Ratio and Proportion
In this post we will learn tricks to solve ratio and proportion questions within seconds. Also we will learn conceptual methods and formulas of ratio and proportion to solve such problems as this is one of the most important chapter which will help in solving questions of other arithmetic chapters too. Later we will learn how to solve Alligation and mixture problems easily using the concept of Ratios.What is a Ratio?
A relationship between two numbers indicating how many times the first number contains the second is known as Ratio. A ratio may be written as "a to b" or “a:b”, or it may be expressed as a quotient of "a and b". The first term of a ratio is known as “Antecedent” and the second term is “Consequent”.When the two quantities are measured with the same unit, their ratio is a dimensionless number. A quotient of two quantities that are measured with different units is called a Rate.
A ratio which is a fraction is not altered by multiplying or dividing both its terms by the same number.
For Example: Ration between 3 & 4.
We can write “3:4” or “3/4” or “the ratio of 3 to 4”or “3 is to 4”.
Here '3' is antecedent and '4' is consequent.
Multiplying any ratio by same number gives the same ratio. For example: multiply 3:4 by 5.
⇒ (3x5/4x5) = 15/20
Clearly 15/20 is same as 3/4 containing a common term ‘5’ between antecedent and consequent.
It means: ‘Ratio is a fraction left after removing the HCF of antecedent and consequent’.
What is Proportion?
The equality of ratios is called Proportion.For example: Consider two ratios: 7:21 and 8:24
Since 7 is one-third of 21 and 8 is one-third of 24, the two ratios are equal. Hence numbers 7, 21, 8 and 24 are said to be in proportion.
We write proportion as: [7:21::8:24] or [7/21 = 8/24]
First and fourth terms (7 & 24) are called ‘Extremes’ while second and third terms (21 & 8) are called ‘Means’.
If four numbers are in proportion then product of extremes is equal to the product of means.
METHODS
Compounded Ratio
Ratios are compounded by multiplying together the antecedents and consequents of different ratios or Fractions.Example: Find the ratio compounded of three ratios: 5:2, 4:11 and 7:9?
Solution: The required ratio = (5x4x7/2x11x9) = 70/99
Duplicate & Sub-Duplicate Ratio
- When any ratio is compounded with itself, the resulting ratio is known as Duplicate ratio.
Solution: Duplicate ratio = (2/3)2 = 4/9
- Root of any Ratio is known as sub-duplicate ratio
Solution: Sub-duplicate ratio = √(4/9) = 2:3
Triplicate & Sub-Triplicate Ratio
- Multiplication of any ratio with itself thrice gives us Triplicate ratio.
Solution: Triplicate ratio = (2/3)3 = 8/27
- Cube root of any ratio is known as sub-triplicate ratio.
Solution: Sub-Triplicate ratio = ∛(8/27) = 2:3
Trick to find a Common Ratio
If the ratio between the first and 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:BDFor given two ratios, it can be written as:
A : B
C : D
Now write B at the space above D and write C at the space below A,
A : B ⃕: B
x C ⃔: C : D
Now multiply: AC : BC : BD
Similarly for three or more ratios like A:B, C:D, E:F, G:H etc we can write a common ratio like this:
A : B : B : B : B
C : C : D : D : D
E : E : E : F : F
G : G : G : G : H
Multiply Column vertically to get Common Ratio:
⇒ ACEG : BCEG : BDEG : BDFG : BDFH
Example 1: A, B and C are three quantities of same kind such that A:B = 2:3, B:C = 5:6. Find A:B:C.
Solution: A:B:C = 2 x 5 : 3 x 5 : 3 x 6 = 10 : 15: 18
Example 2: In 30 litres mixture of acid, the ratio of acid and water is 2:3. What amount of water should be added to the mixtures o that the ratio of acid and water becomes 2:5?
Solution: Method 1: Initial ratio = 2:3 and final ratio = 2:5
Here fixed quantity is acid, so we need to equal the value of fixed quantity in both the ratios. As we have '2' as a value of acid in both the ratios so value is already equal.
Water to be added = (Total quantity of mixture) x (Difference in the ratio of varying quantity)/(Sum of initial ratios)
⇒ 30 x 2/5 = 12 litres
Method 2: Direct formula: If in x litres mixture of milk and water, the ratio of milk and water is a:b, the quantity of water to be added in order to make this ratio c:d = x(ad-bc)/c(a+b)
Putting values in this formula we get = 30(2x5 – 3x2)/ 2(2+3) = 12 litres
Method 3: Initial ratio = 2:3 and Mixture = 30 litres
Acid = 2 x 30/5 = 12 Litres ; Water = 3 x 30/5 = 18 litres
Final ratio = 2:5
Here fixed quantity is acid because we added water to change ratio, as value of acid in final mixture is "2", it means: 2x = 12 litres or x = 6 litres
It means water = 5x = 5x6 = 30 litres
Initially water was 18 litres and now 30 litres, it means water added = 30 – 18 = 12 Litres
Method 4: By Alligation: Acid : Water = 2: 3
Water in the original mixture = 3x100/5 = 60%
Water to be added is pure = 100%
Water in resulting mixture = 5x100/7 = (500/7)%
In 5 litres of mixture water to be added is 2 litres
Therefore in 30 litres of mixture water to be added = 30 x 2/5 = 12 litres
Different types of Proportions
1. Continued Proportion: When the ratio of the first to the second quantities is equal to the ratio of second to the third, such three quantities are said to be in Continued Proportion. For example: 4, 8 and 16 are in continued proportion for 4:8::8:16.2. Mean proportional: The second quantity in a continued proportion is said to be mean proportional of the first and third quantity. For example in proportion ‘4:8::8:16’ the second quantity ‘8’ is called the mean proportional between first and third, and the third quantity ‘16’ is called the Third proportional to the first and second.
Example: Find the fourth proportional to the numbers 6, 9 and 4.
Solution: Let x be the Fourth proportional, then 6:9 = 4:x
⇒ 6/9 = 4/x
⇒ x = 36/6 = 6
Example: Find the third proportional to 8 and 20.
Solution: For third proportional we need to write 8 and 20 like this: 8:20::20:x
⇒ 8/20 = 20/x
⇒ x = 400/8 = 50
Example: Find the mean proportional between 8 and 32.
Solution: Mean proportional = √(8 x 32) = 16
These are few methods for solving ratio and proportion problems. Later we will learn more methods which will be problem specific but concept remains same. That's why always focus on concept than direct formulas. Good Luck 👍
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