Ratio and Proportion Practice Problems: Level 02

Solve the given practice questions based on geometric progression. Also, the answer key and explanations are given for the same.

1.When a sum of money was equally distributed among 49 children, each child received Rs. 20. If the same amount is equally distributed among children, such that each child gets Rs. 3.5, find the number of children.

Option A
Number of children
49= K=49 × 20 ⇒ N = 49 × 20 /3.5 = 280.

2. The cost of a diamond varies as the square of its weight. A diamond weighing 20 decigrams costs Rs. 4,800. Find the cost of a diamond of the same kind weighing 8 decigrams.

Option D
C α W² ⇒ C = kW². 4800 = k(20)² ⇒ k = 12. Now C = 12(8)² ⇒ C =768.

3. The ratio of two numbers is 9 : 5. If 9 is added to the greater number and 5 is subtracted from the smaller number, the greater number becomes thrice the smaller one. Find the numbers.

Option C
Present ratio = 9 : 5. ∴actual values are 9x and 5x.
Hence 9x + 9 : 5x – 5 = 3 : 1 ⇒ x = 4. ∴ numbers are 9(4) = 36 and 5(4) = 20.

4. Find the ratio of the diagonal of a square of side 30 cm, to its side.

Option D
Diagonal of a square = √2 side. Hence this is the required ratio. So the required ratio is √2: 1.

5. The ratio of the first and second-class fares between the two stations is 6 : 4 and the number of passengers traveling by first and second-class is 1 : 30. If Rs. 2100 is collected as fare, what is the amount collected from first class passengers?

Option D
Ratio of the amounts collected from 1^st and 2^nd class = (6 × 1): (4 × 30) = 1 : 20. ∴ Amount collected from 1st class passengers = × 2100 = 100.

6. The ratio of A’s salary to B’s was 4 : 5. A’s salary is increased by 10% and B’s by 20%, what is the ratio of their salaries now?

Option C
Present ratio = 4 : 5. Increase of 10% and 20%. New ratio of salaries will be 4 × 1.1 :5 × 1.2 = 11 : 15.

7. 200 g of 25% sulphuric acid solution was added to 300 g of 40% sulphuric acid solution. Find the concentration of the acid in the mixture.

Option D
Amount of acid in the 1st solution = 0.25 × 200 = 50 g.
Amount of acid in the 2nd solution = 0.4 × 300 = 120 g.
Total amount of acid = 50 + 120 = 170 g. ∴ Required concentration = 170 / (200 + 300) × 100 = 34%

8. In one alloy there is 60% gold in its total mass, while in another alloy it is 35%. 12 kg of the first alloy was melted together with 8 kg of the second one to form a third alloy. Find the percentage of gold in the new alloy.

Option A
Amount of gold in 1^st alloy = 0.6 × 12 = 7.2 kg. Amount of gold in 2^nd alloy = 0.35 × 8 =2.8 kg.
∴ required % age of gold = (7.2 + 2.8) / (12 + 8) × 100 = (10 / 20) × 100 = 50 %.

9. Divide Rs. 390 among 3 persons A, B and C such that 3 times A’s share, 2 times B’s share and 4 times C’s share are all equal. The shares of A, B and C are respectively

Option A
3A = 2B = 4C. Also A + B + C = 390. ∴ A + + = 390
⇒ (4 + 6 + 3)A = 390 × 4 ⇒ A = 120, ∴ B = 180 and C = 90

10.300 coins consists of 1 rupee, 50 paise and 25 paise coins, their values being in the ratio of 10 : 4 : C. Find the number of coins of each types, check.

Option A
Value of rupee coins = 10 i.e. 10 coins. Value of 50 p coins = 4 i.e. 8 coins. Value of 25 p coins = Rs. 3 i.e. 12 coins. ∴ Ratio of coins = 10 : 8 : 12 ⇒ 5 : 4 : 6. ∴ Number of rupee coins = 5/15 × 300 = 100. Number of 50 P coins = 4/15 × 300 = 80 and Number of 25 P coins = 6/15 × 300 = 120