Ratio and Proportion: Definition, Formulas & Examples (2024)

Solved Examples of Ratio and Proportion

Example 1: If A : B = 2 : 3 and B : C = 5 : 7 then what is the ratio A : B : C ?

Solution: A : B = 2 : 3 B : C = 5 : 7

Multiply by 3/5 so as to make the ratio term of B Common, B : C = 5 × 3/5 : 7 × 3/5

⇒ B : C = 3 : 21/5

A : B : C = 2 : 3 : 21/5

=2 × 5 : 3 × 5 : 21/5 × 5

Hence, A : B : C = 10 : 15 : 21

Example 2: What is the equivalent compound ratio of 17 : 23 ∷ 115 : 153 ∷ 18 : 25

Solution: We know, compound ratio of the ratios (a : b), (c : d), (e : f) will be (ace : bdf) Thus, the compound ratio of (17 : 23), (115 : 153), (18 : 25) = (17 × 115 × 18) / (23 × 153 × 25) = 2 : 5

Example 3: If 3 : 27 ∷ 5 : ?

Solution: If 3 : 27 ∷ 5 : ?

3/27 = 5/?

? = 5 × 27/3

? = 45

Example 4: Find the mean proportional between 14 & 15?

Solution: Mean proportional = √(ab)

⇒ √(14 × 15)

⇒ 14.5

So, the mean proportional of 14 and 15 = 14.5

Example 5: Mean proportional of 4 and 36 is a and third proportional of 18 and a is b. Find the fourth proportional of b, 12, 14.

Solution: Given,

Mean proportional of 4 and 36 = a

⇒ a2 = 4 × 36

⇒ a = 12

Third proportional of 18 and 12 = b

⇒ 122 = 18 × b

⇒ b = 8

Fourth proportional of 8, 12 and 14

⇒ 8/12 = 14/?

⇒ ? = 21

Example 6: A bag has coins of Rs. 1, 50 Paise and 25 Paise in ratio of 5 : 9 : 4. What is the worth of the bag if the total number of coins in the bags is 72?

Solution:⇒ Number of Rs. 1 Coins = 5/18 × 72 = 20

⇒ Number of 50 Paise coins = 9/18 × 72 = 36

⇒ Number of 25 Paise coins = 4/18 × 72 = 16

⇒ Total worth of the bag = (20 × 1) + (0.5 × 36) + (0.25 × 16) = 20 + 18 + 4 = Rs. 42

Example 7: If 18 : 13.5 : : 16 : x and (x + y) : y : : 18 : 10, then what is the value of y?

Solution: 18 : 13.5 : : 16 : x x = (16 × 13.5)/18 x = 12

Now,

(x + y) : y : : 18 : 10

(12 + y) : y : : 9 : 5 5(12 + y) = 9y

60 + 5y = 9y

4y = 60

y = 15

Example 8: There are a certain number of Rs.10, Rs.20 and Rs.50 notes available in a box. The ratio of the number of notes of Rs.10, Rs.20 and Rs.50 is 3 ∶ 4 ∶ 6. The total amount available in a box is Rs.2460. The amount of Rs.10 and Rs.50 in a box is –

Solution: Let the number of notes of Rs.10, Rs.20 and Rs.50 be 3a, 4a and 6a respectively. Given,

⇒ 10 × 3a + 20 × 4a + 50 × 6a = 2460

⇒ 410a = 2460

⇒ a = 6

Number of notes of Rs.10 = 3 × 6 = 18

Number of Notes of Rs.20 = 4 × 6 = 24 Number of notes of Rs.50 = 6 × 6 = 36

Required amount = 10 × 18 + 50 × 36 = Rs.1980

Example 9: Mr. Raj divides Rs. 1573 such that 4 times the 1st share, thrice the 2nd share and twice the third share amount to the same. Then the value of the 2nd share is:

Solution: Given:Total amount = Rs. 1573

Calculation:Let the share of A, B and C is 4A : 3B : 2C. A : B : C = 1/4 : 1/3 : 1/2 = 3 : 4 : 6

The value of the 2nd share = (4/13) × 1573 = Rs. 484

Example 10: Wayne wants to use Nitrogen, Potassium, and Phosphorus in his field as fertilizers. When any of them is mixed in the field, their quantity reduces by 1 kg every day due to chemical reactions. He mixed Nitrogen, Potassium, and Phosphorus on 7th November, 9th November, and 15th November, respectively. He spent equal amounts on buying each of the three. What should be the ratio of prices of Nitrogen, Potassium, and Phosphorus, so that there is an equal quantity of each of them in the field on 16th November, and that quantity is 11 kg?

Solution: Given:Quantities of each of Nitrogen, Potassium and Phosphorus in the field on 16th November are 11 kg.

Concept used:If equal amounts are spent on buying the components, the ratio of their prices will be inverse of ratios of their quantities.

Calculation:Nitrogen was mixed on 7th November.

Quantity of Nitrogen when it was mixed = 11 + (16 – 7) = 20 kg

Potassium was mixed on 9th November.

Quantity of Potassium when it was mixed = 11 + (16 – 9) = 18 kg

Phosphorus was mixed on 15th November.

Quantity of Phosphorus when it was mixed = 11 + (16 – 15) = 12 kg

Expenditure = quantity bought × Price per unit

⇒ Ratio of their prices = (1/20):(1/18):(1/12) = 9:10:15

∴ The required ratio is 9 ∶ 10 ∶ 15.

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