Word Problems and Applications

Let x represent the number of rows. Since each row contains x + 3 chairs, the total number of chairs is x(x + 3). The given total therefore produces x(x + 3) = 154, or x² + 3x − 154 = 0. Factoring gives (x + 14)(x − 11) = 0, so x = 11 or x = −14. A number of rows must be positive, so x = 11. The check is 11 rows × (11 + 3) chairs per row = 11 × 14 = 154 chairs. Thus option B is correct. Option C is the number of chairs in each row after finding x, not the number of rows; the other options fail the total or positivity condition.

If the smaller integer is (x), then (x(x+1)=182), which gives (x=13). Writing consecutive integers as (x) and (x+1) is the correct method.

The governing concept is modelling consecutive integers with x and x + 1. Let x be the smaller positive integer. The product condition becomes x(x + 1) = 210, which expands to x² + x − 210 = 0. Factoring gives (x + 15)(x − 14) = 0, so x = −15 or x = 14. The negative value is rejected because the problem specifies positive integers. Hence the smaller integer is 14 and the larger integer is 14 + 1 = 15. Direct checking confirms that 14 × 15 = 210. Therefore option C is correct. Option B is the smaller member of the pair, while 13 and 16 do not form the required consecutive pair with product 210.

Taking breadth as (x), (x(x+6)=135), giving (x=9). For a rectangle, take the product of length and breadth for area.

If breadth is (x), then (x(x+11)=180), giving (x=9) and length (20). Write only the dimension asked in the question.

If the side of the square is \(x\) metres, its area is \(x^2\). Thus, \(x^2=196\), giving \(x=\pm14\). Since a length cannot be negative, \(x=14\) metres. Exam tip: when the area of a square is given, find its side by taking the positive square root.

From (x^2=289), the side is (17) m and perimeter is (4x=68) m. First find the side and then apply perimeter.

Let the smaller number be \(x\). Then the larger number is \(x+8\), so \(x(x+8)=240\). This gives \(x^2+8x-240=0\), or \((x-12)(x+20)=0\). Thus, \(x=12\) or \(x=-20\); since the numbers are positive, \(x=12\) is the valid answer. Check: \(12\times20=240\). Remember that 20 is the larger number, not the smaller one.

(x(x+9)=340) gives (x=11), so the larger number is (20). Check the product of both numbers at the end.

(x^2=3x+70) gives (x^2-3x-70=0), and the positive solution is (10). If a positive number is asked, do not take the negative root.

The governing concept is converting a verbal comparison into a quadratic equation and then selecting the solution allowed by the condition. Let the positive number be x. Its square is 120 more than twice the number, so x² = 2x + 120. Rearranging gives x² − 2x − 120 = 0. Factoring, (x − 12)(x + 10) = 0, so x = 12 or x = −10. The problem specifies a positive number, hence x = 12 is the only admissible solution. Verification gives 12² = 144 and 2(12) + 120 = 144. Thus option C is correct; −10 is rejected because it is not positive, and the other choices do not satisfy the equation.

The direct answer is B: the smaller number is 12. Let the smaller number be x. Since their sum is 29, the other number is 29−x. The product condition gives x(29−x)=204. Expanding and rearranging: 29x−x^2=204, so x^2−29x+204=0. The factors of 204 that add to 29 are 12 and 17, so (x−12)(x−17)=0. The numbers are 12 and 17, and the smaller is 12. Option A, 10, would pair with 19 and give product 190, not 204. Option B is correct. Option C, 17, is the larger number, not the smaller one. Option D, 19, would pair with 10 and again give product 190. Check: 12+17=29 and 12×17=204. Memory cue: find the factor pair whose sum is given, then select the smaller member.

Let one number be \(x\); then the other number is \(31-x\). Thus, \(x(31-x)=238\), which gives \(x^2-31x+238=0\). Factoring, \((x-14)(x-17)=0\), so the two numbers are 14 and 17, and the larger number is 17. Exam tip: When the sum and product are given, look for a pair of numbers satisfying both conditions; the larger member of the pair is the answer.

The governing concept is applying the rectangle area formula to form a quadratic equation. Let the breadth be x metres. Then the length is x + 12 metres, and area = length × breadth gives x(x + 12) = 448. Thus x² + 12x − 448 = 0. Factoring gives (x + 28)(x − 16) = 0, so x = −28 or x = 16. A dimension cannot be negative, so the breadth is 16 m. The corresponding length is 28 m, and 16 × 28 = 448 confirms the result. Therefore option B is correct. The other options do not produce area 448 when their corresponding length is 12 m greater.

The governing concept is using the area of a rectangle to connect its two dimensions through a quadratic equation. Let the breadth be x cm. The length is then x + 13 cm. Since the area is 300 cm², x(x + 13) = 300, giving x² + 13x − 300 = 0. Factoring yields (x + 25)(x − 12) = 0, so x = −25 or x = 12. Breadth must be positive, so the breadth is 12 cm and the requested length is 12 + 13 = 25 cm. A check gives 12 × 25 = 300. Therefore option C is correct. Option A is the breadth, not the requested length; the remaining choices fail the stated difference or area.

For a right triangle, the Pythagorean theorem says that the sum of the squares of the legs equals the square of the hypotenuse. Substituting the given lengths gives \(x^2+(x+5)^2=25^2\u0005. After expansion and simplification, \(x^2+5x-300=0\u0005, which factors as \((x-15)(x+20)=0\u0005.

The algebraic solutions are \(x=15\u0005 and \(x=-20\u0005. A length must be positive, so \(x=15\u0005 cm. The other leg is then 20 cm, and 15 cm is the smaller leg. Therefore option B is correct. The hypotenuse, 25 cm, is indeed the largest side.

By the Pythagorean theorem, \(x^2+(x+4)^2=20^2\). Simplifying gives \(2x^2+8x-384=0\), or \(x^2+4x-192=0\). Factoring yields \((x-12)(x+16)=0\), so \(x=12\) or \(x=-16\). Since a side length cannot be negative, \(x=12\) m is the correct answer. Exam tip: In geometry word problems, reject any negative algebraic root when it represents a length.

The direct answer is B: the original number is 10 under the stated positive-root condition. Let the original number be x. After adding 5, the result is x+5, and its square is 225, so \((x+5)^2=225\). Taking square roots gives x+5=15 or x+5=−15. The positive-root instruction selects x+5=15, hence x=10. Verification: 10+5=15 and 15^2=225. Option A, 8, gives 13^2=169, not 225. Option B is correct. Option C, 12, gives 17^2=289. Option D, 15, gives 20^2=400. The negative-root solution would be x+5=−15, giving x=−20, but it is excluded by the instruction to use the positive root. The wording is therefore sufficient and the supplied answer is consistent. Memory cue: when a squared expression equals 225, first write both roots ±15, then apply any positivity condition.

The governing concept is solving a square equation while respecting the stated positive case. Let the original number be x. After subtracting 7, the result is x − 7, and its square is 100, so (x − 7)² = 100. Taking square roots gives x − 7 = ±10. Thus x = 17 or x = −3. The phrase “in the positive case” selects the positive original number, x = 17. Substitution verifies it: 17 − 7 = 10 and 10² = 100. Therefore option C is correct. Option A is the square-root value rather than the original number, option B does not satisfy the condition, and option D gives a result whose difference from 7 has square 169.

The equation is (x^2-9x=90), or (x^2-9x-90=0), which gives (x=15). Keep the subtraction order correct in difference statements.

(x^2=11x+60) gives (x^2-11x-60=0), whose positive solution is (15). More than means addition in the equation.

The governing concept is converting a verbal condition into a quadratic equation. Let the number be x. “The number plus its square is 156” gives x + x² = 156, or x² + x − 156 = 0. Factoring produces (x + 13)(x − 12) = 0, so the possible roots are x = −13 and x = 12. The question specifically asks for the positive number, so x = 12 must be selected. Verification is immediate: 12 + 12² = 12 + 144 = 156. Thus option C is correct. The value 13 is only the absolute value of the negative root and is not itself a solution; indeed 13 + 13² = 182. The values 10 and 11 also fail the stated equation.

(x^2+x=210) gives (x^2+x-210=0), whose positive solution is (14). While factoring, check factors of (210).

Consecutive odd numbers are (x) and (x+2), so (x(x+2)=255) gives (x=15). Consecutive odd numbers differ by (2).

Consecutive even numbers are (x) and (x+2), then (x(x+2)=360) gives (x=18). So the larger number is (20).