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

Tan 74 = 150/x x = 150/tan 74 = 43.01cm

The word LEADER has 1L 2E 1A 1D and 1R making total of 6! \(\frac{6}{1!2!1!1!1!}\) = \(\frac{6!}{2!}\) = \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\) = 360

The two numbers must be expressed in the same unit. To convert 495g to kg, it will be divided by 1000 495g = \(\frac{495}{1000}\) = 0.495kg To express in percentage, 0.495 will be divided by 16.5 and then multiplied by 100 % will be added to the answer \(\frac{0.4950}{16.5}\) x 100 = 3%

2\sqrt3 - 4) ( 2\sqrt3 + 4) = 12 + 8\sqrt3 - 8\sqrt3 – 16 = 12 – 16 = -4 The two expressions in the bracket are conjugate of each other

log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\) = log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\) = log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\) Based on law of logarithm = 3 log\(_2\) 2 – (-2 log\(_3\) 3) But log\(_2\) 2 = 1, log\(_3\) 3 = 1 So, = 3 + 2 = 5

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

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

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

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

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

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

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

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