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In a college, the total number of students who opted for Biology or Chemistry are 150. Later, 12 students who opted for Biology changed their preference and opted for Chemistry. Now, the number of students in Biology are half the number of students in Chemistry. How many students are in Chemistry now?

This question was previously asked in
SSC Stenographer 2020-21 Previous Year Paper (15-Nov-2021) (Shift 2)
The correct answer is

100

Understanding the College Subject Preference Problem

The problem describes a scenario in a college involving students who have opted for either Biology or Chemistry. We are given the total number of students initially and how some students changed their subject preference. We need to find the final number of students in Chemistry based on a given ratio between the final counts of Biology and Chemistry students.

Initial State of Student Preferences

Let's denote the initial number of students who opted for Biology as \(B_{initial}\) and the initial number of students who opted for Chemistry as \(C_{initial}\).

According to the problem statement, the total number of students who opted for Biology or Chemistry initially is 150.

This can be represented by the equation:

\begin{equation*} B_{initial} + C_{initial} = 150 \end{equation*}

Analyzing the Change in Preferences

The problem states that 12 students who initially opted for Biology changed their preference and opted for Chemistry instead. This change affects the number of students in both subjects.

  • The number of students in Biology decreases by 12.
  • The number of students in Chemistry increases by 12.

Let's denote the number of students in Biology after the change as \(B_{final}\) and the number of students in Chemistry after the change as \(C_{final}\).

The final counts can be expressed as:

  • \(B_{final} = B_{initial} - 12\)
  • \(C_{final} = C_{initial} + 12\)

It's important to note that the total number of students in Biology or Chemistry remains the same after the change, as students only moved between these two subjects: \(B_{final} + C_{final} = (B_{initial} - 12) + (C_{initial} + 12) = B_{initial} + C_{initial} = 150\).

Establishing the Relationship in the Final State

After the students changed their preferences, a new relationship between the number of students in Biology and Chemistry is given:

The number of students in Biology are half the number of students in Chemistry.

This can be written as:

\begin{equation*} B_{final} = \frac{1}{2} C_{final} \end{equation*}

Solving for the Final Number of Chemistry Students

We now have a system of equations. We want to find the value of \(C_{final}\).

We know:

  1. \(B_{initial} + C_{initial} = 150\)
  2. \(B_{final} = B_{initial} - 12\)
  3. \(C_{final} = C_{initial} + 12\)
  4. \(B_{final} = \frac{1}{2} C_{final}\)

Let's use equations (2), (3), and (4) to express the initial counts in terms of the final counts, or substitute final counts into the ratio equation.

From equation (2), \(B_{initial} = B_{final} + 12\).

From equation (3), \(C_{initial} = C_{final} - 12\).

Substitute these into equation (1):

\begin{equation*} (B_{final} + 12) + (C_{final} - 12) = 150 \end{equation*}

\begin{equation*} B_{final} + C_{final} = 150 \end{equation*}

This confirms our earlier check that the total number of students remains 150.

Now substitute equation (4) into this combined equation (\(B_{final} + C_{final} = 150\)):

\begin{equation*} \frac{1}{2} C_{final} + C_{final} = 150 \end{equation*}

Combine the terms involving \(C_{final}\):

\begin{equation*} \left(\frac{1}{2} + 1\right) C_{final} = 150 \end{equation*}

\begin{equation*} \frac{3}{2} C_{final} = 150 \end{equation*}

Now, solve for \(C_{final}\) by multiplying both sides by \(\frac{2}{3}\):

\begin{equation*} C_{final} = 150 \times \frac{2}{3} \end{equation*}

\begin{equation*} C_{final} = \frac{300}{3} \end{equation*}

\begin{equation*} C_{final} = 100 \end{equation*}

So, the number of students in Chemistry now is 100.

Verification of the Solution

Let's check if this value fits all conditions. If \(C_{final} = 100\), then from \(B_{final} = \frac{1}{2} C_{final}\), we get \(B_{final} = \frac{1}{2} \times 100 = 50\).

Using the relationships with initial counts:

  • \(C_{initial} = C_{final} - 12 = 100 - 12 = 88\)
  • \(B_{initial} = B_{final} + 12 = 50 + 12 = 62\)

Let's check the initial total: \(B_{initial} + C_{initial} = 62 + 88 = 150\). This matches the given information.

The final counts are \(B_{final} = 50\) and \(C_{final} = 100\). The condition that Biology students are half of Chemistry students (\(50 = 100/2\)) is satisfied.

The number of students in Chemistry now is indeed 100.

Final Answer Summary

The step-by-step calculation shows that the number of students in Chemistry after 12 Biology students changed their preference is 100.

Here is a summary table:

Initial Count Change Final Count
Biology Students \(B_{initial} = 62\) \(-12\) \(B_{final} = 50\)
Chemistry Students \(C_{initial} = 88\) \(+12\) \(C_{final} = 100\)
Total Students 150 150

Revision Table: Key Concepts

Concept Description How it Applies Here
Algebraic Equations Representing unknown quantities and relationships using variables and equations. Used to set up initial total, preference changes, and the final ratio condition.
Substitution Method Solving a system of equations by expressing one variable in terms of another and substituting it into another equation. Used to solve for the initial/final number of students.
Ratio and Proportion Comparing quantities; specifically, the final relationship between Biology and Chemistry students. The condition \(B_{final} = \frac{1}{2} C_{final}\) is a ratio.

Additional Information: Solving Word Problems

Solving word problems like this involves translating the information given in text into mathematical equations. Here are some general tips:

  • Read Carefully: Understand what is given and what needs to be found. Identify the quantities and their relationships.
  • Define Variables: Assign letters (variables) to the unknown quantities.
  • Formulate Equations: Write down the relationships between the variables as equations based on the problem description.
  • Solve the System: Use algebraic methods (like substitution or elimination) to solve the equations for the unknown variables.
  • Check Your Answer: Plug the values you found back into the original equations or the problem statement to ensure they make sense and satisfy all conditions.

In this problem, clearly defining the 'initial' and 'final' states and tracking the changes was crucial. The total number of students provides one key equation, and the final ratio provides another. Combining these allows us to solve for the unknown numbers.

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