Ratio and Proportion

Subject: Compulsory Maths

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Overview

Ratios are used to compare values. They tell us how much of one thing there is compared to another and Proportions are related to ratios in that they tell you when two ratios are equal to each other.

Ratio and Proportion

 1

Ratio

The ratio of two quantities of the same type ( let a and b) is used to express how many times bigger or smaller, one quantity is compared to other. For example , if a = 3 and b = 4, then we can write \(\frac{a}{b}\) = \(\frac{3}{4}\). Also, we can write a = \(\frac{3}{4}\) b or a is three-fourth of b. The ratios 2:5 and 8:20.

In a ratio a : b or \(\frac{a}{b}\), a is called the antecedent and b is called the consequent. The ratio b : a is the inverse ratio of a:b and vice-versa.

Compound ratio
 If a : b and c : d be any two ratios. then a : b \(\times\) c : d = \(\frac{a}{b}\) \(\times\) \(\frac{c}{d}\) = \(\frac{ac}{bd}\) = ac : bd is called compound ratio.

Duplicate and sub -duplicate ratio
If a : b be a ratio, then the duplicate ratio of \(\frac{a}{b}\) = (\(\frac{a}{b}\))2 = \(\frac{a^2}{b^2}\) = a2: b2
And, the sup-duplicate ratio of \(\frac{a}{b}\) = √ \(\frac{a}{b}\)

Triplicate and sub - triplicate ratio
If a : b be a ratio, then the triplicate ratio of \(\frac{a}{b}\) = (\(\frac{a}{b}\))3 and sub-triplicate vratio of \(\frac{a}{b}\) =3√ \(\frac{a}{b}\)

Proportion

Similarly, if two or more than two ratios are equal, those quantities which make ratios are proportional.
Two ratios a : b and c : d equal or \(\frac{a}{b}\) = \(\frac{c}{d}\), then a, b, c and d are in proportion.
Now, let us study some related examples of proportion.
Example: \(\frac{8}{20}\) = \(\frac{2}{5}\)
Or, \(\frac{20}{8}\) = \(\frac{5 \times 4}{2 \times 4}\) = \(\frac{5}{2}\)
\(\therefore\) \(\frac{20}{8}\) = \(\frac{5}{2}\)

Continued proportion
If a , b and c be any three number such that the ratio of the a and b is equal to the ratio of b and c, then such ratio is known as a compound proportion.

\(\therefore\) \(\frac{a}{b}\) = \(\frac{b}{c}\) is said to be continued proportion. Then, ac= b2

a : b = b : c

Here, a is 1st proportion

b is mean proportion

c is 3rd proportion

Mean proportion (b) =√ac

Properties of proportion

If a, b, c and d are in proportion, then we can verify the following six properties of proportion.

  1. Invertendo
  2. Alternendo
  3. Componendo
  4. Dividendo
  5. Componendo and Dividendo
  6. Addendo

a) Invertendo
If \(\frac{a}{b}\) =\(\frac{c}{d}\), then \(\frac{b}{a}\) =\(\frac{d}{c}\) is known as invertendo properties of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

Then, 1 \(\div\)\(\frac{a}{b}\) = 1\(\div\)\(\frac{c}{d}\) (1 is divided by both ratio)

1 \(\times\)\(\frac{b}{a}\) = 1\(\times\)\(\frac{d}{c}\)

\(\therefore\) \(\frac{b}{a}\) =\(\frac{d}{c}\)

Hence, if \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{b}{a}\) = \(\frac{d}{c}\)

b) Alternendo
If \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{a}{c}\) = \(\frac{b}{d}\) is known as alternendo property of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

Multiplying both by \(\frac{b}{c}\), we get \(\frac{a}{b}\) \(\times\) \(\frac{b}{c}\) = \(\frac{c}{d}\) \(\times\) \(\frac{b}{c}\)

or, \(\frac{a}{c}\) = \(\frac{b}{d}\)

\(\therefore\) \(\frac{a}{c}\) = \(\frac{b}{d}\)

c) Componendo
If \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{a + b}{b}\) = \(\frac{c + d}{d}\) is known as componendo property of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

Then, adding one on both side, we get

\(\frac{a}{b}\) + 1 = \(\frac{c}{d}\) + 1

\(\frac{a + b}{b}\) = \(\frac{c + d}{d}\)

Hence, if \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{a + b}{b}\) = \(\frac{c + d}{d}\)

d) Dividendo
If \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{a - b}{b}\) = \(\frac{c - d}{d}\) is known as dividendo property of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

subtracting 1 from both sides, we get

\(\frac{a}{b}\) - 1 = \(\frac{c}{d}\) - 1

or, \(\frac{a - b}{b}\) = \(\frac{c - d}{d}\)

\(\therefore\) \(\frac{a - b}{b}\) = \(\frac{c - d}{d}\)

e) Componendo and dividendo
If \(\frac{a}{b}\) = \(\frac{c}{d}\), then \(\frac{a + b}{a - b}\) = \(\frac{c + d}{c - d}\) is known as componendo and dividendo property of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

By compendendo we have,

\(\frac{a + b}{b}\) = \(\frac{c + d}{d}\)................. (1)

Again, by dividendo, we have

\(\frac{a - b}{b}\) = \(\frac{c - d}{d}\) ..................... (2)

Now, dividing equation (1) by (2), we get

\(\frac {\frac {a+b}b}{\frac {a-b}b}\) = \(\frac {\frac {c+d}d}{\frac {c-d}d}\)

or, \(\frac{a + b}{b}\) \(\times\) \(\frac{b}{a - b}\) = \(\frac{c + d}{d}\) \(\times\) \(\frac{d}{c - d}\)

or, \(\frac{a + b}{a - b}\) = \(\frac{c + d}{c - d}\)

\(\therefore\) \(\frac{a + b}{a - b}\) = \(\frac{c + d}{c - d}\)

f) Addendo
If \(\frac{a}{b}\) = \(\frac{c}{d}\), then, \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{a + c}{b + d}\) is known as addendo property of proportion.

Proof:

Here, \(\frac{a}{b}\) = \(\frac{c}{d}\)

By alternendo, we get,

\(\frac{a}{c}\) = \(\frac{b}{d}\)

By alternendo we get,

\(\frac{a + c}{c}\) = \(\frac{b + d}{d}\)

Again, by alternendo we get,

\(\frac{a + c}{b + d}\) = \(\frac{c}{d}\)

\(\therefore\)\(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{a + c}{b + d}\)

Hence, if \(\frac{a}{b}\) = \(\frac{c}{d}\) then, \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{a + c}{b + d}\)

Similarly, if \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{e}{f}\) then, \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{e}{f}\) = \(\frac{a + c + e}{b + d + f}\) and so on.

Solution of discontinued and continued proportion (k-method):

Discontinued proportion

If a, b, c and d are in discontinued proportion,

Let, \(\frac ab\) = \(\frac cd\) = k 

Then,

\(\frac ab\) = k,

∴ a = bk....................(i)

\(\frac cd\) = k,

∴ c = dk.....................(ii)

In terms of the denominator with 'k' is a constant number, express the two numerators. We solve the problems related to proportion.

Continued proportion

If a, b, c and d are in a continued proportion, 

Let, \(\frac ab\) = \(\frac bc\) = \(\frac cd\)= k 

or, \(\frac cd\) = k

∴ c = dk...................(i)

or, \(\frac bc\) = k

∴ b = ck = d.k.k = dk2..............(II)

or, \(\frac ab\) = k,

∴ a = bk = dk2.k = dk3.............(iii)

∴ a = dk3, b = dk2 and c = dk

So, if a, b, c and d are in contiuned proportion, we express a, b, c in terms of d with 'k' constant and solve the problem.

Things to remember

A proportion is a name we give to a statement that two ratios are equal. It can be written in following way:

  • using a colon,    a:b = c:d

When two ratios are equal, then the cross products of the ratios are equal.

That is, for the proportion, a:b = c:d ,  a x d = b x c 

  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Ratio and Proportion
Math Antics - Proportions
Ratio and Proportion
Questions and Answers

Here, by chain rule
The work done by Ram in 3 days = the work done by Shyam in 4 days.
The work done by Shyam in 5 days = the work done by Hari in 6 days.
Let, the work done by Hari 16 days = the work done by Ram in x days

Now , 3 \(\times\) 5 \(\times\)16 = 4 \(\times\) 6 \(\times\) x

\(\therefore\) x= \(\frac{3 \times 5 \times 16}{4 \times 6}\) = 10.

Hence, the work done by Hari in 16 days can be done by Ram in 10 days. Ans.

Here , solving the given problem by chain rule ,
The price of 3 ducks = Price of 4 hens
Price of 2 hens= Rs. Rs.750

Let , Rs.x= price of 1 peasants
price of 4 peasants= Price of 7 ducks
Now , 3 \(\times\) 2 \(\times\) x \(\times\) 4 = 4 \(\times\) 750 \(\times\) 1 \(\times\) 7

or , x= \(\frac{4 \times 750\times 1 \times 7}{3 \times 2 \times 4}\) = 875.

The price of 1 peasant = Rs. 875.

Here , solving the given problem by chain rule ,
Food for x horses =food for 153 oxen
Food for 12 oxen = food for 24 sheep
Food for 15 sheep = food for 25 goats
Food for 17 goats= food for 3 baby elephants
Food for 8 baby ele

Let C take x hours to finish the work.
Here , solving the given problem by chain rule ,
The time taken by A to do \(\frac{1}{3}\) work = the time taken by B to do \(\frac{1}{4}\) work.
The time taken by B to do \(\frac{1}{5}\) work = the time taken by C to do \(\frac{1}{2}\) work.
The time taken by C to do 1 work = x hours (suppose)
20 hours = time taken by A to do 1 work

Hence , \(\frac{1}{3}\) \(\times\) \(\frac{1}{5}\) \(\times\) 1 \(\times\) 20 = \(\frac{1}{4}\) \(\times\) \(\frac{1}{2}\) \(\times\) x \(\times\) 1

or , \(\frac{4}{3}\) = \(\frac{x}{8}\)

\(\therefore\) x = \(\frac{32}{3}\) = 10 \(\frac{2}{3}\).

\(\therefore\) C can complete 1 work in 10 \(\frac{2}{3}\) hours.

Let, x eggs of Swan's can be exchanges with two eggs of the hen.
Here, solving the given problem by chain rule,
4 eggs of hen = 3 eggs of duck.
7 eggs of duck= 4 eggs of a swan.

Hence , x × 4 × 7 = 2 × 3 × 4

or , 28x = 24
x = \(\frac{24}{28}\) = \(\frac{6}{7}\)

Now, using unitary method
\(\frac{6}{7}\) eggs of swan = 2 eggs of hen

\(\therefore\) cost of \(\frac{6}{7}\) eggs of swan = Rs. 7.50

or , cost of 1 eggs of swan = Rs. 7.5 \(\times\) \(\frac{7}{6}\) = Rs. 8.75 Ans.

Let , the cost of mixture after 2x kg of rice costing Rs. 15 and 3x kg of rice costing Rs. 20 be Rs. y per kg.

Now , Rs. 15 × 2x + Rs. 20 × 3x = Rs. y (2x + 3x)
or , (30 x + 60x) = 5xy
or , y= \(\frac{90x}{5x}\) = 18

\(\therefore\) The cost of mixture is Rs. 18 per kg.

Here , cost of first type of oil = Rs. 80 per litre
Cost of second type of oil = rS. 90 per litre
Let , cost of mixture = Rs. x per litre.

\(\frac{quantity\;of\;second\;type\;of\;oil}{quantity\;of\;first\;type\;of\;oil}\) = \(\frac{9}{8}\)

By formula ,
\(\frac{quantity \;of \;fist \;type \;of \;oil}{quantity \;of \;second \;type\; of\; oil}\) = \(\frac{cost of second type of oil- cost of mixture}{cost of mixture - cost of first type of oil }\)
or , \(\frac{9}{8}\) = \(\frac{Rs.90 -x }{x - Rs. 80}\)
or , 9x - Rs. 720 = Rs. 720 - 8x
or , 9x + 8x = Rs. 720 + Rs. 720
or , 17x = Rs. 1440
\(\therefore\) x = \(\frac{1440}{17}\) = Rs. 84.70

Let the quantities of three types of tea in kg be 3x , 4x and 7x respectively ,
Total weight of mixture = (3x + 4x + 7x) = 14x kg.
Total cost of mixture = Rs. (3x \(\times\) 80 \(\times\) + 4x \(\times\) 100 + 7x \(\times\) 150)

= Rs. (240x + 400x + 1050x)
= Rs. 1690x.

Here , cost of 14 x kg of mixture = Rs. 1690 x
cost of 1 kg of mixture = \(\frac{1690x}{14x}\)

Cost of 35 kf of mixture = Rs. \(\frac{1690x}{14x}\) \(\times\) 35 = Rs. 4225


Here , the sum of the propertional parts = 56l
Quantity of milk = \(\frac{4}{7}\) \(\times\) 56 l = 4 \(\times\) 8 l= 32l
Quantity of water = \(\frac{3}{7}\) \(\times\) 56 l = 3 \(\times\) 8 l = 24l Ans.

Let , the number of rupee , Mohar and suki be 3x , 4x , and 12 x respectively.
Let us convert the mohor and suki into rupee.

Then , Mohar 4x = Rs. \(\frac{4x}{2}\) = Rs. 2x

and suki 12x = Rs. \(\frac{12x}{4}\) = Rs. 3x

Hence , Rs. 3x + mohar 4x + Suki 12 x = Rs. 3x + 2x + Rs. 3x = Rs. 8x

By question ,
Rs. 8x = Rs. 600

\(\therefore\) x = \(\frac{600}{8}\) = 75.

Hence , the number of suki in bag = 12x = 12 \(\times\) 75 = 900.

Here , the fisrt twot ypes of pulse are mixed in the ratio of 2 : 3 , so let 2x kg and 3x kg be quantity of first , and second type of pulse respectively . Also , let y kg of third type of pulse should be mixed in the mixture so that cost of mixture will be Rs. 78 per kg .

Now , The cost of 2x kg of pulse = Rs. 60 \(\times\) 2x = Rs. 120x
The cost of 3x kg of pulse = 3x \(\times\) 72 = Rs. 216 x
The cost of y kg of pulse = Rs. 90y
Total quantity of mixed pulse = (2x+ 3x + y) = (5x + y)

Total cost of mixed pulse = Rs. 78 (5x + y)
By question , 120x + Rs. 216x + Rs. 90y = Rs. 78 (5x + y)
or , 336x + 90y = 390x + 78y
or , 90y - 78y = 390x - 336x
or , 12y = 54x
or , \(\frac{12}{54}\) = \(\frac{x}{y}\)
or , \(\frac{x}{y}\) = \(\frac{2}{9}\) \(\therefore\) y = 4.5x

Now , in final mixture , the ratio of first , se

Quantity of pure milk = 20 \(\times\) \(\frac{60}{100}\) = 12l
Quantity of water= 20 - 12 = 8l
Let x l of water should be added in 20 l of milk so that pure milk in the mixture is 40& or quantity of milk is 40% in (20 + x ) l of mixture.

\(\therefore\) 12 l = (20 = x) \(\frac{40}{100}\)
or , 1200l = 800 + 40x
or , 400l = 40x
\(\therefore\) x = \(\frac{400}{40}\) = 10l Ans.

Let , the quantity of pure milk = x l
Now , cost of x l of milk costing Rs. 22.50 per litre = 22.5x
Let , y l of water x l of milk be adde in so that cost of per litre of mixture is Rs. 18

Then , quantity of mixture = (x + y) l
Cost of per litre of mixture = Rs. 18

Now , cost of (x + y) l milk = Rs. 18 (x + y)
or , Rs. 22.5x = Rs. 18 (x + y)
or , 4.5x = 18y
or , \(\frac{4.5}{18}\) = \(\frac{y}{x}\)

\(\therefore\) \(\frac{y}{x}\) = \(\frac{1}{4}\)

Hence , the required ratio of water and milk is 1 : 4

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