Word Problems and Applications

The increase in area is ((x+8)^2-x^2=1216). Expanding gives (16x+64=1216), so (16x=1152) and (x=72) metres. Therefore, option B is correct. Exam tip: For a change in the side of a square, use ((x+a)^2-x^2=2ax+a^2); do not incorrectly set ((x+a)^2=1216).

The reduced side is \((x-5)\) cm, so \((x-5)^2=784=28^2\). Since a side length must be positive, take \(x-5=28\), giving \(x=33\) cm. The other algebraic possibility, \(x-5=-28\), gives \(x=-23\) cm, which is not physically meaningful. Exam tip: After finding the reduced side from the area, add back the length that was removed.

If breadth is (x) and length is (x+12), the increase is ((x+6)(x+18)-x(x+12)=480). This gives (x=31).

Let the original breadth be \(x\) m. Then the original length is \(x+9\) m. After the increases, the new length and breadth are \(x+13\) and \(x+3\) m, respectively. Thus, \((x+13)(x+3)=1190\), which gives \(x^2+16x-1151=0\). Using the quadratic formula, \(x=\frac{-16\pm\sqrt{4860}}{2}=-8\pm9\sqrt{15}\). Since a breadth must be positive, the valid value is \(x=-8+9\sqrt{15}\) m; the other root is negative. Exam tip: write the changed dimensions carefully before forming the area equation.

If the side of the square is (x), then (x^2=1296), so (x=36). The wire length is the perimeter (4x=144) cm.

If breadth is (x), then (x(x+16)=1344), giving (x=32) and length (48). The wire length is (2(32+48)=160) cm.

Since the path surrounds the garden, its width is added twice to both the length and the breadth. The outer length is \(34+2(3)=40\) m and the outer breadth is \(22+2(3)=28\) m. Therefore, the total area is \(40\times28=1120\) square metres, so option C is correct. Option B results from an incorrect calculation of the outer dimensions or their product. Exam tip: for a path around all four sides, add twice the path width to each dimension.

The outer dimensions, including the path, are \((42+2x)\) m and \((28+2x)\) m. Thus, \((42+2x)(28+2x)=1848\), which gives \(4x^2+140x-672=0\), or \(x^2+35x-168=0\). Hence, \(x=\frac{-35\pm\sqrt{1897}}{2}\). Since a width must be positive, the valid value is \(x=\frac{-35+\sqrt{1897}}{2}\approx4.28\) m. The negative root is not physically meaningful. Exam tip: when a path surrounds a rectangle externally, add \(2x\) to both its length and breadth.

The strip is removed from both opposite sides, so the inner side is \(52-2x\). Therefore, \(52-2x=36\), giving \(2x=16\) and \(x=8\) m. Using areas gives the equivalent quadratic equation \((52-2x)^2=36^2\); the geometrically valid root is \(x=8\). Exam tip: divide the difference between the outer and inner side lengths by 2. For example, 10 m would reduce the inner side to 32 m, not 36 m.

If the smaller integer is (x), then (x^2+(x+1)^2=1201), giving (x=24). Write consecutive integers as (x) and (x+1).

Let the smaller integer be \(x\). The next consecutive odd integer is therefore \(x+2\). Thus, \(x^2+(x+2)^2=2050\), which gives \(2x^2+4x-2046=0\) or \(x^2+2x-1023=0\). Factoring, \((x-31)(x+33)=0\), so \(x=31\) or \(x=-33\). Since the integers are positive, \(x=31\). Therefore, the correct answer is 31. Exam tip: consecutive odd integers differ by 2, not by 1.

Consecutive even numbers are (x) and (x+2). From (x^2+(x+2)^2=2452), (x=34), so the larger number is (36).

If the smaller number is (x) and the larger is (x+17), then (x(x+17)=1260). This gives smaller number (28) and larger number (45).

Let the numbers be \(x\) and \(80-x\). Then \(x^2+(80-x)^2=3232\), which gives \(2x^2-160x+3168=0\), or \(x^2-80x+1584=0\). Factoring, \((x-36)(x-44)=0\), so the numbers are 36 and 44. Therefore, the smaller number is 36; 44 is the larger number, not the required answer. Exam tip: When the sum and the sum of squares are given, use \(a^2+b^2=(a+b)^2-2ab\) to find the product quickly.

If breadth is (x), then (x(x+21)=1870), giving (x=34). For a rectangle, use the product of length and breadth for area.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+16\) cm. Using the area, \(x(x+16)=2145\), so \(x^2+16x-2145=0\). Factoring gives \((x-39)(x+55)=0\). Since a breadth cannot be negative, \(x=39\) cm. Therefore, the length is \(39+16=55\) cm. Exam tip: reject any negative root when the variable represents a physical dimension.

If breadth is (x), then (x^2+(x+7)^2=97^2), giving (x=65). Use Pythagoras for a rectangle with a diagonal.

From (x(x+15)=2494), breadth is (43) and length is (58). The perimeter is (2(43+58)=202), so the correct option should be (202).

If one side is (x), the other is (93-x), and (x(93-x)=2016). The solutions are (36) and (56), so the shorter side is (36).

Since the areas are equal, \(x(x+28)=48^2=2304\). Thus, \(x^2+28x-2304=0\), which factors as \((x-36)(x+64)=0\). The roots are \(x=36\) and \(x=-64\). Since a breadth cannot be negative, the valid value is \(x=36\). Exam tip: In area-based word problems, form the equality first and reject any negative root that is not physically meaningful.