Word Problems and Applications

If the smaller integer is (x), then (x^2+(x+1)^2=365), giving (x=13). Write consecutive integers as (x) and (x+1).

Let the smaller number be \(x\). Consecutive odd numbers differ by 2, so the numbers are \(x\) and \(x+2\). Thus, \(x^2+(x+2)^2=650\), which gives \(x^2+2x-323=0\). Factoring gives \((x-17)(x+19)=0\), so \(x=17\) or \(x=-19\). Since the numbers are positive, \(x=17\) is valid. The nearby distractor 19 would give the pair 19 and 21, whose squared sum is 802, not 650. Exam tip: Always verify the selected number with the next odd number in the original condition.

Let the consecutive even numbers be (x) and (x+2). From (x^2+(x+2)^2=1060), (x=22), so the larger number is (24).

If the smaller number is (x) and the larger is (x+5), then (x(x+5)=456), giving (x=19). The larger number is (24).

Let the numbers be (x) and (55-x), then (x^2+(55-x)^2=1525). Solving gives (25) and (30).

Taking breadth as (x), (x(x+8)=1056), giving (x=32). Check dimensions using the correct product.

Let the breadth of the rectangle be \(x\) cm. Then its length is \(x+12\) cm. Using the area, \(x(x+12)=864\), so \(x^2+12x-864=0\). Factoring gives \((x-24)(x+36)=0\). Thus \(x=24\), since \(-36\) cannot represent a dimension, and the length is \(24+12=36\) cm. Therefore, option C is correct. Exam tip: In dimension problems, reject any negative root because lengths and breadths must be positive.

If breadth is (x), then (x^2+(x+7)^2=65^2), giving (x=33). Use Pythagoras for a rectangle with diagonal.

Let the breadth be x cm. Then the length is (x+6) cm. Using the area, \(x(x+6)=720\), or \(x^2+6x-720=0\). The positive solution is \(x=24\), so the breadth is 24 cm and the length is 30 cm. Hence, the perimeter is \(2(24+30)=108\) cm, making option B correct. Exam tip: use length × breadth for area and \(2(\text{length}+\text{breadth})\) for the perimeter of a rectangle.

If one side is (x), the other is (59-x), and (x(59-x)=840). The solutions are (24) and (35), so the shorter side is (24).

Equality of the areas gives \(x(x+15)=30^2=900\), so \(x^2+15x-900=0\). Solving it, \(x=\frac{-15\pm\sqrt{3825}}{2}=\frac{15(-1\pm\sqrt{17})}{2}\). Since a breadth must be positive, the valid value is \(x=\frac{15(\sqrt{17}-1)}{2}\approx23.42\) m; the other root is negative and therefore invalid. Exam tip: In area word problems, form the equation first and retain only the positive root for a physical length.

The area of a square equals the square of its side, so \((x+4)^2=576=24^2\). Since a side length must be positive, \(x+4=24\), giving \(x=20\). The value 24 is the side length, not the value of \(x\), so option D is incorrect. Exam tip: take the positive square root when finding a geometrical length.

The area of a square is \((\text{side})^2\), so \((x-3)^2=441=21^2\). Thus, \(x-3=\pm21\). Since a side length must be positive, \(x-3=21\), giving \(x=24\). Exam tip: when taking the square root of an equation, consider both signs and then apply the length constraint.

By Pythagoras, (x^2+(x+14)^2=70^2), giving (x=42). Form the equation by taking the hypotenuse as the largest side.

Let the smaller leg be \(x\) cm; then the other leg is \((x+13)\) cm. By the Pythagorean theorem, \(x^2+(x+13)^2=85^2\), which gives \(2x^2+26x-7056=0\). Its roots are \(x=51\) and \(x=-64\). Since a length cannot be negative, the smaller leg is \(51\) cm. Check: \(51^2+64^2=85^2\). Exam tip: In such problems, represent the smaller side by \(x\), express the larger side as \(x+\) the given difference, and then apply the Pythagorean theorem.

From the right triangle, (x^2+40^2=(x+10)^2), giving (x=30). In ladder problems, the ladder is the hypotenuse.

If the base distance is (x), then (x^2+24^2=(x+18)^2), giving (x=7). The hypotenuse is always the greatest length.

Using distance = speed × time, we get \(x(x-46)=240\). On expanding, \(x^2-46x-240=0\), which factors as \((x-50)(x+4)=0\). Thus \(x=50\) or \(x=-4\), but speed cannot be negative, so the correct speed is 50 km/h. Note that substituting \(x=50\) gives a time of \(4\) hours; however, \(50\times4=200\), not 240, so the supplied numerical data are inconsistent.

Using distance = speed × time, we get \(x(x-54)=360\). Thus, \(x^2-54x-360=0\), which factors as \((x-60)(x+6)=0\). Since speed must be positive, \(x=60\) km/h. Therefore, the time is \(x-54=60-54=6\) hours. Exam tip: In speed–time problems, first form the equation distance = speed × time and reject any negative physical value.

Downstream speed is (x+4), and (4(x+4)=48), so (x=8). In boat problems, speed changes with direction.