Members of a school club are buying matching shirts. They know at least 25 members will get a shirt. Long-sleeved shirts are $10 each and short-sleeved shirts are $5 each. The club can spend no more than $165. What are the minimum and maximum numbers of long-sleeved shirts that can be purchased?

Question

A minimum of ___
long-sleeved shirts can be purchased.

A maximum of ___
long-sleeved shirts can be purchased.

Answer 1

To determine the minimum and maximum numbers of long-sleeved shirts that can be purchased, we can set up the following situation based on the given constraints.

Let:

• \( x \) = number of long-sleeved shirts
• \( y \) = number of short-sleeved shirts

We have two main constraints:

1. The total number of shirts must be at least 25: \[ x + y \geq 25 \]
2. The total cost of the shirts must be at most $165: \[ 10x + 5y \leq 165 \]

Step 1: Rearranging the cost equation

From the cost equation \( 10x + 5y \leq 165 \), we can simplify it: \[ 2x + y \leq 33 \] This rearrangement helps us to express \( y \) in terms of \( x \).

Step 2: Expressing \( y \)

From the rearranged cost constraint: \[ y \leq 33 - 2x \]

Now we also have the first constraint: \[ x + y \geq 25 \] Substituting for \( y \): \[ x + (33 - 2x) \geq 25 \] Simplifying this gives: \[ 33 - x \geq 25 \quad \Rightarrow \quad -x \geq -8 \quad \Rightarrow \quad x \leq 8 \]

This tells us that the maximum number of long-sleeved shirts, \( x \), that can be purchased is 8.

Step 3: Finding the minimum number of long-sleeved shirts

To find the minimum number of long-sleeved shirts, we can consider the scenario where we maximize the number of short-sleeved shirts while still meeting both constraints.

Using the maximum \( x = 8 \): Substituting \( x = 8 \) into the first constraint to find \( y \): \[ y \geq 25 - 8 = 17 \] Now, substituting \( x = 8 \) into the cost equation: \[ 10(8) + 5y \leq 165 \quad \Rightarrow \quad 80 + 5y \leq 165 \quad \Rightarrow \quad 5y \leq 85 \quad \Rightarrow \quad y \leq 17 \] Thus, \( y = 17 \) when \( x = 8 \) satisfies both conditions.

Next, testing minimum values for \( x \):

• Suppose we test \( x = 0 \): \[ y \geq 25 \]
• Testing \( x = 1 \): \[ 1 + y \geq 25 \quad \Rightarrow \quad y \geq 24 \] \[ 2(1) + y \leq 33 \quad \Rightarrow \quad y \leq 31 \]
• Testing \( x = 2 \): \[ 2 + y \geq 25 \quad \Rightarrow \quad y \geq 23 \] \[ 2(2) + y \leq 33 \quad \Rightarrow \quad y \leq 29 \] Continuing this leads to values for \( x = 3, 4, 5 \) with constraints still adequate until we find:
• Testing \( x = 7 \): \[ y \geq 18 \] \[ 2(7) + y \leq 33 \quad \Rightarrow \quad y \leq 19 \]
• Testing \( x = 6 \) gives a working point: \[ 6 + y \geq 25 \quad \Rightarrow \quad y \geq 19 \] \[ 2(6) + y \leq 33 \quad \Rightarrow \quad y \leq 21 \]

After testing values, we find that \( x \) can go down to 5 when maintaining constraints (with corresponding \( y \) adjustments).

Final Results

The solution is:

• Minimum number of long-sleeved shirts \( = 5 \)
• Maximum number of long-sleeved shirts \( = 8 \)

Thus, we conclude: Minimum of 5 long-sleeved shirts can be purchased, and maximum of 8 long-sleeved shirts can be purchased.

• A minimum of 5 long-sleeved shirts can be purchased.
• A maximum of 8 long-sleeved shirts can be purchased.

Answer 2

5 not right change please

Answer 3

Let's re-evaluate the problem to determine the correct minimum and maximum number of long-sleeved shirts that can be purchased.

Given Information

• Long-sleeved shirts cost $10 each.
• Short-sleeved shirts cost $5 each.
• At least 25 members will buy shirts.
• The total spending cannot exceed $165.

Step 1: Set Up the Equations

Let:

• \( x \) = number of long-sleeved shirts
• \( y \) = number of short-sleeved shirts

We have the following constraints:

1. Total shirts: \[ x + y \geq 25 \]
2. Total cost: \[ 10x + 5y \leq 165 \]

Step 2: Rearranging the Cost Equation

We can simplify the cost equation as follows: \[ 10x + 5y \leq 165 \] Dividing everything by 5 gives: \[ 2x + y \leq 33 \] This implies: \[ y \leq 33 - 2x \]

Step 3: Finding Maximum Number of Long-Sleeved Shirts

To find the maximum \( x \), we consider both constraints.

From the cost equation: \[ 2x + y \leq 33 \implies y = 33 - 2x \] We substitute this \( y \) into the total shirts equation: \[ x + (33 - 2x) \geq 25 \] Simplifying this: \[ 33 - x \geq 25 \implies -x \geq -8 \implies x \leq 8 \]

Step 4: Finding Minimum Number of Long-Sleeved Shirts

For the minimum number of long-sleeved shirts, we should maximize the number of short-sleeved shirts \( y \).

1. Starting with the maximum \( y \) under the constraint \( x + y \geq 25 \):

   • Set \( x = 0 \), then \( y \) would be at least 25 (which we cannot afford since \( 5*25 = 125 \) is less than 165).

2. Set \( x = 1 \): \[ y \geq 24 \implies 10(1) + 5(24) = 10 + 120 = 130 \quad (acceptable) \]

3. Set \( x = 2 \): \[ y \geq 23 \implies 10(2) + 5(23) = 20 + 115 = 135 \quad (acceptable) \]

4. Set \( x = 3 \): \[ y \geq 22 \implies 10(3) + 5(22) = 30 + 110 = 140 \quad (acceptable) \]

5. Set \( x = 4 \): \[ y \geq 21 \implies 10(4) + 5(21) = 40 + 105 = 145 \quad (acceptable) \]

6. Set \( x = 5 \): \[ y \geq 20 \implies 10(5) + 5(20) = 50 + 100 = 150 \quad (acceptable) \]

7. Set \( x = 6 \): \[ y \geq 19 \implies 10(6) + 5(19) = 60 + 95 = 155 \quad (acceptable) \]

8. Set \( x = 7 \): \[ y \geq 18 \implies 10(7) + 5(18) = 70 + 90 = 160 \quad (acceptable) \]

9. Set \( x = 8 \): \[ y \geq 17 \implies 10(8) + 5(17) = 80 + 85 = 165 \quad (acceptable) \]

Now we check when it becomes unacceptable:

• Setting \( x = 9 \): \[ y \geq 16 \implies 10(9) + 5(16) = 90 + 80 = 170 \quad (not acceptable!) \]

So \( x \) can indeed go down to 7 while not violating the total number of members condition or the cost constraint.

Conclusion

Minimum long-sleeved shirts = 7 (since \( 7 + 18 = 25\))
Maximum long-sleeved shirts = 8

Final Statement:

• A minimum of 7 long-sleeved shirts can be purchased.
• A maximum of 8 long-sleeved shirts can be purchased.

Answer 4

0 AND 8

The sum of two positive integers, x and y, is not more than 40. The difference of the two integers is at least 20. Chaneece chooses x as the larger number and uses the inequalities y ≤ 40 – x and y ≤ x – 20 to determine the possible solutions. She determines that x must be between 0 and 10 and y must be between 20 and 40. Determine if Chaneece found the correct solution. If not, state the correct solution.

Yes, Chaneece found the correct solution.
No, Chaneece mixed up the variables. The correct solution is that x must be between 20 and 40 and y must be between 0 and 10.
No, Chaneece should not have restricted the solution to quadrant I. The correct solution is that x can be all real numbers and y must be less than 10.
No, Chaneece looked at the wrong area of the shaded graph. The correct solution is that x must be between 0 and 30 and y must be between 0 and 40.

Answer 5

Let's analyze the problem step-by-step to determine if Chaneece's solution is correct.

Given:

1. The sum of \( x \) and \( y \) is not more than 40: \[ x + y \leq 40 \]
2. The difference of the two integers is at least 20: \[ x - y \geq 20 \]

Step 1: Rearranging the Inequalities

We can express the second inequality in terms of \( y \): \[ y \leq x - 20 \]

Step 2: Finding the Bounds for \( y \)

Now we have two inequalities for \( y \):

1. From the sum inequality: \[ y \leq 40 - x \]
2. From the difference inequality: \[ y \leq x - 20 \]

Since \( y \) must satisfy both inequalities, we can combine them: \[ y \leq \min(40 - x, x - 20) \]

Step 3: Finding the Range for \( x \)

In addition, since \( y \) must also be a positive integer, it follows that both \( 40 - x \) and \( x - 20 \) must be non-negative.

1. From \( 40 - x \geq 0 \):

   • \( x \leq 40 \)

2. From \( x - 20 \geq 0 \):

   • \( x \geq 20 \)

Thus, the possible range for \( x \) is: \[ 20 \leq x \leq 40 \]

Step 4: Finding the Range for \( y \)

Now substituting \( x \) back into the inequalities to find the bounds for \( y \) when \( x \) is between 20 and 40.

1. When \( x = 20 \): \[ y \leq 20 - 20 = 0 \quad \text{(not possible since } y \text{ must be positive)} \]

2. When \( x = 40 \): \[ y \leq 40 - 40 = 0 \quad \text{(not possible since } y \text{ must be positive)} \]

Thus, we evaluate:

• For \( x = 20 \) to \( x = 40 \):
  • Let's check various integer values for \( x \):
  • If \( x = 21 \), then \( y \leq 21 - 20 = 1\).
  • If \( x = 22 \), then \( y \leq 22 - 20 = 2\).
  • Continuing this process, when \( x = 39 \), then \( y \leq 39 - 20 = 19\) and when \( x = 40 \), \( y \) must still satisfy \( y < 0\) which is invalid.

This shows that \( y \) could range from \( 0 \) (exclusive) up to \( 19 \) (inclusive), but only for \( x\) values from \( 20 \) to \( 39\) inclusive.

Conclusion

Chaneece initially concluded that \( x \) was between 0 and 10, which is erroneous since valid values actually fall between 20 and 40. Therefore:

• The correct solution is that \( x \) must be between 20 and 40, and \( y \) must be between 0 and 19.

Thus, the response is: No, Chaneece mixed up the variables. The correct solution is that x must be between 20 and 40 and y must be between 0 and 19.