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Prime numbers are numbers that has only two factors (i.e 1 and itself). They are numbers that are only divisible by 1 and their selves. First four Prime numbers greater than 10 are 11, 13, 17 and 19 Average = sum of numbers / number = \(frac{(11 + 13 + 17 + 19)}{4}\) = \(\frac{60}{4}\) = 15

The modal age is the age with the highest frequency, and that is age 5 years with f of 7

Since they are of different base, convert to base 10 243\(_{six}\) = (2 x 62) + (4 x 61) + (3 x 60) = 72 + 24 + 3 = 99 base 10 243\(_{six}\) = 2 x 52 + 4 x 51 +3 x 50 50 + 20 + 3 = 73 base 10 Subtracting them, 99 - 73 = 26

∫\(\frac{5}{x}\) dx = 5 ∫\(\frac{1}{x}\) = 5Inx Since the integral of \(\frac{1}{x}\) is Inx ∫\(^2\) \(_1\)∫ \(\frac{5}{x}\) dx = 5 dx = 5 (In<2 – InIn1) = 3.4657 = 3.47

P( tail on a coin) = \(\frac{1}{2}\) Even numbers on a care 2, 4 and 6 P( even number on a die) = \(\frac{3}{6}\) = \(\frac{1}{2}\) P( tail on a coin and even number on a die) = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)

To form words having 3 consonants and 2 vowels out of 7 consonants and 4 vowels, the number of such words is 7/3C x 4/2C = 35 x 6 = 210

Using Pythagoras theorem PR\(^2\) = 5\(^2\) + 12\(^2\) 25 + 144 = 169 PR = \sqrt(169)= 13km

First we convert the numbers to base ten 23\(_{five}\)= 2 x 51 + 3 x 50 = 10 + 3 = 13 101\(_{five}\) = (1 x 32) + (0 x 31) + (1 x 30) = 9 + 0 + 1 = 10 So, y = 13 + 10 = 23 To convert 23 to base 2 (as in the diagram above) Y = 23 = 10111\(_{five}\) Answer is B

Total number of balls = 2 + 4 = 6 P(of picking a red ball) = \(\frac{2}{6}\) = \(\frac{1}{3}\) P(of picking a blue ball) = \(\frac{4}{6}\) = \(\frac{2}{3}\) With replacement, P( picking two red balls) = \(\frac{1}{3}\) \times \(\frac{1}{3}\) = \(\frac{1}{9}\)

A1 = Elements in the universal set but not in A = {s, w, x, y, z} B = {r, s. t, u} C = {t, u, v, w, x} A1 n B n C = elements common to the three sets = none = empty set = Φ

An equilateral triangle has all sides equal and all angles equal as 600 Area = \(\frac{1}{2}\) absin\theta Area = \(\frac{1}{2}\) x 8 x 8 x sin60 = \(\frac{1}{2}\) x 64 x \(\sqrt{\frac{3}{2}}\) = 16\sqrt3 cm\(^2\)

9 is on one decimal place, the next number to it is 8 which will be rounded up to 1 because it is greater than 5 and then added to 9 to give 10, 10 cannot be written, it will then be rounded up to 1 and added to 0. So the answer is 1.0

Using law of indices 25\(^{(\frac{1}{2}}\) \times 8\(\frac{-2}{3}\) = \(\sqrt{25}\) x (\(\sqrt[3]{8}\))\(^{ -2}\) = 5 x 2\(^{-2}\) = 5 x \(\frac{1}{2^2}\) = \(\frac{5}{4}\) = \(\frac{11}{4}\) The answer is A

Cos \theta = \(\frac{adj}{hyp}\) = \(\frac{300}{600}\) = 0.5 \theta = Cos - 10.5 = 60 ∠ RPQ = ∠ PQs So the bearing of P from Q is 180 + 60 = 240\(^o\) Answer is D

54 \(_{eight}\) and 67\(_{eight}\) = 1438 Starting with normal addition, 4 + 7 gives 11 (it is more than the base, 8) 8 goes in 11 just 1 time, remaining 3, the remainder will be written, and the 1 will be added to the sum of 5 and 6 which gives 12 altogether, 8 goes in 12 one time remaining 4, the remainder 4 was written and then the 1 that was the quotient was then written since nothing to add the 1 to. So answer is 143 in base eight