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No expanded working has been attached for this question yet. The verified answer is Option D.

M = N \(\sqrt{\frac{SL}{T}}\), make T subject of formula square both sides M\(^{2}\) = \(\frac{N^2SL}{T}\) TM\(^{2}\) = N\(^{2}\)SL T = \(\frac{N^2SL}{M^2}\)

3\(^{n - 1}\) \times \(\frac{27^{n + 1}}{81^n}\) = 3\(^{n - 1}\) \times \(\frac{3^{3(n + 1)}}{3^{4n}}\) = 3\(^{n - 1 + 3n + 3 − 4n}\) = 3\(^{4n − 4n − 1 + 3}\) = \(3^{2}\) = 9

The locus of points equidistant from a given straight line PQ consists of two parallel lines, one on each side of PQ, each at the same perpendicular distance from it.

If RST = 60o RXT = 2 × RST (angle at the centre twice angle at the circumference) RXT = 2 × 60 = 120o

Range = Highest Number - Lowest Number Mode is the number with highest occurrence 10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5 Range = 10 − 4 = 6 Mode = 10 Sum of range and mode = range + mode = 6 + 10 = 16

Sum to infinity ∑ = arn − 1 = \(\frac{a}{1}\) − r a = \(\frac{1}{4}\) r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\) r = \(\frac{1}{s}\) × \(\frac{4}{1}\) = \(\frac{1}{2}\) S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\) = \(\frac{1}{4}\) ÷ \(\frac{1}{2}\) = \(\frac{1}{4}\) × \(\frac{2}{1}\) = \(\frac{1}{2}\)

The operation was performed in base 2 because when counting in base 2, there are only two possible digits: 0 and 1.

when x=2, we have \(2 + 2(2^{2})\) = \(2 + 8 = 10\)

2x + x = 180o 3x = 180o x = 60o (exterior angle of the polygon) angle = \(\frac{\text{total angle}}{\text{number of sides}}\) 60 = \(\frac{360}{n}\) n = \(\frac{360}{60}\) n = 6 sides

Capacity = Volume = 3080m3 base diameter = 14m radius = \(\frac{\text{diameter}}{2}\) = 7m Volume of Cylidner = Capacity of cylinder πr2h = 3080 \(\frac{22}{7}\) × 7 × 7 × h = 3080 h = \(\frac{3080}{22 \times 7}\) h = 20m

4\(\sqrt{27}\) + 5\(\sqrt{12}\) − 3\(\sqrt{75}\) = 4\(\sqrt{3}\) × 9 + 5\(\sqrt{3}\) × 4 − 3\(\sqrt{3}\) × 25 = 4 × 3\(\sqrt{3}\) + 5 × 2\(\sqrt{3}\) − 3 × 5\(\sqrt{3}\) = 12\(\sqrt{3}\) + 10\(\sqrt{3}\) − 15\(\sqrt{3}\) = (12 + 10 − 15)\(\sqrt{3}\) = 7\(\sqrt{3}\)

Let the speed of the 1st trip be x miles/hr and the speed of the 2nd trip be 3x miles/hr Speed = distance/time ∴ Time taken to cover a distance of 50 miles on the 1st trip = \(\frac{50}{xhr}\) time taken to cover a distance of 300 miles on the next trip = \(\frac{300}{3xhr}\) = \(\frac{100}{xhr}\) ∴the new time compared with the old time is twice as much

Sum of angle at a point = 360o 2x + 3x + 4x = 360 9x = 360 x = \(\frac{360}{9}\) x = 40o

by method of long division, we get the answer.