Correct Answer: D. 5

Correct Answer: B. 126°

Explanation

In the diagram; 108° + x + x = 180° (sum of angle in a triangle)

108° + 2x = 180°

x = 180° - 108°

= 72°

x = \(\frac{72^o}{2}\)

= 36°

(Angle between tangent and a chord through the point of contact)

Hence, angle PTN = 90 + 36

= 126°

Correct Answer: C. {y : y \(\in\) R, y \(\neq\) 5}

Explanation

(i) \((9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1 = 1\)

(ii) (1) \(10 \otimes m = 2\)

By comparison, \(10 \otimes 9 = 2\).

(2) \(n \otimes n = 4\)

From the table, \(9 \otimes 9 = 4\)

Hence, n = 9.

(b) Let the fraction = \(\frac{m}{n}\)

\(\frac{m}{n} = \frac{3}{4}..... (1)\)

\(\implies n = \frac{4m}{3} ....... (2)\)

\(2m = n + 34 ...... (3)\)

Put (2) in (3),

\(2m = \frac{4m}{3} + 34\)

\(2m - \frac{4m}{3} = 34 \implies \frac{2m}{3} = 34\)

\(m = \frac{34 \times 3}{2} = 51\)

\(n = \frac{4m}{3} = \frac{4 \times 51}{3} = 68\)

Hence, the fraction = \(\frac{51}{68}\).

Explanation

Error in length = 60cm - 50cm = 10cm

Error in breadth = 40cm - 35cm = 5cm

(i) % error in length = \(\frac{10}{50} \times 100% = 20%\)

% error in breadth = \(\frac{5}{40} \times 100% = 12.5%\)

(ii) Area of the actual table = \(50cm \times 40cm = 2000cm^{2}\)

Area of the table in error = \(60cm \times 35cm = 2100cm^{2}\)

Error in area = \(2100cm^{2} - 2000cm^{2} = 100cm^{2}\)

% error in area = \(\frac{100}{2000} \times 100% = 5%\)

(b) Product of the error in length and in breadth = \(20 \times 12.5 = 250%\)

Correct Answer: D. x sec x tan x + secx

Correct Answer: C. 3

Correct Answer: B. 45

Explanation

From the diagram SR//UW given that< RSV = 45^o and< WVT = x^o

< WVT = 45^o(corresponding angle)

x^o = 45

Explanation

No. of farmers = 6

No. of traders = 8

Total = 14

Pr(at least two farmers)

= \(\frac{^6C_2 \times ^8C_2}{^{14}C_4} + \frac{^6C_3 \times ^8C_1}{^{14}}+ \frac{^6C_4 \times ^8C_0}{^{14}C_4}\)

= \(\frac{15 \times 28}{1001} + \frac{20 \times 8}{1001} + \frac{15 \times 1}{1001}\)

= \(\frac{420}{1001} + \frac{160}{1001} + \frac{15}{1001}\)

= \(\frac{420 + 160 + 15}{1001}\)

= \(\frac{595}{1001}\)

(b) P(x) = \(\frac{1}{5}\)

p(x\(^1\)) = 1 - \(\frac{1}{5}\) = \(\frac{4}{5}\)

p(y) = \(\frac{1}{2}\), p(y\(^1\))

= 1 - \(\frac{1}{12}\)

= \(\frac{11}{12}\)

p(2) = \(\frac{1}{6}\)

p(z\(^1\)) = 1 - \(\frac{1}{6}\)

= \(\frac{5}{6}\)

(i) Pr(None)

= \(\frac{4}{5} \times \frac{11}{12} \times \frac{5}{6} = \frac{11}{18}\)

(ii) Pr(all the factories)

= \(\frac{1}{5} \times \frac{1}{12} \times \frac{1}{6} = \frac{1}{360^o}\)

(iii) Pr (at least one factory)

= 1 - Pr (none) = \(\frac{1}{1} - \frac{11}{18}\)

= \(\frac{18 - 11}{18}\)

= \(\frac{7}{18}\)

Explanation

(a)

In the diagram above, area of \(\Delta PSR = \frac{1}{2} \times |PR| \times 6\)

\(36 = \frac{1}{2} \times |PR| \times 6\)

\(36 = 3 \times |PR|\)

\(|PR| = 12 cm\)

But |PQ| = |TS| = 8cm (opp. sides of a parallelogram)

Hence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm.

(b)

In the diagram above, \(\beta = 60°\) (alternate angles)

\(\tan \beta = \frac{10.65}{|QB|}\)

\(|QB| = \frac{10.65}{\tan 60}\)

= \(\frac{10.65}{1.732} = 6.149m\)

|PO| = |QB| = 6.149m (opp. sides of a rectangle)

In \(\Delta POT, \tan 45 = \frac{|TO|}{6.149}\)

\(|TO| = \tan 45 \times 6.149 = 1 \times 6.149 = 6.149m\)

|OB| = |PQ| = 10.65m (opp sides of a rectangle)

\(h = |OB| - |OT| \)

= \(10.65 - 6.149\)

= \(4.501 m \approxeq 4.50m\) (3 sig. figs)

Hence, the height of the flagpole = 4.50m