Word Problems and Applications

Correct answer: A. 13

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.

Correct answer: A. 13

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.

Correct answer: A. 14

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.

Correct answer: A. (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).

Correct answer: A. 10 units

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.

Correct answer: B. \(12\)

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.

Correct answer: A. 8

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.

Correct answer: A. 9

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.

Correct answer: A. 9

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.

Correct answer: A. (9)

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).

Correct answer: A. 14

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.

Correct answer: A. 4√6

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.

Correct answer: A. 12 metres

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.

Correct answer: A. 8

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.

Correct answer: A. 12

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.

Correct answer: A. \(2\sqrt{15}\)

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

Correct answer: A. 30

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.

Correct answer: A. \(\frac{5(1+\sqrt{33})}{2}\) m

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.

Correct answer: A. 50 kilometres per hour

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.

Correct answer: A. 60 kilometres per hour

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.

Correct answer: A. 3

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.

Correct answer: A. (25)

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

Correct answer: A. 36

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.

Correct answer: A. 20

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.

Correct answer: A. 25 वर्ष

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.