Questions: A bookstore sells both the hardback and paperback version of a book. Last month they sold three times as many paperbacks as hardbacks, and they sold a total of 52 copies of the book. How many paperbacks did they sell? 13 14 17 36 38 None of these.

A bookstore sells both the hardback and paperback version of a book. Last month they sold three times as many paperbacks as hardbacks, and they sold a total of 52 copies of the book. How many paperbacks did they sell? 13 14 17 36 38 None of these.
Transcript text: A bookstore sells both the hardback and paperback version of a book. Last month they sold three times as many paperbacks as hardbacks, and they sold a total of 52 copies of the book. How many paperbacks did they sell? 13 14 17 36 38 None of these.
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Solution

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Solution Steps

Solution Approach
  1. Define variables for the number of hardbacks and paperbacks sold.
  2. Set up equations based on the given conditions:
    • Let \( h \) be the number of hardbacks sold.
    • Let \( p \) be the number of paperbacks sold.
    • According to the problem, \( p = 3h \) and \( h + p = 52 \).
  3. Substitute \( p = 3h \) into the second equation to solve for \( h \).
  4. Use the value of \( h \) to find \( p \).
Step 1: Define Variables

Let \( h \) be the number of hardbacks sold and \( p \) be the number of paperbacks sold.

Step 2: Set Up Equations

From the problem, we have the following relationships:

  1. \( p = 3h \) (the number of paperbacks sold is three times the number of hardbacks sold)
  2. \( h + p = 52 \) (the total number of books sold is 52)
Step 3: Substitute and Solve

Substituting \( p = 3h \) into the second equation: \[ h + 3h = 52 \] This simplifies to: \[ 4h = 52 \] Dividing both sides by 4 gives: \[ h = \frac{52}{4} = 13 \]

Step 4: Calculate Paperbacks Sold

Now, substituting \( h \) back into the equation for \( p \): \[ p = 3h = 3 \times 13 = 39 \]

Final Answer

The number of paperbacks sold is \\(\boxed{39}\\).

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