To solve for the maximum marks in the examination, let's first understand the given situation: The passing mark is 300, but the student who scored 225 was said to fail by 10%. This means that the marks required to pass are 10% more than what the student scored.
Let’s denote the maximum marks by \( M \). The percentage required to pass is given by:
\[\frac{300}{M} \times 100\]
The percentage obtained by the student is:
\[\frac{225}{M} \times 100\]
According to the problem, this percentage is 10% below the passing percentage:
\[\frac{300}{M} \times 100 = \left(\frac{225}{M} \times 100\right) + 10\]
We can simplify this equation to find \( M \):
\(\frac{300 \times 100}{M} = \frac{225 \times 100}{M} + 10\)
By clearing the fractions, we derive:
\(30000 = 22500 + 10M\)
Solving for \( M \):
\(30000 - 22500 = 10M\)
\(7500 = 10M\)
Dividing both sides by 10:
\(M = 750\)
Thus, the correct answer is \( 750 \).
Let the maximum marks be \(M\). The student failed by 10%, meaning the student got 90% of the marks required to pass.
The student got 225 marks, and needed 300 to pass. He was short 300 - 225 = 75 marks from passing, therefore, 225 = 300 - 75 marks
Thus the student was declared fail by 75/300 = 25%, instead of 10% as given. We proceed further with the given data: The student needed 300 marks to pass, hence the student achieved 300/M = 1 - 10/100 = 0.9, therefore passing mark percentage can be written as 300/M = 0.9, then
300 = 0.9 * M,
then \(M = \frac{300}{0.9} = \frac{3000}{9} = \frac{1000}{3} \approx 333 \)
This result differs from any options.
Since The student needed 300 marks to pass, the studnet scored 225 marks and was declared fail by 10%, hence this can be written as 225 = 90/100 * Maximum marks i.e. 225 = 0.9 \(\times\) M, hence
then \(M = \frac{225}{0.9} = \frac{2250}{9} = 250\)
This result also differs from any options.
Assuming he needs to pass 300 marks to pass after 225 achieved marks, means he needs 75 more marks, these 75 marks represent 10 % of max marks, hence
75 = 10/100 \(\times\) M M = 750
In a survey, 60 % of 200 students prefer online classes, and 25 % of the remaining prefer hybrid classes. How many students prefer neither?