Percentage Questions Tricks Part 2nd

Let’s continue with Percentage chapter............

Concept 3^rd  :- (Questions Based on Election)

Q9. In an election between two candidates one of them gets 41% of votes and is rejected by a majority of 2412 votes. Find the number of votes drawn in the elections.

Sol.   In these question student should always understand that whatever the voting was held in the election but the total votes which are drawn in the election are always considered as 100% votes. When the number of candidates is two and one of them gets 41% then the votes of the second candidate will be (100% - 41% = 59%) : -

Let the total votes drawn in the election are x then,

         1^st  Candidate got votes = 41% of x.

         2^nd  Candidate got votes = 59% of x.

 According to the concept of these questions the difference between these two percentage values will be equal to the difference of votes given in the questions and the equation will be

             59% of x – 41% of x = 2412

                                18% of x = 2412

                                             x = (2412 × 100) /18    (simplify it.)

                                             x = 13400 votes.

Concept 4^th   :- (Questions Based on Population/Price/Any commodity Percentage increase or decrease)

Concept                 If the present population of a town is x and it increases by p% on the first year and decreases by q% in the second year while again increases by r% in the third year then the population of the town after 3 years will be –          Population after 3 years = x × {(100 + p) / 100)} × {(100 - q) / 100} × {(100 + r) / 100}

Q10.  The population of a town is 10,000. It increases by 10% during 1^st year, while decreases by 20% during 2^nd year and again increases by 30% during the 3^rd year. What will be the population of the town after 3 years.

Sol. 1^st Method :-   Present Population = 10,000

                                 Increase during 1^st year = 10% of 10000 = 1000

                                 Population after 1^st year = 10000 + 1000 = 11000

                                 Decrease during 2^nd  year = 20% of 11000 = 2200

                                 Population after 2^nd  year = 11000 - 2200 = 8800

                                 Increase during 3^rd  year = 30% of 8800 = 2640

                                 Population after 3^rd  year = 8800 + 2640 = 11440

     2^nd Method :- (One line Calculation)

             Required Population = 10000 × {(100 + 10)/100} × {(100 – 20)/100} × {(100 + 30)/100}

                                                    = 10000 × (110/100) × (80/100) × (130/100) (Simplify it)

                                                    = 11440

Q11. The population of a town is 8000. If the number of males increases by 6% and number of females increases by 10% then the population will be 8600. Find the number of males presently present in the town.

Sol. I will show you two conceptual methods of solving these types of questions. Let’s see them.

 1^st Method :-   According to the question the population of the town is 8000. Let the number of male in the town is x therefore the number of female will be (8000 - x) Now

Present Population = 8000

Number of men in the population = x which increases by 6% &

Number of females in the population = (8000 - x) which increases by 10%

After increases the new population will be = 8600.

Now Equation will be :-  x × {(100 + 6)/100} + (8000 - x) × {(100 + 10)/100} = 8600.

                                            x × (106/100) + (8000 - x) × (110/100) = 8600

                                                                         106x + 880000 – 110x = 860000

                                                                                                          4x = 20000

                                                                                                           x = 5000.

                                                      Hence the required number of males = 5000.

Method 2^nd :- (Alligation Method)

                           Present population = 8000

                          The would be population after increase = 8600.

                           Population increase = 600.

   The overall percentage increase in the population will be = (600/8000) × 100 = (15/2)%

 (Always keep this simplification in fraction not in decimal otherwise it may go wrong)   

                                          Males                                               females

                                              6%                                                    10%

                        

                                                                             (15/2)%

                                                 

                                           (10 – 15/2)%                                     (15/2 - 6)%

                                                 (5/2)%                                                (3/2)%

     On simplifying we get the ratio between number of men & females in the town as 5:3

                                          Now number of males = (5/5+3) × 8000

                                                                                   = (5/8) × 8000

                                                    number of males = 5000.

Note you may say that this is so long but dear this is the only way to explain the whole concept of the question to you now its upto you that how fast and short handedly you can solve these type of questions)

Hope you have enjoyed the base concept of this chapter till now. More Conceptual enjoyment will be continued for you soon, just keep visiting this blog regularly and writing me your related queries through comment or forum.

More types of questions will be followed soon on this blog........................