Word Problems and Applications

Let the number be \(x\). Then \(x^2=24x+640\), so \(x^2-24x-640=0\). Factoring gives \((x-40)(x+16)=0\), hence \(x=40\) or \(x=-16\). Therefore, the positive number is 40. Exam tip: Form the quadratic equation first, find both roots, and then select the root matching the condition in the question.

Let the number be \(x\). Then \(x^2-27x=1000\), so \(x^2-27x-1000=0\). Using the quadratic formula, \(x=\frac{27\pm\sqrt{27^2+4(1000)}}{2}=\frac{27\pm\sqrt{4729}}{2}\). The positive root is \(\frac{27+\sqrt{4729}}{2}\); the other root is negative. Therefore, option B is correct. Exam tip: form the quadratic equation carefully and use the stated positivity condition to reject the negative root.

Let the number be \(x\). Then \(x^2+x=1892\), so \(x^2+x-1892=0\). Factoring gives \((x-43)(x+44)=0\), hence \(x=43\) or \(x=-44\). Therefore, the positive number is \(43\). Verification: \(43^2+43=1849+43=1892\); 42 and 44 do not produce this sum. Exam tip: Check the signs of both roots and select only the positive root when the question specifies it.

(x^2+444=1900) gives (x^2=1456), which is not a perfect square. Therefore, none of the options can be correct.

Let the number be \(x\). Then \(x^2-512=1809\), so \(x^2=2321\). Therefore, the positive value is \(x=\sqrt{2321}\), approximately 48.18. Although 48 is close, \(48^2-512=1792\), not 1809. Exam tip: Form the equation first, isolate the square, and then take the positive square root when the question asks for the positive value.

((x+14)^2=2500) and (x+14=50), so (x=36). If the positive case is clear, take the positive square root.

Let the unknown number be x. The phrase “15 less than a number” means x−15, so the given statement becomes (x−15)²=2209. Since 2209=47², the square equation has two algebraic possibilities: x−15=47 or x−15=−47. The question specifies the positive case, so we choose x−15=47. Adding 15 to both sides gives x=62, which is option C. The other branch would give x=15−47=−32 and does not represent the requested positive case. Option A is only the positive square-root value of 2209 and forgets to add 15. Options B and D do not satisfy the equation. Indeed, substituting x=62 gives (62−15)²=47²=2209.

Let the son's age be \(x\) years. Then the father's age is \(x+36\) years. Therefore, \(x(x+36)=1980\), giving \(x^2+36x-1980=0\). Factoring yields \((x-30)(x+66)=0\). Since age cannot be negative, \(x=30\), so the father's age is \(30+36=66\) years. Exam tip: In age problems, reject any negative root of the quadratic equation.

If the daughter's age is (x), then (x(x+34)=1760), giving (x=22). In age problems, ignore the negative root.

((x+9)(x-9)=2119) gives (x^2-81=2119), so (x^2=2200). This is not a perfect square, so none of the given options is correct.

((x+8)(x-8)=1936) gives (x^2-64=1936) and (x^2=2000). Therefore, none of the given options is a correct solution.

The original area is \(x^2\), and the new area is \((x+10)^2\). Hence, \((x+10)^2-x^2=1700\), which gives \(20x+100=1700\). Therefore, \(20x=1600\) and \(x=80\) metres, so option B is correct. Exam tip: represent the increase as the difference of the two areas; it is not merely \(10^2\).

After the reduction, the side is (x − 7) cm, so (x − 7)² = 1296. The reduced side must be positive, and √1296 = 36; therefore x − 7 = 36 and x = 43 cm. The negative square root would give an invalid negative side length.

If breadth is (x) and length is (x+18), the increase is ((x+8)(x+26)-x(x+18)=1056). This gives (x=54).

Let the original breadth be x m. Then the original length is x+15 m. After the increases, the length is x+21 m and the breadth is x+5 m, so \((x+21)(x+5)=1944\). On simplifying, \(x^2+26x-1839=0\), whose positive solution is \(x=-13+2\sqrt{502}\approx31.81\). The negative solution is not valid for a length or breadth. Exam tip: In dimension-change problems, define one dimension as x and equate the product of the changed dimensions to the new area.

If breadth is (x), then (x(x+25)=2100), giving (x=35) and length (60). The wire length is (2(35+60)=190) cm.

Since the path surrounds the garden on all four sides, add twice its width to each garden dimension: the outer length is \(46+2\times4=54\) m and the outer breadth is \(32+2\times4=40\) m. Therefore, the total area is \(54\times40=2160\) square metres, so option B is correct. Option A results from using incorrect outer dimensions. Exam tip: for a path around all four sides, add \(2\times\) the path width to both dimensions before multiplying.

The path adds x metres to each side, so the outer dimensions are (50 + 2x) m and (36 + 2x) m. At x = 5, the area is 60 × 46 = 2760 m². Since the outer area increases as the non-negative path width increases, this is the unique valid width: 5 m.

The border is removed once from each of the two opposite sides, so the inner side is 68 - 2x. Thus, 68 - 2x = 44, giving 2x = 24 and x = 12 metres. The total difference of 24 metres cannot be taken directly as x because it represents two border widths. Exam tip: For a uniform inner border of width x, the difference between the outer and inner sides is always 2x.

This is a speed–time–distance word problem that leads to a quadratic equation. Let the original speed be x km/h, with x>0. The original time is 360/x hours, while the time at the increased speed is 360/(x+10) hours. Their difference is 3, so 360/x − 360/(x+10)=3. On simplifying, 3600/[x(x+10)]=3, giving x²+10x−1200=0. Factoring gives (x−30)(x+40)=0. Since speed cannot be negative, x=30 km/h. Checking: 360/30=12 hours and 360/40=9 hours, a reduction of exactly 3 hours. Therefore option A is correct; the other positive choices do not satisfy the stated time difference.

If there are n students, each handshake corresponds to choosing an unordered pair of distinct students. Therefore the number of handshakes is C(n, 2) = n(n − 1)/2. Setting this equal to 190 gives n(n − 1) = 380, or n^2 − n − 380 = 0. Factoring yields (n − 20)(n + 19) = 0, so n = 20 or n = −19. A number of students cannot be negative, hence n = 20. Option C is correct. The negative root is rejected by the context, while 18, 19, and 21 do not satisfy the pair-count formula: for 20 students, 20 × 19 / 2 = 190 exactly. This is a standard quadratic-equation application based on counting pairs.

The correct answer is C, 8/3 km/h. Let the stream speed be x km/h. Then the still-water speed is x + 8. Downstream speed is (x + 8) + x = 2x + 8, while upstream speed is (x + 8) − x = 8 km/h. Equal times give 40/(2x + 8) = 24/8 = 3. Hence 40 = 3(2x + 8), so 40 = 6x + 24 and 6x = 16, giving x = 8/3 km/h. Option A gives a downstream time of 40/12 and is not equal to 3 hours. Option B gives 40/16 = 2.5 hours. Option C gives downstream speed 40/3, so its time is 3 hours, matching the upstream time. Option D gives a different time. The key idea is to add the stream speed downstream and subtract it upstream.

This is a quadratic word problem involving a fixed total cost and a change in unit price. If the original price is x rupees, the original number of articles bought is 300/x. At the reduced price x − 5, the number is 300/(x − 5). The statement says the latter exceeds the former by 3: 300/(x − 5) − 300/x = 3. Combining the fractions gives 1500/[x(x − 5)] = 3, so x(x − 5) = 500. Therefore x² − 5x − 500 = 0, which factors as (x − 25)(x + 20) = 0. Price must be positive and greater than ₹5, so x = ₹25. Hence option B is correct; ₹20, ₹30 and ₹35 do not make the quantity difference equal to three.

The correct answer is B, 25 m. Let the breadth be x m. The length is then 2x − 5 m. Using area = length × breadth, x(2x − 5) = 1125. Rearranging gives 2x² − 5x − 1125 = 0. Factoring or checking the options, x = 25 works because the length is 2(25) − 5 = 45 m and the area is 25 × 45 = 1125 m². Option A gives length 35 m and area 700 m², so it fails. Option B gives exactly the required area. Option C gives length 55 m and area 1650 m², while option D gives length 65 m and area 2275 m². The negative algebraic root, if obtained, would not represent a physical breadth. Check both the relation and the area before accepting an answer.

For a boat-and-stream problem, if x is the speed in still water, the upstream speed is x − 2 and the downstream speed is x + 2. Therefore, the time equation is 30/(x − 2) + 30/(x + 2) = 8, with x > 2. Multiplying by (x − 2)(x + 2) gives 60x = 8(x² − 4), or x² − 7.5x − 4 = 0; equivalently, 2x² − 15x − 8 = 0. Its positive solution is x = 8 (the other root is negative). Checking: 30/6 = 5 hours upstream and 30/10 = 3 hours downstream, total 8 hours. Thus option B is correct; the other choices fail this time check.