Word Problems and Applications

Let the positive integer be \(x\). Then \(x(x+1)=210\), so \(x^2+x-210=0\). Factoring gives \((x-14)(x+15)=0\), hence \(x=14\) or \(x=-15\). Since the question specifies a positive integer, \(x=14\) is correct, and \(14\times15=210\). Option 15 is incorrect because its next integer is 16, and \(15\times16=240\). Exam tip: Check pairs of consecutive integers whose product equals the given number.

Let the number be (x), so (x^2-11=85). This gives (x^2=96), so the positive number is (4\sqrt{6}).

Let the height reached on the wall be \(x\) metres. Then the distance of the foot of the ladder from the wall is \(x-7\) metres. By the Pythagorean theorem, \(x^2+(x-7)^2=13^2\). This gives \(x^2-7x-60=0\), or \((x-12)(x+5)=0\). Since a length cannot be negative, \(x=12\) metres. The value 5 metres represents the distance from the wall, not the height. Exam tip: in a ladder-wall-ground right triangle, the ladder is the hypotenuse.

Let the shorter leg be \(x\) units. Then the longer leg is \(x+7\) units. By the Pythagorean theorem, \(x^2+(x+7)^2=17^2\), which simplifies to \(x^2+7x-120=0\). Factoring gives \((x+15)(x-8)=0\), so \(x=8\), since a length cannot be negative. Therefore, the correct answer is 8. The value 15 is the longer leg, not the shorter one; 8-15-17 is a Pythagorean triple.

Let the breadth be \(x\) units. Then the length is \(x+8\) units, so \(x(x+8)=240\), giving \(x^2+8x-240=0\). Factoring, \((x-12)(x+20)=0\), so \(x=12\) or \(x=-20\). Since a length cannot be negative, the breadth is 12 units. Exam tip: In rectangle-area problems, assign \(x\) to one dimension, express the other in terms of \(x\), and reject any negative root.

Let the breadth be \(x\) units. Then the length is \(3x\) units. Using the area formula, \(x\times 3x=180\), so \(3x^2=180\). Hence, \(x^2=60\), and since a breadth must be positive, \(x=\sqrt{60}=2\sqrt{15}\) units. Therefore, option A is correct. Exam tip: For dimensions such as length and breadth, select only the positive root of the quadratic equation. Option C, \(15\), does not satisfy the given area because \(3(15)^2\neq 180\).

The equation is (x(50-x)=600). This gives (x^2-50x+600=0), so (x=20) or (30); the larger value is (30).

Let the breadth be \(x\) metres. Then the length is \(x+5\) metres, so \(x(x+5)=200\), or \(x^2+5x-200=0\). Using the quadratic formula, \(x=\frac{-5\pm\sqrt{825}}{2}=\frac{5(-1\pm\sqrt{33})}{2}\). Since breadth must be positive, \(x=\frac{5(\sqrt{33}-1)}{2}\) metres. Therefore, the length is \(x+5=\frac{5(1+\sqrt{33})}{2}\) metres, approximately 16.86 metres. The distractor 20 m is incorrect because it would give a breadth of 15 m and an area of 300 square metres. Exam tip: discard the negative root and verify the positive root using the area condition.

Let the original speed be \(x\) kilometres per hour. The decrease in time gives \(\frac{300}{x}-\frac{300}{x+10}=1\). Simplifying, \(x^2+10x-3000=0\), which factors as \((x-50)(x+60)=0\). Since speed must be positive, \(x=50\) kilometres per hour. In such problems, always reject the negative root as physically meaningless.

Let the original speed be \(x\) kilometres per hour. The original time is \(\frac{240}{x}\) hours, while the time at the reduced speed is \(\frac{240}{x-20}\) hours. Hence, \(\frac{240}{x-20}-\frac{240}{x}=2\). On simplifying, \(x^2-20x-2400=0\), so \((x-60)(x+40)=0\). Since speed must be positive, \(x=60\) kilometres per hour. The root \(x=-40\) is not physically meaningful. Exam tip: In speed-time problems, express the times in both situations first and then form their difference.

The equation is (\frac{36}{15+x}+\frac{24}{15-x}=4). Substituting (x=3) gives time (2+2=4) hours.

If the number of students is (x), then (\frac{1200}{x}-\frac{1200}{x+5}=8). Solving gives (x=25).

Let the original number of students be \(x\). The original share is \(\frac{900}{x}\), while the share after 6 students leave is \(\frac{900}{x-6}\). Thus, \(\frac{900}{x-6}-\frac{900}{x}=5\). Simplifying gives \(x^2-6x-1080=0\), or \((x-36)(x+30)=0\). Therefore, \(x=36\), since the number of students must be positive and greater than 6; \(x=-30\) is inadmissible. Exam tip: In distribution problems, express each person’s share as total amount divided by the number of people.

Let the son's present age be \(x\) years. Then the father's present age is \(x+30\) years. After 5 years, their ages will be \(x+5\) and \(x+35\), respectively. Hence, \((x+5)(x+35)=1375\). For \(x=20\), the product is \(25\times55=1375\), so the son's present age is 20 years. The nearby distractor 25 is incorrect because it gives \(30\times60=1800\). Exam tip: in age problems, add the same number of years to both present ages when calculating their future ages.

Let one friend’s age be \(x\) years. The other friend’s age is then \(40-x\) years. Thus, \(x(40-x)=375\), which gives \(x^2-40x+375=0\). Factoring, \((x-25)(x-15)=0\), so the two ages are 25 years and 15 years. Therefore, the older friend is 25 years old. Exam tip: In age problems, express the second age using the given sum and form a quadratic equation using the product.

If the middle integer is (x), the first and third are (x-1) and (x+1). From ((x-1)(x+1)=143), (x=12).

If the middle one is (x), the first and third are (x-2) and (x+2). From ((x-2)(x+2)=320), (x=18).

Let the breadth be \(x\) units; then the length is \(x+7\) units. By the Pythagorean theorem, \(x^2+(x+7)^2=13^2\). On simplifying, \(x^2+7x-60=0\), or \((x+12)(x-5)=0\). Thus \(x=5\) or \(x=-12\). Since a breadth cannot be negative, the breadth is 5 units. Option 12 is a distractor related to the negative root \(-12\). Exam tip: Reject negative values when a variable represents a physical length or distance.

If the original side is (x), then (x^2=144), so (x=12). The new side is (15), and the new area is (15^2=225).

Let the number be \(x\). Then \(x^2=10x+24\), so \(x^2-10x-24=0\). Factoring gives \((x-12)(x+2)=0\), hence \(x=12\) or \(x=-2\). Since the question asks for the positive number, the correct answer is 12. Exam tip: Find both roots of a quadratic equation and then apply any condition given in the question, such as positivity.