In an army selection process, the ratio of selected candidates to unselected candidates is 3 : 1. If 80 fewer candidates had applied and 40 fewer had been selected, the new ratio of selected to unselected would be 4 : 1. How many candidates originally applied?

Introduction / Context:
This is a word problem involving ratios and algebraic reasoning. It describes an army selection process where candidates are either selected or unselected. We are given an initial ratio of selected to unselected candidates and a hypothetical scenario that changes the total number of applicants and selected candidates, resulting in a new ratio. The goal is to determine how many candidates originally applied.

Given Data / Assumptions:

• Initially, the ratio of selected candidates to unselected candidates is 3 : 1.• Let the original total number of candidates be N.• From this ratio, selected candidates initially are 3N / 4 and unselected candidates are N / 4.• If 80 fewer candidates had applied, the total would have been N − 80.• If 40 fewer candidates had been selected, the number of selected candidates would be 3N / 4 − 40.• Under these new conditions, the ratio of selected to unselected candidates becomes 4 : 1.

Concept / Approach:
This problem is solved through algebra. We express the selected and unselected candidates in terms of N using the given ratios. Then, we apply the changes described and set up a new ratio equation. Ratios of the form selected : unselected can be written as fractions, and we equate that fraction to 4 / 1 in the second scenario. Solving the resulting equation for N gives the total number of candidates who originally applied. Careful handling of the selection and unselection counts is crucial.

Step-by-Step Solution:
Step 1: Initially, selected : unselected = 3 : 1.Step 2: Let total candidates be N. Then selected = 3N / 4 and unselected = N / 4.Step 3: In the hypothetical scenario, 80 fewer apply, so total becomes N − 80.Step 4: Also, 40 fewer are selected, so new selected = 3N / 4 − 40.Step 5: New unselected = total − new selected = (N − 80) − (3N / 4 − 40).Step 6: Simplify new unselected: N − 80 − 3N / 4 + 40 = N / 4 − 40.Step 7: Given that the new ratio of selected to unselected is 4 : 1, we write (3N / 4 − 40) / (N / 4 − 40) = 4.Step 8: Cross multiply: 3N / 4 − 40 = 4 * (N / 4 − 40).Step 9: Expand the right side: 4 * (N / 4) − 4 * 40 = N − 160.Step 10: So the equation is 3N / 4 − 40 = N − 160.Step 11: Bring all terms involving N to one side: 3N / 4 − N = −160 + 40.Step 12: 3N / 4 − N = 3N / 4 − 4N / 4 = −N / 4.Step 13: The right side is −120.Step 14: So −N / 4 = −120, which gives N / 4 = 120.Step 15: Therefore N = 120 * 4 = 480.

Verification / Alternative check:
With N = 480, initial selected = 3 * 480 / 4 = 360 and initial unselected = 480 / 4 = 120, giving ratio 360 : 120 = 3 : 1, which matches the first condition. In the altered scenario, total candidates = 480 − 80 = 400. New selected = 360 − 40 = 320. New unselected = 400 − 320 = 80. The new ratio is 320 : 80 = 4 : 1, which exactly matches the condition given in the problem. This confirms that N = 480 is correct.

Why Other Options Are Wrong:
• 960: This would double all the values and would not satisfy the second condition when the hypothetical changes are applied.• 240: This would give non-integer or inconsistent values for selected and unselected counts under the given ratios.• 1,440: This is too large and does not maintain both ratio conditions when checked.

Common Pitfalls:
A common mistake is to treat 3 : 1 as meaning 3 selected and 1 unselected without relating it to the total N properly. Others incorrectly compute new unselected candidates by subtracting only from the original unselected group rather than from the new total after changing both applied and selected counts. Algebraic manipulation errors, such as misplacing terms when solving the equation, are also frequent. Careful step-by-step reasoning helps avoid such errors.

Final Answer:
The number of candidates who originally applied for the process is 480.