Word Problems and Applications

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.

((x-15)^2=2209) and (x-15=47), so (x=62). After taking square root, choose the sign according to the condition.

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 of the square is \(x-7\) cm. Thus, \((x-7)^2=1296=36^2\). Since a side length is positive, \(x-7=36\), giving \(x=43\) cm. The other algebraic root, \(x-7=-36\), gives \(x=-29\) cm, which is not a valid side length. Exam tip: find the reduced side from the area first, then add back the amount reduced.

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 the side of the square is (x), then (x^2=2025), so (x=45). The wire length is the perimeter (4x=180) cm.

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.

Since the path surrounds the garden on all four sides, the outer dimensions are (50+2x) m and (36+2x) m. Therefore, (50+2x)(36+2x)=2760. Substituting x=5 gives 60×46=2760, so the width of the path is 5 m. The nearby values x=4 and x=6 give areas of 2552 and 2912 square metres, respectively. Exam tip: for a path of width x around all four sides, add 2x to each original dimension.

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.

If the units digit is (x), then the tens digit is (x-2). From (x(x-2)=24), (x=6), so the number is (46).