Percentage Questions Tricks Part 3rd

Let’s continue with this chapter............

Concept 4^th   :- (Questions Based on Population)

Q12. During 1^st year the population of a town increases by 5%, during 2^nd year it decreases by 5%. If the population at the end of 2^nd year was 7980. Find the population of the town at the beginning.

Sol.   Let the population of the town at the beginning was “x”

During 1^st year population of a town increases by 5%

Population of the town after 1^st year = (x) × {(100 + 5)/100}.

During 2^nd year population of a town decreases by 5%

Hence the population of the town after 2^nd year = (x) × (105/100) × {(100 - 5)/100}.

                                                                                  = (x) × (105/100) × (95/100).

But according to the question the population of the town after 2 years = 7980.

Therefore, : - (x) × (105/100) × (95/100) = 7980.

                        x = {7980 × 100 × 100}/{105 × 95}

On simplifying we get:- x = 8000.

Hence the population of the town at the beginning  was 8000.

Concept 5^th   :- (Questions Based on Pass/fail in Examination.)

Q13. In an examination 70% of students passed in English, 80% of the students passed in Mathematics and 10% of the students failed in both.  If the total number of students passed in the examination is 144. Find the number of students appeared in the examination.

Note:- In such type of questions you should analyse the question very carefully and convert the whole given data into either fail percentage or pass percentage and then solve these questions using basic calculation or with the help of formula of Set Theory.

Sol. 1^st Method :-   Analyse the given data :-

 Subject                Student % passed                Student % failed

English                            70%                                Not given

Mathematics                   80%                               Not given

Eng. + Maths              Not given                                10%

Let the total number of students appeared in the examination is “x”.

Re-analyse the given Data in the question:-

Subject                Student % passed                Student % failed

English                            70%                                       30%

Mathematics                   80%                                      20%

Eng. + Maths                  90%                                      10%

Total percentage of students failed in the examination =  (30 + 20 - 10)% = 40%

Hence the total percentage of students who passed the examination = (100 - 40)% = 60%

According to the question number of students passed in the examination = 144.

Therefore, 60% of (x) = 144

                   (60/100) × (x) = 144

                                   x = 144 × (100/60)

                                  x = 240 students. 

Hence the total number of students appeared in the examination = 240 students.

2^nd Method :-   Analyse the given data as :-

Student % passed in Mathematics = n(M) = 80

Student % passed in English = n(E) = 70

Since Student % failed in both subjects = 10% therefore,

Student % passed in Mathematics and English both = n(M ∩ E) = 90

We know from Set theory :- n(M U E) = n(M) + n(E) - n(M ∩ E)  where

n(M U E) = Actual Student % passed in the examination.

Putting the values in the formula we get,

n(M U E) = 80 + 70 – 90

n(M U E) = 60.

→ Student % passed in the examination = 60%

But according to the question number of students passed in the examination = 144.

Therefore, 60% of (x) = 144

                   (60/100) × (x) = 144

                                   x = 144 × (100/60)

                                  x = 240 students. 

Hence the total number of students appeared in the examination = 240 students.

Q14. A candidate scores 20% and fails by 30 marks while another candidate scores 32% and gets 42 marks more than the minimum required marks to pass the examination. Find the percentage of marks required to pass the examination.

Note :- In these type of questions you should be clear that the examination is same, which is faced by two different candidates, therefore the passing percentage of the examination will be same. So you should first of all make both the students just pass the examination and then equate them to get the required result.

Sol. Let the maximum marks of the examination be (x)

Analyse the question for given data:-

1^st candidate scores % marks in the examination = 20%

Marks less for 1^st candidate to pass the examination = 30 marks.

Therefore 1^st candidate passes the examination when he gets = {20% of (x) + 30} marks.

2^nd candidate scores % marks in the examination = 32%

Marks more for 2^nd candidate to pass the examination = 42 marks.

Therefore 2^nd candidate passes the examination when he gets = {32% of (x) - 42} marks.

According to the concept of such question :-

20% of (x) + 30 = 32% of (x) – 42

On simplifying we get ;

12% of (x) = 72

→ x = (72 × 100)/12 = 600 marks

Maximum marks of the examination = 600 marks

Minimum marks required to pass = 20% of (x) + 30 = 20% of 600 + 30 = 150 marks.

So % marks required to pass the examination = (150/600) × 100 = 25%

Hence the required % marks to pass the examination = 25%.

(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........................