Dear Reader,
Below you will find 5 types of percentage problems you can expect in bank exams. Among these, first type is easy and straightforward. For remaining types, you have to remember certain important formulas which you will find below.
Type I: Direct Simplification Type (You Can Skip If You Think This Type Is Easy)
This type is the easiest of all types of percentage problems. This is the same type of problems you would have seen in your school days. If you find this too easy, you can skip to next type.
Below is an example for type 1.
Example Question 1: 20% of 180 + ? = 10% of 350 + 5% of 120
Solution:
In above question, you have to find the missing value. Let us substitute X in place of question mark and start simplification as given below.
20% of 180 + X = 10% of 350 + 5% of 120
20/100 x 180 + X = 10/100 x 350 + 5/100 x 120
X = 10/100 x 350 + 5/100 x 120 – 20/100 x 180
X = 5
Type II: Salary Comparison Percentage Problems:
You will see this type often in bank exams. Though it looks tough, it is very easy if you use simple formula. You should remember the below two formulas.
Assume that there are two persons A and B and A’s salary is X% more than B’s salary.
Then B’s salary is lesser than A’s salary by X/(100+X) x 100%
Variant of the above formula:
If A’s salary is X% lesser than B’s salary,
then B’s salary will be more than A’s salary by X/(100-X) x 100%
Below is an example question using the above formula.
Example Question 2:
Ram’s salary is 30% more than Renu’s salary, by how much percent is Renu’s salary less than Ram?
Solution:
From the question you know that Ram’s salary is 30% more than that of Renu.
Therefore our value of X (to use in formula) = 30
Now, Renu’s formula is lesser than that of Renu by X/(100+X) x 100%
= 30/(100 + 30) x 100%
= 30/130 x 100%
= 23.07%
Type III: Appreciation And Depreciation Based Percentage Problems
As you know, appreciation refers to increase in value and depreciation means decrease in value. When appreciation or depreciation is given in percentage, then we get this type III.
There are two simple formulas you have to remember to solve these problems. They are as follows.
When a value Voriginal increases (i.e., appreciates) by R% per annum (i.e., per year), then the final value Vfinal after N number of years is given by
Vfinal = Voriginal(1 + R/100)N
When a value Voriginal decreases (i.e., depreciates) by R% per annum (i.e., per year), then the final value Vfinal after N years is given by
Vfinal = Voriginal(1 – R/100)N
You will understand this type clearly after seeing the below example.
Example Question 3: The population of the Chennai city in the year 2015 is 3,48,600. If it increases at the rate of 5% per annum, what will be its population in 2017?
Solution:
From the question, you can write down the below values to apply in our formula
Chennai’s population in 2015 = Voriginal = 3,48,600
Rate of increase of population = R = 5%
Now you have to find Vfinal after 2 years.
If you substitue the above values in the Vfinal = Voriginal(1 + R/100)N, you will get
Vfinal = 3,48,600(1 + 5/100)2
= 3,48,600(1 + 5/100)2
= 3,48,600 x 105/100 x 105/100
= 3,84,331.5
But you know population cannot be a fractional number. So we can approximate the above value to 3,84,332.
So you have now find out that the population of Chennai after 2 years (i.e in 2017) will be 3,84,332
Type IV: Price And Consumption Based Percentage Problems
Assume that every month you buy 10 Kgs of onions at 50 per Kg (i.e., you will be paying Rs. 500 in total). Due to lack of rainfall, assume that shopkeeper raises the price of onion to 60 per Kg.
Now, to buy 10 Kgs of onion you have to spend Rs. 600. But if you don’t want to spend the extra 100 rupees, you have to buy lesser than 10 Kg of onions. In other words, you have reduce your consumption (usage). If you are asked to calculate the reduction in consumption, you will get this type IV.
Below are two simple formulas to solve these problems. Though the concept is slightly different, the below formulas look exactly the same as that of type II formulas.
If price of an item increases by X%, then the reduction in consumption (use) so that expenditure (money spent) will not increase is X/(100+X) x 100%
Variant of the above formula:
If price of an item decreases by X%, then the increase in consumption (use) so that expenditure (money spent) will not decrease is X/(100-X) x 100%
Let us now see an example:
Example Question 4: The price of oil increases by 10%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase?
Solution:
From the question, you know that the increase in price of oil = X = 10%
If you apply the above value in the formula (that we saw above), you will get:
Reduction in consumption = [X / (100+X)] x 100%
= [10/(100+10)] x 100%
= 10/110 x 100%
= 9.09%
Type V: Set Theory Formula Based Percentage Problems
In school days when learning Venn Diagrams, you would have come across a diagram and formula as shown below:
n(A∪B) = n(A) + n(B)- n(A∩B)
Now, there can be some percentage problems which use the above formula. Below is an example:
Example Question 5: In an examination 30% of total students failed in Maths, 15% in Hindi and 5% in both. Find the percentage of those who failed in both the subjects.
Solution:
If you draw a simple venn diagram and substitute the values in question, you will get
In the above diagram,
n(A) = percentage of students who failed in maths = 30%
n(B) = percentage of students who failed in hindi = 15%
The shaded region represents the number of students failed in both.
Therefore, n(A∩B) = percentage of students who failed in both = 5%
n(A∪B) represents the number of students who failed at least in 1 subject. (i.e., n(A∪B) denotes students who failed in maths or hindi or both)
You know the formula that n(A∪B) = n(A) + n(B)- n(A∩B)
If you substitute values for n(A), n(B) and n(A∩B), you will get
n(A∪B) = 30 + 15 -5 = 40%
Therefore, you have found that 40% of students have failed in at least 1 subject.
Therefore 100 – 40 = 60% students would have passed in both the subjects.