Ratio and Proportion

Q1: Express each of the following ratios in simplest form:

i. \(18 : 30\)

Step 1: Find the greatest common divisor (GCD) of 18 and 30.
GCD of 18 and 30 is 6.
Step 2: Divide both terms by 6. \[ \frac{18}{6} : \frac{30}{6} = 3 : 5 \] Answer: \(3 : 5\)

ii. \(7.5 : 9\)

Step 1: Express 7.5 and 9 as fractions.
\(7.5 = \frac{15}{2}\), \(9 = \frac{9}{1}\)
Step 2: Write ratio as fractions: \[ \frac{15}{2} : \frac{9}{1} = \frac{15}{2} : \frac{9}{1} \] Step 3: Convert ratio of fractions to fraction by dividing: \[ \frac{\frac{15}{2}}{\frac{9}{1}} = \frac{15}{2} \times \frac{1}{9} = \frac{15}{18} = \frac{5}{6} \] So the ratio is \(5 : 6\).
Answer: \(5 : 6\)

iii. \(6\frac{2}{3} : 7\frac{1}{2}\)

Step 1: Convert mixed numbers to improper fractions. \[ 6\frac{2}{3} = \frac{20}{3}, \quad 7\frac{1}{2} = \frac{15}{2} \] Step 2: Write ratio: \[ \frac{20}{3} : \frac{15}{2} \] Step 3: Convert ratio to fraction: \[ \frac{\frac{20}{3}}{\frac{15}{2}} = \frac{20}{3} \times \frac{2}{15} = \frac{40}{45} = \frac{8}{9} \] So, ratio = \(8 : 9\).
Answer: \(8 : 9\)

iv. \(\frac{1}{6} : \frac{1}{9} : \frac{1}{12}\)

Step 1: Find the LCM of denominators 6, 9, and 12.
LCM of 6, 9, 12 = 36.
Step 2: Express each fraction with denominator 36: \[ \frac{1}{6} = \frac{6}{36}, \quad \frac{1}{9} = \frac{4}{36}, \quad \frac{1}{12} = \frac{3}{36} \] Step 3: Write ratio using numerators:
\(6 : 4 : 3\)
Step 4: Simplify by dividing by GCD 1 (already simplest).
Ratio = \(6 : 4 : 3\).
Answer: \(6 : 4 : 3\)

v. \(5 : 7 : \frac{9}{2}\)

Step 1: Express all terms with a common denominator.
\(5 = \frac{10}{2}, \quad 7 = \frac{14}{2}, \quad \frac{9}{2} = \frac{9}{2}\)
Step 2: Write ratio with numerators:
\(10 : 14 : 9\)
Step 3: Simplify by dividing by GCD 1 (already simplest).
Ratio = \(10 : 14 : 9\).
Answer: \(10 : 14 : 9\)

vi. \(3\frac{1}{5} : 5\frac{1}{3} : 6\frac{2}{3}\)

Step 1: Convert mixed numbers to improper fractions. \[ 3\frac{1}{5} = \frac{16}{5}, \quad 5\frac{1}{3} = \frac{16}{3}, \quad 6\frac{2}{3} = \frac{20}{3} \] Step 2: Find LCM of denominators 5 and 3.
LCM(5, 3) = 15.
Step 3: Express all fractions with denominator 15: \[ \frac{16}{5} = \frac{48}{15}, \quad \frac{16}{3} = \frac{80}{15}, \quad \frac{20}{3} = \frac{100}{15} \] Step 4: Write ratio as numerators:
\(48 : 80 : 100\)
Step 5: Simplify by dividing by GCD 4: \[ \frac{48}{4} : \frac{80}{4} : \frac{100}{4} = 12 : 20 : 25 \] Answer: \(12 : 20 : 25\)

Q2: Express each of the following ratios in simplest form:

i. 75 paise : 4 rupees

Step 1: Convert both amounts to the same unit.
Since 1 rupee = 100 paise,
\(4 \text{ rupees} = 4 \times 100 = 400 \text{ paise}\)
Step 2: Write ratio in paise:
\(75 : 400\)
Step 3: Find GCD of 75 and 400.
GCD(75, 400) = 25.
Step 4: Divide both terms by 25: \[ \frac{75}{25} : \frac{400}{25} = 3 : 16 \] Answer: \(3 : 16\)

ii. 1 m 8 cm : 72 cm

Step 1: Convert meters and centimeters to the same unit (cm).
\(1 \text{ m } = 100 \text{ cm}\)
So, \(1 \text{ m } 8 \text{ cm } = 100 + 8 = 108 \text{ cm}\)
Step 2: Write ratio:
\(108 : 72\)
Step 3: Find GCD of 108 and 72.
GCD(108, 72) = 36.
Step 4: Divide both terms by 36: \[ \frac{108}{36} : \frac{72}{36} = 3 : 2 \] Answer: \(3 : 2\)

iii. 1 hour 15 minutes : 45 minutes

Step 1: Convert hours and minutes to the same unit (minutes).
\(1 \text{ hour} = 60 \text{ minutes}\)
So, \(1 \text{ hour } 15 \text{ minutes } = 60 + 15 = 75 \text{ minutes}\)
Step 2: Write ratio:
\(75 : 45\)
Step 3: Find GCD of 75 and 45.
GCD(75, 45) = 15.
Step 4: Divide both terms by 15: \[ \frac{75}{15} : \frac{45}{15} = 5 : 3 \] Answer: \(5 : 3\)

iv. 2 kg 750 g : 3 kg

Step 1: Convert kilograms and grams to the same unit (grams).
\(1 \text{ kg} = 1000 \text{ g}\)
So, \(2 \text{ kg } 750 \text{ g } = 2 \times 1000 + 750 = 2750 \text{ g}\)
\(3 \text{ kg} = 3 \times 1000 = 3000 \text{ g}\)
Step 2: Write ratio:
\(2750 : 3000\)
Step 3: Find GCD of 2750 and 3000.
GCD(2750, 3000) = 250.
Step 4: Divide both terms by 250: \[ \frac{2750}{250} : \frac{3000}{250} = 11 : 12 \] Answer: \(11 : 12\)

v. 1 year 9 months : 2 years 4 months

Step 1: Convert years and months to months.
\(1 \text{ year} = 12 \text{ months}\)
So,
\(1 \text{ year } 9 \text{ months } = 12 + 9 = 21 \text{ months}\)
\(2 \text{ years } 4 \text{ months } = 2 \times 12 + 4 = 28 \text{ months}\)
Step 2: Write ratio:
\(21 : 28\)
Step 3: Find GCD of 21 and 28.
GCD(21, 28) = 7.
Step 4: Divide both terms by 7: \[ \frac{21}{7} : \frac{28}{7} = 3 : 4 \] Answer: \(3 : 4\)

Q3: Which ratio is greater?

i. \(4 : 9\) or \(3 : 7\)

Step 1: Convert each ratio to fraction form. \[ \frac{4}{9} \quad \text{and} \quad \frac{3}{7} \] Step 2: Cross multiply to compare:
\(4 \times 7 = 28\)
\(3 \times 9 = 27\)
Since \(28 > 27\), \[ \frac{4}{9} > \frac{3}{7} \] Answer: \(4 : 9\) is greater than \(3 : 7\)

ii. \(2\frac{1}{3} : 3\frac{1}{3}\) or \(3.6 : 4.8\)

Step 1: Convert mixed numbers to improper fractions. \[ 2\frac{1}{3} = \frac{7}{3}, \quad 3\frac{1}{3} = \frac{10}{3} \] Step 2: Write ratio 1 as a fraction: \[ \frac{7/3}{10/3} = \frac{7}{3} \times \frac{3}{10} = \frac{7}{10} = 0.7 \] Step 3: Calculate ratio 2: \[ \frac{3.6}{4.8} = 0.75 \] Step 4: Compare decimal values: \[ 0.7 < 0.75 \] Answer: \(3.6 : 4.8\) is greater than \(2\frac{1}{3} : 3\frac{1}{3}\)

iii. \(\frac{1}{2} : \frac{1}{3}\) or \(\frac{1}{6} : \frac{1}{4}\)

Step 1: Convert ratios to fractions: \[ \frac{\frac{1}{2}}{\frac{1}{3}} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2} = 1.5 \\ \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{1}{6} \times \frac{4}{1} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \] Step 2: Compare decimal values: \[ 1.5 > 0.6667 \] Answer: \(\frac{1}{2} : \frac{1}{3}\) is greater than \(\frac{1}{6} : \frac{1}{4}\)

iv. \(3\frac{1}{3} : 4\frac{1}{6}\) or \(0.9 : 1\)

Step 1: Convert mixed numbers to improper fractions. \[ 3\frac{1}{3} = \frac{10}{3}, \quad 4\frac{1}{6} = \frac{25}{6} \] Step 2: Write ratio 1 as a fraction: \[ \frac{10/3}{25/6} = \frac{10}{3} \times \frac{6}{25} = \frac{60}{75} = \frac{4}{5} = 0.8 \] Step 3: Calculate ratio 2: \[ \frac{0.9}{1} = 0.9 \] Step 4: Compare decimal values: \[ 0.8 < 0.9 \] Answer: \(0.9 : 1\) is greater than \(3\frac{1}{3} : 4\frac{1}{6}\)

Q4: Arrange the following ratios in ascending order of magnitude:

i. \(2 : 3\), \(5 : 9\) and \(11 : 15\)

Step 1: Convert each ratio to decimal form. \[ \frac{2}{3} = 0.6667, \quad \frac{5}{9} = 0.5556, \quad \frac{11}{15} = 0.7333 \] Step 2: Arrange decimals in ascending order:
\(0.5556 < 0.6667 < 0.7333\)
Step 3: Write corresponding ratios: \[ 5 : 9 < 2 : 3 < 11 : 15 \] Answer: \(5 : 9\), \(2 : 3\), \(11 : 15\)

ii. \(5 : 7\), \(9 : 14\), \(20 : 21\) and \(3 : 8\)

Step 1: Convert each ratio to decimal form. \[ \frac{5}{7} \approx 0.7143, \quad \frac{9}{14} = 0.6429, \quad \frac{20}{21} \approx 0.9524, \quad \frac{3}{8} = 0.375 \] Step 2: Arrange decimals in ascending order:
\(0.375 < 0.6429 < 0.7143 < 0.9524\)
Step 3: Write corresponding ratios: \[ 3 : 8 < 9 : 14 < 5 : 7 < 20 : 21 \] Answer: \(3 : 8\), \(9 : 14\), \(5 : 7\), \(20 : 21\)

Q5: Divide ₹142.20 between Gagan and Mukesh in the ratio \(\frac{1}{4} : \frac{1}{5}\).

Step 1: Given ratio is \(\frac{1}{4} : \frac{1}{5}\). To work easily, express both fractions with a common denominator.
LCM of 4 and 5 is 20.
Convert each fraction: \[ \frac{1}{4} = \frac{5}{20}, \quad \frac{1}{5} = \frac{4}{20} \] So the ratio becomes \(5 : 4\).
Step 2: Total parts in the ratio = \(5 + 4 = 9\) parts.
Step 3: Value of one part: \[ \text{One part} = \frac{₹142.20}{9} = ₹15.80 \]Step 4: Amount for Gagan (5 parts): \[ 5 \times ₹15.80 = ₹79.00 \]Step 5: Amount for Mukesh (4 parts): \[ 4 \times ₹15.80 = ₹63.20 \]Answer: Gagan gets ₹79.00 and Mukesh gets ₹63.20.

Q6: Divide ₹3726 among A, B, C in the ratio \(\frac{1}{3} : \frac{1}{4} : \frac{1}{6}\).

Step 1: Express all fractions with a common denominator.
Denominators are 3, 4, and 6.
LCM of 3, 4, and 6 is 12.
Convert each fraction: \[ \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{6} = \frac{2}{12} \] So the ratio becomes \(4 : 3 : 2\).
Step 2: Find total parts: \[ 4 + 3 + 2 = 9 \]Step 3: Find the value of one part: \[ \text{One part} = \frac{₹3726}{9} = ₹414 \]Step 4: Amount for A (4 parts): \[ 4 \times ₹414 = ₹1656 \]Step 5: Amount for B (3 parts): \[ 3 \times ₹414 = ₹1242 \]Step 6: Amount for C (2 parts): \[ 2 \times ₹414 = ₹828 \]Answer: A gets ₹1656, B gets ₹1242, and C gets ₹828.

Q7: Divide ₹810 among A, B and C in the ratio \(\frac{1}{4} : \frac{2}{5} : 1\frac{3}{8}\).

Step 1: Convert mixed number \(1\frac{3}{8}\) to improper fraction: \[ 1\frac{3}{8} = \frac{8 \times 1 + 3}{8} = \frac{11}{8} \]Step 2: Write the ratio as: \[ \frac{1}{4} : \frac{2}{5} : \frac{11}{8} \]Step 3: Find LCM of denominators 4, 5, and 8: \[ \text{LCM}(4,5,8) = 40 \]Step 4: Convert each fraction to have denominator 40: \[ \frac{1}{4} = \frac{10}{40}, \quad \frac{2}{5} = \frac{16}{40}, \quad \frac{11}{8} = \frac{55}{40} \]Step 5: Now ratio becomes: \[ 10 : 16 : 55 \]Step 6: Total parts: \[ 10 + 16 + 55 = 81 \]Step 7: Value of one part: \[ \text{One part} = \frac{₹810}{81} = ₹10 \]Step 8: Amount for A (10 parts): \[ 10 \times ₹10 = ₹100 \]Step 9: Amount for B (16 parts): \[ 16 \times ₹10 = ₹160 \]Step 10: Amount for C (55 parts): \[ 55 \times ₹10 = ₹550 \]Answer: A gets ₹100, B gets ₹160, and C gets ₹550.

Q8: Divide ₹1050 between Geeta and Renu in the ratio \(2\frac{2}{3} : 6\frac{2}{3}\).

Step 1: Convert mixed numbers to improper fractions: \[ 2\frac{2}{3} = \frac{3 \times 2 + 2}{3} = \frac{8}{3}, \quad 6\frac{2}{3} = \frac{3 \times 6 + 2}{3} = \frac{20}{3} \]Step 2: Write the ratio as: \[ \frac{8}{3} : \frac{20}{3} \]Step 3: Since denominators are the same, ratio simplifies to: \[ 8 : 20 \]Step 4: Simplify the ratio by dividing both terms by 4: \[ \frac{8}{4} : \frac{20}{4} = 2 : 5 \]Step 5: Total parts in the ratio: \[ 2 + 5 = 7 \]Step 6: Value of one part: \[ \text{One part} = \frac{₹1050}{7} = ₹150 \]Step 7: Amount for Geeta (2 parts): \[ 2 \times ₹150 = ₹300 \]Step 8: Amount for Renu (5 parts): \[ 5 \times ₹150 = ₹750 \]Answer: Geeta gets ₹300 and Renu gets ₹750.

Q9: Divide ₹747 among A, B, C such that \(4A = 5B = 7C\).

Step 1: Let the common value be \(k\). Then, \[ 4A = 5B = 7C = k \]Step 2: Express A, B, and C in terms of \(k\): \[ A = \frac{k}{4}, \quad B = \frac{k}{5}, \quad C = \frac{k}{7} \]Step 3: Total amount divided among A, B, C: \[ A + B + C = \frac{k}{4} + \frac{k}{5} + \frac{k}{7} = ₹747 \]Step 4: Find LCM of denominators 4, 5, and 7: \[ \text{LCM}(4, 5, 7) = 140 \]Step 5: Express sum as: \[ k \left( \frac{1}{4} + \frac{1}{5} + \frac{1}{7} \right) = k \left( \frac{35}{140} + \frac{28}{140} + \frac{20}{140} \right) = k \times \frac{83}{140} = 747 \]Step 6: Solve for \(k\): \[ k = \frac{747 \times 140}{83} = \frac{104580}{83} = 1260 \]Step 7: Find amounts for A, B, and C: \[ A = \frac{1260}{4} = 315, \quad B = \frac{1260}{5} = 252, \quad C = \frac{1260}{7} = 180 \]Answer: A gets ₹315, B gets ₹252, and C gets ₹180.

Q10: A bag contains one rupee, 50 p and 25 p coins in the ratio 5 : 6 : 8 amounting to ₹210. Find the number of coins of each type.

Step 1: Let the number of one rupee coins = \(5x\), 50 paise coins = \(6x\), and 25 paise coins = \(8x\).
Step 2: Find the total amount contributed by each type of coin.
– One rupee coins contribute: \[ 5x \times ₹1 = 5x \, \text{rupees} \] – 50 paise coins contribute: \[ 6x \times ₹0.50 = 3x \, \text{rupees} \] – 25 paise coins contribute: \[ 8x \times ₹0.25 = 2x \, \text{rupees} \]Step 3: Total amount is ₹210: \[ 5x + 3x + 2x = 210 \\ 10x = 210 \\ x = \frac{210}{10} = 21 \]Step 4: Calculate the number of coins of each type: \[ \text{One rupee coins} = 5x = 5 \times 21 = 105 \\ \text{50 paise coins} = 6x = 6 \times 21 = 126 \\ \text{25 paise coins} = 8x = 8 \times 21 = 168 \]Answer: Number of one rupee coins = 105, 50 paise coins = 126, and 25 paise coins = 168.

Q11: Find \(A : B : C\) when:

i. \(A : B = 2 : 5\) and \(B : C = 7 : 9\)

Step 1: Find the common term \(B\) in both ratios.
Ratios: \(2 : 5\) and \(7 : 9\)
\(B\) in first ratio = 5, \(B\) in second ratio = 7.
Find LCM of 5 and 7 = 35.
Step 2: Adjust both ratios to have the same \(B\) value: \[ A : B = 2 \times 7 : 5 \times 7 = 14 : 35 \\ B : C = 7 \times 5 : 9 \times 5 = 35 : 45 \]Step 3: Combine to get \(A : B : C\): \[ 14 : 35 : 45 \]Answer: \(A : B : C = 14 : 35 : 45\)

ii. \(A : B = 3 : 4\) and \(B : C = 6 : 11\)

Step 1: Find the common term \(B\) in both ratios.
\(B\) in first ratio = 4, \(B\) in second ratio = 6.
LCM of 4 and 6 = 12.
Step 2: Adjust both ratios: \[ A : B = 3 \times 3 : 4 \times 3 = 9 : 12 \\ B : C = 6 \times 2 : 11 \times 2 = 12 : 22 \]Step 3: Combine: \[ 9 : 12 : 22 \]Answer: \(A : B : C = 9 : 12 : 22\)

iii. \(A : B = \frac{1}{2} : \frac{1}{3}\) and \(B : C = \frac{1}{4} : \frac{1}{5}\)

Step 1: Convert ratios to fractions: \[ A : B = \frac{1}{2} : \frac{1}{3}, \quad B : C = \frac{1}{4} : \frac{1}{5} \]Step 2: Express \(A : B\) as a fraction: \[ \frac{\frac{1}{2}}{\frac{1}{3}} = \frac{1}{2} \times \frac{3}{1} = \frac{3}{2} \] So, \(A : B = 3 : 2\).
Step 3: Express \(B : C\) as a fraction: \[ \frac{\frac{1}{4}}{\frac{1}{5}} = \frac{1}{4} \times \frac{5}{1} = \frac{5}{4} \] So, \(B : C = 5 : 4\).
Step 4: Find common term \(B\) in \(3 : 2\) and \(5 : 4\):
\(B\) in first ratio = 2, \(B\) in second ratio = 5.
LCM of 2 and 5 = 10.
Step 5: Adjust ratios: \[ A : B = 3 \times 5 : 2 \times 5 = 15 : 10 \\ B : C = 5 \times 2 : 4 \times 2 = 10 : 8 \]Step 6: Combine: \[ A : B : C = 15 : 10 : 8 \]Answer: \(A : B : C = 15 : 10 : 8\)

Q12: If \(A : B = 4 : 9\) and \(A : C = 2 : 3\), find \(B : C\) and \(A : B : C\).

Step 1: Given ratios: \[ A : B = 4 : 9 \] and \[ A : C = 2 : 3 \]Step 2: Find a common value of \(A\) to combine both ratios.
The values of \(A\) are 4 (in first ratio) and 2 (in second ratio).
LCM of 4 and 2 is 4.
Step 3: Adjust the ratios to have the same \(A\) value: \[ A : B = 4 : 9 \] (no change) \[ A : C = 2 \times 2 : 3 \times 2 = 4 : 6 \]Step 4: Now write \(A : B : C\) as: \[ 4 : 9 : 6 \]Step 5: Find \(B : C\): \[ B : C = 9 : 6 = 3 : 2 \]Answer: \(B : C = 3 : 2\) and \(A : B : C = 4 : 9 : 6\).

Q13: If \(A : C = 5 : 8\) and \(B : C = 5 : 6\), find \(A : B\) and \(A : B : C\).

Step 1: Given ratios: \[ A : C = 5 : 8 \] and \[ B : C = 5 : 6 \]Step 2: Find common value of \(C\) in both ratios.
\(C\) in first ratio = 8, \(C\) in second ratio = 6.
LCM of 8 and 6 is 24.
Step 3: Adjust both ratios to have same \(C\) value: \[ A : C = 5 \times 3 : 8 \times 3 = 15 : 24 \\ B : C = 5 \times 4 : 6 \times 4 = 20 : 24 \]Step 4: Now write \(A : B : C\) as: \[ 15 : 20 : 24 \]Step 5: Find \(A : B\): \[ A : B = 15 : 20 = 3 : 4 \]Answer: \(A : B = 3 : 4\) and \(A : B : C = 15 : 20 : 24\).

Q14: Two numbers are in the ratio 6 : 11. On adding 2 to the first and 7 to the second, their ratio becomes 8 : 15. Find the numbers.

Step 1: Let the two numbers be \(6x\) and \(11x\) according to the ratio \(6 : 11\).
Step 2: After adding 2 to the first number and 7 to the second, the new ratio is: \[ \frac{6x + 2}{11x + 7} = \frac{8}{15} \]Step 3: Cross multiply: \[ 15(6x + 2) = 8(11x + 7) \\ 90x + 30 = 88x + 56 \]Step 4: Rearranging to solve for \(x\): \[ 90x – 88x = 56 – 30 \\ 2x = 26 \\ x = 13 \]Step 5: Find the two numbers: \[ 6x = 6 \times 13 = 78 \\ 11x = 11 \times 13 = 143 \]Answer: The two numbers are 78 and 143.

Q15: A sum of money is divided between Mohan and Sohan in the ratio 5 : 7. If Sohan’s share is ₹665, find the total amount.

Step 1: Let the shares of Mohan and Sohan be \(5x\) and \(7x\) respectively.
Step 2: Given Sohan’s share is ₹665: \[ 7x = 665 \]Step 3: Solve for \(x\): \[ x = \frac{665}{7} = 95 \]Step 4: Calculate Mohan’s share: \[ 5x = 5 \times 95 = 475 \]Step 5: Total amount: \[ 5x + 7x = 12x = 12 \times 95 = 1140 \]Answer: The total amount is ₹1140.

Q16: Two numbers are in the ratio 7 : 4. If their difference is 72, find the numbers.

Step 1: Let the two numbers be \(7x\) and \(4x\), based on the given ratio \(7 : 4\).
Step 2: The difference of the numbers is given as 72. \[ 7x – 4x = 72 \\ 3x = 72 \]Step 3: Solve for \(x\): \[ x = \frac{72}{3} = 24 \]Step 4: Calculate the numbers:
First number = \(7x = 7 \times 24 = 168\)
Second number = \(4x = 4 \times 24 = 96\)
Answer: The two numbers are 168 and 96.

Q17: A certain sum of money is divided among A, B, C in the ratio 5 : 6 : 7. If A’s share is ₹175, find the total amount and the share of each one of B and C.

Step 1: Let the shares of A, B, and C be \(5x\), \(6x\), and \(7x\) respectively.
Step 2: It is given that A’s share is ₹175: \[ 5x = 175 \]Step 3: Solve for \(x\): \[ x = \frac{175}{5} = 35 \]Step 4: Calculate shares of B and C: \[ B = 6x = 6 \times 35 = ₹210 \\ C = 7x = 7 \times 35 = ₹245 \]Step 5: Total amount: \[ A + B + C = 175 + 210 + 245 = ₹630 \]Answer: Total amount = ₹630, B’s share = ₹210, C’s share = ₹245.

Q18: The ratio of number of boys to the number of girls in a school of 1440 students is 7 : 5. If 40 new boys are admitted, find how many new girls may be admitted to make this ratio 4 : 3.

Step 1: Let the number of boys be \(7x\) and the number of girls be \(5x\).
Given total students = 1440 \[ 7x + 5x = 12x = 1440 \\ x = \frac{1440}{12} = 120 \]Step 2: Find actual number of boys and girls: \[ \text{Boys} = 7x = 7 \times 120 = 840 \\ \text{Girls} = 5x = 5 \times 120 = 600 \]Step 3: After 40 new boys are admitted: \[ \text{New number of boys} = 840 + 40 = 880 \] Let the number of new girls admitted be \(g\).
Then, new number of girls = \(600 + g\)
Step 4: New ratio is to be \(4 : 3\): \[ \frac{880}{600 + g} = \frac{4}{3} \]Step 5: Cross-multiply and solve: \[ 3 \times 880 = 4 \times (600 + g) \\ 2640 = 2400 + 4g \\ 2640 – 2400 = 4g \\ \Rightarrow 240 = 4g \\ \Rightarrow g = \frac{240}{4} = 60 \]Answer: 60 new girls must be admitted.

Q19: The sum of three numbers is 212. If the ratio of the first to the second is 13 : 16 and that of the second to the third is 2 : 3, then find the numbers.

Step 1: Let the three numbers be A, B, and C.
Given: \[ A : B = 13 : 16,\quad B : C = 2 : 3 \]Step 2: Make the B terms same in both ratios.
LCM of 16 and 2 = 16.
Adjust both ratios: – \(A : B = 13 : 16\) (already in desired form)
– \(B : C = 2 \times 8 : 3 \times 8 = 16 : 24\)
Step 3: Combine into one ratio: \[ A : B : C = 13 : 16 : 24 \]Step 4: Add the parts: \[ 13 + 16 + 24 = 53 \]Step 5: Let one part be \(x\). Then, total sum = \(53x = 212\) \[ x = \frac{212}{53} = 4 \]Step 6: Multiply each ratio term by 4: \[ A = 13 \times 4 = 52,\quad B = 16 \times 4 = 64,\quad C = 24 \times 4 = 96 \]Answer: The numbers are 52, 64, and 96.

Q20: Find the number which when added to each term of the ratio 27 : 35, changes the ratio to 4 : 5.

Step 1: Let the number to be added be \(x\).
Then, the new ratio becomes: \[ \frac{27 + x}{35 + x} = \frac{4}{5} \]Step 2: Cross-multiply: \[ 5(27 + x) = 4(35 + x) \\ 135 + 5x = 140 + 4x \]Step 3: Solve for \(x\): \[ 5x – 4x = 140 – 135 \\ \Rightarrow x = 5 \]Answer: The number is 5.

Q21: The present ages of Mr. Sen and his son are in the ratio 17 : 9. If the ratio of their ages, 9 years ago was 7 : 3, then find their present ages.

Step 1: Let the present ages of Mr. Sen and his son be \(17x\) and \(9x\) respectively.
Step 2: 9 years ago, their ages would have been:
Mr. Sen: \(17x – 9\)
Son: \(9x – 9\)
According to the question: \[ \frac{17x – 9}{9x – 9} = \frac{7}{3} \]Step 3: Cross-multiply: \[ 3(17x – 9) = 7(9x – 9) \\ 51x – 27 = 63x – 63 \]Step 4: Rearranging terms: \[ 51x – 63x = -63 + 27 \\ \Rightarrow -12x = -36 \\ \Rightarrow x = 3 \]Step 5: Find actual ages:
Mr. Sen’s age = \(17x = 17 \times 3 = 51\) years
Son’s age = \(9x = 9 \times 3 = 27\) years
Answer: Mr. Sen is 51 years old and his son is 27 years old.

Q22: A sample of vinegar was prepared by dissolving 705 ml of acetic acid in 4.7 litres of water. Find the ratio by volume of acetic acid to water in this sample.

Step 1: Convert all quantities to the same unit.
Water = 4.7 litres = \(4.7 \times 1000 = 4700\) ml
Step 2: Acetic acid = 705 ml, Water = 4700 ml
Ratio by volume = Acetic acid : Water \[ = 705 : 4700 \]Step 3: Simplify the ratio.
Find HCF of 705 and 4700 = 235 \[ \frac{705}{235} : \frac{4700}{235} = 3 : 20 \]Answer: The ratio by volume of acetic acid to water is 3 : 20.

Q23: The salaries of A, B and C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be the new ratio of their salaries?

Step 1: Let the original salaries of A, B and C be \(2x, 3x,\) and \(5x\) respectively.
Step 2: Apply the percentage increment to each salary:
– A gets 15% increase → \[ \text{New salary of A} = 2x + \frac{15}{100} \times 2x = 2x(1 + \frac{15}{100}) = 2x \times \frac{115}{100} \]– B gets 10% increase → \[ \text{New salary of B} = 3x \times \frac{110}{100} \]– C gets 20% increase → \[ \text{New salary of C} = 5x \times \frac{120}{100} \]Step 3: Simplify each expression:
– A: \[ 2x \times \frac{115}{100} = \frac{230x}{100} \]– B: \[ 3x \times \frac{110}{100} = \frac{330x}{100} \]– C: \[ 5x \times \frac{120}{100} = \frac{600x}{100} \]Step 4: Cancel out common denominator 100 and factor \(x\): \[ \text{New ratio} = 230 : 330 : 600 \]Step 5: Simplify the ratio:
Divide each term by 10: \[ 23 : 33 : 60 \]Answer: The new ratio of their salaries is 23 : 33 : 60.

Q24: Anuj got 58 marks out of 75 in Physics and 97 marks out of 120 in Maths. In which of the two subjects did he perform better?

Step 1: Find the percentage of marks obtained in each subject.
Step 2: Physics: \[ \frac{58}{75} \times 100 = 77.33\% \]Step 3: Maths: \[ \frac{97}{120} \times 100 = 80.83\% \]Step 4: Compare the percentages:
Physics = 77.33%
Maths = 80.83%
Answer: Anuj performed better in Maths.

Q25: A mixture has only two components A and B. In 60 litres of this mixture, the components A and B are present in the ratio 2 : 1. What quantity of component B has to be added to this mixture so that the new ratio 1 : 2?

Step 1: Let the original quantities of A and B be in the ratio 2 : 1 in 60 litres.
Total parts = \(2 + 1 = 3\) parts
Each part = \(60 \div 3 = 20\) litres
So,
A = \(2 \times 20 = 40\) litres
B = \(1 \times 20 = 20\) litres
Step 2: Let \(x\) litres of B be added.
New amount of B = \(20 + x\)
Amount of A remains the same = 40 litres
New ratio of A : B is 1 : 2, so: \[ \frac{40}{20 + x} = \frac{1}{2} \]Step 3: Cross-multiply and solve: \[ 2 \times 40 = 1 \times (20 + x) \\ \Rightarrow 80 = 20 + x \\ \Rightarrow x = 60 \]Answer: 60 litres of component B must be added.