Word Problems and Applications

Let the smaller integer be \(x\). Then the larger integer is \(x+1\), so \(x(x+1)=156\), giving \(x^2+x-156=0\). Factoring, \((x-12)(x+13)=0\). Since the integers are positive, \(x=12\), and the larger integer is therefore \(13\). Option B is the smaller integer, not the larger one. Exam tip: For consecutive-integer problems, represent the numbers as \(x\) and \(x+1\) before forming the quadratic equation.

This is a word problem modeled by a quadratic equation. Let the smaller positive odd integer be n. The next consecutive odd integer is n + 2, so n(n + 2) = 195, giving n² + 2n − 195 = 0. Factoring, (n + 15)(n − 13) = 0, so n = 13 or n = −15. Because the question requires positive integers, n = −15 must be rejected. Direct verification confirms that 13 × 15 = 195, and 13 and 15 are consecutive odd integers. Thus the smaller integer is 13. Option B is the larger member of the pair, while 11 and 17 do not form a consecutive odd pair whose product is 195.

Let the smaller even integer be x. The next consecutive even integer is x + 2. Thus, x(x + 2) = 168, giving x² + 2x − 168 = 0 = (x − 12)(x + 14). Since the integers are positive, x = 12, so the larger integer is 12 + 2 = 14. Exam tip: consecutive even integers always differ by 2.

Let the number be (x), so (x+\frac{1}{x}=\frac{5}{2}). This gives (2x^2-5x+2=0), so the larger number is (2).

Let the breadth be \(x\) units. Then the length is \(x+3\) units, so \(x(x+3)=70\), or \(x^2+3x-70=0\). Factoring gives \((x+10)(x-7)=0\). Since a breadth must be positive, \(x=7\), and the length is \(7+3=10\) units. The root \(-10\) is not physically meaningful. Exam tip: In geometry word problems, reject roots that make a length or breadth negative.

Let the sides be \(l\) and \(w\) cm. Perimeter \(34\) gives \(2(l+w)=34\Rightarrow l+w=17\). Area \(60\) gives \(lw=60\). Substitute \(w=17-l\): \(l(17-l)=60\Rightarrow 17l-l^2=60\Rightarrow l^2-17l+60=0\). Factoring: \((l-5)(l-12)=0\), so \(l=5\) or \(l=12\). The larger side is \(12\) cm.

Let the breadth be \(x\) units. Then the length is \(x+4\) units. Using the area, \(x(x+4)=96\), so \(x^2+4x-96=0\). Factoring gives \((x-8)(x+12)=0\), yielding \(x=8\) or \(x=-12\). Since a breadth cannot be negative, the correct answer is 8. Exam tip: In geometry word problems, reject any negative root because lengths must be positive.

Let the side of the square be \(x\). Its area is \(x^2\) and its perimeter is \(4x\). Therefore, \(x^2=4x+45\), giving \(x^2-4x-45=0\). Factoring, \((x-9)(x+5)=0\), so \(x=9\) or \(x=-5\). Since a length cannot be negative, the valid answer is 9. Exam tip: For a geometrical length, reject any negative root of the quadratic equation.

Use the sum-and-product relationship for a quadratic equation. Let one number be y; then the other is 15 − y. Their product gives y(15 − y) = 54, so y² − 15y + 54 = 0. Factoring yields (y − 9)(y − 6) = 0, hence the two numbers are 9 and 6. The larger number is therefore 9. Substitution verifies both conditions: 9 + 6 = 15 and 9 × 6 = 54. Option B is the smaller number, not the requested larger one. Options C and D do not produce the required product with a partner having sum 15: 12 would pair with 3 and give product 36, while 15 would pair with 0 and give product 0.

If the smaller number is (x), the larger is (x+2). From (x^2+(x+2)^2=130), (x=7), so the larger number is (9).

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 unknown number be x. The statement translates directly to x² − 11 = 85. Adding 11 to both sides gives x² = 96. Since 96 = 16 × 6, its square root is √96 = 4√6. The question asks for the positive number, so the positive root is selected; mathematically the equation also has the negative root −4√6, but it is not an allowed answer here. Option B simplifies to 12, whose square is 144 and does not satisfy the equation. Option C gives 81 − 11 = 70, and option D gives 144 − 11 = 133. Therefore option A is the only value that satisfies the original condition.

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 \(3(15)^2\neq 180\).

Set the stated revenue equal to 600: x(50 − x) = 600. Expanding and rearranging gives 50x − x² = 600, so x² − 50x + 600 = 0. The quadratic factors as (x − 20)(x − 30) = 0. Therefore the two possible values are x = 20 and x = 30, and the larger one is 30. Both can be verified in the original model: 20(50 − 20) = 20 × 30 = 600, while 30(50 − 30) = 30 × 20 = 600. Hence option A is correct. Option B is the smaller valid value, option C gives revenue 625, and option D gives revenue 400. The wording asks specifically for the larger solution.

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.

For a boat and stream, downstream speed is 15 + x and upstream speed is 15 − x. Therefore the total-time equation is 36/(15 + x) + 24/(15 − x) = 4, with 0 < x < 15. Testing the proposed value x = 3 gives downstream speed 18 and upstream speed 12. The corresponding times are 36/18 = 2 hours and 24/12 = 2 hours, whose sum is 4 hours. Thus x = 3 satisfies the original condition. For completeness, clearing denominators produces a quadratic whose admissible root is 3; any root violating x < 15 would be physically invalid. Options 2, 4, and 5 do not make the two travel times add to exactly four 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.