Correct Answer: B. x = 1 and y = 1

Explanation

x + y = 2........... (i)

3x - 2y = 1......(ii)

From (i): x = 2 - y ..........(iii)

subt. (2 - y) for x in (ii)

3(2 - y) 2y = 1

6 - 3y - 2y = 1

-5y = 1 - 6 = -5

y = \(\frac{-5}{-5}\) = 1

subst. (i) for y in (iii)

x = 2 - 1 = 1

x = 1, y = 1

Correct Answer: A. 1 - x

Explanation

If p is the probability of an event happening(occurring) and the probability of an event not occurring is p¹ then p + p¹ = 1

p¹ = 1 - p

x¹ = 1 - x

Correct Answer: D. \(\frac{1}{3}\)

Explanation

8^x-1 = \(\frac{1}{4}\)

2^{3(x - 1} = 2^-2

3x - 3 = -2

3x = -2 + 3

3x = 1

x = \(\frac{1}{3}\)

Correct Answer: C. \(9\pi\)

Explanation

Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Given that \(x^{2} + y^{2} - 4x + 8y + 11 = 0\)

Expanding the equation of a circle, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Comparing this expansion with the given equation, we have

\(2a = 4 \implies a = 2\)

\(-2b = 8 \implies b = -4\)

\(r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9\)

\(r = 3\)

\(Area = \pi r^{2} = \pi \times 3^{2}\)

= \(9\pi\)

Correct Answer: D. 20

Explanation

\(\frac{(n - 2)180}{n}\) = 162 (where n = no. of sides)

162n = 180n - 360

162n - 180n = -360

-18n = - 360

n = \(\frac{360}{18}\)

n = 20

Correct Answer: A. least value of x is -3.2

Explanation

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Correct Answer: D. 4.0cm

Explanation

Distance(d) = d² = 5² - 3²

= 25 - 9

= 16

d = \(\sqrt{16}\)

= 4cm

Correct Answer: D. 8.66 N

Explanation

\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)

\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)

\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)

\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)

= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)

= 8.66 N

Correct Answer: D. 0.976

Explanation

\(P(\text{at least one solves the problem}) = 1 - P(\text{none solving the problem})\)

= 1 - (0.4)(0.3)(0.2)

= 1 - 0.024

= 0.976

Explanation

P(at most two winning prizes) = 1 - [P(all winning) + P(three winning)]

P(three winning) = \(P(ABCD') + P(ABC'D) + P(AB'CD) + P(A'BCD)

P(A) = \(\frac{1}{5}\), P(A') = \(\frac{4}{5}\)

P(B) = \(\frac{1}{4}\), P(B') = \(\frac{3}{4}\)

P(C) = \(\frac{1}{3}\), P(C') = \(\frac{2}{3}\)

P(D) = \(\frac{1}{2}\), P(D') = \(\frac{1}{2}\)

P(three winning) = \((\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2}) + (\frac{1}{5} \times \frac{3}{4} \times \frac{1}{3} \times \frac{1}{2}) + (\frac{4}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2})\)

= \(\frac{1}{120} + \frac{2}{120} + \frac{3}{120} + \frac{4}{120}\)

= \(\frac{10}{120}\)

P(all winning) = \(\frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2}\)

= \(\frac{1}{120}\)

P(at most 2 winning) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)

= \(\frac{109}{120}\)