“Percentage Tricks for Competitive Exams 2025 – A Complete Easy Guide”

Introduction

Using percentage tricks, you can solve problems in exams very quickly. In the exam hall, no one checks your step-by-step explanation — the final goal is the answer. The examiner doesn’t care about the process; he only cares about the correct answer. So, always use percentage tricks in exams. We can also use these tricks in other concepts like profit and loss, partnership, and other percentage-related problems. Let’s start.

Learn Basics

Basics are very important when using percentage tricks, so always learn the basics first.

→ 25% means 25/100. The symbol % means 1/100.

→ Example: If 100% is 360, then 10% of 360 is 36 (for 10%, place a decimal point one position from the right: 36.0 = 36).

Now, 1% of 360 is 3.6 (for 1%, place the decimal point two positions from the right: 3.60 = 3.6).

→ Therefore: 100% = 360, 10% = 36, 1% = 3.6.

“Now, let’s learn fractions of percentages.”

Fractions of Percentages

“We use fractions of percentages to simplify calculations, and we apply these fractional values in percentage tricks.”

—> Example if we have to calculate 7.14% of 14, it’s very hard but if we know the fraction value of 7.14%, then it is easy. The fraction of 7.14% is 1/14. Now (1/14) × 14 = 1. See it is very easy. So let’s learn remaining fraction values also.

Remember, for 9 and 11 their values are like this: for 9 it’s 11 multiples, and for 11 it’s 9 multiples.

—> Example: 1/9 = 11.11%, 2/9 = 22.22%, and so on. Now, 1/11 = 9.09%, 2/11 = 18.18%, and so on.

—> Now Fractions to Percentages

—> Example: 7/4 – we solve by separating or rewriting them with known fractions like:
7/4 = (4/4) + (3/4) = 100% + 75% = 175%.

—> Now 8/3 – using division to solve this: 3 divides 8, the quotient is 2, remainder is 2 (because 3 × 2 = 6, and we still need 2 for 8). Now (quotient) + (remainder/divisor) = 2 + (2/3). That is 200% + 66.66% = 266.66%. Therefore, 8/3 = 266.66%. Like that we have to do for other fractions.

Always revise these methods to understand them easily.

Percentages Change Problems

We also use percentage tricks in percentage problems. These types of questions are also often asked in competitive exams. So let’s see a question and solve it by using percentage tricks. This question is made by me.

—> If a company got profit 24 rs in past and at present the company got 15 rs profit. What is the % change in the profit made by the company?

Sol: Always take the old value in the denominator because we are comparing on the old value, here 24. First we need to find the difference between past and present, that is 24 – 15 = 9. Now profit is decreased because in past company got 24 but now it is 15. So it is decrease in percentage. So we compare 9 on 24, that is (9/24) × 100 = 37.5%. Therefore the % decrease is 37.5%.

This is how we have to solve these problems by using percentage tricks.

Effective Change Problems

Here effective change means total change. Let me explain with:

—> Example: If A = B × C. If B is increased by 20% and C is increased by 25% then the total change is ? That is A. There is a formula for this, that is x + y + (xy / 100). x = 20%, y = 25% here.

By substituting we get the answer.

Other percentage trick is ratio method:

We know that 20% means 1/5, that at first it is 5 parts, now one part is increased, so now 5 + 1 = 6 parts. Now the ratio is 5:6. This is B.

We know that 25% means 1/4, that at first it is 4 parts, now one part is increased, so now 4 + 1 = 5 parts. Now the ratio is 4:5. This is C.

Now Ainitial / Anow = (Binitial × Cinitial) / (Bnow × Cnow) = (5 × 4) / (4 × 5) = 2/3. That is Ainitial is 2 parts, Anow is 3 parts. Now we can tell that one part is increased. Therefore (1/2) × 100 = 50%.

Therefore A is increased by 50%.

This is how we have to solve this type of problems using percentage tricks.

Pulp and Water Percentage change Problems

We can solve this type of problems with a few percentage tricks and it also has a formula. Lets understand by solving a question

—> Fresh fruits contains 68% water and dry Fruits contains 20% water. How much dry fruits can be obtained from 100kg of fresh fruits?

Sol: In question they have given fresh fruits contains 68% water that means remaining 32% is pulp and they also given dry fruits contains 20% water that means 80% is pulp. Lets equal the pulp.

—> 32% of f.f = 80% of d.f

—> By solving we get f.f/d.f ratio that is 5/2

—> they question they have given 100 kg fresh fruits that means 5parts is 100 then 1 part is 20 then 2 parts is 40kg. Therefore the answer is 40kg. Now methood 2 with formula.

Method 2

—> Formula is pulp constant = % change x total quantity

—> (32/80) is 2/5 now inverse this we get 5/2 that is total quantity. Therefore 5 is 100kg and 2 is 40kg. So direct answer is 40kg.

This is how you have to solve this type of problems.

Tax and Income Problems

If we know the formula we can easily solve the question. Let me explain with a question.

—> Formula: Tax = Change of tax × Income. —- Tax is constant

—> Income of a person decreased by Rs. 36,000 and rate of income tax increased from 15% to 18%. The ratio of paid tax is 3:2. 26.5% income is tax free. Find initial income?

Sol: Equal the tax 3:2 —> 3×2 : 2×3 —> 6:6 (because tax should be constant).

—> Change of tax 15/18 —> 5/6. Now multiply this with 2/3 because we multiplied the ratios to make them equal. So here also we have to do.

—> After multiplying we get 5/9. Now inverse this 9/5, difference is 4 parts, and 4 parts is 36,000, then 9 parts is 81,000. Therefore 81,000 is the answer.

This is how we have to solve this type of questions. Now I will give some similar formulas:

—> Married people = percentage change × total population

—> Expenditure = price × consumption

—> Revenue = price × sales

—> Expenditure = price × quantity

These are some you should always remember.

Student Marks Probelms

If you understand it, it will be easy. Let me explain the concept:

—> Two students A and B. If A got 42% of marks and still failed by 12 marks, and if B got 54% of marks and passed by 36 marks, what are the passing marks?

Sol: Now take the difference of percentages, that is 54 – 42 = 12%.

—> Now take the difference of marks: 36 – (–12) (because less than pass marks) = 48 marks.

—> Now 12 parts equal to 48 marks, then 1 part is 4 marks. Now passing marks is: Let’s take A’s marks, that is 42 parts × 4 = 168 marks, and A failed by 12 marks, so add those 12 marks also.

—> Therefore 168 + 12 = 180. So 180 is the passing marks.

—> If they give x% less marks and y% more marks, what is the maximum marks? Then the formula is:

m.m = a+by−x×100\frac{a + b}{y – x} \times 100y−xa+b​×100

—> a, b are marks.

This is how you have to solve this type of problems.

Conclusion

These are some percentage tricks to solve problems in less time. Always revise daily and try new questions to solve. Now I have covered only a few topics; in the future, I will write another article on percentage tricks. So prepare smart and work smart daily. All the best for your future. Remember, consistency is the key to mastering any topic, and the more you practice, the faster you’ll improve. Keep challenging yourself with new problems every day to stay ahead.

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