# Formula - Ratio & Proportion

**RATIO**

Ratio is strictly a mathematical term to compare two similar quantities expressed in the same units.

The ratio of two terms ‘x’ and ‘y’ is denoted by x : y.

In general, the ratio of a number x to a number y is defined as the quotient of the numbers x and y.

The numerator of the ratio is called the antecedent (x) and the denominator is called the consequent (y) of the ratio.

**COMPARISON OF TWO OR MORE RATIOS**

Two or more ratios may be compared by reducing the equivalent fractions to a common denominator and then comparing the magnitudes of their numerator.Thus, suppose 2 : 5, 4 : 3 and 4 : 5 are three ratios to be compared then the fractions ⅖ , 4/3 and ⅘ are reduced to equivalent fractions with a common denominator.For this, the denominator of each is changed to 15 equal to the L.C.M. their denominators Hence the given ratios are expressed 6/15 , 20/15 and 12/15 or 2 : 5, 4 : 3, 4 : 5 according to magnitude.

**Example 1**

**1. Which of the ratios 2 :3 and 5 :9 is greater ?**

**Sol.** In the form of fractions, the given ratios are ⅔ and &frac59; Reducing them to fractions with a common denominator they are written as &frac69; and &frac59;.

Hence the greater ratio is &frac69; or 2:3.

**Example 2**

**2. Are the ratios 3 to 4 and 6:8 equal ?**

**Sol.** The ratios are equal if 3/4 = 6/8. These are equal if their cross products are equal; that is,if 3 × 8 = 4 × 6. Since both of these products equal 24, the answer is yes, the ratios are equal.

Remember to be careful! Order matters!

A ratio of 1 : 7 is not the same as a ratio of 7:1.

✦ The two quantities must be of the same kind and in same unit.

✦ The ratio is a pure number, i.e., without any unit of measurement.

✦ The ratio would stay unaltered even if both the antecedent and the consequent are multiplied or divided by the same number.

**Compound ratio :** Ratios are compounded by multiplying together the antecedents for a new antecedent and the consequents for a new consequent.The compound of a : b and c : d is \(\frac{a \times c}{b \times d}\) , i.e., ac : bd.

**Example 3**

**3. Find the compound ratio of the four ratios : 4 : 5, 15 : 13, 26 : 3 and 6 : 17**

**Sol.** The required ratio \(= \frac{4 \times 15 \times 26 \times 6}{15 \times 13 \times 3 \times 17} = \frac{48}{17}\)

or 48 : 17

The duplicate ratio of x : y is x^{2} : y^{2}.

The triplicate ratio of x : y is x^{3} : y^{3}.

The subduplicate ratio of x : y is √x : √y

The subtriplicate ratio of x : y is \(\sqrt[3]{x} : \sqrt[3]{y}\)

Reciprocal ratio of a : b is 1/a : 1/b or b : a

**Inverse ratio**

Inverse ratio of x : y is y : x.

**PROPERTIES**

**1**. If \(\frac{a}{b} : \frac{c}{d} \;\; then \;\; \frac{b}{a} : \frac{d}{c}\) , i.e., the inverse ratios of two equal ratios are equal. The property is called Invertendo.

**2**. If \(\frac{a}{b} : \frac{c}{d} \;\; then \;\; \frac{a}{c} : \frac{b}{d}\) , i.e., the ratio of antecedents and consequents of two equal ratios are equal. This property is called Alternendo.

**3**. If \(\frac{a}{b} : \frac{c}{d} \;\; then \;\; \frac{a + b}{b} : \frac{c + d}{d}\). This property is called Componendo.

**4**. If \(\frac{a}{b} : \frac{c}{d} \;\; then \;\; \frac{a ? b}{b} : \frac{c ? d}{d}\). This property is called Dividendo.

**5**. If \(\frac{a}{b} : \frac{c}{d} \;\; then \;\; \frac{a + b}{a ? b} : \frac{c + d}{c ? d}\). This property is called Componendo – Dividendo.

**6**. If \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = ?..\; Then,\)

Each ratio = \(\frac{Sum \; of \; Numerators}{Sum \; of \; denominators}\)

i.e. \(\frac{a}{b} = \frac{c}{d} = \frac{a + c + e + ?}{b + d + f + ?}\)

**7**. If we have two equations containing three unknowns as

a_{1}x + b_{1}y + c_{1}z = 0 and… (i)

a_{2}x + b_{2}y + c_{2}z = 0… (ii)

then, the values of x, y and z cannot be resolved without having a third equation.

However, in the absence of a third equation, we can find the proportion x : y : z.

This will be given by

b_{1}c_{2} – b_{2}c_{1} : c_{1}a_{2} – c_{2}a_{1} : a_{1}b_{2} – a_{2}b_{1}.

**8**. To find the ratio of the two variables of a homogeneous equation of second degree.For this all the terms of the homogeneous equation are taken on one side and factorized into linear factoA linear equation is formed from each of the factors and the ratio of the variables obtained

If the ratio between the first and the second quantities is a : b and the ratio between the second and third quantities is c : d, then the ratio among first, second and third quantities is given by ac : bc : bd

If the ratio between the first and the second quantities is a : b and the ratio between the second and third quantities is c : d and the ratio between the third and fourth quantities is e : f, then the ratio among first, second, third and fourth quantities is given by ace : bce : bde : bdf

#### To divide a given quantity into a given ratio.

Suppose any given quantity a, is to be divided in the ratio m : n.

Let one part of the given quantity be x then the other part will be a – x.

∴ \(\frac{x}{a - x} = \frac{m}{n}\) nx = ma – mx or (m + n) x = ma

∴ one part is \(\frac{ma}{m + n}\) and the other part will be

\(a - \frac{ma}{m + n} = \frac{na}{m + n}\)**Example 4**

**4. Divide 70 in the ratio 3 : 7.
**

**Sol.**Let one part be x

then the other part = 70 – x

∴ \(\frac{x}{70 ? x} = \frac{3}{7}\) or 7x = 210 – 3x

or x = 21 and 70 – x = 49

Hence the two required parts of 70 are 21 and 49.

**Example 5**

**5. What is the least integer which when subtracted from both the numerator and denominator of 60/70 will give a ratio equal to 16/21 ?**

**Sol.** Let x be the required integer. Then,

=> 1260 – 21x = 1120 – 16x

=> 5x = 140 => x = 28.

**Example 6**

**6. If \(\mathbf{\frac{x}{y} = \frac{4}{5}}\) , find the value of \(\mathbf{\frac{3x + 4y}{4x + 3y}}\).**

**Sol.** \(\frac{3x + 4y}{4x + 3y} = \frac{\frac{3x}{y} + 4}{\frac{4x}{y} + 3} = \frac{3 \times \frac{4}{5} + 4}{4 \times \frac{4}{5} + 3} = \frac{32}{31}.\)

**Example 7**

**7.Find the value of \(\mathbf{\frac{x + a}{x ? a} + \frac{x + b}{x ? b}}\) , if x = \(\mathbf{\frac{2ab}{a + b}}\).**

**Sol.** \(x = \frac{2ab}{a + b} \Rightarrow \frac{x}{a} = \frac{2b}{a + b}\)

By componendo – dividendo,

\(\frac{x + a}{x ? a} = \frac{3b + a}{b ? a}\)Similarly, \(\frac{x}{b} = \frac{2a}{a + b}\)

=> \(\frac{x + b}{x ? b} = \frac{3a + b}{a ? b}\)

∴ \(\frac{x + a}{x ? a} + \frac{x + b}{x ? b} = \frac{3b + a}{b ? a} + \frac{3a + b}{a ? b}\)

= \(\frac{-(3b + a)}{a ? b} + \frac{3a + b}{a ? b} = \frac{2a ? 2b}{a ? b} = 2.\)

**Example 8**

**8. Divide 581 among A, B and C such that four time A’s share is equal to 5 times B’s share which is equal to seven times C’s share.**

**Sol.** 4 times A’s share = 5 times B’s share

= 7 times C’s share.

∴ A :B :C =35 : 28 :20

∴ Share of A = \(\frac{35}{35 + 28 + 20} \times 581\) = 245.

Share of B = \(\frac{28}{83} \times 581\) = 196.

Share of C = \(\frac{20}{83} \times 581\) = 140.

In any 2-dimensional figures, if the corresponding sides are in the ratio x : y,then their areas are in the ratio x^{2} : y^{2}.

**Example 9**

**9. The ratio of the radius of two circles is 2 : 5.Find the ratio of their areas.**

**Sol.** Ratio of their areas = 2^{2}: 5^{2}= 4 : 25

In any two 3-dimensional figures, if the corresponding sides are in the ratio x : y, then their volumes are in the ratio x^{3} : y^{3}. If the ratio between two numbers is a : b and if each number is increased by x, the ratio becomes c : d.Then

Sum of the two numbers = \(\frac{x\left ( a + b \right )\left ( c ? d \right )}{ad ? bc}\)

Difference of the two numbers = \(\frac{x\left ( a ? b \right )\left ( c ? d \right )}{ad ? bc}\)

Two numbers are given as \(\frac{xa\left ( c ? d \right )}{ad ? bc} \; and \; \frac{xb\left ( c ? d \right )}{ad ? bc}\)

**Example 10**

**10. The ratio between two numbers is 3: 4. If each number be increased by 2, the ratio becomes 7 : 9. Find the number.**

**Sol.** Numbers are \(\frac{2 \times 3\left ( 7 ? 9 \right )}{3 \times 9 ? 4 \times 7}\; and \; \frac{2 \times 4\left ( 7 ? 9 \right )}{3 \times 9 ? 4 \times 7}\) or 12 and 16.

If the sum of two numbers is A and their difference is a, then the ratio of numbers is given by A + a : A – a.

**Example 11**

**11. The sum of two numbers is 60 and their difference is 6. What is the ratio of the two numbers ?**

**Sol.** The required ratio of the numbers

= \(\frac{60 + 6}{60 ? 6} = \frac{66}{54} = \frac{11}{9}\) or 11 : 9

**Example 12**

**12.Three persons A, B, C whose salaries together amount to 14400, spend 80, 85 and 75 percent of their salaries respectively. If their savings are in the ratio 8 : 9 : 20, find their respective salaries.**

**Sol.** A, B and C spend 80 %, 85 % and 75 % respectively of their salaries.

=> A, B and C save 20 %, 15 % and 25 % respectively of their salaries.

So, 20 %of A’s salary : 15 % of B’s salary :

25 % of C’s salary = 8 : 9 : 20

=> 1/5 of A’s salary : 3/20 of B’s salary :

1/4 of C’s salary = 8 : 9 : 20

Now \(\frac{\frac{1}{5} \; of \; A?s \; salary}{\frac{3}{20} \; of \; B?s \; salary} = \frac{8}{9}\)

=> \(\frac{A?s \; salary}{B?s \; salary} = \frac{3}{20} \times 8 \times \frac{5}{9} = \frac{2}{3}\)

∴ A’s salary : B’s salary = 2: 3 …(ii)

Similarly, B’s salary : C’s salary = 3 :4 …(iii)

From (ii) and (iii)

A’s salary : B’s salary : C’s salary = 2 :3 :4.

∴ A’s salary = \(\frac{2}{2 + 3 + 4} \times 14400 = 3200\)

B’s salary = \(\frac{3}{2 + 3 + 4} \times 14400 = 4800\)

C’s salary = \(\frac{4}{2 + 3 + 4} \times 14400 = 6400\)

**PROPORTION**

When two ratios are equal, the four quantities composing them are said to be in proportion.

If a/b = c/d, then a, b, c, d are in proportions.

This is expressed by saying that ‘a’ is to ‘b’ as ‘c’ is to ‘d’ and the proportion is written as

a :b :: c: d or a : b = c : d

The terms a and d are called the extremes while the terms b and c are called the means.

Let a, b, c, d be in proportion, then \(\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\)

✦ If three quantities a, b and c are in continued proportion,then a : b = b : c

∴ ac = b^{2}

b is called mean proportional.

✦ If three quantities are proportionals, then first is to the third is the duplicate ratio of the first is to the second.

✦ If a : b:: b : c then a : c = a^{2} : b^{2}

**To find the mean proportional.**

**Example 13**

13.Find the mean proportional between 3 and 75.

13.Find the mean proportional between 3 and 75.

**Sol.** Let x be the required mean proportional. Then,

3 : x :: x : 75

∴ x = \(\sqrt{3 \times 75} = 15\)

**Example 14**

**14. What must be added to each of the four numbers 10, 18, 22, 38 so that they become in proportion ?**

**Sol.** Let the number to be added to each of the four numbers be x

By the given condition, we get

(10 + x) : (18 + x) : : (22 + x) : (38 + x)

=> (10 + x) (38 + x) = (18 + x) (22 + x)

=> 380 + 48x + x^{2} = 396 + 40x + x^{2}

Cancelling x² from both sides, we get

380 + 48x = 396 + 40x

=> 48x – 40x = 396 – 380

=> 8x = 16 => x = 16/8 = 2

Therefore, 2 should be added to each of the four given number

**To find the fourth proportional**

**Example 15**

**15. Find the fourth proportional to
p ^{2} – pq + q^{2}, p^{3} + q^{3}, p – q**

**Sol.** Let x be the fourth proportional

∴ (p^{2} – pq + q^{2}) : (p^{3} + q^{3}) = (p – q) : x

=> (p^{2} – pq + q^{2}) × x = (p^{3} + q^{3}) (p – q)

∴ \(x = \frac{\left ( p^{3} + q^{3} \right )\left ( p ? q \right )}{\left ( p^{2} ? pq ? q^{2} \right )}\)

=> \(x = \frac{\left ( p + q \right )\left ( p^{2} ? pq + q^{2} \right )\left ( p ? q \right )}{\left ( p^{2} ? pq ? q^{2} \right )}\)

=> x = (p + q)(p – q) = p^{2} – q^{2}

∴ The required fourth proportional is p^{2} – q^{2}

**To find the third proportional**

**Example 16**

**16. Find third proportional to a² – b² and a + b.**

**Sol.** Let x be the required third proportional

Then a^{2} – b^{2} : a + b = a + b : x

∴ (a^{2} – b^{2}) x = (a + b) (a + b)

∴ \(x = \frac{\left ( a + b \right )\left ( a + b \right )}{\left ( a^{2} ? b^{2} \right )} = \frac{a + b}{a ? b}\)

**Direct Proportion :**

If on the increase of one quantity, the other quantity increases to the same extent or on the decrease of one, the other decreases to the same extent, then we say that the given two quantities are directly proportional. If A and B are directly proportional then we denote it by A ∝ B.

Also, A = kB, k is constant

=> A/B = k

If b_{1} and b_{2} are the values of B corresponding to the values a_{1}, a_{2} of A respectively, then

**Some Examples :**

1. Work done ∝ number of men

2. Cost ∝ number of Articles

3. Work ∝ wages

4. Working hour of a machine ∝ fuel consumed

5. Speed ∝ distance to be covered

**Indirect Proportion (or inverse proportion) :**

If on the increase of one quantity, the other quantity decreases to the same extent or vice versa, then we say that the given two quantities are indirectly proportional. If A and B are indirectly proportional then we denote it by A ∝ 1/B.

Also, A = k/B(k is constant)

=> AB = k

If b_{1}, b_{2} are the values of B corresponding to the values a_{1}, a_{2} of A respectively, then

a_{1}b_{1} = a_{2}b_{2}

**Some Examples :**

1. More men, less time

2. Less men, more hours

3. More speed, less taken time to be covered distance

**Example 17**

**17. A garrison of 3300 men had provision for 32 days, when given at the rate of 850 gm per head. At the end of 7 days, a reinforcement arrived and it was found that the provision would last 17 days more, when given at the rate of 825 gm per head. What was the strength of the reinforcement ?**

(a) 1500

(b) 1700

(c) 1800

(d) 1600

**Sol.(b)** There is a provision for 2805 × 32 kg for 3300 men for 32 days @ 850 gm per head per day.

In 7 days, 3300 men consumed

Let the strength of the reinforcement arrived after 7 days be x.

∴ (3300 + x) men had provision of 2805 × 25 kg for 17 days @ 825 gm per head per day, i.e.

∴ \(\frac{\left ( 3300 + x \right ) \times 825 \times 17}{1000} \times 7 = 2805 \times 25\)

=> (3300 + x) = \(\frac{1000 \times 2805 \times 25}{825 \times 17} = 5000\)

=> x = 1700

∴ Strength of the reinforcement arrived after 7 days= 1700.

**RULE OF THREE**

In a problem on simple proportion, usually three terms are given and we have to find the fourth term, which we can solve by using Rule of three. In such problems, two of given terms are of same kind and the third term is of same kind as the required fourth term.First of all we have to find whether given problem is a case of direct proportion or indirect proportion.

For this, write the given quantities under their respective headings and then mark the arrow in increasing direction. If both arrows are in same direction then the relation between them is direct otherwise it is indirect or inverse proportion. Proportion will be made by either head to tail or tail to head.

The complete procedure can be understand by the examples.

**Example 18**

**18. A man completes 5/8 of a job in 10 days. At this rate, how many more days will it take him to finish the job ?**

(a) 5

(b) 6

(c) 7

(d) 7½

**Sol.(b)** Work done = ⅝. Balance work = (1 – ⅝) = ⅜.

Less work, Less days (Direct Proportion)

Let the required number of days be x. Then,

Then, ⅝ : ⅜ :: 10 : x =>; ⅝ × x = ⅜ × 10

=> x = (⅜ × 10 × 8/5) = 6.

**Example 19**

**19. A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. The number of days for which the remaining food will last, is :**

(a) 29⅕

(b) 37¼

(c) 42

(d) 54

**Sol.(c)** After 10 days : 150 men had food for 35 days. Suppose 125 men had food for x days. Now,**Less men,More days (Indirect Proportion)** Then,

∴ 125 : 150 :: 35 : x => 125 × x = 150 × 35

=> x = \(\frac{150 \times 35}{125} \Rightarrow x = 42.\)

Hence, the remaining food will last for 42 days.

**Compound Proportion or Double Rule of Three
**

In the compound proportion, number of ratios are more than two.

**Example 20**

**20. If the cost of printing a book of 320 leaves with 21 lines on each page and on an average 11 words in each line is 19, find the cost of printing a book with 297 leaves, 28 lines on each page and 10 words in each line.**

(a) 22⅜

(b) 20⅜

(c) 21⅜

(d) 21¾

**Sol.(c)**

Less leaves, less cost (Direct Proportion)

More lines, more cost (Direct Proportion)

Less words, less cost (Direct Proportion)

∴ 320 × 21 × 11 × x = 297 × 28 × 10 × 19

=> x = 171/8 = 21⅜

**PARTNERSHIP**

A partnership is an association of two or more persons who invest their money in order to carry on a certain business.

A partner who manages the business is called the **working partner** and the one who simply invests the money is called the **sleeping partner**.

Partnership is of two kinds :

(i) Simple

(ii) Compound.

**Simple partnership :** If the capitals is of the partners are invested for the same period, the partnership is called simple.

**Compound partnership :** If the capitals of the partners are invested for different lengths of time, the partnership is called compound.

If the period of investment is the same for each partner,then the profit or loss is divided in the ratio of their investments.

If A and B are partners in a business, then

\(\frac{Investment \; of \; A}{Investment \; of \; B} = \frac{Profit \; of \; A}{Profit \; of \; B} \; or = \frac{Loss \; of \; A}{Loss \; of \; B}\)If A, B and C are partners in a business, then Investment of A: Investment of B :Investment of C

= Profit of A : Profit of B : Profit of C, or

= Loss of A : Loss of B : Loss of C

**Example 21**

**21. Three partner Rahul, Puneet and Chandan invest 1600, 1800 and 2300 respectively in a business.How should they divide a profit of 399 ?**

**Sol.** Profit is to be divided in the ratio 16 :18 : 23

Rahul’s share of profit = \(\frac{16}{16 + 23 + 18} \times 399\)

= \(\frac{16}{57} \times 399\) = 112

Puneet’s share of profit = \(\frac{18}{57} \times 399\) = 126

Chandan’s share of profit = \(\frac{23}{57} \times 399\) = 161

**Example 22**

**22. A and B invested in the ratio 3 : 2 in a business. If 5% of the total profit goes to charity and A’s share is855, find the total profit.**

**Sol.** Let the total profit be 100.

Then, 5 goes to charity.

Now, 95 is divided in the ratio 3 : 2.

∴ A’s share = \(\frac{95}{3 + 2} \times 3\) = 57

But A’s actual share is 855.

∴ Actual total profit = \(855\left ( \frac{100}{57} \right )\) = 1500

**MONTHLY EQUIVALENT INVESTMENT**

It is the product of the capital invested and the period for which it is invested.

If the period of investment is different, then the profit or loss is divided in the ratio of their Monthly Equivalent Investment.

= \(\frac{Profit \; of \; A}{Profit \; of \; B} \; or \; \frac{Loss \; of \; A}{Loss \; of \; B}\)

i.e., \(\frac{Investment \; of \; A \times Period \; of \; Investment \; of \; A}{Investment \; of \; B \times Period \; of \; Investment \; of \; B}\)

= \(\frac{Profit \; of \; A}{Profit \; of \; B} \; or \; \frac{Loss \; of \; A}{Loss \; of \; B}\)

If A, B and C are partners in a business, then

Monthly Equivalent Investment of A : Monthly Equivalent Investment of B : Monthly Equivalent Investment of C

= Profit of A : Profit of B : Profit of C.

= Loss of A : Loss of B : Loss of C.

**Example 23**

**23. A and B start a business. A invests 600 more than B for 4 months and B for 5 months. A’s share is48 more than that of B, out of a total profit of 528. Find the capital contributed by each.**

**Sol.** B’s profit = \(\frac{528 ? 48}{2}\) = 240

A’s profit = 528 – 240 = 288

\(\frac{A?s \; capital \times 4}{B?s \; capital \times 5} = \frac{288}{240} = \frac{6}{5}\)∴ \(\frac{A?s \; capital}{B?s \; capital} = \frac{6}{5} \times \frac{5}{4} = \frac{3}{2}\)

=> \(\frac{B?s \; capital + 600}{B?s \; capital} = \frac{3}{2}\)

=> B’s capital = 1200 and A’s capital =1800

**Example 24**

**24. Three persons A, B, C rent the grazing of apark for 570. A puts in 126 oxen in the park for 3 months, B puts in 162 oxen for 5 months and C puts in 216 oxen for 4 months. What part of the rent should each person pay ?**

**Sol.** Monthly equivalent rent of A = 126 × 3 = 378

Monthly equivalent rent of B = 162 × 5 = 810

Monthly equivalent rent of C = 216 × 4 = 864

∴ Rent is to be divided in the ratio

378 : 810 : 864, ie. 7 : 15 : 16

∴ A would have to pay \(\frac{7}{7 + 15 + 16}\) of the rent

= 7/38 of the rent = 7/38 × 570 = 105

B would have to pay 15/38 of the rent = 15/38 × 570

= 225

and C would have to pay 16/38, i.e. 8/19 of the rent

\(\frac{8}{19} \times 570\) = 240**MIXTURE**

**Simple Mixture:** When two different ingredients are mixed together,it is known as a simple mixture.

**Compound Mixture :** When two or more simple mixtures are mixed together to form another mixture, it is known as a compound mixture.

**Alligation :** Alligation is nothing but a faster technique of solving problems based on the weighted average situation as applied to the case of two groups being mixed together.The word ‘Alligation’ literally means ‘linking’.

**Alligation rule :** It states that when different quantities of the same or different ingredients of different costs are mixed together to produce a mixture of a mean cost, the ratio of their quantities is inversely proportional to the difference in their cost from the mean cost.

**Graphical representation of Alligation Rule :**

Applications of Alligation Rule :

(i) To find the mean value of a mixture when the prices of two or more ingredients, which are mixed together and the proportion in which they are mixed are given.

(ii) To find the proportion in which the ingredients at given prices must be mixed to produce a mixture at a given price.

**Example 25**

**Example 26**

**26. A mixture of a certain quantity of milk with 16 litres of water is worth 90 P per litre. If pure milk be worth 1.08 per litre, how much milk is there in the mixture ?**

**Sol.** The mean value is 90P and the price of water is 0 P.

By the Alligation Rule, milk and water are in the ratio of 5 : 1.

∴ Quantity of milk in the mixture = 5 × 16 = 80 litres.

**Price of the Mixture :**

When quantities Q_{i} of ingredients M_{i}’s with the cost C_{i}’s are mixed then cost of the mixture C_{m} is given by

**Straight line approach of Alligation**

Let Q_{1} and Q_{2} be the two quantities, and n_{1} and n_{2} are the number of elements present in the two quantities respectively,

where Av is the average of the new group formed then n_{1} corresponds to Q_{2} – Av, n_{2} corresponds to Av – Q_{1} and (n_{1} + n_{2}) corresponds to Q_{2} – Q_{1}.Let us consider the previous example.

**Example 27**