Aptitude - Percentage - Discussion
Discussion Forum : Percentage - General Questions (Q.No. 7)
7.
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Answer: Option
Explanation:
Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
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2 | of 48 |
3 |
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80 | x = 80 |
100 |
x = 100.
Video Explanation: https://youtu.be/yPfocU6DA2M
Discussion:
135 comments Page 6 of 14.
Ramesh said:
9 years ago
It clearly indicates that 80% of students is 100 but 100% is 120 must be the answer.
Sachin Dubey said:
9 years ago
x = total number of students.
Below 8 year = 20%x.
Exact 8 year = 48.
Above 8 year= (2/3) * 48 = 32.
Then the equation becomes (we have to find x):
x = (20%x) + 48 + 32.
x - (20%x) = 80.
100x - 20x = 80 * 100.
x = 100.
Hope you will understand it.
Below 8 year = 20%x.
Exact 8 year = 48.
Above 8 year= (2/3) * 48 = 32.
Then the equation becomes (we have to find x):
x = (20%x) + 48 + 32.
x - (20%x) = 80.
100x - 20x = 80 * 100.
x = 100.
Hope you will understand it.
Neeraj said:
9 years ago
Why we are doing 2/3*48? Please, anyone explain me.
Asad said:
9 years ago
Nice one @Rumman.
Ayesha said:
9 years ago
Nice explanation @Ramesh.
SAMEER said:
9 years ago
80% of 80 = ?
Liyakat said:
9 years ago
8 years old students are exact no = 48.
Above 8 years old students are=2/3 * 48 = 32.
8years students+above 8year students = 80.
20% no of below 8 years students = 20/100 * 80 = 96.
Above 8 years old students are=2/3 * 48 = 32.
8years students+above 8year students = 80.
20% no of below 8 years students = 20/100 * 80 = 96.
RAJIB KUMAR KOHAR said:
9 years ago
Students at the age of 8 = 48,
Students above age of 8 = 2/3 of 48 = 32,
Students at the age of 8 = 20,
Total students = (48 + 32 + 20) = 100.
Students above age of 8 = 2/3 of 48 = 32,
Students at the age of 8 = 20,
Total students = (48 + 32 + 20) = 100.
Anshra Khan said:
8 years ago
Good and easy method @Ramesh. Thanks.
Rehan said:
8 years ago
Thanks for explaining the solution.
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