Solving the Examination Pass-Fail Ratio Problem

The problem involves ratios of students who passed and failed an examination under two different scenarios. We are given the initial ratio and a changed ratio based on specific conditions, and we need to find the difference between the initially passed and failed students.

Understanding the Initial Student Ratio

Initially, the ratio of passed students to failed students was 25:4. Let's denote the number of students who initially passed as $P$ and the number of students who initially failed as $F$.

We can write this ratio as:

$\frac{P}{F} = \frac{25}{4}$

This implies that $P$ and $F$ are multiples of 25 and 4, respectively, for some common factor. Let this common factor be $k$. Since the number of students must be a whole number, $k$ must be a positive integer.

• Initial Passed Students ($P$) $= 25k$
• Initial Failed Students ($F$) $= 4k$

The total number of students initially was $P + F = 25k + 4k = 29k$.

Analyzing the Changes and the New Ratio

The problem describes a hypothetical scenario:

"If one more student had appeared and passed and the number of failed students was 3 less than earlier..."

Let's break this down to find the new number of passed and failed students:

1. "one more student had appeared": This suggests the total number of students considered in this new scenario is 1 more than the initial total, i.e., $(P+F)+1$.
2. "...and passed": This additional student is specifically counted among the passed students.
3. "...and the number of failed students was 3 less than earlier": The number of failed students in this new scenario is $F-3$.

Let the new number of passed students be $P_{new}$ and the new number of failed students be $F_{new}$.

• New Failed Students ($F_{new}$) $= F - 3 = 4k - 3$.
• The total students in this new scenario are $(P+F)+1$.
• The new passed students must be the new total minus the new failed students: $P_{new} = (P+F+1) - (F-3)$ $P_{new} = P + F + 1 - F + 3$ $P_{new} = P + 4$ Substituting $P=25k$: $P_{new} = 25k + 4$.

In this new scenario, the ratio of passed students to failed students is 22:3.

$\frac{P_{new}}{F_{new}} = \frac{25k+4}{4k-3} = \frac{22}{3}$

Setting Up and Solving the Equation

Now we have an equation with one variable, $k$. We can solve for $k$ by cross-multiplying:

$3 \times (25k+4) = 22 \times (4k-3)$

Distribute on both sides:

$75k + 12 = 88k - 66$

Gather the $k$ terms on one side and the constant terms on the other side:

$12 + 66 = 88k - 75k$

$78 = 13k$

Divide by 13 to find $k$:

$k = \frac{78}{13}$

$k = 6$

Calculating the Initial Number of Students

Now that we have the value of $k$, we can find the initial number of passed and failed students:

• Initial Passed Students ($P$) $= 25k = 25 \times 6 = 150$
• Initial Failed Students ($F$) $= 4k = 4 \times 6 = 24$

The total number of students initially was $150 + 24 = 174$.

Finding the Difference

The question asks for the difference between the number of students who initially passed the examination and the number of students who initially failed the examination.

Difference $= P - F = 150 - 24$

Difference $= 126$

Verification

Let's check if these numbers satisfy the conditions of the hypothetical scenario:

• Initial Passed: 150, Failed: 24. Initial Ratio: $150:24$. Dividing both by 6 gives $25:4$. Correct.
• Hypothetical total students: Initial Total + 1 = $174 + 1 = 175$.
• Hypothetical failed students: Initial Failed - 3 = $24 - 3 = 21$.
• Hypothetical passed students: Hypothetical Total - Hypothetical Failed = $175 - 21 = 154$.
• Hypothetical Ratio: Passed : Failed = $154 : 21$. Dividing both by 7 gives $22:3$. Correct.

The calculated initial numbers are consistent with both the initial and the new ratio conditions.

Revision Table: Key Concepts

Additional Information: Ratio Problems

Ratio problems often involve setting up proportions and solving for an unknown quantity. Here are some common types and approaches:

• Direct Ratios: Given $a:b = c:d$, this means $\frac{a}{b} = \frac{c}{d}$. This is solved by cross-multiplication $ad=bc$.
• Changes in Ratios: When quantities change and form a new ratio, represent the initial quantities using a variable (like $k$ in our problem), apply the changes to these expressions, and set up a new equation with the new ratio.
• Total Quantity: The sum of the parts in a ratio often represents the total quantity. If $a:b = m:n$, the total is proportional to $m+n$. For example, if a mixture has components A and B in the ratio 3:2, the total is in parts $3+2=5$. If the total volume is 100 units, then 1 part is $100/5=20$, so A is $3 \times 20 = 60$ and B is $2 \times 20 = 40$.
• Consistency Check: Always check if your final numbers make sense in the context of the original problem statement and conditions.

Solving ratio problems requires careful reading, setting up correct mathematical relationships, and applying algebraic techniques.