Let the length of the shortest piece be x cm. The second piece is x + 4 cm, and the third piece is 3x cm. The sum of the lengths cannot exceed the total wire length of 84 cm, giving x + (x + 4) + 3x ≤ 84.

Let the length of the shortest piece be x cm. The second piece is x + 4 cm, and the third piece is 3x cm. The sum of the lengths cannot exceed the total wire length of 84 cm, giving x + (x + 4) + 3x ≤ 84.

From the total length constraint, we obtain 5x ≤ 80, which simplifies to x ≤ 16. Additionally, the third piece is at least 6 cm longer than the second piece, meaning 3x ≥ (x + 4) + 6, which gives 2x ≥ 10, so x ≥ 5. Combining these inequalities yields 5 ≤ x ≤ 16.

From the total length constraint, we obtain 5x ≤ 80, which simplifies to x ≤ 16. Additionally, the third piece is at least 6 cm longer than the second piece, meaning 3x ≥ (x + 4) + 6, which gives 2x ≥ 10, so x ≥ 5. Combining these inequalities yields 5 ≤ x ≤ 16.

Let the length of the shortest piece in Problem 2 be y cm. The second piece is y + 5 cm, and the third piece is 2y cm. The sum of the three pieces cannot exceed the length of the wooden plank, 120 cm, giving y + (y + 5) + 2y ≤ 120.

Let the length of the shortest piece in Problem 2 be y cm. The second piece is y + 5 cm, and the third piece is 2y cm. The sum of the three pieces cannot exceed the length of the wooden plank, 120 cm, giving y + (y + 5) + 2y ≤ 120.

From the plank length constraint, 4y ≤ 115 yields y ≤ 28.75 cm (or y ≤ (115)/(4) cm). From the condition that the third piece is at least 8 cm longer than the second piece, we have 2y ≥ (y + 5) + 8, which simplifies to y ≥ 13. Combining these inequalities gives 13 ≤ y ≤ 28.75.

From the plank length constraint, 4y ≤ 115 yields y ≤ 28.75 cm (or y ≤ (115)/(4) cm). From the condition that the third piece is at least 8 cm longer than the second piece, we have 2y ≥ (y + 5) + 8, which simplifies to y ≥ 13. Combining these inequalities gives 13 ≤ y ≤ 28.75.

$\text{\underline{Problem 1 :}}$ A wire of length $84\text{ cm}$ is to be cut into three pieces. The second piece is to be $4\text{ cm}$ longer than the shortest piece, and the third piece is to be three times as long as the shortest piece. If the third piece is at least $6\text{ cm}$ longer than the second piece, what are the possible lengths of the shortest piece? $\text{\underline{Problem 2 :}}$ A man has a $120\text{ cm}$ wooden plank and wants to cut it into three pieces. The second piece is to be $5\text{ cm}$ longer than the shortest piece, and the third piece is to be twice as long as the shortest piece. If the third piece is at least $8\text{ cm}$ longer than the second piece, what are the possible lengths of the shortest piece?

Answer: Problem 1: The possible length of the shortest piece is between $5\text{ cm}$ and $16\text{ cm}$ (inclusive), i.e., $5\text{ cm} \le x \le 16\text{ cm}$ . Problem 2: The possible length of the shortest piece is between $13\text{ cm}$ and $28.75\text{ cm}$ (inclusive), i.e., $13\text{ cm} \le y \le 28.75\text{ cm}$ .

Step-by-step solution

Step 1: Set up conditions for Problem 1

Let the length of the shortest piece be $x\text{ cm}$ . The second piece is $x + 4\text{ cm}$ , and the third piece is $3x\text{ cm}$ . The sum of the lengths cannot exceed the total wire length of $84\text{ cm}$ , giving $x + (x + 4) + 3x \le 84$ .

Step 2: Solve inequalities for Problem 1

From the total length constraint, we obtain $5x \le 80$ , which simplifies to $x \le 16$ . Additionally, the third piece is at least $6\text{ cm}$ longer than the second piece, meaning $3x \ge (x + 4) + 6$ , which gives $2x \ge 10$ , so $x \ge 5$ . Combining these inequalities yields $5 \le x \le 16$ .

Step 3: Set up conditions for Problem 2

Let the length of the shortest piece in Problem 2 be $y\text{ cm}$ . The second piece is $y + 5\text{ cm}$ , and the third piece is $2y\text{ cm}$ . The sum of the three pieces cannot exceed the length of the wooden plank, $120\text{ cm}$ , giving $y + (y + 5) + 2y \le 120$ .

Step 4: Solve inequalities for Problem 2

From the plank length constraint, $4y \le 115$ yields $y \le 28.75\text{ cm}$ (or $y \le \frac{115}{4}\text{ cm}$ ). From the condition that the third piece is at least $8\text{ cm}$ longer than the second piece, we have $2y \ge (y + 5) + 8$ , which simplifies to $y \ge 13$ . Combining these inequalities gives $13 \le y \le 28.75$ .

Solve your own maths question free →