Question:
David bought greater than 10 paperback books that cost 8 each and greater than 8 hardcover books that cost 20 each. If the total cost of all the books that he bought was between 240 and $300, exclusive, how many total books could he buy?
Indicate all such answers.
[Note: Select one or more answer choices]
David bought greater than 10 paperback books that cost 8 each and greater than 8 hardcover books that cost 20 each. If the total cost of all the books that he bought was between 240 and $300, exclusive, how many total books could he buy?
Indicate all such answers.
[Note: Select one or more answer choices]
Show Hint
When solving Diophantine inequalities like this, it is most efficient to iterate on the variable with the larger coefficient (in this case, \(h\)). This narrows down the possibilities for the other variable more quickly.
Updated On: Oct 3, 2025
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
This is a problem that requires setting up and solving a system of linear inequalities with integer constraints. We need to find the possible total number of items based on constraints on the quantity of each item and the total cost.
Step 2: Key Formula or Approach:
1. Define variables: Let \(p\) be the number of paperbacks and \(h\) be the number of hardcovers. 2. Translate the given information into mathematical inequalities: - \(p>10 \implies p \ge 11\) (since \(p\) must be an integer). - \(h>8 \implies h \ge 9\) (since \(h\) must be an integer). - Total Cost: \(C = 8p + 20h\). - \(240<C<300 \implies 240<8p + 20h<300\). 3. Simplify the cost inequality and test possible integer values for \(h\) (starting from its minimum value) to find the corresponding possible integer values for \(p\). 4. For each valid pair \((p, h)\), calculate the total number of books, \(T = p+h\). 5. Collect all possible values of \(T\) and match them with the options.
Step 3: Detailed Explanation:
The cost inequality is \(240<8p + 20h<300\). Let's simplify by dividing all parts by 4, the greatest common divisor of 8 and 20. \[ 60<2p + 5h<75 \] We know \(h \ge 9\). Let's test values for \(h\) starting from 9. Case 1: h = 9 \[ 60<2p + 5(9)<75 \] \[ 60<2p + 45<75 \] Subtract 45 from all parts: \[ 15<2p<30 \] Divide by 2: \[ 7.5<p<15 \] Since \(p\) must be an integer and \(p \ge 11\), the possible values for \(p\) are \{11, 12, 13, 14\}. The possible total books \(T = p+h = p+9\) are: \(11+9=20\), \(12+9=21\), \(13+9=22\), \(14+9=23\). Case 2: h = 10 \[ 60<2p + 5(10)<75 \] \[ 60<2p + 50<75 \] Subtract 50: \[ 10<2p<25 \] Divide by 2: \[ 5<p<12.5 \] Since \(p \ge 11\), the possible values for \(p\) are \{11, 12\}. The possible total books \(T = p+h = p+10\) are: \(11+10=21\), \(12+10=22\). Case 3: h = 11 \[ 60<2p + 5(11)<75 \] \[ 60<2p + 55<75 \] Subtract 55: \[ 5<2p<20 \] Divide by 2: \[ 2.5<p<10 \] There are no possible values for \(p\), since we require \(p \ge 11\). For any \(h>11\), the lower bound for \(p\) will become even smaller, so there will be no solutions. The set of all possible values for the total number of books \(T\) is the union of the results from the valid cases: \(\{20, 21, 22, 23\} \cup \{21, 22\} = \{20, 21, 22, 23\}\).
Step 4: Final Answer:
Matching our findings with the options, the possible total number of books are 20, 21, 22, and 23. These correspond to options (D), (E), (F), and (G).
This is a problem that requires setting up and solving a system of linear inequalities with integer constraints. We need to find the possible total number of items based on constraints on the quantity of each item and the total cost.
Step 2: Key Formula or Approach:
1. Define variables: Let \(p\) be the number of paperbacks and \(h\) be the number of hardcovers. 2. Translate the given information into mathematical inequalities: - \(p>10 \implies p \ge 11\) (since \(p\) must be an integer). - \(h>8 \implies h \ge 9\) (since \(h\) must be an integer). - Total Cost: \(C = 8p + 20h\). - \(240<C<300 \implies 240<8p + 20h<300\). 3. Simplify the cost inequality and test possible integer values for \(h\) (starting from its minimum value) to find the corresponding possible integer values for \(p\). 4. For each valid pair \((p, h)\), calculate the total number of books, \(T = p+h\). 5. Collect all possible values of \(T\) and match them with the options.
Step 3: Detailed Explanation:
The cost inequality is \(240<8p + 20h<300\). Let's simplify by dividing all parts by 4, the greatest common divisor of 8 and 20. \[ 60<2p + 5h<75 \] We know \(h \ge 9\). Let's test values for \(h\) starting from 9. Case 1: h = 9 \[ 60<2p + 5(9)<75 \] \[ 60<2p + 45<75 \] Subtract 45 from all parts: \[ 15<2p<30 \] Divide by 2: \[ 7.5<p<15 \] Since \(p\) must be an integer and \(p \ge 11\), the possible values for \(p\) are \{11, 12, 13, 14\}. The possible total books \(T = p+h = p+9\) are: \(11+9=20\), \(12+9=21\), \(13+9=22\), \(14+9=23\). Case 2: h = 10 \[ 60<2p + 5(10)<75 \] \[ 60<2p + 50<75 \] Subtract 50: \[ 10<2p<25 \] Divide by 2: \[ 5<p<12.5 \] Since \(p \ge 11\), the possible values for \(p\) are \{11, 12\}. The possible total books \(T = p+h = p+10\) are: \(11+10=21\), \(12+10=22\). Case 3: h = 11 \[ 60<2p + 5(11)<75 \] \[ 60<2p + 55<75 \] Subtract 55: \[ 5<2p<20 \] Divide by 2: \[ 2.5<p<10 \] There are no possible values for \(p\), since we require \(p \ge 11\). For any \(h>11\), the lower bound for \(p\) will become even smaller, so there will be no solutions. The set of all possible values for the total number of books \(T\) is the union of the results from the valid cases: \(\{20, 21, 22, 23\} \cup \{21, 22\} = \{20, 21, 22, 23\}\).
Step 4: Final Answer:
Matching our findings with the options, the possible total number of books are 20, 21, 22, and 23. These correspond to options (D), (E), (F), and (G).
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