Word Problems and Applications

Correct answer: C. 15 cm

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.

Correct answer: B. 13

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.

Correct answer: B. 30 km/h

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.

Correct answer: B. 45

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.

Correct answer: C. 12 m

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.

Correct answer: B. 12

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.

Correct answer: A. 10 cm and 15 cm

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.

Correct answer: C. 6 s

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.

Correct answer: C. 16

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.