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).

Let the smaller positive number be x. Since the difference is 5, the larger number is x + 5. The product condition gives x(x + 5) = 456, which becomes x² + 5x − 456 = 0. Factoring yields (x − 19)(x + 24) = 0, so x = 19 or x = −24. The negative root is invalid because both numbers are positive. Thus the smaller number is 19 and the larger is 19 + 5 = 24. Direct verification gives 19 × 24 = 456 and 24 − 19 = 5. Therefore option C is correct. Option A identifies the smaller number, while 21 and 27 do not satisfy both the difference and product conditions simultaneously.

Let one number be x; because the sum is 55, the other is 55 − x. The sum-of-squares condition becomes x² + (55 − x)² = 1525. Expanding gives 2x² − 110x + 3025 = 1525, hence 2x² − 110x + 1500 = 0. Dividing by 2 gives x² − 55x + 750 = 0, which factors as (x − 25)(x − 30) = 0. Thus the two numbers are 25 and 30. Both are positive, and the smaller is 25, so option C is correct. Substitution confirms 25 + 30 = 55 and 25² + 30² = 625 + 900 = 1525. Option D is the larger number, while 20 and 22 fail the square-sum condition.

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).

Equal areas give x(x + 15) = 30² = 900, so x² + 15x − 900 = 0. Solving gives x = 15(−1 ± √17)/2. Since a breadth must be positive, x = 15(√17 − 1)/2 ≈ 23.42 m; the other root is negative and cannot represent a breadth.

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.

The direct answer is option A, 42 cm. In a right triangle, the Pythagorean theorem says leg1^2+leg2^2=hypotenuse^2. Here, x^2+(x+14)^2=70^2. Expanding gives 2x^2+28x+196=4900, or x^2+14x-2352=0. Factoring gives (x-42)(x+56)=0. Length cannot be negative, so x=42. The other leg is 56, and 42^2+56^2=70^2, confirming the result. Option A is correct. Option B, 44, gives the other leg 58 and does not satisfy the theorem. Option C, 46, gives 60, whose squares do not total 70^2. Option D, 48, gives 62 and also fails. The hypotenuse must be the longest side, so it is correctly used as 70. Exam cue: write the square of each leg, add them, and equate the result to the square of the hypotenuse; reject any negative algebraic root.

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.

Direct answer: The distance of the foot from the wall is 7 m, so option A is correct. Let this horizontal distance be x metres. The ladder is 18 m longer, so its length is x+18. The wall, ground and ladder form a right triangle. By Pythagoras, x^2+24^2=(x+18)^2. Expanding gives x^2+576=x^2+36x+324, so 252=36x and x=7. The ladder length is therefore 25 m, which is sensible because it is the hypotenuse and is longer than both legs. Option A is correct. Options B, C and D fail the equation: for x=8, 10 or 12, the corresponding ladder lengths 26, 28 and 30 do not satisfy the Pythagorean relation with height 24. Also, the negative algebraic possibility is not a physical distance. Memory cue: identify the right angle and use hypotenuse² = base² + height².

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.

Using distance = speed × time, we get \(x(x+5)=84\). Thus, \(x^2+5x-84=0\), which factors as \((x+12)(x-7)=0\). Therefore, \(x=-12\) or \(x=7\). Since time cannot be negative, the valid answer is 7 hours. Exam tip: Reject any negative root when the variable represents time or distance.

The governing concept is the work relation: total work equals the number of days multiplied by the work completed each day. Here, the total work is 180 units, the number of days is x, and the daily work is x + 3. Therefore, x(x + 3) = 180, which gives x² + 3x − 180 = 0. Factoring, (x + 15)(x − 12) = 0, so x = −15 or x = 12. A duration cannot be negative, so x = 12 is the meaningful solution. The worker therefore takes 12 days and completes 15 units per day; 12 × 15 = 180 verifies the answer. The other options do not satisfy the complete work equation.

The total number of questions is x(x+4). Hence, x(x+4)=320, giving x²+4x−320=0. Factoring, (x+20)(x−16)=0, so x=−20 or x=16. Since the number of days cannot be negative, x=16 is valid. Exam tip: In word problems, calculate both roots of the quadratic and retain only the value that fits the real-world condition.

The total number of books gives the equation \(x(x+9)=682\). Thus, \(x^2+9x-682=0\), which factors as \((x-22)(x+31)=0\). Hence, \(x=22\) or \(x=-31\). Since the number of shelves cannot be negative, the valid answer is 22. Exam tip: In word problems, retain only the positive root that fits the real-life context.

The total-chair condition gives \(x(x+15)=1000\), so \(x^2+15x-1000=0\). Factoring, \((x-25)(x+40)=0\), which gives \(x=25\) or \(x=-40\). Since the number of rows cannot be negative, the valid answer is 25. In such word problems, always reject the negative root on the basis of the context.