Word Problems and Applications

Let the original numerator be \(x\). Then the denominator is \(x+4\). According to the condition, \(\frac{x+2}{x+6}=\frac{3}{5}\). Cross-multiplication gives \(5x+10=3x+18\), so \(2x=8\) and \(x=4\). Check: the original fraction is \(\frac{4}{8}\), and adding 2 to both terms gives \(\frac{6}{10}=\frac{3}{5}\). Exam tip: in such problems, represent the numerator by \(x\) and express the denominator using the stated difference.

The fraction is (\frac{x}{x+3}). From (\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}), (x=2) or (x=15), and among the options (2) is correct.

Let the current speed be (x). Then (\frac{30}{12+x}+\frac{20}{12-x}=5). Solving gives (5x^2+10x-60=0), so (x=2).

Let the current speed be (x). Then (\frac{24}{10+x}+\frac{16}{10-x}=4). This gives (4x^2+8x-16=0), so (x=2).

If breadth is (x), length is (100-2x) and area is (A=x(100-2x)). For maximum area, (x=\frac{100}{4}=25).

Let the breadth be b m and the length be l m. From the three-sided fencing condition, l+2b=96, so l=96−2b. Hence, the area is A=b(96−2b)=−2b²+96b. This quadratic reaches its maximum at its vertex, where b=−96/(2×−2)=24 m. The corresponding length is 48 m and the maximum area is 1152 square m. Exam tip: For ax²+bx+c, the x-coordinate of the vertex is −b/(2a).

Let the breadth be \(x\) cm. Then the length is \(x+7\) cm. Using the area, \(x(x+7)=330\), so \(x^2+7x-330=0\). Factoring gives \((x-15)(x+22)=0\). Thus \(x=15\) cm, since the negative root \(-22\) is not physically meaningful, and the length is \(22\) cm. Therefore, the perimeter is \(2(15+22)=74\) cm. Exam tip: In geometry word problems, reject any negative root because a length cannot be negative.

Let the base be \(x\) cm. Then the perpendicular is \(x+7\) cm. By the Pythagorean theorem, \(x^2+(x+7)^2=25^2\), which simplifies to \(x^2+7x-288=0\). Its positive root is \(x=\frac{-7+\sqrt{1201}}{2}\approx13.82\) cm, so option A is correct. Option C is incorrect because a base of 15 cm would make the perpendicular 22 cm, giving \(15^2+22^2=709\ne625\). Exam tip: For a length, select only the positive root of the quadratic equation.

By the Pythagorean theorem, x² + (x+2)² = 10². Therefore, 2x² + 4x − 96 = 0, or x² + 2x − 48 = 0. Factoring gives (x−6)(x+8)=0, so x=6 or x=−8. Since a length cannot be negative, x=6 and the other shorter side is x+2=8 cm. Hence, the correct answer is 8 cm. Exam tip: In a quadratic equation representing a length, reject any negative root.

The correct answer is C, (7 + √1201)/2 m, approximately 20.33 m. Let the initial horizontal distance be x and the initial height be h. Since the ladder is 25 m, x² + h² = 625. After moving the foot 7 m away and lowering the top 7 m, (x + 7)² + (h − 7)² = 625. Subtracting the first equation gives 14x − 14h + 98 = 0, so h = x + 7. Substituting gives x² + (x + 7)² = 625, or x² + 7x − 288 = 0. The positive root is x = (−7 + √1201)/2, hence h = (7 + √1201)/2 ≈ 20.33 m. A, B and D do not satisfy both right-triangle equations. In particular, the familiar 7-24-25 triangle cannot be used automatically because the stated 7 m vertical drop must also be satisfied.

The first height is (\sqrt{13^2-5^2}=12) m. With new distance (12) m, height is (5) m, so the fall is (12-5=7) m.

For a height of 15 metres, set \(20t-5t^2=15\). Rearranging gives \(t^2-4t+3=0\), or \((t-1)(t-3)=0\). Thus, \(t=1\) s and \(t=3\) s. The first time occurs while the object is rising, and the second while it is falling. Exam tip: check both positive roots because a projectile can reach the same height twice.

Let the number of students be (n). Then (\frac{n(n-1)}{2}=276). This gives (n^2-n-552=0), so (n=24).

Let the number of teams be \(n\). Each match is played between one distinct pair of teams, so the total number of matches is \(\frac{n(n-1)}{2}\). Therefore, \(\frac{n(n-1)}{2}=153\), giving \(n^2-n-306=0\). Factoring, \((n-18)(n+17)=0\), so \(n=18\) or \(n=-17\). Since the number of teams cannot be negative, the answer is 18. Exam tip: Use \(\frac{n(n-1)}{2}\) when every unordered pair participates in exactly one match.

Let the smaller page number be \(x\); then the next page number is \(x+1\). Thus, \(x(x+1)=930\), or \(x^2+x-930=0\). Factoring gives \((x-30)(x+31)=0\), so \(x=30\) or \(x=-31\). Since a page number must be positive, the smaller page number is 30. Exam tip: For consecutive positive integers, look for factor pairs of 930 that differ by 1—\(30\) and \(31\).

If the tens digit is (x), the units digit is (11-x). Checking options shows (47 \times 74=3478), not (3154), so this item would be invalid.

The reverse of (48) is (84), and (48 \times 84=4032). In digit problems, checking the reversed number is essential.

Let the positive number be x. Its reciprocal is 1/x, so the statement becomes x−1/x=15/4, with x≠0. Multiplying by 4x gives 4x²−4=15x, or 4x²−15x−4=0. Factoring gives (4x+1)(x−4)=0. The roots are x=−1/4 and x=4. Since the question asks for the positive number, only x=4 is acceptable.

Option B is therefore correct. A quick check confirms it: 4−1/4=16/4−1/4=15/4. The negative root also satisfies the algebraic equation, but it is excluded by the word positive. The supplied explanation correctly forms the equation and selects the positive root; the factorization provides the missing clear verification.

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).