Word Problems and Applications

If breadth is (x), then (x(x+21)=1870), giving (x=34). For a rectangle, use the product of length and breadth for area.

Let the breadth of the rectangle be \(x\) cm. Its length is then \(x+16\) cm. Using the area, \(x(x+16)=2145\), so \(x^2+16x-2145=0\). Factoring gives \((x-39)(x+55)=0\). Since a breadth cannot be negative, \(x=39\) cm. Therefore, the length is \(39+16=55\) cm. Exam tip: reject any negative root when the variable represents a physical dimension.

If breadth is (x), then (x^2+(x+7)^2=97^2), giving (x=65). Use Pythagoras for a rectangle with a diagonal.

From (x(x+15)=2494), breadth is (43) and length is (58). The perimeter is (2(43+58)=202), so the correct option should be (202).

If one side is (x), the other is (93-x), and (x(93-x)=2016). The solutions are (36) and (56), so the shorter side is (36).

Since the areas are equal, \(x(x+28)=48^2=2304\). Thus, \(x^2+28x-2304=0\), which factors as \((x-36)(x+64)=0\). The roots are \(x=36\) and \(x=-64\). Since a breadth cannot be negative, the valid value is \(x=36\). Exam tip: In area-based word problems, form the equality first and reject any negative root that is not physically meaningful.

The area of a square is \((\text{side})^2\), so \((x+12)^2=1764=42^2\). Since a side length must be positive, \(x+12=42\), giving \(x=30\). Option D is the side length, not the value of \(x\). Exam tip: take the positive square root when finding a geometrical length.

A square’s area is (side)², so (x − 13)² = 1600. Taking square roots gives x − 13 = 40 or −40. Since a side length must be positive, x − 13 = 40, so x = 53.

The direct answer is B, 72 cm. Let the smaller leg be x cm; the other leg is x+24 and the hypotenuse is 120. Pythagoras gives \\(x^2+(x+24)^2=120^2\\). Expanding: \\(2x^2+48x+576=14400\\), so \\(x^2+24x-6912=0\\). This factors as \\( (x-72)(x+96)=0\\). Since a length is positive, x=72. Option B is correct. Option A, 66, gives legs 66 and 90, whose squares do not total \\(120^2\\). Option C, 80, gives legs 80 and 104 and fails the equation. Option D, 96, gives legs 96 and 120; a second leg cannot equal the hypotenuse in a non-degenerate right triangle. The smaller leg is the requested quantity.

Let the shorter leg be \(x\) cm; then the longer leg is \(x+17\) cm. By the Pythagorean theorem, \(x^2+(x+17)^2=145^2\), which simplifies to \(2x^2+34x-20736=0\). Applying the quadratic formula gives \(x=\frac{-17\pm\sqrt{41761}}{2}\). Since a length must be positive, the valid answer is \(\frac{\sqrt{41761}-17}{2}\) cm, approximately 93.68 cm. Option B results from using the wrong sign and is not a root of the equation. Exam tip: define the shorter leg first, use the Pythagorean theorem, and reject any negative length.

From the right triangle, (x^2+80^2=(x+20)^2), giving (x=150). In ladder problems, the ladder is the hypotenuse.

The direct answer is B, 128 m. Let the ground distance be x metres. The height is 96 m and the ladder length is x+32 m, the hypotenuse. Thus \\(x^2+96^2=(x+32)^2\\). Expanding gives \\(x^2+9216=x^2+64x+1024\\), so \\(64x=8192\\) and x=128. Option B is correct; the ladder is 160 m long. Option A, 112, would give a ladder of 144 and does not satisfy Pythagoras. Option C, 144, would give a ladder of 176 and also fails. Option D, 160, is the resulting ladder length, not the base distance; using it as x would make the ladder 192. Remember to assign the hypotenuse as x plus the stated extra length.

Using distance = speed × time, we get \(x(x-72)=720\), so \(x^2-72x-720=0\). Applying the quadratic formula gives \(x=36\pm12\sqrt{14}\). The negative root \(36-12\sqrt{14}\) is not a valid speed, so the speed is \(36+12\sqrt{14}\) km per hour, approximately 80.9 km per hour. Option 80 is incorrect because it would give a distance of \(80\times8=640\) km. Exam tip: Form the distance = speed × time equation first, then retain only the positive root that fits the context.

Using distance = speed × time, we get \(x(x-89)=990\). Thus, \(x^2-89x-990=0\), which factors as \((x-99)(x+10)=0\). Since speed must be positive, \(x=99\) km/h; the root \(x=-10\) is not physically meaningful. Therefore, the time is \(x-89=99-89=10\) hours. Exam tip: In speed–time problems, first use \(D=S\times T\), then reject any negative value that is not physically meaningful.

Downstream speed is (x+4), and (7(x+4)=154), so (x=18). In boat problems, speed changes with direction.

The width is \(x\) and the length is \(x+5\). Hence, \(x(x+5)=84\), which gives \(x^2+5x-84=0\). Option B uses the wrong sign for the added length. Exam tip: write both dimensions before forming the area equation.

The direct answer is option B: 28 days. Let the number of days be x. The worker completes x+14 units each day, so total work equals days × daily work: x(x+14)=1176. Hence x²+14x−1176=0. Factoring gives (x−28)(x+42)=0, so x=28 or x=−42. A number of days cannot be negative, so x=28. Check: daily work is 28+14=42 units, and 28×42=1176 units. Option A, 24, gives 24×38=912, not enough. Option B, 28, gives exactly 1176 and is correct. Option C, 32, gives 32×46=1472, too much. Option D, 42, gives 42×56=2352, also too much. The important model is total work = number of days × work per day. Memory cue: identify the two quantities being multiplied before forming the quadratic.

For the rectangle, area gives \(x(x+3)=54\), or \(x^2+3x-54=0\), which is quadratic. In option A, the perimeter gives only \(4x=48\), a linear equation. Exam tip: a product of two expressions containing \(x\) often produces a quadratic equation.

The total-number condition gives \(x(x+21)=1530\), so \(x^2+21x-1530=0\). Factoring gives \((x-30)(x+51)=0\), hence \(x=30\) or \(x=-51\). Since the number of shelves cannot be negative, the valid answer is 30. Exam tip: In word problems, reject any algebraic root that is not meaningful in the given context.

The total-chair condition gives \(x(x+23)=1674\). Thus, \(x^2+23x-1674=0\), which factors as \((x-31)(x+54)=0\). Therefore, \(x=31\) or \(x=-54\). Since the number of rows cannot be negative, the valid answer is 31. Exam tip: In arrangement problems, express the total as ‘number of rows × items in each row’.

From the total number of plants, \(x(x−12)=1408\). Thus, \(x^2−12x−1408=0\). Factoring gives \((x−44)(x+32)=0\), so \(x=44\) or \(x=−32\). A number of rows cannot be negative, so \(x=44\) is valid. Checking: \(44(44−12)=44×32=1408\). For option 40, the total would be \(40×28=1120\), not 1408. Exam tip: In word problems, retain only roots that make sense in the stated situation.

The total number of questions gives \(x(x-11)=1092\). Thus, \(x^2-11x-1092=0\), which factors as \((x-39)(x+28)=0\). Therefore, \(x=39\) or \(x=-28\). Since the number of students cannot be negative, the valid answer is 39. Check: \(39\times(39-11)=39\times28=1092\). In such word problems, always reject a negative root when the variable represents a count.

Let the original number be (x). Then (x+6)(x+13)=1394. On expanding, (x^2+19x-1394=0, which factors as (x+47)(x-28)=0. Thus (x=28 or x=-47). Among the given options, the correct answer is 28. The negative root (-47 also satisfies the equation, but it is not listed. Exam tip: In word problems, represent the unknown by (x), translate each condition algebraically, and then form the quadratic equation.

Let the original number be \(x\). Then \((x-9)(x+16)=1650\), which gives \(x^2+7x-1650=0\), or \((x-39)(x+46)=0\). Thus, \(x=39\) or \(x=-46\); the positive original number is \(39\). In such questions, translate ‘9 less’ as \(x-9\) and ‘16 more’ as \(x+16\) before expanding.

Let the original number be \(x\). Then \((x-8)(x+19)=2368\). Expanding gives \(x^2+11x-2520=0\), or \((x-45)(x+56)=0\). Thus \(x=45\) or \(x=-56\); since the number is positive, the answer is 45. Exam tip: Translate each phrase into an algebraic expression first, solve the resulting quadratic, and then apply any condition such as positivity.