Aptitude Ratio & Proportion Theory

Ratio and Proportion - Important Formulas

1.Ratio:

The ratio of two quantities a and b in the same units, is the fraction $\dfrac{a}{b}$ and we write it as a : b.

In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.

Eg. The ratio 5 : 9 represents $\dfrac{5}{9}$ with antecedent = 5, consequent = 9.

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.

Eg. 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.

2. Proportion:

The equality of two ratios is called proportion.

If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.

Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a : b :: c : d <=> (b x c) = (a x d).

3.Fourth Proportional:

If a : b = c : d, then d is called the fourth proportional to a, b, c.

a : b = c : d, then c is called the third proportion to a and b.

Mean proportional between a and b is $\sqrt{ab}$.

4.Comparison of Ratios:

We say that (a : b) > (c : d) <=> $\dfrac{a}{b}>\dfrac{c}{d}$

The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).

5.Duplicate Ratios:

Duplicate ratio of (a : b) is (a² : b²).

Sub-duplicate ratio of (a : b) is ( $\sqrt{a} : \sqrt{b}$).

Triplicate ratio of (a : b) is (a³ : b³).

Sub-triplicate ratio of (a : b) is (a^1/3 : b^1/3).

If $\dfrac{a}{c}= \dfrac{c}{d}$ , then $\dfrac{a+b}{a-b}= \dfrac{c+d}{c-d}$. [componendo and dividendo]

6.Variations:

We say that x is directly proportional to y, if x = ky for some constant k and we write, x $\propto$ y.

We say that x is inversely proportional to y, if xy = k for some constant k and we write, x $\propto$1/y

Ratio And Proportion :

4 Types Of Ratio And Proportion Problems Are Important For Competitive Exams (3rd Type Is Interesting)

As always, do not forget to attend the short practice test after this tutorial. This will help you to test if you understood well.

Type 1: Combined Ratio Based On Individual Ratios

This type is very easy to solve. Below is an example, to understand this type clearly.

Example Question 1:

1: If a:b = 5:8 and b:c = 6:7, Find a:b:c

Solution 1:
To solve this type, first you have to identify the common term appearing in both the ratios.

In this question, b is common in both the ratios. The value of b in first ratio is 8 and in second ratio is 6.
Now, you have to find the LCM of 8 and 6, which is 24.

Then, you have to transform a:b and b:c so that b becomes 24 in both the cases.
Consider first ratio a:b
You know that a:b = 5:8
To transform b to 24, you have to multiply both the terms by 3.
Therefore, a:b = 5×3:8×3 = 15:24

Consider second ratio b:c
You know that b:c = 6:7
To transform b from 6 to 24, you have to multiply both the terms by 4.
Therefore, b:c = 6×4:7×4 = 24:28
After transformation, a:b becomes 15:24 and b:c becomes 24:28

Now, you can spot that b is equal (24) in both the ratios.
Now you to combine both the transformed ratios by writing b value only once.
Therefore, you will get a:b:c = 15:24:28

Exercise:

Type 2: Distributing Any Quantity Based On Ratios

In this type, you will find that a particular quantity (e.g .,Amount in rupees, Mixture in liters) is to be shared among individuals based on ratios. You will understand this type after the below example.

Example Question 2:

Ram, Gita and Anu shared Rs.5400 among themselves in the ratio 2:3:4. Find the amounts received by each of them.

Solution:
To solve this type of problems, you have to remember a simple formula shown below:

Amount received by a person = (Ratio value of that person / Sum of the ratio values) x Total amount

Based on the above formula, you can easily derive the below 3 formulas:
Amount received by Ram = (Ram’s ratio value / Sum of the ratio values) x Total amount
Amount received by Gita = (Gita’s ratio value / Sum of the ratio values) x Total amount
Amount received by Anu = (Anu’s ratio value / Sum of the ratio values) x Total amount

You know that Ram’s ratio value = 2 , Gita’s value = 3 and Anu’s value = 4
Sum of the ratio values = 2+3+4 = 9
And total amount = 5400

Therefore, you can find individual amounts as shown below
Ram’s amount = 2/9 x 5400 = 1200
Gita’s amount = 3/9 x 5400 = 1800
Anu’s amount = 4/9 x 5400 = 2400

Exercise:

Type 3: Coins Based Ratio Problems (This type is interesting)

This is a special type of ratio problems is very interesting. If you have not seen this before, below example will help you.

Example Question 3:

A bag contains 50p, 20p and 10p coins in the ratio 4 : 8 : 6, amounting to Rs. 210. Find the number of coins of each type.

Solution:
You know that the given ratio of the number 50, 20 and 10 paisa coins is 4:8:6

To make calculations easier, you have to assume number of coins based on their ratio values. For example, the ratio value of 50p coins is 4. Therefore, you have to assume that there are 4X number of 50p coins. (Here X is the unknown quantity, which you will solve)

Similarly, you have to assume that there are 8X number of 20p coins and 6X number of 10p coins.
You know that in total value of all the coins is Rs. 210.

You also know that, two 50p coins make 1 rupee, five 20p coins make 1 rupee and ten 10 paisa coins make 1 rupee. Therefore, we can write the below 3 equations:
Amount in rupees corresponding to 4x number of 50p coins = 4X x (1/2)
Amount in rupees corresponding to 8x number of 20p coins = 8X x (1/5)
Amount in rupees corresponding to 6x number of 10p coins = 6X x (1/10)

Adding all the above three amounts in rupees, you should get Rs. 210. Therefore, you can write,
4X/2 + 8X/5 + 6X/10 = 210
Or (20X + 16X + 6X) / 10 = 210
42X = 2100
X = 50

Number of 50p coins = 4X = 4 x (50) = 200
Number of 20p coins = 8X = 8 x (50) = 400
Number of 10p coins = 6X = 6 x (50) = 300

Exercise:

Type 4: Mixtures & Replacement Based Ratio Problems

You may see problems that involve replacement of a liquid in a mixture of two different liquids. Now, let us see an example.

Example Question 4:

A 15 liters of mixture contains water and milk in the ratio 2 : 4. If 3 liters of this mixture is replaced by 3 liters of water, the ratio of water to milk in the new mixture would be?

Solution:
After 3 liters of mixture is taken out, the remaining mixture will be12 liters.

First you find the amount of water in 12 liters of mixture by using the below formula

Amount of water in 12 liters of mixture = (Ratio value of water / Sum of ratios ) x Total Quantity

Note: Above formula is the same as that we used in example 2.

So, Amount of water in 12 liters of mixture = (2/6) x 12 = 4 liters … equation 1

After 3 liters of mixture is taken out, 3 liters of water is added.
Therefore, Amount of water in 15 liters of new mixture = 3 liters of water + Amount of water in 12 liters of mixture
= 3 + 4 = 7 liters of water …. equation 2

(Note: If you doubt from where 4 appeared refer to equation 1)
Therefore, quantity of milk in the mixture = 15 liters of mixture – 7 liters of water
= 8 liters of milk … equation 3 From equations 1 and 2, you can conclude that the ratio of water and milk in the new mixture = 7 : 8

Exercise: