Ratio and proportion are basic mathematical concepts that are very important in solving a wide range of real-world and academic issues. In the United States, knowledge of ratios and proportions is especially useful for students who are preparing for different competitive exams, such as the SAT, ACT, GRE, GMAT, and ASVAB. For SAT Math, problems relating to ratios and proportions commonly manifest in problem-solving and data analysis, for which a good conceptual understanding of such is required in order to be able to read relationships among quantities and answer word problems in the most effective way. 

• In addition, such ideas are equally relevant in applications related to finance, science, and engineering wherein it is necessary to compare values, scale quantities, and make equivalencies. 
• In this article, we will look at the essential concepts of ratio and proportion, proportion types, helpful formulas, tips, and tricks to get problems solved and in-depth solved examples that will help candidates to prepare for exams.

What are Ratio and Proportion?

Ratios are used for comparing two quantities of an identical style whereas when two or more such ratios are identical, they are declared to be in proportion.

Ratios are used when we are required to express one number as a fraction of another. If we have two quantities, say x and y, then the ratio of x to y is calculated as \(\frac{x}{y}\) and is written as \(x : y\). The first term of the ratio is called antecedent and the second term is called the consequent.

Compound ratio is the ratio obtained if two or more ratios are given and the antecedent of one is multiplied by the antecedent of others and consequents are multiplied by the consequences of others. Compounded ratio of the ratios \((a : b), (c : d), (e : f) \: will \: be \: (ace : bdf)\)

Proportion is an equation that specifies that the two given ratios are identical to one another. We can say that the proportion states the equivalency of the two fractions or the ratios i.e Equivalent Ratios. Proportions are represented by the symbol \(( : : )\) or equal to \(( = )\).

That is the proportion is signified by double colons. For example, ratio \(6 : 8\) is the same as ratio \(3 : 4\). This can be written as \(6 : 8 :: 3 : 4\).

Product of means = Product of extremes

Thus, \(a : b ∷ c : d ⇔ (b × c) = (a × d)\)

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Types of Proportion

The proportion can be categorized into the following types:

• Direct Proportion
• Inverse Proportion
• Continued Proportion

Direct Proportion: In proportion, if two sets of given numerals are rising or falling in the same ratio, then the ratios are considered to be directly proportional to each other. That is we can say that direct proportion illustrates the relationship between two portions wherein the gains in one there is a growth in the other quantity too. Likewise, if one quantity drops, the other also decreases.

Therefore, if “X” and “Y” are two quantities, then the direction proportion is composed as \( x ∝ y \).

Inverse Proportion: The inverse proportion as the name outlines are in contrast to the direct one; where the relationship between two quantities is defined such that growth in one leads to a decline in the other quantity. Likewise, if there is a drop in one portion, there is an expansion in the other portion.

Accordingly, the inverse proportion of two quantities, say “p” and “q” is represented by \( p ∝ (\frac{1}{q}) \).

Continued Proportion: If we assume two ratios to be p: q and r: s and we are interested in determining the continued proportion for the given ratio. Then we transform the means to a single term/digit. That is we will find the LCM of means.

• For the provided ratio, the LCM of q & r will be qr.
• The next step is to multiply the first ratio by r and the second ratio by q as shown:
• First ratio- \(rp : rq\)
• Second ratio- \(rq : qs\)
• Thus, the continued proportion can be documented in the form of \(rp : qr : qs[/ latex]

Ratio and Proportion Formula

Ratio Formula: For any 2 given quantities say x and y, the formula for the ratio is: \(\)x : y ⇒ \frac{x}{y} \)

where

• x that is the first term is also called the antecedent.
• y which is the second term is also called the consequent.

For instance, ratio \(7 : 17 \) is depicted by \(\frac{7}{17}\), where 7 is antecedent and 17 is consequent.

Proportion Formula: Now for the proportion formula consider two ratios, \( p : q \)and \( r : s \). Then, \(p:q:: r:s⟶\frac{p}{q}=\frac{r}{s}\)

• Here, the two terms q and r are named mean terms.
• Whereas the other two terms i.e. p and s are understood as extreme terms.

Difference Between Ratio and Proportion

The difference between ratio and proportion is given in the table below:

Tips and Tricks on Ratio and Proportion

Students can find different tips and tricks for solving questions related to ratio and proportion below:

Tip 1: In ratio, if both the antecedent and the consequent are multiplied or divided by the same number (except 0) then the ratio will remain the same.

Tip 2:If a proportion is such as a:x::x:b then x is called the mean proportional or second proportional of a and b. And if a proportion is such that a:b::b:x then x is called the third proportional of a and b.

Tip 3: Componendo rule: If \( \frac{a}{b} = \frac{c}{d} \: then \: \frac{a+b}{b} = \frac{c+d}{d}\)

Tip 4: Dividendo rule: If \( \frac{a}{b} = \frac{c}{d} \: then \: \frac{a-b}{b} = \frac{c-d}{d}\)

Tip 5: Componendo & Dividendo rule: If \( \frac{a}{b} = \frac{c}{d} \: then \: \frac{a+b}{a-b} = \frac{c+d}{c-d}\)

Tip 6: Invertendo rule: If \( \frac{a}{b} = \frac{c}{d} \: then \: \frac{b}{a} = \frac{d}{c}  \)

Tip 7: Alternendo rule: If \( \frac{a}{b} = \frac{c}{d} \: then \: \frac{a}{c} = \frac{b}{d} \)

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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.

In summary, it is important for American students to master the principles of ratio and proportion if they are going to take competitive exams like the SAT, ACT, GRE, GMAT, and ASVAB. The principles are not only critical to problem-solving as well as the analysis of data, but they also come in handy in real-world applications in sciences, finance, and engineering. By learning ratio and proportion formulas, types, and problem-solving methods, students can enhance their mathematical skills to a great extent and solve exam questions confidently. In addition, practicing solved examples and using helpful tips and tricks will guarantee a better conceptual understanding and greater accuracy in solving intricate problems.  You can also download the Testbook App, which is absolutely free and start preparing for any government competitive examination by taking the mock tests before the examination to boost your preparation.