Word Problems and Applications

By Pythagoras, (x^2+(x+8)^2=40^2), giving (x=24). In a right triangle, forming the correct equation is most important.

If the smaller integer is (x), then (x(x+1)=306), giving (x=17). Write consecutive integers as (x) and (x+1).

Let the smaller integer be \(x\), so the larger integer is \(x+1\). Thus, \(x(x+1)=380\), giving \(x^2+x-380=0\). Factoring gives \((x-19)(x+20)=0\). Since the integers are positive, \(x=19\), so the larger integer is \(20\). Option 19 is the smaller integer, not the required larger one. Exam tip: For consecutive integers, look for factor pairs whose difference is 1.

Let the smaller number be \(x\). The next consecutive odd number is \(x+2\). Thus, \(x(x+2)=575\), so \(x^2+2x-575=0\). Factoring gives \((x-23)(x+25)=0\), yielding \(x=23\) or \(x=-25\). Since the numbers are positive, the smaller number is 23. Check: \(23\times25=575\). Exam tip: consecutive odd numbers differ by 2, not by 1.

Let the smaller consecutive even number be x. The next consecutive even number is x + 2, not x + 1, because even numbers differ by 2. The product condition becomes x(x + 2) = 728, or x² + 2x − 728 = 0. Factoring gives (x − 26)(x + 28) = 0, so x = 26 or x = −28. Since the numbers are positive, x = 26. The larger number is therefore x + 2 = 28, making option C correct. Verification is 26 × 28 = 728. Options A, B, and D do not represent the larger member of a positive consecutive-even pair whose product is 728; in particular, 24 × 26 = 624 and 28 × 30 = 840.

Taking breadth as (x), (x(x+16)=576), giving (x=18). For a rectangle, use the product of length and breadth.

If breadth is (x), then (x(x+19)=680), but this does not match the options, so checking options gives (17 \times 40=680). The correct equation would be formed when length is (x+23).

If breadth is (x), then (x(x+17)=468), giving (x=9) and length (26). Always choose the dimension asked in the question.

Taking breadth as (x), (x(x+22)=864), giving (x=18). Avoid wrong checking and verify the correct product.

If the side of the square is \(x\) metres, its area is \(x^2\). Thus, \(x^2=529\), so \(x=\sqrt{529}=23\), since \(23^2=529\). A side length cannot be negative, so the positive root is selected. Exam tip: To find the side of a square from its area, take the square root of the area.

From (x^2=625), the side is (25) m and perimeter is (4x=100) m. Keep area and perimeter formulas separate.

Take the numbers as (x) and (43-x). From (x(43-x)=460), the numbers are (20) and (23), so the smaller number is (20).

If the numbers are (x) and (47-x), then (x(47-x)=540). The solutions are (20) and (27), so the larger number is (27).

If the smaller number is (x) and larger is (x+9), then (x(x+9)=486), giving (x=18). Keep the difference between larger and smaller correct.

Let the smaller number be x. Because the difference is 11, the larger number is x + 11. Their product condition gives x(x + 11) = 672, so x² + 11x − 672 = 0. This factors as (x − 21)(x + 32) = 0. Thus x = 21 or x = −32. Under the intended positive-number interpretation, the smaller number is 21 and the larger number is 21 + 11 = 32. Hence option D is correct. Substitution verifies the answer: 21 × 32 = 672 and 32 − 21 = 11. Option A is the smaller number, not the larger one; options B and C do not form a pair with difference 11 and product 672. The governing skill is choosing variables according to the stated difference and rejecting the irrelevant negative root.

Let the smaller number be \(x\). Then the larger number is \(x+14\), so \(x(x+14)=560\). Thus, \(x^2+14x-560=0\), which factors as \((x-16)(x+30)=0\). Hence, \(x=16\) or \(x=-30\); since the numbers are positive, \(x=16\) is the valid answer, and the other number is 30. Option 20 is incorrect because \(20\times34=680\). Exam tip: In such problems, represent the smaller number by \(x\) and the larger one by \(x+14\).

Let the positive number be \(x\). The other number is \(x+18\), so \(x(x+18)=703\), giving \(x^2+18x-703=0\). Factoring, \((x-19)(x+37)=0\), and the positive condition gives \(x=19\). Hence, the larger number is \(19+18=37\). Exam tip: In product word problems, represent the smaller number by \(x\) and express the other number in terms of \(x\).

Let the larger number be \(x\). The other number is then \(x-10\), so \(x(x-10)=375\). This gives \(x^2-10x-375=0\), or \((x-25)(x+15)=0\). Since the numbers are positive, \(x=25\), and the other number is 15; indeed, \(25\times15=375\). Therefore, option C is correct. Exam tip: Translate “10 less than a number” as \(x-10\), not \(10-x\).

If the larger number is (x), then (x(x-13)=414), giving (x=23). The smaller number is (10), so the listed options reveal a mismatch.

(x^2=6x+187) gives (x^2-6x-187=0), whose positive solution is (17). If a positive number is asked, ignore the negative root.

Let the positive number be x. The phrase “the square is 384 more than 8 times the number” means x² = 8x + 384. Bringing all terms to one side gives x² − 8x − 384 = 0. Factoring, (x − 24)(x + 16) = 0, so x = 24 or x = −16. The positive-number condition selects x = 24, hence option C is correct. Verification is 24² = 576, while 8 × 24 + 384 = 192 + 384 = 576. For comparison, 18² is not 8 × 18 + 384, and the other listed values also fail the equation. The governing concept is converting a comparative sentence into a quadratic equation, solving it, and applying the restriction that the required number must be positive.

The equation is (x^2-12x=589), or (x^2-12x-589=0), giving (x=31). In a difference statement, keep the subtraction order correct.

Let the number be x. Then x² + x = 342, so x² + x − 342 = 0. Factoring gives (x − 18)(x + 19) = 0, so x is 18 or −19. The positive number is 18; indeed, 18 + 18² = 18 + 324 = 342.

(x^2+x=506) gives (x^2+x-506=0), whose positive solution is (22). While factoring, check factors of (506).

Let the original number be \(x\). Then \((x+9)^2=676\). Taking the positive square root gives \(x+9=26\), so \(x=26-9=17\). Option 26 is the resulting number, not the original number. In exams, remember that a square equation generally has positive and negative roots, but here the question specifically asks for the positive root.