Word Problems and Applications

The equation is (x-\frac{1}{x}=\frac{15}{4}). This gives (4x^2-15x-4=0), so the positive root is (x=4).

From (x+\frac{1}{x}=\frac{17}{4}), we get (4x^2-17x+4=0). The roots are (4) and (\frac{1}{4}), so the larger value is (4).

Let the breadth of the rectangle be \(x\) cm. Then its length is \(x+17\) cm. By the Pythagorean theorem, \(x^2+(x+17)^2=25^2\). On simplifying, we get \(x^2+17x-168=0\), or \((x-7)(x+24)=0\). Thus, \(x=7\) or \(x=-24\). Since a length cannot be negative, the breadth is 7 cm. Exam tip: Treat the diagonal, length, and breadth as the sides of a right triangle and reject any negative root obtained from the quadratic equation.

The length, breadth and diagonal of a rectangle form a right-angled triangle. By the Pythagorean theorem, \(x^2+12^2=37^2\), which gives \(x^2+144=1369\). Option C has the incorrect sign before the square of the breadth. Exam tip: The diagonal of a rectangle is the hypotenuse of the right triangle formed by its length and breadth.

Let the shorter perpendicular side be \(x\) cm. Then the other side is \(x+5\) cm. Using the area formula, \(\frac{1}{2}x(x+5)=84\), which gives \(x^2+5x-168=0\). Applying the quadratic formula gives \(x=\frac{-5\pm\sqrt{697}}{2}\). Since a length must be positive, the valid answer is \(x=\frac{\sqrt{697}-5}{2}\) cm, approximately 10.69 cm. The 12 cm option is incorrect because the corresponding sides would be 12 cm and 17 cm, giving an area of 102 square cm. In an exam, reject the negative root as a physical length.

Let the inner radius be r metres. Then the outer radius is (r+3) metres. The area of the path is π[(r+3)^2−r^2] = 75π. Cancelling π gives (r+3)^2−r^2 = 75, so 6r+9=75 and r=11. Therefore, option B, 11 m, is correct. Exam tip: For a circular path, use π[(outer radius)^2−(inner radius)^2], not the area of the outer circle alone.

The border adds 5 cm on each side, so the side of the outer square is \(x+5+5=x+10\) cm. Therefore, \((x+10)^2=900\), giving \(x+10=\pm 30\). Since a side length must be positive, \(x+10=30\), and hence \(x=20\) cm. The negative root is not physically valid. Exam tip: for a uniform border around a square, add twice the border width to the original side.

The photo breadth is (x) and length is (x+8). With frame, dimensions are (x+4) and (x+12), and ((x+4)(x+12)=480) gives (x=12).

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.

If the original number of rows is (x), then (\frac{240}{x-4}-\frac{240}{x}=5). Solving gives (x=16).