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Percentage | Elementary Mathematics | Aptitude | IAS, SSC, Bank, etc Competitive Exams

Concept of Percentage and Technique Based on Problems

Percent can be translated as per 100 or out of 100. Percentage is a dimensionless number that is used to indicate the number of parts per hundred that are being considered. The percent sign (%) percent, is frequently used to represent it. For percentage calculations, 100 is used as the starting point.

Let’s suppose that, in a school of 80 students passed out of 100 students in the exam then the passing percentage of students who passed in the exam is 80%.

If the number of students is increased to 200, the percentage will drop to 40%. This is due to the fact that just 40 students out of every 100 passed the exam.

It can be computed mathematically as (80/200)*100=40 percent.

On the other hand, if 40 percent of 200 students passed the exam, the total number of students who passed the exam is 80.

It can be computed mathematically as (40/100)*200=80.

Generally, we have to face three types of problems regarding calculating of percentage(%).

So, we take three variables and formulate for easily calculation;

X% of Y = Z

Case-I If Y and Z are given and X is missing

Then, X =(Z/Y) *100

Example X% of 500 = 300

Then X = 300/500*100 = 60.

Hence, we can say that 60% of 500 is 3000.

Case-II If X and Z are given and Y is missing

Then, Y = (Z/X) *100

Example 60% of Y = 300

Then Y = 300/60*100 = 500.

Hence, we can say that 300 is 60% of 500.

Case-III If X and Y are given and Z is missing

Then, Z = (X*Y)/100

Example 60% of 500 = Z

Then Z = 60*500/100 = 300.

Hence, we can say that 60% of 500 is 300.

 

 

Now we discuss problems based on percentage for competitive exams

Convert x% into decimal =  x/100

7% = 7/100 = 0.07 ;    12% = 12/100 = 0.12

Convert fraction into percent

Fraction X/Y =  (X/Y*100)%

 Some Fast Track Trick for solving questions of examinations

Technique 1. If X% of A is equal to Y% of B then Z% of A = (Y*Z/X) % of B.

Ex: If 20% of A is equal to 12% of B then 15% of A is equal to what percent of B?

Sol. Percent of B = (12*15/20) %

                              = 180/20% = 9%

Technique 2. If there is a percentage rise then the effect of its change can be nullified by 100*x/(100 + x) %

Ex: If the cost of a shirt went up by 20% by what percentage should the cost be reduced to make the price even? 

Percentage reduction required= (100 × 20)/(100 + 20) = 2000/120 = 16.67%

Successive Percentage Change

Technique 3. If a value X increases by a%, b%, c% and so on upto n%, the total rise in % is given by the formula,

Final output = X(1 + a/100)(1 + b/100)(1 + c/100)……….(1 + n/100)

Technique 4. If a value X decreases by a%, b%, c% and so on upto n%, the total decline in % is given by the formula,

Final output = X(1 – a/100)(1 – b/100)(1 – c/100)……….(1 – n/100)

Technique 5. (a) If A is X% more than Y, then Y is (X/100+X * 100) % less than X.
                            (b) If A is X% less than Y, then Y is (X/100-X * 100) % more than X.

Ex. If income of Ram is 10% more than that of Shyam, then income of Shyam is how much per cent less than of Ram?

Required percentage = (10/100+10 * 100)%

                                           =(10/110 * 100)%

                                              = 9.09%

Ex. If income of Ram is 10% less than that of Shyam, then income of Shyam is how much per cent more than of Ram?

Required percentage = (10/100-10 * 100)%

                                           =(10/90 * 100)%

                                               = 11.11%

Technique 6. If the value of a number is first increased by X% and later decreased by X% then the net effect is always a decrease which is equal to X%  of X and is written as X2/100 %.

Ex.  The salary of Ram is first increased by 10% and then it is Decreased by 10%. What is the change in his salary?

Sol. Required change = 102 /100 %

                                             =  100/100% = 1%

Technique 6. If a number X is increased by Y%, then the new number will be   

                           100+Y/100 * X

Technique 6. If a number X is deceased by Y%, then the new number will be           

                            100-Y/100 * X

Technique 7. If the passing marks in an examination is P%. If a candidate scores S marks and fails by F marks then


Example: – Pankaj Sharma has to score 40% marks to get through. If he gets 40 marks and fails by 40, then find the total marks set for the examination?

Technique 8. If a candidate scores X% marks and fails by a marks while an another candidate scores y% marks and gets b marks more than minimum passing marks, then


Example:- A candidate score 25% and fails by 60 marks, while another candidate who score 50% marks, gets 40 marks more than the minimum required marks to pass the examination. find the maximum marks for the examination.

Sol:  Maximum marks = (60 + 40)/50-25 *100

                                          = 100/25 *100

                                         = 4*100 = 400

 

 

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