Word Problems and Applications

((x+10)(x-10)=1500) gives (x^2-100=1500), so (x=40). The difference of squares idea helps in such questions.

The original area is \(x^2\), and the new area is \((x+12)^2\). Hence, \((x+12)^2-x^2=2448\), which gives \(24x+144=2448\). Therefore, \(24x=2304\) and \(x=96\) metres. Thus, 96 is correct. Exam tip: For such questions, use \((a+b)^2-a^2=2ab+b^2\) to simplify the calculation quickly.

The reduced side is \((x-9)\) cm, so \((x-9)^2=1681=41^2\). Since a side length is positive and the reduced side must also be positive, \(x-9=41\), not \(-41\). Hence, \(x=50\) cm. Exam tip: In quadratic equations involving lengths, retain only the positive root that fits the situation.

If breadth is (x) and length is (x+22), the increase is ((x+10)(x+32)-x(x+22)=1520). This gives (x=60).

Let the original breadth be (x) m. Then the original length is (x+20) m. After the increases, the dimensions become (x+28) m and (x+6) m. Thus, (x+28)(x+6)=3360
, which gives x^2+34x-3192=0. Its roots are x=42 and x=-76. Since a rectangle cannot have a negative dimension, the original breadth is 42 m. Exam tip: In geometrical word problems, reject any negative root because dimensions must be positive.

If the side of the square is (x), then (x^2=3025), so (x=55). The wire length is the perimeter (4x=220) cm.

If breadth is (x), then (x(x+30)=2800), giving (x=40) and length (70). The wire length is (2(40+70)=220) cm.

Since the path surrounds the garden on all four sides, 5 + 5 = 10 m must be added to both dimensions. Thus, the outer length is 60 + 10 = 70 m and the outer width is 42 + 10 = 52 m. The total area is \(70 \times 52 = 3640\) square metres, so option C is correct. Exam tip: for a path around all sides, add twice the path width to each original dimension.

The outer dimensions, including the path, are \(64+2x\) metres and \(44+2x\) metres. Thus, \((64+2x)(44+2x)=3996\). On simplification, \(x^2+54x-295=0\), which factors as \((x-5)(x+59)=0\). The roots are \(5\) and \(-59\), but a width cannot be negative, so \(x=5\) metres. Exam tip: for a uniform path around all four sides, add \(2x\) to each original dimension.

The border occupies width (x) on both opposite sides, so the inner side is (92-2x) metres. Thus, (92-2x=60) , giving (2x=32) and (x=16) metres. Hence, option C is correct. Exam tip: for a border around a square, the difference between the outer and inner sides is (2x) , not (x) ; therefore, 32 m is the total reduction, not the border width.

Let the breadth be (x\text{ m}), so (x(x+5)=300), giving (x=15). In exams, reject the negative dimension.

Let the original speed be (x), then (\frac{360}{x}-\frac{360}{x+10}=3). This gives (x^2+10x-1200=0), so (x=30).

Let the faster pipe take (x\text{ h}), then (\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}). This gives (x^2-7x-30=0), so (x=10).

Let the number of students be (n), then (\frac{n(n-1)}{2}=190). This gives (n^2-n-380=0), so (n=20).

Let the shorter leg be \(x\text{ cm}\); then the longer leg is \((x+5)\text{ cm}\). By the Pythagorean theorem, \(x^2+(x+5)^2=25^2\), which simplifies to \(x^2+5x-300=0\). Factoring gives \((x-15)(x+20)=0\), so \(x=15\) or \(-20\). Since a length cannot be negative, the shorter leg is \(15\text{ cm}\). Exam tip: verify the result using \(15^2+20^2=25^2\).

Let the side of the field be (x\text{ m}), so the outer side is (x+4) and ((x+4)^2=196). Hence (x=10).

Let the digits be (x) and (11-x), then (x(11-x)=30), giving (x=5) or (x=6). The possible numbers are (56) and (65), so the larger one is (65).

From (20t-5t^2=15), we get (t^2-4t+3=0). Hence (t=1) and (t=3), representing upward and downward motion.

Let the stream speed be (x), so still-water speed is (x+8). From (\frac{40}{2x+8}=\frac{24}{8}), we get (x=4).

If the length is (x), the breadth is (23-x) and (x(23-x)=120). The roots are (8) and (15), so the length is (15\text{ cm}).