Word Problems and Applications

Let the original contribution per student be x rupees. The total amount is therefore 30x. When 5 students were absent, the remaining 25 students each paid (x+20) rupees, so 30x = 25(x+20). Thus, 30x = 25x + 500, giving x = 100. Hence, the total amount is 30 × 100 = 3000 rupees, so option B is correct. Exam tip: In such problems, equate the total amount in the original and changed situations before solving for the per-student contribution.

Let the number of students be (n). Then (\frac{540}{n-3}-\frac{540}{n}=30). This gives (30n^2-90n-1620=0), so (n=9).

Let initial workers be (w). The time is (\frac{60}{w}). From (\frac{60}{w}-\frac{60}{w+2}=3), (w=5).

Let the faster pipe take (x) hours, so the slower takes (x+5). From (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}), (x=10).

The combined filling rate is \(\frac{1}{4}\) tank per hour, so \(\frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}\). On simplification, \(x^2-2x-24=0\), or \((x-6)(x+4)=0\). Since time cannot be negative, \(x=6\), making the other pipe’s time 12 hours. The pipe with the smaller individual time is faster, so the correct answer is 6 hours. Exam tip: In work-rate problems, express each rate as \(1\/\text{time}\) and reject any negative time value.

The original cost per item is \(\frac{720}{x}\), and with 3 additional items it would be \(\frac{720}{x+3}\). Therefore, \(\frac{720}{x}-\frac{720}{x+3}=40\). On simplification, \(x^2+3x-54=0\), or \((x-6)(x+9)=0\). Thus \(x=6\) or \(x=-9\); since the number of items cannot be negative, \(x=6\) is the valid answer. Exam tip: In word problems involving quantities, reject any negative root.

Let the original number be (n). Then (\frac{1200}{n}-\frac{1200}{n+4}=10). This gives (10n^2+40n-4800=0), so (n=20).

Let the number be \(x\). Then \((x-5)(x+5)=119\). Using the difference of squares, \(x^2-25=119\), so \(x^2=144\) and \(x=\pm 12\). Since the question asks for the positive number, the answer is \(12\). Exam tip: For expressions of the form \((x-a)(x+a)\), apply \(x^2-a^2\) directly.

Let the number be \(x\). Then \(x^2=11x+60\), so \(x^2-11x-60=0\). Factoring gives \((x-15)(x+4)=0\), hence \(x=15\) or \(x=-4\). Since the number is positive, \(-4\) must be rejected, making 15 the correct answer. Exam tip: Always check the sign or other condition stated in the question before selecting a root.

If the sides are \((x+2)\) and \((x-1)\), then the area is \((x+2)(x-1)=x^2+x-2\), which gives a quadratic equation. Distance at constant speed is linear. Exam tip: look for a product of variable expressions.

Let the smaller number be \(x\); then the larger number is \(x+9\). Thus, \(x^2+(x+9)^2=521\), which gives \(2x^2+18x-440=0\), or \(x^2+9x-220=0\). Factoring gives \((x-11)(x+20)=0\). Since the numbers are positive, \(x=11\). Therefore, option B is correct. As a quick check, \(11^2+20^2=121+400=521\); choosing 10 would give 461 instead.