Sunday, 18 December 2016

Tricks and Shortcuts to solve Aptitude Problems on Percentage within 30 seconds

                        In Aptitude Test most important section is Percentage problems, because to solve aptitude problems on Percentage, you should know tricks and shortcuts otherwise percentage problems section will take more time. Percentage problems are somewhat difficult than Aptitude problems on Ages, Aptitude problems on Averages, Aptitude problems on Trains. But If you follow below tricks and shortcuts methods then percentage problems are also become easy to you. To solve aptitude problems on averages within 30 seconds you should to more practice also.

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                      Now come back to our “Tricks and Shortcuts to solve Aptitude Problems on Percentage” section. Here we are giving percentage basic formula's, tricks shortcuts and some examples.

Percentage problems tricks and shortcuts 

Basic Formula's to remember:               

1.  x% = x/100

      Example: 25% means 25/100 value

2. To solve Percentage problems on Population, you should remember the below formula

        After n years, The population = P[1+(R/100)]^n
         n years ago the population = P/[1+(R/100)]^n
       Note: In the above formulae P = Number of population in the city
                                                                 R = % increase per year
                                                                  n = years

3. The above formulae is for R% increase per year, Now the reverse R% decrease per year

Note: Here P may be ‘Machine’, ‘Mobile’ or something else (Not Population)
                After n years, The value of Mobile = P[1-(R/100)]^n
         n years ago, The value of Mobile = P/[1 - (R/100)]^n
       Note: In the above formulae P = Number of population in the city
                                                                 R = % increase per year
                                                                  n = years

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4. If X is R% greater than Y, then Y is less than X by (R/(100+R))*100]%.

5. Now Reverse, If X is R% less than Y, then Y is greater than X by (R/(100-R))*100]%.


- The first formula is very basic, everyone knew.
- If you remember second formula then third one is bonus.
- If you remember fourth formula then fifth one is bonus.

                So you need to remember only Two formulas to solve aptitude problems on percentage.
Now we will see some examples, tricks and shortcuts.

Example 1:

                 What is the fraction of 60%, 6% and 0.6%?

Percentage tricks for competitive exams

Ans: (a) 60% = 60/100 = 3/5,
          (b) 6% = 6/100 = 3/50,
           (c) 0.6% = 0.6/100 = 6/1000 = 3/500

 Example 2:

                    What is the value of 40%, 4%, 0.4% in decimal?

Shortcut tricks of percentage problems

Ans:  (a) 40% = 40/ 100 = 0.4,
          (b) 4% = 4/100 = 0.04,
           (c) 0.4% = 0.4/100 = 0.004

Example 3:

                 Solve 25% of 300 + 30% of 250

Percentage shortcut tricks

Answer: (25/100)*300 + (30/100)*250
             = 75 + 75 = 150

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Example 4:

                 Sehwag Scored 240 runs in a match which includes 9 sixes and 13 boundaries. What percentage of score did he make through running between the wickets?

Shortcuts for percentage problems in Aptitude

A) Total Runs = 240
   Through Boundaries (13) and sixes (9) = 13*4 + 9*6 = 106 runs
   Through running = 240-106=134
Percentage of score he did through running = (134/240)*100 = 335/6 = 55 5/6% or 55.83%

Example 5:

                A fruit seller had some Mangos. He sells 30% mangos and still has 350 mangos. The total number of mangos he had initially?

Tricks to solve percentage Aptitude problems

Answer : for example he had ‘x’ mango's initially.
         Then, (100 - 30)% of x = 350
(70/100)*x = 350
X = 350*100/70 = 500

Note: example 5 type of problems frequently asked question in Aptitude problems on Percentage

Example 6:

                        The Above question is Very frequently asked percentage aptitude problem with exact options also.

Solve percentage problems with shortcuts and tricks

Answer: take x=10 and y = 100 as examples
Now A = x % of y = 10% of 100 = (10/100)*100 = 10
 B = y% of x = (100/100)*10 = 10
A = B
So, The Answer is None of these (Option E)

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Example 7:

                       In one Election, 2 candidates are participated. 64% of voters are used their votes. In this 50% votes 2% votes are declared as Invalid. One candidate got 6272 votes which were 60% of the total valid votes. Find out how many voters are there in that election.

Percentage solve shortcut tricks in Aptitude


Let us Total no. of voters in the election = x
Number of votes cast = 64% of x
In that above 50% votes 2% votes are invalid.
So, Number of votes cast = 50% of x = 98% of (64% of x)
64% of (98% of (50% of x)) = 6272
(64/100)*(98/100)*(50/100)*X = 6272
X = (6272*100*100*100)/(64*98*50) = 20000

Example 8:

                The Population of a town is from 3,50,000 to 5,25,000in a decade (in last 10 years). What is the average percent increase of population per year?

Percentage Aptitude problems with solutions

In the last 10 years, Population increase is  5,25,000 - 3,50,000  = 1,75,000
Percentage increased in last 10 years = (1,75,000/3,50,000)*100 % = 50%
Per year 50/10 = 5% increase

Example 9:

                 In a Town present population is 3,52,800. If it is increase 5% every year, What is the population of a town after 2 years and before 2 years

Tricks shortcuts for population percentage problems

Here You need to use the second formula as I mentioned above

        After n years, The population = P[1+(R/100)]^n
         n years ago the population = P/[1+(R/100)]^n
So, After 2 years, The population = 352800[1+(5/100)]^2 = 352800*(105/100)*(105/100) = 388962
      2 years ago the population = 352800/[1+(5/100)]^2 = 352800/[(105/100)*(105/100)] = 320000

Example 10:

               In a School, 30% of the candidates failed in Telugu, 40% of the candidates failed in Hindi and 25% in both. What percentage of candidates are passed in both Telugu and Hindi.

Tricks to solve percentage problems with solutions

Candidates failed in Telugu n(A) = 30
Candidates failed in Hindi n(B) = 40
Candidates failed in both Telugu and Hindi n(A intersection B) = 25
n(A Union B) = n(A) + n(B) - n(A intersection B) = 30 + 40 - 25 = 45
So, the answer is 45%

                  I hope, you got better understand on percentage problems with all the above different types of problems. In most of competitive Aptitude exams, problems from percentage should not go beyond these types. Ensure you should have good practice on all the above types of percentage problems.

                 Here the tricks and tips to solve the percentage problems is Better understanding of problem and apply the simple formula. You will get the proper answer within 30 seconds. But you should be very fast in mathematics calculations.

              Bookmark this page and practice regularly with these type of percentage problems apply some tricks means respective formula.

                 If you want to get tricks and tips for other type of aptitude problems, then Drop your mail Here. I will send pdf Aptitude problems with examples.

                If you have any other type of percentage problems in aptitude, then comment here. I will include those type of aptitude problems here and that will become help for others also. “Share knowledge and get Success”


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