Word Problems and Applications

Let the smaller integer be \(x\). Then the larger integer is \(x+1\), so \(x(x+1)=306\), or \(x^2+x-306=0\). Factoring gives \((x-17)(x+18)=0\). Since the integers are positive, \(x=17\), and the larger integer is \(x+1=18\). Therefore, 18 is correct. Exam tip: For consecutive integers, represent them as \(x\) and \(x+1\) before forming the quadratic equation.

The direct answer is A, 19. Consecutive positive odd integers differ by 2, not by 1. Let the smaller integer be x; the next is x+2. Their product gives \\(x(x+2)=399\\), or \\(x^2+2x-399=0\\). We can check the factor pair \\(19\\times21=399\\), so x=19. Option A is correct. Option B, 21, is the larger integer, not the smaller one. Option C, 17, would be paired with 19 and gives 323, not 399. Option D, 23, would be paired with 25 and gives 575. The positivity condition also rules out any negative pair. Memory cue: consecutive odd numbers are written x and x+2.

Let the smaller even integer be \(x\). The next consecutive even integer is \(x+2\). Thus, \(x(x+2)=288\), giving \(x^2+2x-288=0\). The positive solution is \(x=16\), so the larger integer is \(16+2=18\). Exam tip: consecutive even integers differ by 2, not 1.

Let the number be (x), so (x+\frac{1}{x}=\frac{10}{3}). This gives (3x^2-10x+3=0), so the larger number is (3).

Let the breadth be \(x\) units. Then the length is \(x+5\) units. From the area, \(x(x+5)=204\), so \(x^2+5x-204=0\). Factoring gives \((x-12)(x+17)=0\); hence the positive breadth is \(12\) units and the length is \(12+5=17\) units. Thus, 12 units is the breadth, not the length. Exam tip: reject the negative root because a rectangle’s dimensions must be positive.

Let the sides of the rectangle be \(x\) cm and \(y\) cm. From the perimeter, \(2(x+y)=46\), so \(x+y=23\). The area gives \(xy=126\). Therefore, \(x(23-x)=126\), which becomes \(x^2-23x+126=0\), or \((x-9)(x-14)=0\). Thus, the sides are 9 cm and 14 cm, so the larger side is 14 cm. Exam tip: divide the perimeter by 2 first to obtain the sum of the two sides.

Let the breadth be \(x\) units. Then the length is \(x+6\) units. Using the area, \(x(x+6)=216\), so \(x^2+6x-216=0\). Factoring gives \((x-12)(x+18)=0\), leading to \(x=12\) or \(x=-18\). Since a length cannot be negative, the breadth is \(12\) units. Exam tip: In rectangle word problems, assign \(x\) to the smaller dimension and form the area equation first.

Let the side of the square be x. Its area is x² and its perimeter is 4x. According to the question, x² = 4x + 96, so x² − 4x − 96 = 0. Factoring gives (x − 12)(x + 8) = 0. Thus x = 12 or −8; a side length cannot be negative, so the correct answer is 12. Exam tip: In measurement problems, retain only the positive root.

Let the number be \(x\). From the question, \(x^2=11x+42\). Therefore, \(x^2-11x-42=0\), which factors as \((x-14)(x+3)=0\). Hence, \(x=14\) or \(x=-3\). The positive number is \(14\). Option 3 is incorrect because the other root is \(-3\), not \(3\). Exam tip: After finding both roots of a quadratic equation, always check the required condition, such as positive or negative.

Let the numbers be \(a\) and \(b\). Then \(a+b=29\) and \(ab=204\). So \(a\) and \(b\) are roots of \(t^2-29t+204=0\). Factor: \(t^2-29t+204=(t-12)(t-17)\). Hence the numbers are \(12\) and \(17\), so the larger number is \(17\).

The direct answer is A, 21. Let the smaller number be x. Since the difference is 8, the larger number is x+8. Their product gives \\(x(x+8)=273\\), so \\(x^2+8x-273=0\\). The factorisation is \\( (x-13)(x+21)=0\\), giving x=13 or x=-21. For the intended positive numbers, x=13, and the larger number is \\(13+8=21\\). Option A is correct. Option B, 13, is the smaller number. Option C, 19, differs from 21 by only 2 and gives product 399 with 21. Option D, 24, differs from 16 by 8, but their product is 384. Always identify which number the question asks for after solving.

If the smaller number is (x), the larger is (x+4). From (x^2+(x+4)^2=250), (x=9), so the larger number is (13).

Let the positive integer be \(x\). Its product with the next integer is \(x(x+1)=552\), so \(x^2+x-552=0\). Factoring gives \((x-23)(x+24)=0\), yielding \(x=23\) or \(x=-24\). Since the number is specified as positive, \(x=23\) is correct. Check: \(23\times24=552\). Exam tip: Represent the number and its next integer as \(x\) and \(x+1\) before forming the quadratic equation.

Let the number be \(x\). Then \(x^2-17=127\), so \(x^2=144\) and \(x=\pm 12\). Since the question asks for the positive number, the answer is 12. The values 10 and 14 are incorrect because their squares are 100 and 196, respectively. Exam tip: after taking a square root, consider both signs and then select the value required by the question.

Let the height reached on the wall be \(x\) units. Then the distance of the foot of the ladder from the wall is \(x-17\) units. By the Pythagorean theorem, \(x^2+(x-17)^2=25^2\). This gives \(x^2-17x-168=0\), or \((x-24)(x+7)=0\). Thus \(x=24\) units, since a physical length cannot be negative. Exam tip: reject any negative root when solving a word problem involving length or distance.

Let the shorter leg be \(x\). Then the other leg is \(x+31\). By the Pythagorean theorem, \(x^2+(x+31)^2=41^2\), which simplifies to \(x^2+31x-360=0\). Factoring gives \((x-9)(x+40)=0\), so the roots are 9 and −40. Since a length cannot be negative, the shorter leg is 9. Exam tip: always reject a negative root when solving for a physical length.

Let the breadth be \(x\) units. Then the length is \(x+4\) units, so the area gives \(x(x+4)=360\). Thus, \(x^2+4x-360=0\), which factors as \((x-18)(x+20)=0\). Therefore, \(x=18\) or \(x=-20\). Since a length cannot be negative, the breadth is 18 units. Exam tip: Always reject a negative root when solving a geometrical measurement problem.

Let the breadth be \(x\) units. Then the length is \(2x\) units. Using area, \(x\times 2x=242\), so \(2x^2=242\), giving \(x^2=121\) and \(x=\pm 11\). Since length and breadth must be positive, the breadth is \(11\) units. The value \(22\) represents the length, not the breadth. Exam tip: reject the negative root when solving measurement problems.

We have \(x(70-x)=1200\). Rearranging gives \(x^2-70x+1200=0\), which factors as \((x-30)(x-40)=0\). Thus, \(x=30\) or \(x=40\), and the larger value is 40. Exam tip: In such problems, form the quadratic equation first, find both roots, and then select the one requested.

Let the breadth of the field be \(x\) metres. Then its length is \(x+8\) metres. Using the area, \(x(x+8)=384\), so \(x^2+8x-384=0\). Factoring gives \((x-16)(x+24)=0\), hence \(x=16\) or \(x=-24\). Since a length cannot be negative, the breadth is 16 m. Option B (24 m) is the corresponding length, not the breadth. Exam tip: Always reject negative roots when solving geometrical measurement problems.

Let the original speed be \(x\) km/h. Then \(\frac{360}{x}-\frac{360}{x+15}=2\). On simplifying, \(x(x+15)=2700\), so \(x^2+15x-2700=0\). Factoring gives \((x-45)(x+60)=0\), hence \(x=45\) or \(x=-60\). Since speed cannot be negative, the original speed is \(45\) km/h. Check: at 45 km/h the time is 8 hours, and at 60 km/h it is 6 hours, a decrease of exactly 2 hours. Exam tip: reject any negative root when the variable represents speed, distance, or time.

Let the original speed be \(x\) km/h. The original travel time is \(\frac{300}{x}\) hours, while the time at the reduced speed is \(\frac{300}{x-10}\) hours. Therefore, \(\frac{300}{x-10}-\frac{300}{x}=1\). On simplifying, \(x^2-10x-3000=0\), giving \(x=60\) or \(x=-50\). Since speed cannot be negative, the original speed is \(60\) km/h. Check: at 60 km/h the time is 5 hours, and at 50 km/h it is 6 hours, an increase of exactly 1 hour. Exam tip: reject physically impossible negative roots and verify the remaining value in the original condition.

The governing concept is forming a rational equation from speed, distance, and time. Downstream speed is 20+x and upstream speed is 20-x, with 0<x<20. Therefore the total-time equation is 48/(20+x)+32/(20-x)=4. Testing x=4 gives 48/24+32/16=2+2=4, so option A satisfies the conditions. For completeness, clearing denominators gives 48(20-x)+32(20+x)=4(400-x^2), which simplifies to x^2-4x-80=0, so x=10 or x=-8; wait, this indicates the stated numerical data do not support x=4. Direct substitution confirms that x=4 works only because 48/24+32/16=4, while the cleared equation calculation must be expanded carefully: 1600-16x=1600-4x^2, hence x(x-4)=0. The positive stream speed is x=4. Distractors do not satisfy the total-time equation.

If the number of students is (x), then (\frac{1800}{x}-\frac{1800}{x+10}=6). Solving gives (x=50).

Let x be the original number of students. The original share is 1680/x, while with four fewer students the share would be 1680/(x-4). The condition says 1680/(x-4)-1680/x=10, with x>4. Combining fractions gives 6720/[x(x-4)]=10, so x(x-4)=672. Hence x^2-4x-672=0, which factors as (x-28)(x+24)=0. The mathematically possible roots are x=28 and x=-24, but a number of students must be positive, so x=28. Option A is correct. Options B, C, and D do not make the increase in each share equal to 10 rupees when checked in the original condition.