Notes on Ratio and Proportion | Grade 9 > Compulsory Maths > Algebra | KULLABS.COM

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### 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

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, $$\frac{a + b}{a - b}$$ = $$\frac{c + d}{c - d}$$

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

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.

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

.

### Very Short Questions

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, 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 , 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.

0%

1:3
1:4
1:2
1:5

Rs.15

Rs.10

Rs.30

Rs.20

7

4

6

8

Rs.250

Rs.500

Rs.600

Rs.300

13

10

12

11

1

2

6

3

3

4

6

2

• ### If(frac{3m-5n}{3m+5n}) =(frac{1}{4}),what is the value of (frac{m}{n})?

(frac{24}{9})

(frac{25}{9})

(frac{15}{9})

(frac{21}{9})

• ### If (frac{p}{q})=(frac{3}{4}), What is value of (frac{2p-3q}{2p+3q})?

-(frac{1}{3})

-(frac{1}{7})

-(frac{1}{4})

-(frac{1}{8})

• ### If (frac{a}{b})=(frac{3}{4}), what is the value of (frac{2a-3b}{2a+3b})?

(frac{-1}{5})

(frac{-1}{3})

(frac{-1}{6})

(frac{-1}{3})

## ASK ANY QUESTION ON Ratio and Proportion

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(nk pk)³÷(n p)3

##### Aarjan

When rs 105 is divided in the ratio of 2:5,how much money will be?

##### Sworup

When rs 105 is divided in the ratio of 2:5,how much money will be in the first part