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In a classroom, if every student has either 2 or 3 pencils, and the total number

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Question: In a classroom, if every student has either 2 or 3 pencils, and the total number of pencils is 30, which of the following could be the number of students with 2 pencils?

Options:

  1. 10
  2. 5
  3. 15
  4. 20

Correct Answer: 10

Solution: 

If there are 10 students with 2 pencils, then there are 10 students with 3 pencils, totaling 30 pencils.

In a classroom, if every student has either 2 or 3 pencils, and the total number

Practice Questions

Q1
In a classroom, if every student has either 2 or 3 pencils, and the total number of pencils is 30, which of the following could be the number of students with 2 pencils?
  1. 10
  2. 5
  3. 15
  4. 20

Questions & Step-by-Step Solutions

In a classroom, if every student has either 2 or 3 pencils, and the total number of pencils is 30, which of the following could be the number of students with 2 pencils?
  • Step 1: Let the number of students with 2 pencils be represented as 'x'.
  • Step 2: Let the number of students with 3 pencils be represented as 'y'.
  • Step 3: We know that the total number of pencils is 30.
  • Step 4: The equation for the total number of pencils can be written as: 2x + 3y = 30.
  • Step 5: We also know that the total number of students is x + y.
  • Step 6: To find possible values for 'x', we can try different values for 'y' and see if 'x' remains a whole number.
  • Step 7: For example, if y = 0 (no students with 3 pencils), then 2x = 30, so x = 15.
  • Step 8: If y = 1, then 2x + 3(1) = 30, which simplifies to 2x = 27, so x = 13.5 (not a whole number).
  • Step 9: If y = 2, then 2x + 3(2) = 30, which simplifies to 2x = 24, so x = 12.
  • Step 10: Continue this process until you find all possible whole number values for 'x'.
  • System of Equations – The problem involves setting up a system of equations to represent the relationship between the number of students with 2 pencils and 3 pencils, and the total number of pencils.
  • Integer Solutions – The solution must consider that the number of students must be a whole number, leading to constraints on possible values.
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