Word Problems and Applications

Let the breadth be \(x\) cm. Then the length is \((x+5)\) cm. Using the area, \(x(x+5)=300\), so \(x^2+5x-300=0\). Factoring gives \((x-15)(x+20)=0\), hence \(x=15\) or \(x=-20\). Since a length cannot be negative, the breadth is \(15\) cm. In such word problems, always reject the negative root as physically meaningless.

Let the smaller number be \(x\). The next consecutive odd number is \(x+2\), so \(x^2+(x+2)^2=394\). Simplifying gives \(2x^2+4x-390=0\), or \(x^2+2x-195=0=(x+15)(x-13)\). Thus, \(x=13\) or \(-15\); since the number must be positive, the smaller number is \(13\). Exam tip: consecutive odd numbers differ by 2, not by 1.

Let the original speed be \(x\) km/h. The decrease in travel time gives \(\frac{240}{x}-\frac{240}{x+10}=2\). Simplifying, \(x^2+10x-1200=0\), or \((x-30)(x+40)=0\). Thus \(x=30\) or \(-40\); the negative value cannot represent a speed. Therefore, the original speed was 30 km/h. Exam tip: For a fixed distance, use \(\text{time}=\frac{\text{distance}}{\text{speed}}\) and reject any negative root.

Let the tens digit be \(x\). Then the units digit is \(x+1\). Therefore, \(x^2+(x+1)^2=41\), which gives \(2x^2+2x-40=0\) or \((x-4)(x+5)=0\). Since a digit cannot be negative, \(x=4\); hence the units digit is 5 and the number is 45. Although 56 also has digits differing by 1, its digit-square sum is \(25+36=61\), not 41. Exam tip: reject any negative root when the variable represents a digit.

Let the breadth of the hall be \(x\) metres. Its length will then be \(x+2\) metres. Using the area, \(x(x+2)=168\), so \(x^2+2x-168=0\). Factoring gives \((x+14)(x-12)=0\), yielding \(x=12\) or \(x=-14\). Since a length cannot be negative, the breadth is 12 metres. For example, 10 metres would give an area of \(10\times12=120\) square metres, so it is not correct. Exam tip: reject any negative root when solving for a physical dimension.

Let the number of students be \(x\). Each student donated \(x+2\) rupees, so \(x(x+2)=168\). Thus, \(x^2+2x-168=0\), which factors as \((x+14)(x-12)=0\). Therefore, \(x=12\) or \(x=-14\); the number of students cannot be negative, so the correct answer is 12, option B. Exam tip: In word problems, reject roots that are impossible in the given context, such as a negative number of students.

The governing relations are perimeter = 2(l + b) and area = lb. From 2(l + b) = 50, we obtain l + b = 25, while the area condition gives lb = 150. Let either dimension be t; then the two dimensions are roots of t² − 25t + 150 = 0. Factoring gives (t − 10)(t − 15) = 0, so the dimensions are 10 cm and 15 cm. Option A is correct. Verification is important: 2(10 + 15) = 50 cm and 10 × 15 = 150 cm². Option B has the correct perimeter but area 144, option C has area 156, and option D has area 136, so none of them satisfies both conditions.

The ball is on the ground when its height h is zero. Set the given quadratic expression equal to zero: 30t − 5t² = 0. Factoring gives 5t(6 − t) = 0, so t = 0 or t = 6 seconds. The value t = 0 represents the instant of launch, when the ball is initially at ground level; the question asks when it returns, so the relevant time is t = 6 seconds. Therefore option C is correct. The quadratic model also shows that the height is positive between t = 0 and t = 6 and becomes zero again at t = 6. The other choices are not roots of the height equation and therefore do not represent ground contact.

Let the original number of rows be x. Then each row initially contains 240/x students. After reducing the rows by 4, each row contains 240/(x − 4) students. The condition gives 240/(x − 4) − 240/x = 5, where x > 4. Simplifying, 960/[x(x − 4)] = 5, so x(x − 4) = 192 and x² − 4x − 192 = 0. Factorisation gives (x − 16)(x + 12) = 0. Since the number of rows must be positive, x = 16. Option C is correct. Checking, 240/16 = 15 students per row; with 12 rows, 240/12 = 20, which is exactly 5 more. The other options fail this comparison.