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


Dear Reader, below are 4 types of ratio problems you can expect in SBI and IBPS exams. Among these, 3rd type is really interesting and may be new to you. Each type is explained with example.

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

You can expect this type of problems not only in bank but also in other government exams. This type is very easy to solve. Below is an example, to understand this type clearly.

Example Question 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

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 litres) 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.

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

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.

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

This type is interesting, isn’t it? Share your views on comments section below. Now let us move on to our final type.

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

After 3 litres of mixture is taken out, the remaining mixture will be12 litres.

First you find the amount of water in 12 litres of mixture by using the below formula
Amount of water in 12 litres 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 litres of mixture = (2/6) x 12 = 4 litres … equation 1

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

(Note: If you doubt from where 4 appeared refer to equation 1)
Therefore, quantity of milk in the mixture = 15 litres of mixture – 7 litres of water
= 8 litres 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

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