Word Problems and Applications

Let the original number be \(x\). Then \((x-4)(x+5)=360\), which gives \(x^2+x-380=0\). Factoring, \((x-19)(x+20)=0\), so \(x=19\) or \(x=-20\). Since the question asks for the positive number, the answer is \(19\). In such problems, write “4 less” as \(x-4\) and “5 more” as \(x+5\).

The governing concept is the square-area formula A=s². Let the original side be s units. From s²=196 and the fact that a length is positive, s=14 units. Increasing the side by 4 units gives a new side of 14+4=18 units. Therefore the new area is 18²=324 square units, so option A is correct. The answer can also be checked by expansion: (14+4)²=14²+2(14)(4)+4²=196+112+16=324. Option B is 16² and would use an increase of only 2 units; option C uses a side of 20, and option D uses a side of 17. Neither matches the stated four-unit increase.

Let the units digit be x. Then the tens digit is x+3. The product condition becomes x(x+3)=18, so x²+3x−18=0. Factoring gives (x+6)(x−3)=0, leading to x=−6 or x=3. A digit cannot be negative, so the units digit is 3 and the tens digit is 6. The number is therefore 63, making option A correct. Direct checking gives 6−3=3 and 6×3=18. Option B reverses the digits and has the tens digit smaller, not larger. In option C the product is 18 but the digit difference is not 3, while option D has a difference of 3 but product 54, so neither satisfies both conditions.

When a wire is bent into a square, the wire length becomes the perimeter of the square. Let the side be s cm. The area condition gives s² = 1296, so s = 36 because a length must be positive. The perimeter, and hence the length of the wire, is 4s = 4×36 = 144 cm. Therefore option C is correct. Option A could result from an incorrect side length of 30 cm, while options B and D do not equal four times the valid side. The negative square-root value −36 is rejected because geometric length cannot be negative. A direct check also confirms that a square of side 36 cm has area 36² = 1296 square cm and perimeter 144 cm.