Question

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of heartbeats per minute	50-52	53-55	56-58	59-61	62-64
Number of boxs	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer 1

(i) System of equations:
1. \(x + y = 14\)
2. \(x − y = 4\)
From equation (1), \(x = 14 − y\) (3).
Substitute (3) into (2): \((14-y) -y =4\).
This simplifies to \(14 - 2y = 4\), so \(10 = 2y\), which gives \(y = 5\).
Substitute \(y = 5\) into equation (3): \(x = 14 - 5 = 9\).
Solution: \(x = 9, y = 5\).

(ii) System of equations:
1. \(s-t =3\)
2. \(\frac{s}{3} + \frac{t}{2} = 6\)
From equation (1), \(s = t + 3\) (3).
Substitute (3) into (2): \(\frac{t+3}{3} + \frac{t}{2} = 6\).
Multiply by 6 to clear denominators: \(2(t+3) + 3t = 36\).
This simplifies to \(2t + 6 + 3t = 36\), so \(5t = 30\), which gives \(t = 6\).
Substitute \(t = 6\) into equation (3): \(s = 6 + 3 = 9\).
Solution: \(s = 9, t = 6\).

(iii) System of equations:
1. \(3x − y = 3\)
2. \(9x − 3y = 9\)
From equation (1), \(y = 3x − 3\) (3).
Substitute (3) into (2): \(9x - 3(3x - 3) = 9\).
This simplifies to \(9x - 9x + 9 = 9\), resulting in \(9 = 9\).
This identity indicates that the system has infinite solutions. The relationship between the variables is given by \(y = 3x - 3\).
An example solution is \(x = 1, y = 0\).

(iv) System of equations:
1. \(0.2x + 0.3y = 1.3\)
2. \(0.4x + 0.5y = 2.3\)
From equation (1), \(x = \frac{1.3 - 0.3y}{0.2}\) (3).
Substitute (3) into (2): \(0.4(\frac{1.3 - 0.3y}{0.2}) + 0.5y = 2.3\).
This simplifies to \(2(1.3 - 0.3y) + 0.5y = 2.3\), so \(2.6 - 0.6y + 0.5y = 2.3\).
Rearranging gives \(0.3 = 0.1y\), so \(y = 3\).
Substitute \(y = 3\) into equation (3): \(x = \frac{1.3 - 0.3 \times 3}{0.2} = \frac{1.3 - 0.9}{0.2} = \frac{0.4}{0.2} = 2\).
Solution: \(x = 2, y = 3\).

(v) System of equations:
1. \(\sqrt{2}x + \sqrt{3}y = 0\)
2. \(\sqrt{3}x - \sqrt{8}y = 0\)
From equation (1), \(x = -\frac{\sqrt{3}y}{\sqrt{2}}\) (3).
Substitute (3) into (2): \(\sqrt{3}(-\frac{\sqrt{3}y}{\sqrt{2}}) - \sqrt{8}y = 0\).
This simplifies to \(-\frac{3y}{\sqrt{2}} - 2\sqrt{2}y = 0\).
Factoring out \(y\): \(y(-\frac{3}{\sqrt{2}} - 2\sqrt{2}) = 0\).
This implies \(y = 0\).
Substitute \(y = 0\) into equation (3): \(x = -\frac{\sqrt{3}(0)}{\sqrt{2}} = 0\).
Solution: \(x = 0, y = 0\).

(vi) System of equations:
1. \(\frac{3}{2}x - \frac{5}{3}y = -2\)
2. \(\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\)
From equation (1), multiply by 6 to get \(9x - 10y = -12\). Then, \(x = \frac{-12 + 10y}{9}\) (3).
Substitute (3) into (2): \(\frac{1}{3}(\frac{-12 + 10y}{9}) + \frac{y}{2} = \frac{13}{6}\).
This simplifies to \(\frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6}\).
Multiply by the least common multiple of the denominators (54): \(2(-12 + 10y) + 27y = 9 \times 13\).
So, \(-24 + 20y + 27y = 117\).
This gives \(47y = 141\), so \(y = 3\).
Substitute \(y = 3\) into equation (3): \(x = \frac{-12 + 10 \times 3}{9} = \frac{-12 + 30}{9} = \frac{18}{9} = 2\).
Solution: \(x = 2, y = 3\).