Correct Answer: B. 20

Explanation

\(T_{n} = a + (n - 1) d\) (terms of an A.P)

\(T_{1} = a\)

\(T_{2} = a + d\)

\(T_{10} = a + 9d\)

\(a + a + d = 2a + d = 4 ... (i)\)

\(a + 9d = 19 ... (ii)\)

(ii) x 2: \(2a + 18d = 38 ... (iii)\)

(iii) - (i) : \(17d = 34 \implies d = 2\)

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

\(a = 1\)

\(T_{5} + T_{6}\)

= \((a + 4d) + (a + 5d)\)

= \(2a + 9d\)

= \(2(1) + 9(2)\)

= 20

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) = \(1 - [\frac{1}{120} + \frac{10}{120}]\)

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

Correct Answer: B. \(\beta\)

Explanation

From the diagram, \(g(3) = s; h(s) = \beta\).

Correct Answer: C. 3.5

Explanation

\(U_{n} = a + (n - 1)d\)

\(U_{4} = 2 + (4 - 1) \times 0.5\)

= \(2 + 1.5\)

= 3.5

Correct Answer: C. N1,640

Explanation

(a) \(T_{n} = a + (n - 1)d\) (terms of an A.P)

3rd term = -9 ; 8th term = 26

\(\implies T_{3} = a + 2d = -9 ... (1)\)

\(\implies T_{8} = a + 7d = 26 ..... (2)\)

(i) Solving (2) - (1), we have

\(7d - 2d = 26 - (-9)\)

\(5d = 35 \implies d = 7\)

(ii) Putting d = 7 in (1) above, we have

\(a + 2(7) = -9\)

\(a + 14 = -9\)

\(a = - 9 - 14 = -23\)

(b) \(\frac{XP}{XY} = \frac{2}{5}\) (similar triangles)

\(\frac{24}{\text{Area of \Delta XYZ} = \frac{2^{2}}{5^{2}}\)

\(\text{Area of } \Delta XYZ = \frac{24 \times 25}{4} = 150 cm^{2}\)

\(\therefore \text{Area of trapezium PQZY} = 150 - 24 = 126 cm^{2}\)

Correct Answer: C. \(1\)

Correct Answer: C. 0.00799

Correct Answer: C. 360

Correct Answer: B. 2

Explanation

P ∩ Q = {f, e} = 2