 60^o}\)) + (\(^{P \cos 30^o}_{P \sin 30^o}\)) + (\(^{Q \cos 60^o}_{-Q \sin 60^o}\)) + (\(^{3 \cos 90^o}_{-3 \sin 90^o}\)) + (\(^{-5 \cos 0^o}_{5 \sin 0^o}\)) = (\(^0_0\))

which simplifies to;

(\(^{0.866P + 0.5Q}_{0.5P - 0.866Q}\)) = (\(^{2.5}_{-1.33}\))

This will yield two equations to be solved, the equations are 0.866P + 0.5Q = 2.5 and 0.5P - 0.866Q = -1.33

When these equations are solved, they will give P = 1.5 and Q = 2.4019