Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row, there would be 3 rows more. Find the total number of students in the class.

• A) 55
• B) 30
• C) 115
• D) 60

To solve the problem of how many students are in the class, we can set up a couple of equations based on the information provided. Let's break it down step by step.

Understanding the Problem

We have a certain number of students, which we'll denote as S, and a certain number of rows, which we'll denote as R. The number of students per row can be expressed as P. Thus, we can say:

• S = R × P (Total students equals rows times students per row)

Setting Up the Equations

From the problem, we have two scenarios:

• If one student is added to each row, the number of rows decreases by 2.
• If one student is removed from each row, the number of rows increases by 3.

We can express these scenarios mathematically:

• When one student is added: S = (R - 2) × (P + 1)
• When one student is removed: S = (R + 3) × (P - 1)

Formulating the Equations

Now, we can set up our equations based on the above expressions:

• From the first scenario: S = (R - 2)(P + 1)
  Expanding this gives us:
  S = RP + R - 2P - 2
• From the second scenario: S = (R + 3)(P - 1)
  Expanding this gives us:
  S = RP - R + 3P - 3

Equating the Two Expressions

Since both expressions equal S, we can set them equal to each other:

RP + R - 2P - 2 = RP - R + 3P - 3

Simplifying the Equation

Now, let's simplify this equation:

• Cancel RP from both sides:
• R - 2P - 2 = -R + 3P - 3

Rearranging gives us:

• 2R = 5P - 1

Finding the Total Number of Students

Now we have a relationship between R and P. We can express R in terms of P:

R = (5P - 1)/2

Now, substituting this back into the equation for S:

S = R × P = ((5P - 1)/2) × P

This simplifies to:

S = (5P^2 - P)/2

Testing Possible Values

Now, let's test the answer choices provided to find a suitable integer value for S:

• A) 55: 2S = 110 → 5P^2 - P = 110 → 5P^2 - P - 110 = 0 (not a perfect square)
• B) 30: 2S = 60 → 5P^2 - P = 60 → 5P^2 - P - 60 = 0 (not a perfect square)
• C) 115: 2S = 230 → 5P^2 - P = 230 → 5P^2 - P - 230 = 0 (not a perfect square)
• D) 60: 2S = 120 → 5P^2 - P = 120 → 5P^2 - P - 120 = 0

Solving for P in option D:

P = 5 gives us R = (5*5 - 1)/2 = 12 and S = 60. This fits all conditions.

Final Answer

The total number of students in the class is 60.