Fractions

I. A man spends \(\frac{2}{5}\) of his salary on house-rent, \(\frac{3}{10}\) of his salary on food and \(\frac{1}{8}\) of his salary on conveyance. He saves the remaining ₹10500.

Q1: The total salary of the man is:

Step 1: Let the total salary = ₹\(S\).Step 2: Sum of fractions spent: \[ \frac{2}{5} + \frac{3}{10} + \frac{1}{8} = ? \]Step 3: Find LCM of denominators 5, 10, and 8: \[ LCM(5,10,8) = 40 \]Step 4: Convert to common denominator: \[ \frac{2}{5} = \frac{2 \times 8}{5 \times 8} = \frac{16}{40} \\ \frac{3}{10} = \frac{3 \times 4}{10 \times 4} = \frac{12}{40} \\ \frac{1}{8} = \frac{1 \times 5}{8 \times 5} = \frac{5}{40} \]Step 5: Add the fractions: \[ \frac{16}{40} + \frac{12}{40} + \frac{5}{40} = \frac{33}{40} \]Step 6: Fraction of salary saved: \[ 1 – \frac{33}{40} = \frac{40}{40} – \frac{33}{40} = \frac{7}{40} \]Step 7: Given savings = ₹10500, so: \[ \frac{7}{40} \times S = 10500 \]Step 8: Solve for \(S\): \[ S = \frac{10500 \times 40}{7} = 10500 \times \frac{40}{7} = 1500 \times 40 = 60000 \]Answer: d. ₹60000

Q2: The amount spent on house-rent is:

Step 1: House-rent = \(\frac{2}{5}\) of salary: \[ \frac{2}{5} \times 60000 = \frac{2 \times 60000}{5} = 2 \times 12000 = 24000 \]Answer: c. ₹24000

Q3: The amount spent on food is:

Step 1: Food = \(\frac{3}{10}\) of salary: \[ \frac{3}{10} \times 60000 = \frac{3 \times 60000}{10} = 3 \times 6000 = 18000 \]Answer: a. ₹18000

Q4: The amount spent on conveyance is:

Step 1: Conveyance = \(\frac{1}{8}\) of salary: \[ \frac{1}{8} \times 60000 = \frac{60000}{8} = 7500 \]Answer: c. ₹7500

II. A drum of kerosene is \(\frac{3}{4}\) full. When 30 litres of kerosene is drawn from it, it remains \(\frac{7}{12}\) full.

Q1: The capacity of the drum is:

Step 1: Initial quantity of kerosene: \[ \frac{3}{4}C \]Step 2: Quantity after drawing 30 litres: \[ \frac{3}{4}C – 30 \]Step 3: After drawing 30 litres, quantity is \(\frac{7}{12}C\): \[ \frac{3}{4}C – 30 = \frac{7}{12}C \]Step 4: Rearrange equation: \[ \frac{3}{4}C – \frac{7}{12}C = 30 \]Step 5: Find common denominator (12): \[ \frac{9}{12}C – \frac{7}{12}C = 30 \\ \frac{2}{12}C = 30 \\ \frac{1}{6}C = 30 \\ \]Step 6: Multiply both sides by 6: \[ C = 30 \times 6 = 180 \]Answer: c. 180 litres

Q2: The quantity of kerosene in the drum is:

Step 1: Initial quantity: \[ \frac{3}{4} \times 180 = \frac{3 \times 180}{4} = 135 \]Answer: a. 135 litres

Q3: If 15 l of kerosene is added to the drum, what fraction of the capacity would be filled?

Step 1: Quantity after addition: \[ 135 + 15 = 150 \text{ litres} \]Step 2: Fraction filled: \[ \frac{150}{180} = \frac{5}{6} \]Answer: c. \(\frac{5}{6}\)

Q4: If 35 l of kerosene is drawn from the drum, What fraction of the capacity would be filled?

Step 1: Quantity after drawing 35 litres: \[ 135 – 35 = 100 \text{ litres} \]Step 2: Fraction filled: \[ \frac{100}{180} = \frac{10}{18} = \frac{5}{9} \]Answer: a. \(\frac{5}{9}\)