Word Problems and Applications

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.

Consecutive even numbers are (x) and (x+2). From (x(x+2)=728), (x=26), so the larger number is (28).

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.

If the smaller number is (x), the larger is (x+11). From (x(x+11)=672), (x=21), so the larger number is (32).

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.

(x^2=8x+384) gives (x^2-8x-384=0), giving (x=24). Convert the sentence into an equation in the correct order.