In this article, we simply deal with the different types of problem based on Percentages. This lesson does not have any problems of its own but simply is based on the concepts we have studied so far. This article is an extension of the previous article. You can read it here.

Problem Type-5: Problems based on students and marks

Shortcut-1 for the problem type:
A candidate scores x% is an examination fails by ‘a’ marks, while another candidate who scores y% marks and get b marks more than the maximum required passing marks, then the maximum marks for the examination is given by Shortcut-2 for the problem type:
In an examination a% of total number of candidates failed in a subject X and b% of total number of candidates failed in subject Y and c% failed in both subjects, then percentage of candidates, who passed in both the subjects, is [100 – (a + b – c)]%.

Example: In a quarterly examination a student secured 30% marks and failed by 12 marks. In the same examination another student secured 40% marks and got 28 marks more than minimum marks to pass. The maximum marks in the examination is:
(a) 300
(b) 500
(c) 700
(d) 400 Example: In an examination, 52% of the candidates failed in English and 42% failed in Mathematics. If17% failed in both the subjects, then the percentage of candidates who passed in both the subjects, was:
(a) 23%
(b) 21%
(c) 25%
(d) 22%

Solution: (a)
In an examination a% of total number of candidates failed in a subject X and b% of total number of
candidates failed in subject Y and c% failed in both subjects.
Then, percentage of candidates who passed in both the subjects: [100–(a + b– c)%].
Why so?
We need to subtract failed candidate from total to calculate pass candidates, but in this process, we subtract people who were fail in both of subjects twice, so we added c again.
Now equation can be read as [ 100—a-b+c]
Here, a = 52, b = 42 and c = 17, then Percentage of candidates who passed in both the subjects
= [100 –(52 +42–17)]
= 100–77=23%

Problem Type-2: Problems based on percentage in excess or in short

Example:
lf A has 4/5 of the number of books that shelf B has. If 25% of the books A are transferred to B and then 25 % of the books from B are transferred to A then the percentage of the total number of books that on shelf A is:
(a) 25%
(b) 50%
(c) 75%
(d) 100%

Solution: (b)

Let the number of books in shelf B be 100.
∴ Number of books in shelf A = {(100 x 4)/5} = 80
On transferring 25% i.e.,of books of shelf A to shelf B, the books on
shelf B = {100 + (80 x 25)/100}
B = 100 + 20 = 120
Books left in shelf A = 80-20 = 60
Again, on transferring 1/4th books of shelf B to shelf A, the books on
shelf A = {60 + (120/4)} = 90
Total no of books in A and B = 120 +60 = 180
Required percentage of books on shelf A = (90/180) x 100 = 50%

Problem Type-3: Problems based on percentages and number system

Example: A number, on subtracting 15 from it, reduces to its 80%. What is 40% of the number?
(a) 75
(b) 60
(c) 30
(d) 90

Solution: (c)
Let the number be p.
According to the question,
p – 15 = p(80/100)
p -15 = 4p/5
⇒ 5p – 75 = 4p ⇒ p = 75
Now, 40% of 75 = (75 x 40)/ 100 = 30