Percentage All Competence Aptitude

To find the combined percentage, we need to first find the total marks obtained and the total maximum marks.

For the first paper:
- Scored: \(72\%\) of \(900\) marks
- \(72\% \text{ of } 900 = 0.72 \times 900 = 648\) marks

For the second paper:
- Scored: \(80\%\) of \(700\) marks
- \(80\% \text{ of } 700 = 0.80 \times 700 = 560\) marks

Now, find the total marks obtained and the total maximum marks:

- Total marks obtained: \(648 + 560 = 1208\) marks
- Total maximum marks: \(900 + 700 = 1600\) marks

Now, find the combined percentage:

\[ \text{Combined percentage} = \dfrac{\text{Total marks obtained}}{\text{Total maximum marks}} \times 100 \]

\[ \text{Combined percentage} = \dfrac{1208}{1600} \times 100 \]

\[ \text{Combined percentage} \approx 75.5\% \]

Therefore, the combined percentage is approximately \( \boxed{75.5\%} \).

Let's denote the total number of students in the school as \(S\).

The problem states that 10% of the students have less than 75% attendance and are not allowed to sit in the exam. Therefore, the number of students with less than 75% attendance and not allowed to sit is \(0.10 \times S\).

It also states that 20% of the students who have less than 75% attendance are allowed to sit in the exam. Therefore, the number of students with less than 75% attendance and allowed to sit is \(0.20 \times (0.10 \times S)\).

Now, let's find the total number of students with less than 75% attendance:

\[ \text{Total with less than 75% attendance} = 0.10 \times S + 0.20 \times (0.10 \times S) \]

\[ \text{Total with less than 75% attendance} = 0.10 \times S + 0.02 \times S \]

\[ \text{Total with less than 75% attendance} = 0.12 \times S \]

Now, we want to find the percentage of students with less than 75% attendance:

\[ \text{Percentage with less than 75% attendance} = \dfrac{\text{Total with less than 75\% attendance}}{S} \times 100% \]

\[ \text{Percentage with less than 75% attendance} = \dfrac{0.12 \times S}{S} \times 100\% \]

\[ \text{Percentage with less than 75% attendance} = 12\% \]

Therefore, \( \boxed{12%} \) of the students in the school have less than 75% attendance.

Let \( T \) be the total number of marks in the examination.

The passing marks are \( 36\% \) of \( T \), which is \( 0.36 \times T \).

According to the given information, the student who gets 190 marks failed by 35 marks. So, the passing marks would be \( 190 + 35 \).

The passing marks are equal to \( 36\% \) of the total marks:

\[ 190 + 35 = 0.36 \times T \]

Solve for \( T \):

\[ 225 = 0.36 \times T \]

\[ T = \dfrac{225}{0.36} \]

\[ T = 625 \]

Therefore, the total number of marks in the examination is \( \boxed{625} \).

Let \( E \) be the set of students who passed in English.
Let \( M \) be the set of students who passed in Mathematics.
Let \( B \) be the set of students who passed in both subjects.

Given:\[ |E| = 0.90 \times T \]\[ |M| = 0.85 \times T \]\[ |B| = 150 \]

We know that the total number of students is the union of students who passed in English and those who passed in Mathematics:

\[ |E \cup M| = |E| + |M| - |B| \]

Substitute the given values:

\[ T = 0.90 \times T + 0.85 \times T - 150 \]

Simplify:

\[ T = 1.75 \times T - 150 \]

Now solve for \( T \):

\[ 0.75 \times T = 150 \]

\[ T = \dfrac{150}{0.75} \]

\[ T = 200 \]

Therefore, the total number of students is \( \boxed{200} \), consistent with the previous solution.

1. According to the question:
   \[    \dfrac{30x + 5}{100} = \dfrac{40x - 10}{100}
   \]

2. Simplify the equation:
   \[    30x + 5 = 40x - 10
   \]

3. Rearrange terms:
   \[    10x = 15 \implies x = 1.5
   \]

So, the full marks (\(x\)) would be 1.5, not 15.

Now, let's use this corrected value to find the minimum pass marks:

\[ \text{Minimum pass marks} = \dfrac{30 \times 1.5 + 5}{100} = \dfrac{50}{100} = 0.5\]

So, the minimum pass marks are 0.5 which is 50 percent.

Let's denote the maximum marks as "M" and the minimum passing marks as "P."

1. The candidate who secured 30% marks failed by 6 marks, so their actual marks are \(P - 6\).
2. The candidate who secured 40% marks got 6 marks more than the minimum passing marks, so their actual marks are \(P + 6\).

Now, let's use the percentage information:

1. For the candidate scoring 30%: \((P - 6) = 0.3M\)
2. For the candidate scoring 40%: \((P + 6) = 0.4M\)

Now, we have two equations to solve for "P" and "M":

1. \(P - 6 = 0.3M\)
2. \(P + 6 = 0.4M\)

Let's solve these equations:

From equation (1):\[P = 0.3M + 6\]

Substitute this into equation (2):\[(0.3M + 6) + 6 = 0.4M\]

Combine like terms:\[0.3M + 12 = 0.4M\]

Subtract \(0.3M\) from both sides:\[12 = 0.1M\]

Divide by 0.1:\[M = 120\]

So, the maximum marks are 120.

Among the options provided, the correct one is 120.

Let's denote the total number of candidates who appeared in the examination as "T."

According to the given information:

1. 60% of the candidates passed in English, which means 40% failed in English.
2. 70% of the candidates passed in Mathematics, which means 30% failed in Mathematics.
3. 20% failed in both subjects.

Now, let's express the number of candidates who failed in English, failed in Mathematics, and failed in both subjects in terms of "T."

1. Number of candidates who failed in English = 40% of T
2. Number of candidates who failed in Mathematics = 30% of T
3. Number of candidates who failed in both subjects = 20% of T

Since 20% failed in both subjects, we can write an equation:

\[ \text{Number of candidates who failed in English} + \text{Number of candidates who failed in Mathematics} - \text{Number of candidates who failed in both subjects} = T \]

\[ (40\%T + 30\%T - 20\%T) = T \]

Now, simplify the equation:

\[ 50\%T = T \]

Divide both sides by 50% (or 0.5):

\[ T = \dfrac{T}{0.5} = 2T \]

So, the total number of candidates who appeared in the examination (\(T\)) is twice the number of candidates who failed in both subjects. Given that 2500 candidates passed in both subjects:

\[ T = 2 \times 2500 = 5000 \]

Therefore, the correct answer is 5000 candidates who appeared at the examination.

Let's denote the percentage of students who pass in English as \( A \), the percentage who pass in Hindi as \( B \), and the percentage who pass in both subjects as \( C \).

According to the given information:\[ A = 60\% \]\[ B = 70\% \]\[ C = 40\% \]

The percentage of students who fail in both English and Hindi can be calculated using the formula:\[ \text{Failure in both} = 100\% - (A + B - C) \]

Substitute the given values:\[ \text{Failure in both} = 100\% - (60\% + 70\% - 40\%) \]\[ \text{Failure in both} = 100\% - (130\% - 40\%) \]\[ \text{Failure in both} = 100\% - 90\% \]\[ \text{Failure in both} = 10\% \]

Therefore, the percentage of students who fail in both English and Hindi is \( 10\% \). Hence, the correct answer is \( 10\% \).

Let's denote the marks scored in the science subjects as \( S \) (out of 300) and the marks to be scored in language papers as \( L \) (out of 200).

The overall percentage is given by the formula:\[ \text{Overall percentage} = \left( \dfrac{\text{Total marks obtained}}{\text{Total marks possible}} \right) \times 100 \]

The student's overall percentage is given as 46%. We can express this as:\[ 46\% = \left( \dfrac{S + L}{300 + 200} \right) \times 100 \]

Now, we know \( S = 32\% \) of 300, so \( S = 0.32 \times 300 = 96 \).

Substitute these values into the overall percentage equation:\[ 46\% = \left( \dfrac{96 + L}{500} \right) \times 100 \]

Solve for \( L \):\[ \dfrac{46}{100} \times 500 = 96 + L \]\[ 230 = 96 + L \]\[ L = 230 - 96 \]\[ L = 134 \]

Now, the student should score 134 out of 200 in language papers.

To find the percentage, use the formula:\[ \text{Percentage} = \left( \dfrac{\text{Marks obtained}}{\text{Total marks possible}} \right) \times 100 \]

Substitute the values:\[ \text{Percentage} = \left( \dfrac{134}{200} \right) \times 100 \]\[ \text{Percentage} = 67\% \]

Therefore, the student should score 67% in language papers to get an overall percentage of 46%. Hence, the correct answer is 67%.

Let's assume Mahesh's income is \( M \). 

According to the given information, Mohan's income is 150% higher than Mahesh's income. This means Mohan's income (\( P \)) is given by:

\[ P = M + \dfrac{150}{100} \times M \]

Simplify this expression:

\[ P = M + \dfrac{3}{2}M = \dfrac{5}{2}M \]

Now, we want to find the percentage by which Mahesh's income is less than Mohan's income. This percentage can be calculated using the formula:

\[ \text{Percentage decrease} = \left( \dfrac{\text{Difference}}{\text{Original}} \right) \times 100 \]

The difference between Mohan's income (\( \dfrac{5}{2}M \)) and Mahesh's income (\( M \)) is \( \dfrac{3}{2}M \).

\[ \text{Percentage decrease} = \left( \dfrac{\dfrac{3}{2}M}{\dfrac{5}{2}M} \right) \times 100 \]

Cancel out \( \dfrac{M}{2} \) from the numerator and denominator:

\[ \text{Percentage decrease} = \left( \dfrac{3}{5} \right) \times 100 \]

\[ \text{Percentage decrease} = 60\% \]

Therefore, the income of Mahesh is 60% less than that of Mohan. Hence, the correct answer is 60%.