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A = (p, q, r, t, u, v) B = (r, s, t, u) A U B = Elements in both A and B = (p, q, r, s, t, u, v) (A U B)1 = elements in the universal set E but not in (A U B)= (w, x, y, z) n(A U B) 1 = number of the elements in (A U B)1 = 4

Locus is the path traced at by a point which moves in accordance with a certain law. It is also the set of all possible position occupied by an object The path traced from all possible location of 4cm from a given point P form a circle of radius 4cm with centre P.

Sin(5x - 28) = Cos(3x - 50)………..i But Sin\alpha = Cos(90 - \alpha) So Sin(5x - 28) = Cos(90 - [5x - 28]) Sin(5x - 28) = Cos(90 - 5x + 28) Sin(5x - 28) = Cos(118 - 5x)………ii Combining i and ii Cos(3x - 50) = Cos(118 - 5x) 3x - 50 = 118 - 5x Collecting the like terms 3x + 5x = 118 + 50 8x = 168 x = \(\frac{168}{8}\) x = 21\(^o\) Answer is B

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