Mathematics Questions and Answers

Total number of students = total frequency = 3 + 6 + 9 + 1 + 12 + 7 + 5 + 10 = 53
Since the failed mark was 4,
Total number of students that passed = 12 + 7 + 5 + 10 = 34

Pr( passed) = \(\frac{34}{53}\) = 0. 64

\(8 + 2x - x^2 = 0\)
\(8 + 4x - 2x - x^2 = 0\)
4(2 + x) - x(2 + x) = 0
(4 - x)(2 + x) = 0
4 - x = 0 or 2 + x = 0
x = 4 or x =- 2
So, p = 4 and q = -2
∴ p + q = 2

1st bag → 4 W, 3 B = 7 marbles
2nd bag → 5 R, 6 B = 11 marbles
The only possible way of picking marbles of the same colour is to pick a blue marble from each bag

therefore, Pr( same colour) = \(\frac{3}{7}\times \frac{6}{11} = \frac{18}{77}\)
 

(7, 4) and (-3, 9) = (\( y_1 , x_1)(y_2 , x_2)\)

slope\( m_1\) of the points(7, 4) and (-3, 9) = \(\frac{ y_2 - y_1}{ x_2 - x_1}\)

 \( m_1 = \frac{ 9 - 4}{ -3 - 7} = \frac{5}{-10} = \frac{-1}{2}\)

\( m_1 = m_2\) ( condition for parallelism)
Since the line L passes through the point (6, -13) and is parallel to the points (7, 4) and (-3, 9)

\(\frac{-1}{2} = \frac{ y - (- 13)}{x - 6}\)

= \(\frac{-1}{2} = \frac{ y + 13}{x - 6}\)

= \(\frac{-1}{2}( x - 6) = y + 13\)

\(\frac{-1}{2}x + 3 = y + 13\)

= \(\frac{-1}{2}x + 3 - 13 = y\)

∴ y = \(\frac{-1}{2}x - 10\)

Consider |NS|
θ = 180° - (90° + 35°) (sum of angles on a straight line is 180°)
= 180° - 90° - 35°
= 90° - 35°
= 55°
∴ His new bearing is N55°E.

Total numbers from 40 to 50 inclusive = 11 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50).

Prime numbers between 40 and 50 (inclusive) are 41, 43, and 47.

There are 3 prime numbers in this range.

therefore, Pr(prime numbers) = \(\frac{3}{11}\)

Converting each number to base 10
\(110_{two} = 1 × 2^2 + 1 × 2^1 + 0 × 2^0\)
= 1 × 4 + 1 × 2 + 0 × 1
= 4 + 2 + 0
= \(6_{ten}\)
\(31_{eight} = 3 × 8^1 + 1 × 8^0\)
= 3 × 8 + 1 × 1
= 24 + 1
= \(25_{ten}\)
\(42_{five} = 4 × 5^1 + 2 × 5^0\)
= 4 × 5 + 2 × 1
= 20 + 2
= \(22_{ten}\)
Hence, \(31_{eight} > 42_{five} > 110_{two}\)
In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)

given that diameter = 140cm ⇒radius = 70cm, volume of cylinder = 200litres = \(200,000cm^3\)

volume of cylinder = base area times height = \(\pi r^2h\)

\( 200,000cm^3 = \frac{22}{7} \times 70 \times 70 \times h\)

\(200,000 = \frac{107,800h}{7}\)

cross multiply

1,400,000 = 107,800h 

\( h = \frac{1,400,000}{107,800}\)

   = 13cm (to the nearest cm).

\(y = \frac{ax^3 - b}{3z}\)

cross multiply

\(ax^3 - b\) = 3yz

\(ax^3\) = 3yz + b 

divide both sides by a 

\(x^3 = \frac{3yz + b}{a}\)

take cube root of both sides

therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)

m:n = \(2\frac{1}{3} : 1\frac{1}{5}\) = m : n = \(\frac{7}{3} : \frac{6}{5}\)

\(\frac{7}{3} : \frac{6}{5}\) =  \(\frac{7}{3} \div \frac{6}{5}\)

\(\frac{m}{n}\) =  \(\frac{7}{3} \times \frac{5}{6}\)

\(\frac{m}{n}\) =  \(\frac{35}{18}\) = m =  \(\frac{35n}{18}\)

n : q = \(1\frac{1}{2} : 1\frac{1}{3}\) =  \(\frac{3}{2} : \frac{4}{3}\)

\(\frac{n}{q}\) =   \(\frac{3}{2} \times\frac{3}{4}\)

\(\frac{n}{q}\) =  \(\frac{9}{8}\) = q =  \(\frac{8n}{9}\) 

q : m =   \(\frac{8n}{9}\) :   \(\frac{35n}{18}\)

\(\frac{q}{m}\) =  \(\frac{8n}{9} \div \frac{35n}{18}\)

\(\frac{q}{m}\) =  \(\frac{8n}{9}\times\frac{18}{35n}\)

 =\(\frac{q}{m} = \frac{16}{35}\) = q : m = 16 : 35