Subject: Compulsory Maths

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.

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**

**Duplicate and** **sub -****duplicate** **ratio**If a : b be a ratio, then the 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 \(\frac{a}{b}\) = (\(\frac{a}{b}\))

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= b^{2}

a : b = b : c

Here, a is 1^{st }proportion

b is mean proportion

c is 3^{rd }proportion

Mean proportion (b) =√ac

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

- Invertendo
- Alternendo
- Componendo
- Dividendo
- Componendo and Dividendo
- 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.

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.

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 = dk^{2..............(II)}

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

∴ a = bk = dk^{2}.k = dk^{3}.............(iii)

∴ a = dk^{3}, b = dk^{2} 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.

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.

The work done by Ram is 3 days can be done by Shyam in 4 days. Again, the work done by Shyam in 5 Days can be done by Hari in 6 days. In how many days the work done by Hari in 16 days can be done by Ram?

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.

The price of 3 ducks is equal to the price of 4 hens. The price of 4 peasants and the price of 7 ducks are equal. If the price of hens is Rs. 750 then what is the price of one peasant ?

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.

12 oxen can eat as much as 24 sheep where 15 sheep can eat as much as food as 25 goats. Again the food for 17 goats is equal to the food for baby elephants. Similarly , the food eaten by 8 baby elephants is equal to the food eatn by 12 horses. How many horses can eat the food required for 153 oxen ?

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

When A does \(\frac{1}{3}\) of a work , B does \(\frac{1}{4}\) of the same work. Similarly , when B does \(\frac{1}{5}\) of work , C does \(\frac{1}{2}\) of work . If A finishes the work in 20 hours then in how many days C finishes the work ?

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.

3 ducks eggs can be exchanged for 4 hen eggs. Similarly, with 4 eggs of the swan , 7 eggs of duck can be exchanged. If the cost of 2 hens eggs is Rs. 7.50 then what is the cost of a swan's egg?

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.

Two types of Rice costing Rs. 15 and Rs. 20 per kg are mixed in the ratio 2:3 . What is the cost of mixture of rice per kg ?

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.

When two types of oil costinf Rs. 80 and Rs. 90 per l are mixed in the ratio 9 : 8 , what if the cost of mixture of oil ?

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

Rahul mixed three types of tea in the ratio 3: 4: 7. If the price of tea per kg is Rs. 80, Rs. 100 And Rs. 150 respectively, find the price of 35 kg of mixed tea.

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

In a mixture of 56 l of milk and water , the ratio of milk and water is 4 : 3 , Find the quantity of milk and water in the mixture.

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.

There is Rs. 600 in a bag which consists of rupee , mohar and suki in the ratio od 3 : 4 : 12. Find the number of suki inside the bag.

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.

A shopkeeper has three types of black pulse costing Rs. 60 , Rs. 71 and Rs. 90 per kg respectively. Out of them the first two types of pulse are mixed in the ratio of 2 : 3 . In what ratio the third type of pulse should be mixed so that the cost f mixture will be Rs. 78 per kg ?

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

In 20 litres of milk , 69% is pure milk and the remaining is water . What amount of water should be added in the milk so that pure milk in the ratio mixture is 40% ?

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.

The milk previoulsy sold for Rs. 22.50 per litre is to be sold for Rs. 18. What is the ratio of water and milk in the mixture ?

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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