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 speed be (x). Then (\frac{360}{x}-\frac{360}{x+5}=1). This gives (x^2+5x-1800=0), so (x=40).

Let the speed be (x). Then (\frac{300}{x-10}-\frac{300}{x}=1). This gives (x^2-10x-3000=0), so (x=60).

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.

Let initial distance be (x). Then (x^2+h^2=25^2) and ((x+7)^2+(h-7)^2=25^2). Solving gives (h=24).

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.

On the ground, (h=0). From (30t-5t^2=0), (t=0) or (t=6), so the return time is (6) seconds.

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.