# Concept: Ratio, Proportion, Variation and Partnerships

CONTENTS

**INTRODUCTION**

The Concept of Ratio, Proportion and variation is very important for MBA Entrance exams. Basic fundamentals of this topic are extensively used in the areas like Data Interpretation, Time & Work, Time, Speed & Distance etc.

**RATIO**

Comparison of two numbers or quantities having the same units is known as a ratio. Sometimes we need to find out how many times one number (say a) is compared to another number (say b). The ratio of a to b is written as:

a : b = $\frac{\mathrm{a}}{\mathrm{b}}$

In the ratio a : b, a is known as the antecedent and b is consequent.

For **Example**:

Gautam and Virendra have Rs. 56 and Rs. 14 with them respectively.

Then, we can say that Gautam has Rs. 42 more than Virendra or alternatively Gautam has 4 times as much money as Virendra. The latter way of comparing is by finding ratios between the amount with them

$\frac{\mathrm{Amount\; with\; Gautam}}{\mathrm{Amount\; with\; Virendra}}$ = $\frac{56}{14}=\frac{4}{1}$

- Ratio is used to compare similar type of numbers (i.e. their units should be same).
- Ratio is generally expressed in simplest form (i.e. by cancelling out any common factor between the numbers).
- Ratio does not have any unit.
- The order of terms is important, i.e. a : b is not same as b : a.

**Example**: Calculate the ratio of 15 and 25?

**Solution**:

Required ratio is $\frac{15}{25}$.

This can further be simplified by eliminating the common factor from both the numbers. Here, the common factor is 5.

Hence, the simplified ratio = $\frac{3}{5}$

**Example** : 7 is what part of 8?

**Solution**:

Required part is the ratio of 7 and 8.

∴ part = $\frac{7}{8}$.

It means 7 is $\frac{7}{8}$^{th} of 8.

**Example** : $\frac{3}{48}$ is what part of $\frac{1}{12}$?

**Solution**:

Required part is the ratio of $\frac{3}{48}$ and $\frac{1}{12}$

∴ part = $\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$48$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}}=\frac{3}{48}\times \frac{12}{1}=\frac{3}{4}$ part.

It means $\frac{3}{48}$ is $\frac{3}{4}$^{th} of $\frac{1}{12}$.

**Example**:

In a ratio which is equal to 3 : 7, if the antecedent is 33, what is the consequent?

**Solution**:

$\frac{3}{7}=\frac{33}{x}$

⇒ x = 77

**Example**:

Find a fraction which bears the same ratio to $\frac{3}{7}$ that $\frac{1}{5}$ does to $\frac{7}{15}$?

**Solution**:

Let the number be x

Then, x ∶ 3/7 = 1/5 ∶ 7/15

⇒ $\frac{x}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$15$}\right.}}$

⇒ $x\times \frac{7}{15}=\frac{1}{5}\times \frac{3}{7}$

⇒ $x=\frac{1}{5}\times \frac{3}{7}\times \frac{15}{7}$ = $\frac{9}{49}$

Let’s take the ratio of amount of money with Gautam and Virendra again.

This can also be interpreted in the following manner: For every Rs. 4 that Gautam has, Virendra has Rs. 1. Therefore, out of total Rs. 5, Gautam has Rs. 4 and Virendra has Rs. 1, i.e. Gautam has $\frac{4}{5}$^{th} of the total and Virendra has $\frac{1}{5}$^{th} of the total.

Hence, if the ratio of two numbers is a : b, it means for every ‘a’ parts of the first numbers there are ‘b’ parts of the second number.

**Example** : Divide Rs. 72 in the ratio 4 : 5.

**Solution**:

First part = $\frac{\mathrm{First}\mathrm{ratio}\mathrm{term}}{\mathrm{Sum}\mathrm{of}\mathrm{ratio}\mathrm{terms}}$ × Total amount = $\frac{4}{4+5}\times 72$ = Rs. 32

and, second part = $\frac{\mathrm{Second}\mathrm{ratio}\mathrm{term}}{\mathrm{Sum}\mathrm{of}\mathrm{ratio}\mathrm{terms}}$ × Total amount = $\frac{5}{4+5}\times 72$ = Rs. 40

As mentioned earlier, ratio is expressed in simplest form. Hence, if the ratio of two numbers is a : b, the numbers need not be exactly ‘a’ and ‘b’. The numbers would be a × x and b × x, where x is the common factor which would’ve been eliminated while simplifying the ratio.

**Example** : Amit and Atul have money in the ratio of 3 : 5. What is the amount with them respectively.

**Solution**:

Here, we know the ratio of the amounts with them i.e. 3 : 5.

Hence, the amount with Amit = 3x

and the amount with Atul = 5x.

Now, there is no additional information given to calculate the value of x. Hence, we cannot determine the amount with each of them.

Hence, the answer cannot be determined.

Alternately,

Amount with Amit = $\frac{3}{8}$^{th} of the total amount.

Since we do not know the total amount, we cannot figure out the exact value of the amount with Amit.

**Example** : Amit and Atul have money in the ratio of 3 : 5. The total amount with them is Rs. 160. What is the amount with them respectively.

**Solution**:

Here, we know the ratio of the amounts with them i.e. 3 : 5.

Hence, the amount with Amit = 3x

and the amount with Atul = 5x.

Now, we know the total amount with them is Rs. 160.

Hence, 3x + 5x = 160

⇒ x = 20

Now, we can calculate the amounts with each of them.

Amount with Amit = 3x = 60.

Amount with Atul = 5x = 100.

- The value of a ratio remains unchanged if each one of its terms is multiplied or divided by a same non-zero number.

i.e. a : b ⇔ $\frac{a}{b}=\frac{ma}{mb}$

and a : b ⇔ $\frac{a}{b}=\frac{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}}{{\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}}$

whatever m may be except zero.

a

^{2}∶ b^{2}is called duplicate ratio of a : b.a

^{3 }: b^{3}is called triplicate ratio of a : b.√a ∶ √b is called sub-duplicate ratio of a : b.

∛a ∶ ∛b is called sub-triplicate ratio of a : b.

If a : b and c : d are two ratios, then product of ratios i.e. ac : bd is called the compound ratio.

If the ratio is less than 1 then it is said to be less in equality. i.e., 1/5,6/7 etc.

If the ratio is equal to unit, then it is said to be equality ratio i.e., 5/5,7/7 etc.

If the ratio is greater than 1, then it is said to be greater in equality i.e., 5/2,17/7 etc.

In a ratio of greater inequality, if the same number is added to both numerator and denominator, the ratio decreases, i.e. If a > b then, $\frac{a+x}{b+x}<\frac{a}{b}$

In a ratio of lesser inequality, if the same number is added to both numerator and denominator, the ratio increases, i.e. if a < b then, $\frac{a+x}{b+x}>\frac{a}{b}$ where x is any natural number.

If a < b then, $\frac{a-x}{b-x}<\frac{a}{b}$ and if a > b then, $\frac{a-x}{b-x}>\frac{a}{b}$.

If a : b : c = x : y : z ⇒ $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$

a : b > x : y if ay > bx (where a, b, c, x, y, z are positive numbers)

a : b < x : y if ay < bx (where a, b, c, x, y, z are positive numbers)

**Example** : The ratio between two numbers is 12 : 13. If each number is reduced by 20, the ratio becomes 2 : 3. Find the numbers.

**Solution**:

Let the numbers be 12x and 13x

∴ $\frac{12x-20}{13x-20}=\frac{2}{3}$

⇒ x = 2

∴ Numbers are 24, 26.

**Example** : If bc : ac : ab = 1 : 2 : 3, find a/bc ∶b/ca

**Solution**: $\frac{bc}{ac}=\frac{1}{2}$⇒ a ∶ b = 2 ∶ 1.

∴ $\frac{a}{bc}:\frac{b}{ca}=\frac{a}{bc}\times \frac{ca}{b}=\frac{{a}^{2}}{{b}^{2}}=\frac{4}{1}$

When the ratio is expressed in terms of fractions, we multiply all the terms with LCM of denominators to simplify the ratio.

**Example** : A person distributes his pens among four friends A, B, C and D in the ratio
$\frac{1}{3}$ : $\frac{1}{4}$ ∶ $\frac{1}{5}$ ∶ $\frac{1}{6}$.
What is the minimum number of pens that the person should have?

**Solution**:

LCM of 3, 4, 5 and 6 is 60. Hence, we multiply all terms with 60.

So, the pens are distributed among A, B, C and D in the ratio 1/3 × 60 ∶ 1/4 × 60 : 1/5 × 60 ∶ 1/6 × 60

i.e. 20 : 15 : 12 : 10

∴ Total number of pens = 20x + 15x + 12x + 10x = 57x

For minimum number of pens, x = 1

∴ The person should have at least 57 pens.

**Example** : A bag contains 25 paise, 10 paise and 5 paise coins in the ratio 1 : 2 : 3. If their total value is Rs. 30, the number of 5 paise coins is :

**Solution**:

Let the number of coins be x, 2x and 3x

Value of 'x' 25 paise coins = 25x, that of '2x' 10 paise coins = 2x × 10 and that of '3x' 5 paise coins = 3x × 5.

Since, the total value is Rs. 30.

⇒ x × 25 + 2x × 10 + 3x × 5 = 30 × 100 (converting all the amounts in paise)

⇒ 60x = 3000

i.e. x = 50

∴ Number of 5 paise coins are 3x = 150.

**Example** : Two numbers are in the ratio 3 : 7 and if 6 be added to each of them, they are in the ratio 5 : 9, then find the numbers.

**Solution**:

Let the numbers be 3x and 7x

∴ $\frac{3x+6}{7x+6}=\frac{5}{9}$ or 27x + 54 = 35x + 30

or 8x = 24 i.e., x = 3

Numbers are 3 × 3; 7 × 3 i.e., 9 and 21.

**Example** : A profit of Rs. 84 is divided between A and B in the ratio of $\frac{1}{3}$ ∶ $\frac{1}{4}$. What will be the difference of their profits?

**Solution**:

$\frac{1}{3}$ ∶ $\frac{1}{4}$ is the same as $\frac{1}{3}$ × 12 ∶ $\frac{1}{4}$ × 12 i.e., 4 : 3

Share of A = 4/7 × Rs. 84 = Rs. 48

Share of B = 3/7 × Rs. 84 = Rs. 36

Difference = Rs. 48 – Rs. 36 = Rs. 12

**Example** : The ratio of ages of Meera and Meena is 4 : 3. The sum of their ages is 28 years. The ratio of their ages after 8 years will be :

**Solution**:

Age of Meena = 4/7 × 28 = 16 years

Age of Meera = 3/7 × 28 = 12 years

After 8 years their respective ages will be 24 years and 20 years which shall be in the ratio 24 : 20 or 6 : 5

**Example** : A man divided Rs. 2,60,000 amongst his three daughters such that four times the amount received by the first, thrice the amount received by the second and twice the amount received the third are all equal. Find the amount received by the third daughter.

**Solution**:

Let the amounts received by first, second and third daughter by x, y and z respectively.

Hence, 4x = 3y = 2z.

We divided all the expression by LCM of coefficients i.e., LCM(4, 3, 2) = 12

⇒ x/3 = y/4 = z/6

Now assuming, x/3 = y/4 = z/6 = k

⇒ x = 3k, y = 4k and z = 6k

∴ x : y : z = 3k : 4k : 6k = 3 : 4 : 6

∴ Third daughter’s share = 6/13 × 2,60,000 = 12,000.

When two or more ratios are equal then ratio of numerators is same as the ratio of denominators.

If $\frac{\mathrm{a}}{\mathrm{p}}$ = $\frac{\mathrm{b}}{\mathrm{q}}$ = $\frac{\mathrm{c}}{\mathrm{r}}$

then, a : b : c = p : q : r

There may be situations where there are more than two quantities and they are not in the same ratio. The ratios can be scaled to find a common ratio.

**Example** : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ and $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{3}{4}$. Find a : b : c.

**Solution**:

Here, calculating combined ratio is very simple.

For every 2 parts of a, there are 3 parts of b and for every 3 parts of b, there are 4 parts of a.

Hence, for every 2 parts of a, there are 3 parts of b and 4 parts of c.

∴ a : b : c = 2 : 3 : 4

In the previous **Example**, the value of the common term ‘b’ was same hence, it was easy to calculate the combined ratio.

What if the value of the common term is not same :

**Example** : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ and $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$. Find a : b : c.

**Solution**:

Here, since the common term ‘b’ has different values, we will first have to make its value same in both the ratios,

i.e. is equal to LCM of the two values.

⇒ $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$ = $\frac{8}{12}$ (Multiplying 4 in both numerator and denominator)

⇒ $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$ = $\frac{12}{15}$ (Multiplying 3 in both numerator and denominator)

Now, for every 8 parts of a, there are 12 parts of b and 15 parts of c.

∴ a : b : c = 8 : 12 : 15

We can extend this logic to combine more than 2 ratios also but that would require a lot of effort and time.

There exists an easier and quicker method of solving the same.

If $\frac{a}{b}=\frac{{n}_{1}}{{d}_{1}}$, $\frac{b}{c}=\frac{{n}_{2}}{{d}_{2}}$, $\frac{c}{d}=\frac{{n}_{3}}{{d}_{3}}$ and $\frac{d}{e}=\frac{{n}_{4}}{{d}_{4}}$

Then, a : b : c : d : e = (n_{1} × n_{2} × n_{3} × n_{4}) : (d_{1} × n_{2} × n_{3} × n_{4}) : (d_{1} × d_{2} × n_{3} × n_{4}) : (d_{1} × d_{2} × d_{3} × n_{4}) : (d_{1} × d_{2} × d_{3} × d_{4})

**Example** : $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{2}{3}$, $\frac{\mathrm{b}}{\mathrm{c}}$ = $\frac{4}{5}$, $\frac{\mathrm{c}}{\mathrm{d}}$ = $\frac{1}{2}$ and $\frac{\mathrm{d}}{\mathrm{e}}$ = $\frac{2}{1}$. Find a : b : c : d : e.

**Solution**: Here, a : b : c : d : e = (n_{1} × n_{2} × n_{3} × n_{4}) : (d_{1} × n_{2} × n_{3} × n_{4}) : (d_{1} × d_{2} × n_{3} × n_{4}) : (d_{1} × d_{2} × d_{3} × n_{4}) : (d_{1} × d_{2} × d_{3} × d_{4})

⇒ a : b : c : d : e = 2 × 4 × 1 × 2 : 3 × 4 × 1 × 2 : 3 × 5 × 1 × 2 : 3 × 5 × 2 × 2 : 3 × 5 × 2 × 1.

⇒ a : b : c : d : e = 16 : 24 : 30 : 60 : 30.

⇒ a : b : c : d : e = 8 : 12 : 15 : 30 : 15

Duplicate ratio of $\frac{\mathrm{a}}{\mathrm{b}}$ = ${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{2}$

Triplicate ratio of $\frac{\mathrm{a}}{\mathrm{b}}$ = ${\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{3}$

Sub-Duplicate ratio of $\frac{\mathrm{a}}{\mathrm{b}}$ = $\sqrt{\frac{\mathrm{a}}{\mathrm{b}}}$

Sub-Triplicate ratio of $\frac{\mathrm{a}}{\mathrm{b}}$ = $\sqrt[3]{\frac{\mathrm{a}}{\mathrm{b}}}$

Inverse/Reciprocal ratio of a : b = $\frac{1}{\mathrm{a}}$ : $\frac{1}{\mathrm{b}}$

**PROPORTION**

Four quantities are said to be proportional, when the ratio of the first to the second is equal to the ratio of the third to the fourth.

Thus, a, b, c, d are proportional if a : b = c : d. This is often expressed as a : b :: c : d and is read ‘a’ is to ‘b’ as ‘c’ is to ‘d’.

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

The term d is called fourth proportional to a, b, c

**Example** : If 0.75 : x :: 5 : 8 then find x.

**Solution**:

Product of extremes = product of mean terms

0.75 × 8 = x × 5

⇒ x = 1.2

**Example** : The ratio of number of boys and girls in a school in 4 : 3. If there are 480 boys in the school, find the number of girls in the school.

**Solution**:

4 : 3 = 480 : x

⇒ (3 × 480)/4 = 360 girls

**CONTINUED PROPORTION**

Three or more quantities are said to be in continued proportion, when the first is to the second as the second is to the third, as the third is to the fourth; and so on. Thus, a, b, c, d are in continued proportion when

a : b = b : c = c : d

If three quantities a, b, c are in continued proportion (a : b :: b : c) then b is called the mean proportional between a and c.

i.e. $\frac{a}{b}=\frac{b}{c}$ ⇒ b = √ac

and c is called third proportional to a and b.

**Example** : The mean proportion of 0.32 and 0.02 is what?

**Solution**:

If x be the required mean proportional then,

0.32 : x :: x : 0.02

⇒ x^{2 }= 0.32 × 0.02

⇒ x = 0.08

**Example** : Find the third proportional to 16 and 24.

**Solution**:

16 : 24 :: 24 : x

⇒ x = (24 × 24)/16 = 36

**Example** : The sum of the squares of 3 numbers is 532 and the ratio of the first to the second as also of the second to the third is 3 : 2. What is the second number?

**Solution**: $\frac{Firstnumber}{Secondnumber}=\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$ and $\frac{Secondnumber}{Thirdnumber}=\frac{3}{2}\times \frac{2}{2}=\frac{6}{4}$

[Make the second number same in both the ratios, i.e. 6]

∴ First : Second : Third = 9 : 6 : 4

∴ (9x)^{2} + (6x)^{2} + (4x)^{2} = 532 (given)

⇒ 133 x 2 = 532

⇒ x = 2

Second number is 6x = 12

If a : b :: c : d, then

ad = bc

**Invertendo**b : a :: d : c, This operation is called Invertendo or b/a = d/c**Alternendo**a : c :: b : d, This operation is called Alternando or $\frac{\mathrm{a}}{\mathrm{c}}$ = $\frac{\mathrm{b}}{\mathrm{d}}$**Componendo**(a + b) : b :: (c + d) : d. This operation is called Componendo or $\frac{a+b}{b}=\frac{c+d}{d}$**Dividendo**(a – b) : b :: (c – d) : d. This operation is called Dividendo or $\frac{a-b}{b}=\frac{c-d}{d}$**Componendo & Dividendo**a + b : a – b :: c + d : c – d This operation is called Componendo and Dividendo. or $\frac{a+b}{a-b}=\frac{c+d}{c-d}$

k = $\frac{a+c+e}{b+d+f}$

k = ${\left(\frac{p{a}^{n}+q{c}^{n}+r{e}^{n}}{p{b}^{n}+q{d}^{n}+r{f}^{n}}\right)}^{1/n}$ where p, q, r, n are any real numbers.

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

In any two dimensional figure (which are similar), the corresponding sides are in the ratio a : b, then their areas are in the ratio a

^{2}: b^{2}.In any three dimensional figure (which are similar), the corresponding sides or other measuring lengths are in the ratio a : b, then their volumes are in the ratio a

^{3}: b^{3}.If the ratio of any quantities be a : b : c : d, then the ratio of other quantities which are inversely proportional to them is given by $\frac{1}{\mathrm{a}}$ : $\frac{1}{\mathrm{b}}$ : $\frac{1}{\mathrm{c}}$ : $\frac{1}{\mathrm{d}}$

If the sum of two numbers is S and their difference is D, then the ratio of numbers is given by S + D : S – D

**Example** : Rs. 3000 are divided amongst A, B and C so that if Rs. 20, Rs. 40 and Rs. 60 be taken from their shares respectively, they will have money in the ratio 3 : 4 : 5. Find the share of C.

**Solution**:

Total decrease = Rs. 20 + Rs. 40 + Rs. 60 = Rs. 120

Remaining sum = Rs. 3000 – Rs. 120 = Rs. 2880

C’s share = 5/12 × Rs. 2880 = Rs. 5 × 240 = Rs. 1200

∴ Share of C out of Rs. 3000 = Rs. 1200 + Rs. 60 = Rs. 1260

**Example** : A fort has provisions for 50 days. If after 10 days they are strengthened by 500 men and the food lasts 35 days long, how many men are there in the fort?

**Solution**:

Let there be x men in the beginning so that after 10 days the food for them is left for 40 days. Now (x + 500) men can have the same food for 35 days.

∴ 40 × x = (x + 500) × 35

or 40x = 35x + 35 × 500

or 5x = 35 × 500

or x = 3500 men

**Example** : Rs. 4400 are divided among A, B and C so that A receives 3/8 as much as B and C together. What is the share of A?

**Solution**:

when (B + C) get Rs. 1, A gets Rs. 3/8

∴ A’s share : (B + C)’s share = 3/8 : 1 = 3∶8

∴ A’s share = Rs. $\frac{3}{11}$ × 4400 = Rs. 1200

**Example** : The ratio of the first and second class fares between the two stations is 4 : 1 and that the number of passengers travelling by first and second class is 1 : 40. If Rs. 1100 is collected as fare, the amount collected from first class passengers is :

**Solution**:

Ratio of amounts collected from 1st and 2nd class = (4 × 1) : (1 × 40) = 1 : 10

∴ Amount collected from 1st class passengers = $\frac{1}{11}$ × Rs. 1100 = Rs. 100

**Example** : One year ago the ratio between Laxman’s and Gopal’s salary was 3 : 4. The ratio of their individual salaries between last year’s and this year’s salaries are 4 : 5 and 2 : 3 respectively. At present the total of their salary is Rs. 4160. The salary of Laxman now is :

**Solution**:

Let the salaries of Laxman and Gopal one year before be a, b respectively and now c, d respectively.

⇒ a : b = 3 : 4

a : c = 4 : 5

b : d = 2 : 3

It is given that c + d = 4160

⇒ $\frac{\mathrm{a}}{\mathrm{b}}$ = $\frac{3}{4}$; $\frac{\mathrm{a}}{\mathrm{c}}$ = $\frac{4}{5}$ and $\frac{\mathrm{b}}{\mathrm{d}}$ = 2/3

Let a be 3x and b be 4x.

Hence, c = 15x/4 and d = 6x

⇒ c ∶ d = $\frac{{\displaystyle \raisebox{1ex}{$15x$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{6x}=\frac{5}{8}$

∵ c + d = 4160

∴ c = $\frac{5}{13}$ × 4160 = Rs. 1600.

**VARIATION**

When the two (or more) quantities are so related that if one of them be changed the other is changed with respect to it, this relation is called variation.

There are two types of variations

**DIRECT VARIATION**

One quantity is said to vary directly as another when the two quantities are so related that if one of them be increased (or decreased) the other increases (or decreases) in the same ratio, i.e., if x varies directly as y, then

If x ∝ y then, x = ky, where k is a constant.

Also, $\frac{{x}_{1}}{{y}_{1}}=\frac{{x}_{2}}{{y}_{2}}$

Note : The sign ‘∝’ is called the sign of variation.

**Example** : If y varies as x, and y = 5 when x = 12, find the value of y, when x = 18.

**Solution**:

By supposition, y = mx, where m is constant.

Putting y = 5 and x = 12, we have 5 = m × 12

∴ m =5/12

Hence, x and y are connected by the relation y = 5/12 x

Hence, when x = 18, we have y = 5/12 × 18 = 15/2 = 7.5

Alternately,

Since, y varies as x ⇒ y_{1}/x_{1} = y_{2}/x_{2}

⇒ 5/12 =y_{2}/18

∴ y_{2} = 15/2 = 7.5

**Example** : If z varies as px + y and if z = 3 when x = 1, y = 2 and z = 5, when x = 2 and y = 3, find p.

**Solution**:

By supposition, z = m × (px + y), where m is constant.

Putting z = 3, x = 1, y = 2, we have 3 = m(p + 2) …(i)

Again, putting z = 5, x = 2, y = 3, we have 5 = m(2p + 3)

Hence, from (1) and (2), by division, $\frac{3}{5}$ = (p + 2)/(2p + 3)

Hence p = 1

**INVERSE VARIATION**

One quantity is said to be vary inversely as another when the two quantities are so related that if one of them be increased (or decreased) the other is decreased (or increased) in the same ratio.

i.e., If x varies inversely as y, then x ∝ 1/y or x = k/y where k is any constant.

Also, x_{1 }× y_{1 }= x_{2 }× y_{2}

**Example** : If y varies inversely as x, and y = 12 when x = 5, find the value of y, when x = 20.

**Solution**:

By supposition, y = m/x, where m is constant.

Putting y = 12 and x = 5, we have 12 = m/5

∴ m = 60

Hence, x and y are connected by the relation y = 60/x

Hence, when x = 20, we have y = 60/20 = 3

Alternately,

Since, y varies inversely as x ⇒ y_{1 }× x_{1} = y_{2} × x_{2}

⇒ 12 × 5 = y_{2} × 20

∴ y_{2} = 60/20 = 3

**Example** : Given that the illumination from a source of light varies inversely as the square of the distance. How much farther from a candle must a book, which is now three meters off, be removed so as to receive just half as much light?

**Solution**:

I α 1/d^{2} or I = k/d_{2}

Here I_{1 }= k/3^{2} ⇒ k = 9I_{1}

Again, I_{2 }= I_{1}/2 = (9I_{1})/(d_{2}^{2} )

⇒ d_{2}^{2 }= 18 ⇒ d_{2 }= 4.242

Hence, book is 4.242 – 3 = 1.242 m farther apart

If A ∝ B and B ∝ C, then A ∝ C

A ∝ C and B ∝ C, then (A + B) ∝ C and also, (A – B) ∝ C

If A ∝ B and C ∝ D, then AC ∝ BD

If A ∝ BC, then B ∝ A/C and C ∝ A/B

If A ∝ B and A ∝ C, then A ∝ (B – C) and also A ∝ (B + C)

If A ∝ B when C is constant, and A ∝ C when B is constant, then A ∝ BC when both B and C vary.

If A ∝ B, then A

^{n}∝ B^{n}.If A ∝ B, then A × P ∝ B × P, where P is any quantity, variable or constant.

**Example** : Assuming that the quantity of work done varies as the cube root of the number of agents, when the time is the same, and varies as the square root of the time, when the number of agents is the same; find how long 3 men would take to do one-fifth of the work which 24 men can do in 25 hours.

**Solution**:

Let x denotes the quantity of work done by y men in z hours

Then, by supposition, x ∝ y^{1/3}, when z and ∴ z^{1/2} is constant, and also, x ∝ z^{1/2}, when y and ∴ y^{1/3} is constant.

Hence, when both y and z, and ∴ y^{1/3} and z^{1/2} are variable.

x ∝ y^{1/3} z^{1/2}

i.e. x = k × y^{1/3} z^{1/2}, when k is constant.

Now, since by the problem, x = 1, when y = 24 and z = 25.

∴1 = k × ∛24 × √25 …(i)

Also, if Z_{1} be the required number of hours, since the corresponding values of x and y are respectively 1/5 and

3, we have

1/5 = k^{3} × ∛3 √z_{1} …(ii)

Hence, dividing (i) by (ii), $5=\frac{\sqrt[3]{24}\times 5}{\sqrt[3]{3}\times \sqrt{{z}_{1}}}=\frac{\sqrt[3]{8}\times 5}{\sqrt{{z}_{1}}}$

∴ √z_{1 }= 2 and ∴ z_{1} = 4;

i.e., the required time = 4 hours

**Example** : If 15 men or 24 women or 36 boys can do a piece of work in 12 days, working 8 hours a day. How many men must be associated with 12 women and 6 boys to do $2\frac{1}{4}$ times the same work in 30 days working 6 hours a day?

**Solution**:

24 women = 15 men ⇒ 12 women = 7 1/2 men

36 boys = 15 men ⇒ 6 boys = 2 1/2 men

∴ 12 women and 6 boys = ($7\frac{1}{2}$ + $2\frac{1}{2}$) men i.e. 10 men

More work ⇒ more men

More days ⇒ less men

Less hours ⇒ more men

∴ Number of men = 15 × 9/4 × 8/6 × $\frac{12}{30}$ = 18

∴ Number of men associated with 12 women and 6 boys = 18 – 10 i.e. 8 men.

**Example** : A sphere of metal is known to have a hollow space about its center in the form of a concentric sphere, and its weight is $\frac{7}{8}$ of the weight of a solid sphere of the same substance and radius; compare the inner and outer radii, having given that the weights of spheres of the same substance ∝ (radii)3.

**Solution**:

Let R be the outer radius and W the weight of a solid sphere of the given metal of radius R; also let r be the inner radius (i.e., radius of the spherical cavity), and w the weight of a solid sphere of the given metal of radius r.

Then, by hypothesis, W = KR^{3} and w = Kr^{3}, where K is constant.

Now, since (W – w) is the weight of the given sphere, we have, by the question, W – w = $\frac{7}{8}$ W, hence we must have

K(R^{3 }- r^{3}) = $\frac{7}{8}$ KR^{3}

∴ 1/8 R^{3} = r^{3},

hence r/R = 1/2

**Example** : If y be equal to the sum of 3 quantities, of which the 1st ∝ x^{2}, the 2nd ∝ x, and the third is constant; and when x = 1, 2, 3; y = 6, 11, 18 respectively. Find the equation between x and y.

**Solution**:

By supposition y = mx^{2 }+ nx + p, where m, n, p are constants.

Now, since y = 6, when x = 1, we have

6 = m + n + p …(i)

Similarly, 11 = 4m + 2n + p …(ii)

and 18 = 9m + 3n + p …(iii)

From (1) and (2), by subtraction, 3m + n = 5, …(iv)

Similarly, from (ii) and (3), 5m + n = 7 …(v)

Now, subtracting (4) and (5), we have 2m = 2

∴ m = 1;

Hence, from (4), n = 2,

∴ from (1), p = 3

Hence, the equation between x and y is y = x^{2 }+ 2x + 3

**Example** : If 5 persons make 15 kilometers road in a certain time, how many kilometers road will be made by 20 persons in the same time?

**Solution**:

Let the required length of the road be x km.

Hence, arrangement of the terms in the proportion will be as follows :

x : 15 :: 20 : 5

⇒ 5x = 15 × 20

∴ x = (15 × 20)/5 = 60

**Example** : Two clerks A and B whose salaries were Rs. 75 and Rs. 145 received an increase in their salaries. If their salaries increase in the same ratio, what increase B secures when A’s salary reached Rs. 90?

**Solution**:

A’s old salary = Rs. 75

A’s new salary = Rs. 90

Increase in A’s salary = 90 – 75 = Rs. 15

Ratio of increase in A’s salary = 15/75

Let the ratio of increase in B’s salary = x/145

Then x/145 = 15/75

⇒ x × 75 = 15 × 145

∴ x = (15 × 145)/75 = 29

Increase in B’s salary is Rs. 29.

**Example** : Two brothers have their annual income in the ratio 8 : 5 and their spending in the ratio 5 : 3. If they save Rs. 2,400 and Rs. 2,000 p.a.. Find their incomes.

**Solution**:

Since the ratio of incomes of two brothers is 8 : 5, so we can assume that their incomes are 8x and 5x respectively.

∴ Expenditure of one brother= 8x – 2,400

Expenditure of second brother = 5x – 2,000

Ratio of their expenditures (8x - 2400)/(5x - 2000) = 5/3

⇒ 3(8x – 2400) = 5(5x – 2000)

⇒ 24x – 7200 = 25x – 10000

⇒ 24x – 25x = -10000 + 7200

⇒ -x = - 2800

i.e. x = 2800

Hence, the income of one brother = 8 × 2800 = Rs. 22,400

And the income of other brother = 5 × 2800 = Rs. 14,000

**Example** : In a cash box there are coins worth Rs. 64. The ratio between 1 rupee and 50 paise and 25 paise coins as per their numbers is 4 : 5 : 6. Find the total number of coins of each denominations.

**Solution**:

1 rupee 50 paise 25 paise

Ratio of the

number of coins 4 : 5 : 6

Ratio of the

values of the coins

(in paise) 400 : 250 : 150

Or 8 : 5 : 3

Sum of the terms in the ratio = 8 + 5 + 3 = 16

Number of 1 rupee coins = 8/16 × 6400/100 = 32

Number of 50 paise coins = 5/16 × 6400/50 = 40

Number of 25 paise coins = 3/16 × 6400/25 = 48

**Example** : A bag contains an equal number of one rupee, 50 paise and 25 paise coins respectively. If the total value is Rs. 35, how many coins of each type are there?

**Solution**:

Here number of each type of coin is same. Hence, we may write,

Number of each type of coin = $\frac{Totalamount}{Sumofvalueofeachcoin}$

∴ Number of each type coin = $\frac{35}{1+0.5+0.25}$ = 20 coins of each type

**PARTNERSHIP**

If two or more persons invest their money (capital) in a joint business, their association is called partnership.

A partner who simply invests money but does not attend to the business is called a silent / sleeping partner. One who invests money as well as attends to the business is a working partner.

Such type of questions require us to distribute profit among partners.

Profit is directly proportional to investment when everyone invests for same amount of time.

Profit is also directly proportional to time invested when everyone invests for the same amount.

Hence, **Profit ∝ Investment × Time**

Generally, the gain or loss is divided among the partners in a partnership on the basis of the following rules:

- Time invested is same : In a partnership, the gain or loss is distributed among the partners in the ratio of their capital investments when the investments of all the partners are for the same period.
- Amount invested is same : In a partnership, the gain or loss is distributed among the partners in the ratio of their invested time when the investments of all the partners are same.
- Amount and Time invested both are different : In a partnership, the gain or loss is distributed among the partners in the ratio of their equivalent capital investments for a unit of time. The equivalent capital investments for a unit of time are calculated by taking [Capital × Number of units of time]

**Example** : Ram and Shyam enter into a partnership. Ram invests Rs. 24,000 and Shyam invests Rs. 20,000. Find the share of each in a Profit of Rs. 66,000 after one year.

**Solution**:

Ratio of investments of Ram and Shyam = 24,000 : 20,000 = 6 : 5

Ram’s share in a profit of Rs. 66,000 = Rs. $\frac{6}{11}$ × 66000 = Rs. 36,000

Shyam’s share = Rs. $\frac{5}{11}$ × 66000 = Rs. 30,000

**Example** : Ram and Shyam enter into a partnership with same investment. Ram invests for 8 months and leaves the business. Find the ratio of share of each after one year.

**Solution**:

Ratio of investments of Ram and Shyam = I × 8 : I × 12 = 2 : 3

**Example** : Ram, Mohinder and Jatinder enter into a partnership. Ram invests Rs. 24,000, Mohinder Rs. 20,000 and Jatinder Rs. 30,000. Find the share of each in a Profit of Rs. 74,000 after one year.

**Solution**:

Ratio of equivalent investments of Ram, Mohinder & Jatinder for one month = 24,000 : 20,000 : 30,000 = 12 : 10 : 15

Ram’s share in a profit of Rs. 74,000 = Rs. 12/37 × 74000 = Rs. 24,000

Mohinder’s share = Rs. 10/37 × 74000 = Rs. 20,000

Jatinder’s share = 15/37 × 74000 = Rs. 30,000.

**Example** : Ram, Mohinder and Jatinder enter into a partnership. Ram invests Rs. 24,000 for 8 months, Mohinder Rs. 20,000 for 12 months and Jatinder Rs. 30,000 for 6 months. Find the share of each in a Profit of Rs. 51,000 after one year.

**Solution**:

Ratio of equivalent investments of Ram, Mohinder & Jatinder for one month = 24,000 × 8 : 20,000 × 12 : 30,000 × 6 = 16 : 20 : 15

Ram’s share in a profit of Rs. 51,000 = Rs. 16/51 × 51000 = Rs. 16,000

Mohinder’s share = Rs. 20/51 × 51000 = Rs. 20,000

Jatinder’s share = 15/51 × 51000 = Rs. 15,000.

**Example** : Ram, Mohinder and Jatinder enter into a partnership. Ram invests Rs. 24,000 for 8 months, Mohinder Rs. 20,000 for 12 months and Jatinder Rs. 30,000 for 6 months. Find the share of each in a Profit of Rs. 51,000 after one year.

**Solution**:

Ratio of equivalent investments of Ram, Mohinder & Jatinder for one month = 24,000 × 8 : 20,000 × 12 : 30,000 × 6 = 16 : 20 : 15

Ram’s share in a profit of Rs. 51,000 = Rs. 16/51 × 51000= Rs. 16,000

Mohinder’s share = Rs. 20/51 × 51000 = Rs. 20,000

Jatinder’s share = 15/51 × 51000 = Rs. 15,000.

**Example** : Rajkumar and Sandeep entered into a partnership investing Rs. 48,000 and Rs. 54,000 respectively. After 3 months Jugal Kishore also joined the business with a capital of Rs. 40,000. Find the share of each in a profit of Rs. 55,000 after a year.

**Solution**:

Ratio of equivalent investments of Raj Kumar, Sandeep and Jugal Kishore for one month :

48000 × 12 : 54000 × 12 : 40000 × 9 = 8 : 9 : 5

Raj Kumar’s share in the profit = Rs. $\frac{8}{22}$ × 55,000 = Rs. 20,000

Sandeep’s share in the profit = Rs. $\frac{9}{22}$ × 55,000 = Rs. 22,500

Jugal Kishore’s share = Rs. $\frac{5}{22}$ × 55000 = Rs. 12,500

**Example** : A, B and C hire a grassland for Rs. 2,934.60. A puts in 10 cows for 20 days; B, 30 cows for 8 days and C, 16 cows for 9 days. What is the rent paid by each?

**Solution**:

Ratio of rents to be paid by A, B and C = (10 × 20 : 30 × 8 : 16 × 9) = 25 : 30 : 18

∴ A’s share of rent = Rs. $\frac{25}{73}$ × 2934.60 = Rs. 1005

B’s share of rent = Rs. $\frac{70}{73}$ × 2934.60 = Rs.1206

C’s share of rent = Rs. [2934.60 – (1005 + 1206)] = Rs. 723.60

**Example** : A and B invest Rs. 12,500 and Rs. 8,500 respectively in a business. They agree that 40% of the profit shall be divided equally between them and the rest divided in proportion to their investments. If A’s share of profit is Rs. 1950, what is B’s share of profit?

**Solution**:

Let the total profit be Rs. 100. Rs. 40 is divided equally between A and B and Rs. 60 is divided in the ratio of 12500 : 8500 i.e. 25 : 17.

A’s share out of Rs. 40 = Rs. 20. A’s share out of Rs. 60 = 25/42 × 60 = Rs. 250/7

A’s total profit =(20 + 250/7) = Rs. $\frac{390}{7}$

If A’s profit is Rs. $\frac{390}{7}$ total profit is Rs. 100.

If A’s profit is Rs. 1950, total profit is Rs. $\frac{100}{390}$ × 7 × 1950 i.e. Rs. 3500.

∴ B’s share of profit = Rs. 3500 – Rs. 1950 = Rs. 1550.

**Example** : P and Q make investment in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and P’s share is Rs. 912. Find the total profit.

**Solution**:

Let the total profit be P.

∴ Resultant total profit ⇒ p - $\frac{\mathrm{5p}}{100}$ = $\frac{\mathrm{19p}}{20}$

or $\frac{\mathrm{19p}}{20}$ × $\frac{3}{5}$ = 912

⇒ 57p = 912 × 100

∴ Total profit = Rs. 1600.

**Example** : P, Q and R contract a work for Rs. 1100. P and Q together are to do $\frac{7}{1}$ of the work. Find R’s share of the amount contracted for.

**Solution**:

R’s share of work = 1 - $\frac{7}{1}$ = $\frac{4}{1}$ of the total work.

∴ R’s share in total money = $\frac{4}{11}$ × 1100 = Rs. 400.

**Example** : A is working partner and B is sleeping partner in a business. A puts in Rs. 5,000 and B puts in Rs. 6,000. A receives 12$\frac{1}{2}$ per cent of profit for managing the business and the rest is divided in proportionof their capitals. How much does each get out of a profit of Rs. 880?

**Solution**:

A receives 12$\frac{1}{2}$% of profit for managing the business = $\frac{25}{2\times 100}$× 880 = Rs. 110

Remaining profit = 880 – 110 = Rs. 770

Dividing Rs. 770 in 5000 : 6000 or 5 : 6 we get

A’s profit on capital = 770 × $\frac{5}{11}$ = Rs. 350

A’s total gain = 350 + 110 = Rs. 460

B’s profit on capital = 770 × $\frac{6}{11}$ = Rs. 420

B’s gain = Rs. 420

**Example** : What amount of money is divided between A, B and C if B and C together get Rs. 100, A gets twice as much as B, and C with A gets Rs. 150?

**Solution**:

Given B + C = 100 and A + C = 150

∴ A = 2B, ∴ 2B + C = 150

⇒ B + (B + C) = 150 (Since B + C = 100)

∴ B = 150 – 100 = 50

∴ A + B + C = (A + C) + B = 150 + 50 = Rs. 200

**Example** : Three students A, B and C hired a computer for a month. ‘A’ runs 27 floppy disks for 19 days, B runs 21 for 17 days and C runs 24 for 23 days. If at the end of the month, the rent amounts to Rs. 23,700, How much ought to be paid by each?

**Solution**:

Floppy-days = No. of floppies × days run

Here, A’s floppy-days : B’s floppy days : C’s floppy days

= A’s payment for rent : B’s payment for rent : C’s payment for rent

= 27 × 19 : 21 × 17 : 24 × 23

= A’s rent : B’s rent : C’s rent

= (171) : (119) : (184)

∴ Payment for rent by A = $\frac{171}{171+119+84}$ × 23,700 = Rs. 8,550

Payment for rent by B = $\frac{119}{171+119+84}$ × 23,700 = Rs. 5,950

Payment for rent by C = $\frac{184}{171+119+84}$ × 23,700 = Rs. 9,200

**Example** : A, B and C invest Rs. 4,000, 5,000 and 6,000 respectively in a business and A gets 25% profit for managing the business, the rest of the profit is divided by A, B and C in proportion to their investment. If in a year, A gets Rs. 100 less than B and C together, what was total profit for that year?

**Solution**:

After giving 25% of the total profit amount to A for managing the business, the rest 75% of total profit is divided among A, B and C in proportion to their investments.

In 75% of total profit, A’s share : B’s share : C’s share

= 4,000 : 5,000 : 6,000 = 4 : 5 : 6

∴ 75% of total profit = 4x + 5x + 6x

∴Total profit = 15x/(75%) = 20x

∴ Share of A = 4x + 25% of 20x = 9x

Share of B = 5x

Share of C = 6x

Given, (5x + 6x) – 9x = 100 ⇒ x = 50

∴ Total profit = 20x = 20 × 50 = Rs. 1,000

**Example** : Two partners invested Rs. 1,250 and Rs. 850 respectively in a business. Both the partners distribute 60% of the profit equally and distribute the rest 40% as the interest on their capitals. If one partner received Rs. 30 more than the other. Find the total profit.

**Solution**:

Since 60% of the profit is distributed equally. So, one partner receives Rs. 30 more than the other only due to distribution of rest 40% of the profit on the basis of their invested capitals.

∴ $\frac{A\text{'}ssharein40\%profit}{B\text{'}ssharein40\%profit}=\frac{1250}{850}=\frac{25}{17}$

⇒ $\frac{A\text{'}s40\%profit+B\text{'}s40\%profit}{A\text{'}s40\%profit-B\text{'}s40\%profit}=\frac{25+17}{25-17}$ = $\frac{42}{8}$

⇒ (40% profit)/30 = $\frac{42}{8}$

⇒ Profit = $\frac{42}{8}$ × 30 × 100/40 = Rs. 393.75

∴ Total profit is Rs. 393.75

**Example** : In a partnership, A invests 1/6 of the capital for 1/6 of the time, B invests 1/3 of the capital for 1/3 of the time and C, the rest of the capital for whole time. Find A’s share of the total profit of Rs. 2,300.

**Solution**:

Capital of C = 1 - 1/6 - 1/3 = 1/2

Let the total time be 12 months

∴ A’s profit : B’s profit : C’s profit

= $\frac{1}{6}\times \left(\frac{1}{6}\times 12\right)$ : $\frac{1}{3}\times \left(\frac{1}{3}\times 12\right)$ : $\left(\frac{1}{2}\times 12\right)$ = $\frac{1}{3}:\frac{4}{3}:6$ = 1 : 4 : 18

∴ Share of A = $\frac{1}{1+4+8}$ × 2300 = Rs. 100.

**Example** : A, B and C enter into a partnership. A invests Rs. 8,000 for the whole year, B puts in Rs. 12,000 at the first and increasing to Rs. 16,000 at the end of 4 months, whilst C puts in at first Rs. 16,000 but withdraw Rs. 4,000 at the end of 9 months. Find the profit of A at the end of year, if the total profit is Rs. 22,600.

(a) Rs. 4,800 (b) Rs. 4,600 (c) Rs. 4,750 (d) Rs. 4,300 (e) None of these

**Solution**:

Capital of C = 1 - 1/6 - 1/3 = 1/2

Ratio of profit:

A : B : C = (8000 × 12) : (12000 × 4) + (16000 × 8) : (16000 × 9) + (12000 × 3) = 24 : 44 : 45

Share of A = $\frac{24}{113}$ × 22600 = Rs. 4800.