Word Problems and Applications

The cost price is (1200-200=1000), and profit percentage is (\frac{200}{1000}\times100=20). This does not match (\frac{1000}{20}=50), so the statement is inconsistent.

Let the height be \(x\text{ cm}\); then the base is \((x+6)\text{ cm}\). Using the triangle-area formula, \(\frac{1}{2}x(x+6)=140\), which gives \(x^2+6x-280=0\). Factoring, \((x-14)(x+20)=0\), so \(x=14\) or \(x=-20\). Since a length cannot be negative, the height is \(14\text{ cm}\). Exam tip: after solving a quadratic word problem, reject any root that is not physically meaningful.

Let the fraction be (\frac{x}{x+4}), then (\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}). This gives (x=5), so the fraction is (\frac{5}{9}).

The original side of the square is \(\sqrt{225}=15\text{ cm}\), and the new side is \(\sqrt{400}=20\text{ cm}\). Thus, \(15+x=20\), giving \(x=5\text{ cm}\). Option B is incorrect because a side of \(15+4=19\text{ cm}\) would produce an area of \(361\text{ cm}^2\), not \(400\text{ cm}^2\). Exam tip: take the square root of each area to find the corresponding side, then subtract the original side from the new side.

With perimeter (20), (l+b=10) and (lb=96). The equation (x(10-x)=96) has discriminant (100-384<0), so the rectangle is impossible.

Let the actual speed be (x), then (\frac{240}{x-20}-\frac{240}{x}=1). This gives (x^2-20x-4800=0), so (x=80).

Any two points determine one line segment, so the number of segments is \(\frac{n(n-1)}{2}\). Thus, \(\frac{n(n-1)}{2}=45\), which gives \(n^2-n-90=0\). Factoring, \((n-10)(n+9)=0\), so n = 10 or -9. Since the number of points cannot be negative, n = 10 is the valid answer. In such problems, always reject the negative root because it has no practical meaning.

Let breadth be (x), then (x(x+4)=96), so (x=8) and length is (12). The perimeter is (2(8+12)=40\text{ m}).

Let the original number be \(x\). Then \(x(x+6)=135\), giving \(x^2+6x-135=0\). Factoring, \((x-9)(x+15)=0\), so \(x=9\) or \(x=-15\). Since the question asks for the positive number, the correct answer is 9. Exam tip: Find both roots of a quadratic equation and then apply the condition stated in the question, such as positivity.

By the Pythagorean theorem, \(x^2+(x+7)^2=(x+8)^2\). On simplifying, \(x^2-2x-15=0\), or \((x-5)(x+3)=0\). Thus, \(x=5\) or \(x=-3\); since a length cannot be negative, the shorter side is 5 cm. In such problems, always reject roots that give a non-positive length.

Let original price be (x), then (\frac{300}{x-5}-\frac{300}{x}=3). This gives (x^2-5x-500=0), so (x=25).

Let breadth be (x), then increase is ((x+5)(x+15)-x(x+10)=350). This gives (10x+75=350), so (x=27.5), hence no listed option is exact.

Let students per room be (x), then (\frac{60}{x}-\frac{60}{x+2}=5). This gives (5x^2+10x-120=0), so (x=4).

Let the number be \(x\). Then \(x^2+x=156\), so \(x^2+x-156=0\). Factoring gives \((x+13)(x-12)=0\), hence \(x=-13\) or \(x=12\). Since the number is positive, the correct answer is 12. The nearby option 13 is incorrect because \(13+13^2=182\), not 156. Exam tip: Form the quadratic equation first, then select the root that satisfies the given condition.

Let breadth be (x), then length is (2x-5), and (x(2x-5)=875). This gives (2x^2-5x-875=0), so (x=25).

Let the stream speed be (x), then (\frac{48}{10+x}+\frac{32}{10-x}=10). This gives (x^2+4x-12=0), so (x=2).

Let original breadth be (x), then the box base is ((x-4)(x+4)) and height is (2). From (2(x-4)(x+4)=240), (x^2=136), so none of the options is exact.

Let the actual speed be (x\text{ km/h}), then (\frac{300}{x}-\frac{300}{x+10}=1). This gives (x^2+10x-3000=0), so the positive root is (x=50).

Let the breadth be (x\text{ m}), then (x(x+8)=273). This gives (x=13) and length (21), so the perimeter is (2(13+21)=68\text{ m}).

Let the first worker's time be (x\text{ days}), then (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}). This gives (x^2-7x-30=0), so (x=10).