Word Problems and Applications

((x+4)(x-4)=609) gives (x^2-16=609) and (x=25). The difference of squares idea helps in such questions.

The original area is \(x^2\), while the new area is \((x+5)^2\). Thus, \((x+5)^2-x^2=425\), which simplifies to \(10x+25=425\). Hence, \(10x=400\) and \(x=40\) metres. Therefore, option C is correct. Exam tip: For such problems, equate the difference of the two areas before solving the resulting linear equation.

The reduced side is \((x-3)\) cm, so \((x-3)^2=225\). This gives \(x-3=\pm15\). Since a side length must be positive and the reduced side is 15 cm, \(x-3=15\), hence \(x=18\) cm. The negative possibility would give \(x=-12\) cm, which is not physically valid. Exam tip: reject any root that gives a negative length in a geometric word problem.

If breadth is (x) and length is (x+10), the increase is ((x+4)(x+14)-x(x+10)=224). This gives (x=20).

Let the original breadth be \(x\) m. Then the original length is \(x+5\) m. After the increase, the length becomes \(x+8\) m and the breadth becomes \(x+2\) m. Hence, \((x+8)(x+2)=462\), which gives \(x^2+10x-446=0\). Using the quadratic formula, \(x=-5\pm\sqrt{471}\). Since a breadth must be positive, \(x=-5+\sqrt{471}\approx16.70\) m. The distractor 18 m is incorrect because it would give a new area of \(26\times20=520\) square metres. Exam tip: discard the negative root and verify the remaining root in the original area equation.

If the side of the square is (x), then (x^2=625), so (x=25). The wire length is the perimeter (4x=100) cm.

If breadth is (x), then (x(x+9)=700), giving (x=25) and length (34). The wire length is (2(25+34)=118), so the correct option is (118).

Since the path surrounds all four sides, its width is added twice to both dimensions. Outer length = 20 + 2 + 2 = 24 m and outer width = 14 + 2 + 2 = 18 m. Therefore, total outer area = 24 × 18 = 432 square metres. Values such as 384 or 408 result from failing to add the path width correctly in one or both directions. Exam tip: For a path around all four sides, add twice the path width to each original dimension.

The consecutive integers are \(x\) and \(x+1\). Thus, \(x(x+1)=56\), which gives \(x^2+x-56=0\). Option B incorrectly represents the second integer as \(x-1\). Exam tip: translate each word condition into algebra first.

The border reduces the outer side by \(x\) on both opposite sides, so the inner side is \(28-2x\). Thus, \(28-2x=20\), giving \(2x=8\) and \(x=4\) m. Therefore, option B is correct. Exam tip: for a uniform border around a square, subtract twice the border width from the outer side.

Let the smaller integer be \(x\). Then the larger integer is \(x+1\), so \(x(x+1)=306\), or \(x^2+x-306=0\). Factoring gives \((x-17)(x+18)=0\). Since the integers are positive, \(x=17\), and the larger integer is \(x+1=18\). Therefore, 18 is correct. Exam tip: For consecutive integers, represent them as \(x\) and \(x+1\) before forming the quadratic equation.

If the smaller odd integer is (x), then (x(x+2)=399). Since (19\times21=399), the smaller integer is (19).

Let the smaller even integer be \(x\). The next consecutive even integer is \(x+2\). Thus, \(x(x+2)=288\), giving \(x^2+2x-288=0\). The positive solution is \(x=16\), so the larger integer is \(16+2=18\). Exam tip: consecutive even integers differ by 2, not 1.

Let the number be (x), so (x+\frac{1}{x}=\frac{10}{3}). This gives (3x^2-10x+3=0), so the larger number is (3).

Let the breadth be \(x\) units. Then the length is \(x+5\) units. From the area, \(x(x+5)=204\), so \(x^2+5x-204=0\). Factoring gives \((x-12)(x+17)=0\); hence the positive breadth is \(12\) units and the length is \(12+5=17\) units. Thus, 12 units is the breadth, not the length. Exam tip: reject the negative root because a rectangle’s dimensions must be positive.

Let the sides of the rectangle be \(x\) cm and \(y\) cm. From the perimeter, \(2(x+y)=46\), so \(x+y=23\). The area gives \(xy=126\). Therefore, \(x(23-x)=126\), which becomes \(x^2-23x+126=0\), or \((x-9)(x-14)=0\). Thus, the sides are 9 cm and 14 cm, so the larger side is 14 cm. Exam tip: divide the perimeter by 2 first to obtain the sum of the two sides.

Let the breadth be \(x\) units. Then the length is \(x+6\) units. Using the area, \(x(x+6)=216\), so \(x^2+6x-216=0\). Factoring gives \((x-12)(x+18)=0\), leading to \(x=12\) or \(x=-18\). Since a length cannot be negative, the breadth is \(12\) units. Exam tip: In rectangle word problems, assign \(x\) to the smaller dimension and form the area equation first.

Let the side of the square be x. Its area is x² and its perimeter is 4x. According to the question, x² = 4x + 96, so x² − 4x − 96 = 0. Factoring gives (x − 12)(x + 8) = 0. Thus x = 12 or −8; a side length cannot be negative, so the correct answer is 12. Exam tip: In measurement problems, retain only the positive root.

Let the number be \(x\). From the question, \(x^2=11x+42\). Therefore, \(x^2-11x-42=0\), which factors as \((x-14)(x+3)=0\). Hence, \(x=14\) or \(x=-3\). The positive number is \(14\). Option 3 is incorrect because the other root is \(-3\), not \(3\). Exam tip: After finding both roots of a quadratic equation, always check the required condition, such as positive or negative.

If one number is (x), the other is (29-x). From (x(29-x)=204), the numbers are (17) and (12).