Correct Answer: B. 5.50

Explanation

Mean \(\bar{x}\) = \(\frac{3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10}{10} \)

= \(\frac{55}{10} = 5.50\)

Correct Answer: B. \(7x^{2} - 4x + 5 = 0\)

Explanation

Let the roots of the equation be \(\alpha\) and \(\beta\).

\(\alpha + \beta = \frac{-b}{a} = \frac{4}{7}\)

\(\alpha \beta = \frac{c}{a} = \frac{5}{7}\)

\(\implies a = 7, b = -4, c = 5\)

Equation: \(ax^{2} + bx + c = 0 \)

= \(7x^{2} - 4x + 5 = 0\)

Correct Answer: C. 60.255

Explanation

\((1.98)^{6} = (1 + 0.98)^{6} = 1 + 6(0.98) + 15(0.98)^{2} + 20(0.98)^{3} + 15(0.98)^{4} + 6(0.98)^{5} + (0.98)^{6}\)

\(\approxeq 1 + 5.88 + 14.406 + 18.823 + 13.836 + 5.424 + 0.886 \)

= \(60.255\)

Explanation

\(U = 25\)

\(\therefore (2x + 1) + x + (x - 2)^{2} = 25\)

\(x^{2} - 4x + 4 + x + 2x + 1 = 25\)

\(x^{2} - x + 5 - 25 = 0\)

\(x^{2} - x - 20 = 0\)

\(\implies x^{2} - 5x + 4x - 20 = 0\)

\(x(x - 5) + 4(x - 5) = 0\)

\(\implies \text{x = -4 or 5}\)

But x cannot be negative in this case, hence, x = 5.

(b) Number of guests that selected just one meal = \(2x + 1 + x^{2} - 4x + 4\)

\(x^{2} - 2x + 5 = 5^{2} - 2(5) + 5 = 25 - 10 + 5 = 20\)

Probability of selecting a person who picked one meal = \(\frac{20}{25} = \frac{4}{5}\).

Correct Answer: C. \(\sim q \implies \sim p\)

Explanation

(a) \(^{n + 1}C_{3} - ^{n - 1}C_{3}\)

= \(\frac{(n + 1)!}{3! (n + 1 - 3)!} - \frac{(n - 1)!}{3! (n - 1 - 3)!}\)

= \(\frac{(n + 1)!}{3! (n - 2)!} - \frac{(n - 1)!}{3! (n - 4)!}\)

= \(\frac{(n + 1)(n)(n - 1)(n - 2)!}{3! (n - 2)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)!}{3! (n - 4)!}\)

= \(\frac{(n + 1)(n)(n - 1)}{6} - \frac{(n - 1)(n - 2)(n - 3)}{6}\)

= \(\frac{(n - 1)[(n + 1)(n) - (n - 2)(n - 3)]}{6}\)

= \(\frac{(n - 1)[n^{2} + n - (n^{2} - 5n + 6)}{6}\)

= \(\frac{(n - 1)[6n - 6]}{6}\)

= \(\frac{6(n - 1)^{2}}{6}\)

= \((n - 1)^{2}\)

(b) p(six) = p = \(\frac{1}{6}\), p(not six) = q = \(\frac{5}{6}\).

\((p + q)^{5} = p^{5} + 5p^{4} q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)

(i) p( at most 2 sixes) = \(q^{5} + 5pq^{4} + 10p^{2} q^{3}\)

= \((\frac{5}{6})^{5} + 5(\frac{1}{6})(\frac{5}{6})^{4} + 10(\frac{1}{6})^{2} (\frac{5}{6})^{3}\)

= \(\frac{3125}{7776} + \frac{3125}{7776} + \frac{1250}{7776}\)

= \(\frac{7500}{7776} = 0.9645 \approxeq 0.965\) (3 d.p)

(ii) p(exactly 3 sixes) = \(10p^{3} q^{2}\)

= \(10(\frac{1}{6})^{3} (\frac{5}{6})^{2}\)

= \(\frac{250}{7776}\)

= \(0.032\) (3 d.p)

Correct Answer: B. 22

Explanation

\(\text{Total frequency} = 50\)

Median occurs at \(\frac{50}{2} = \text{25th position}\)

= Class 20 - 24 with classmark \(\frac{20 + 24}{2} = 22\)

Explanation

\((x - a)^{2} + (y - b)^{2} = r^{2}\)

\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Using A(3, 2) :

\(9 - 6a + a^{2} + 4 - 4b + b^{2} = r^{2}\)

\(13 - 6a - 4b + a^{2} + b^{2} = r^{2} .... (1)\)

Using B(-1, -2) :

\(1 + 2a + a^{2} + 4 + 4b + b^{2} = r^{2}\)

\(5 + 2a + 4b + a^{2} + b^{2} = r^{2} .... (2)\)

Using C(5, -4) :

\(25 - 10a + a^{2} + 16 + 8b + b^{2} = r^{2}\)

\(41 - 10a + 8b + a^{2} + b^{2} = r^{2} ... (3)\)

(1) - (2) :

\(8 - 8a - 8b = 0 \implies 1 - a - b = 0 .... (4)\)

(3) - (2) :

\(24 - 12a + 12 + 4b = 0 \implies 36 - 12a + 4b = 0 \)

\(\equiv 9 - 3a + b = 0... (5)\)

Solving (4) and (5), we have :

\(a = 2\frac{1}{2} ; b = -1\frac{1}{2}\)

Coordinates of the centre : \(C(2\frac{1}{2}, -1\frac{1}{2})\).

(b) To get the radius, we can make use of any point on the circle as well as the centre.

Using point \(A(3, 2) & C(2\frac{1}{2}, -1\frac{1}{2})\)

Radius = \(\sqrt{(3 - 2\frac{1}{2})^{2} + (2 + 1\frac{1}{2})^{2}}\)

= \(\sqrt{\frac{1}{4} + \frac{49}{4}}\)

= \(\sqrt{\frac{50}{4}}\)

= \(\frac{5\sqrt{2}}{2} cm\)

(c) Equation of the circle :

\((x - 2\frac{1}{2})^{2} + (y + 1\frac{1}{2})^{2} = (\frac{5\sqrt{2}}{2})^{2}\)

\(x^{2} - 11x + \frac{25}{4} + y^{2} + 3y + \frac{9}{4} = \frac{50}{4}\)

\(x^{2} + y^{2} - 11x + 3y + \frac{34}{4} - \frac{50}{4} = 0\)

\(x^{2} + y^{2} - 11x + 3y - 4 = 0\)

Correct Answer: C. p = 6, k = 6

Explanation

\(\frac{P}{2} + \frac{k}{3}\) = 5

\(\frac{3p + 2k}{6}\) = 5

2p + 3k = 30

- 2p - k = 6

\(\overline{\frac{4k}{4} = \frac{24}{4}}\)

k = 6

From 2p - k = 6

2p - 6 = 6

\(\frac{2p}{2} = \frac{12}{2}\)

p = 6

Explanation

(a)

(b)(i) Lower quartile, \(Q_{1} = 32.5\)

Upper quartile, \(Q_{3} = 53.0\)

Semi-interquartile range = \(\frac{1}{2} (Q_{3} - Q_{1})\)

= \(\frac{1}{2} (53.0 - 32.5) = \frac{1}{2} (20.5)\)

= \(10.25\)

(ii) Number of students that passed with distinction = (80 - 74)

= 6 students

Percentage = \(\frac{6}{80} \times 100% = 7.5%\)