Correct Answer: C. 5/19

Explanation

Probability that the student is against the proposal

= \(\frac{32 + 48}{128 + 32 + 96 + 48}\)

= \(\frac{80}{304}\)

= \(\frac{5}{19}\)

Explanation

(a) Let S = sample space = integers from 5 to 25 inclusive.

F = multiple of 5 ; T = multiple of 3.

S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}

F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24}

P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\)

(b) Let B = blue balls and R = red balls

n(B) = 4 ; n(R) = 6.

(i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\)

= \(\frac{9}{25}\)

(ii) P(both are same colour) = P(both are red) or P(both are blue)

= \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\)

= \(\frac{9}{25} + \frac{4}{25}\)

= \(\frac{13}{25}\)

Explanation

Put the value of d in (1), so we have

\(a + 0.75 = 2 \implies a = 1.25\)

Therefore, the length of the poles in metres = \(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\)

Explanation

(a) Formula : Volume = \(\frac{1}{3} \pi r^{2} h\)

\(\pi = 3.142; r = 4cm; h = 6cm\)

\(Volume = \frac{1}{3} \times 3.142 \times 4^{2} \times 6 = 100.544 cm^{3}\)

(b) Volume of hemisphere = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}\)

= \(\frac{2}{3} \times \frac{22}{7} \times 5^{3}\)

= \(\frac{5500}{21} cm^{3} = 261.905 cm^{3}\)

Volume of water discharged per second = \(\pi r^{2} h\)

= \(\frac{22}{7} \times 0.2m \times 0.2m \times 2m = \frac{1.76}{7}\)

= \(0.2514 m^{3} / sec\)

\(\therefore \text{The time to fill the tank = } \frac{261.905}{0.2514 \times 60}\)

= \(\frac{261.905}{15.084} = 17.363 mins\)

\(\approxeq 17.4 mins\)

Correct Answer: D. 3.6258

Explanation

log(0.65)² = 2log(0.65)

but log0.65 = 1.8129

2 x 1.8129 = 3.6258

Correct Answer: B. 1.5

Explanation

3^2x = 27

3^2x = 3³

2x = 3

x = 1.5

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: B. 63°

Explanation

tan θ - 7/3.5 = 2

tan θ= 2.0000

θ = 63°

Explanation

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

Table of value for \(y = x + 6\)

(b)

(d) (i) Least value of y = -2 occurs at x = 1

(ii) Solution of \(2x^{2} - 5x + 1 = x + 6\) are x = -0.7 and x = 3.7.