Finding Minimum Percentage of Students Failing All Subjects

The problem provides the passing percentages of students in four subjects: Physics, Chemistry, Mathematics, and Biology. We are asked to find the minimum percentage of students who failed in all four subjects.

First, let's determine the percentage of students who failed in each individual subject:

• Percentage failed in Physics: $100\% - 70\% = 30\%$
• Percentage failed in Chemistry: $100\% - 80\% = 20\%$
• Percentage failed in Mathematics: $100\% - 75\% = 25\%$
• Percentage failed in Biology: $100\% - 85\% = 15\%$

The sum of the percentages of students who failed in individual subjects is $30\% + 20\% + 25\% + 15\% = 90\%$.

Let $N_k$ represent the percentage of students who failed in exactly $k$ subjects, for $k = 0, 1, 2, 3, 4$. The total percentage of students is 100, so the sum of percentages failing in exactly $k$ subjects must be 100:

$\qquad N_0 + N_1 + N_2 + N_3 + N_4 = 100$

where $N_0$ is the percentage of students who failed in exactly 0 subjects (i.e., passed all subjects), and $N_4$ is the percentage of students who failed in exactly 4 subjects (i.e., failed all subjects). We are looking for the minimum value of $x = N_4$.

The sum of the individual failure percentages ($90\%$) can also be expressed as the sum of percentages failing in exactly $k$ subjects, weighted by $k$ (since a student failing in $k$ subjects is counted in the failure percentage of each of those $k$ subjects):

$\qquad 1 \times N_1 + 2 \times N_2 + 3 \times N_3 + 4 \times N_4 = 90$

We have a system of two equations with five variables ($N_0, N_1, N_2, N_3, N_4$):

1. $N_0 + N_1 + N_2 + N_3 + N_4 = 100$
2. $N_1 + 2N_2 + 3N_3 + 4N_4 = 90$

We want to find the minimum possible value of $N_4 = x$, given that $N_0, N_1, N_2, N_3, N_4$ must all be non-negative percentages ($\ge 0$).

From equation (1), we can express $N_1 + N_2 + N_3$ as $100 - N_0 - N_4$. Substitute this into equation (2):

$\qquad (N_1 + N_2 + N_3) + N_2 + 2N_3 + 3N_4 = 90$

$\qquad (100 - N_0 - N_4) + N_2 + 2N_3 + 3N_4 = 90$

$\qquad 100 - N_0 + N_2 + 2N_3 + 2N_4 = 90$

Rearranging this equation to isolate $N_0$:

$\qquad N_0 = 100 - 90 + N_2 + 2N_3 + 2N_4$

$\qquad N_0 = 10 + N_2 + 2N_3 + 2N_4$

We know that $N_0$ must be non-negative ($N_0 \ge 0$). Since $N_2 \ge 0$, $N_3 \ge 0$, and $N_4 \ge 0$, the equation $N_0 = 10 + N_2 + 2N_3 + 2N_4$ implies $N_0 \ge 10$. The minimum percentage of students passing all subjects ($N_0$) is $10\%$. This minimum occurs when $N_2=0$, $N_3=0$, and $N_4=0$. In this specific case:

• If $N_0 = 10$, then $10 = 10 + N_2 + 2N_3 + 2N_4$.
• This simplifies to $N_2 + 2N_3 + 2N_4 = 0$.

Since $N_2, N_3,$ and $N_4$ are percentages and must be non-negative, the only way their weighted sum can be zero is if each term is zero. Thus, $N_2 = 0$, $N_3 = 0$, and $2N_4 = 0 \implies N_4 = 0$.

This shows that if $N_0 = 10$, then $N_2, N_3, N_4$ must all be 0. Let's check if a valid distribution exists for $N_4 = x = 0$.

If $x = N_4 = 0$, $N_2 = 0$, $N_3 = 0$:

• From $N_0 = 10 + N_2 + 2N_3 + 2N_4$, $N_0 = 10 + 0 + 2(0) + 2(0) = 10$.
• From $N_0 + N_1 + N_2 + N_3 + N_4 = 100$, $10 + N_1 + 0 + 0 + 0 = 100$, so $N_1 = 90$.
• Let's check if this distribution satisfies the second equation ($N_1 + 2N_2 + 3N_3 + 4N_4 = 90$): $90 + 2(0) + 3(0) + 4(0) = 90$. This is true.

So, a valid distribution of students exists where $N_0 = 10\%$ (passed all), $N_1 = 90\%$ (failed exactly one), $N_2 = 0\%$ (failed exactly two), $N_3 = 0\%$ (failed exactly three), and $N_4 = 0\%$ (failed exactly four). In this scenario, the percentage of students who failed in all four subjects ($x$) is $0\%$.

Since we found a valid scenario where $x=0$, and $x$ must be non-negative, the minimum value of $x$ is $0\%$.

The given options are 10, 12, 15, and None of the above. Since the minimum value we found is 0, which is not among the numerical options, the correct answer is "None of the above".

Explanation Summary

By setting up equations based on the percentages of students failing exactly $k$ subjects and the total given failure percentages, we derived a relationship for the percentage passing all subjects ($N_0$). This relationship showed that $N_0 \ge 10$. The minimum value of $N_0 = 10$ implies that the percentages of students failing exactly 2, 3, or 4 subjects ($N_2, N_3, N_4$) must all be 0. Thus, the minimum percentage failing all four subjects ($N_4$ or $x$) is 0. A valid student distribution matching this scenario (10% pass all, 90% fail exactly one, 0% fail 2, 3, or 4 subjects) exists and is consistent with the problem's data.

Revision Table: Key Concepts

Additional Information on Overlapping Percentages

When dealing with percentages of groups that can overlap (like students passing or failing multiple subjects), the sum of individual percentages often exceeds $100\%$. This excess represents the overlap. For instance, the sum of pass percentages is $70+80+75+85 = 310\%$. This clearly shows significant overlap among the students who passed different subjects.

Similarly, the sum of failure percentages is $30+20+25+15 = 90\%$. While this sum is less than $100\%$, it doesn't mean there's no overlap among students who failed. A student failing two subjects contributes to two separate failure percentages. The formulas used with $N_k$ (percentage failing exactly $k$ subjects) correctly account for these overlaps to relate the sum of individual percentages to the distribution of students based on how many subjects they failed.

The minimum percentage failing all subjects occurs when the failures are distributed among the students in a way that minimizes the number of students failing multiple subjects, especially failing 4 subjects. As shown in the solution, it is possible for 90% of the students to fail exactly one subject, and the remaining 10% pass all subjects. In this scenario, no student fails more than one subject, and importantly, no student fails all four subjects. Thus, the minimum percentage failing all four subjects is 0.