Word Problems and Applications

If the smaller integer is (x), then (x^2+(x+1)^2=1513), giving (x=27). Write consecutive integers as (x) and (x+1).

Let the smaller number be x. The next consecutive odd number is x+2. Thus, \(x^2+(x+2)^2=2314\), which simplifies to \(2x^2+4x-2310=0\) or \(x^2+2x-1155=0\). Factoring gives \((x-33)(x+35)=0\). Since the numbers are positive, \(x=33\) is valid, while \(x=-35\) is not. Therefore, the correct answer is 33. Exam tip: For consecutive odd numbers, use a difference of 2 and reject any non-positive root.

The direct answer is C, 40. Consecutive positive even numbers can be written as x and x+2. Their squares satisfy x²+(x+2)²=3044, so 2x²+4x−3036=0, or x²+2x−1518=0. This factors as (x−38)(x+40)=0. The positive value is x=38, making the larger number 40. Option A, 36, would pair with 38 and give 1296+1444=2740. Option B, 38, is the smaller number, not the requested larger one. Option C works because 38²+40²=1444+1600=3044. Option D, 42, would pair with 40 and give 3364. Remember: for consecutive even numbers, add 2, then identify which member the question asks for.

If the smaller number is (x) and the larger is (x+19), then (x(x+19)=1632). This gives smaller number (32) and larger number (51).

Let the smaller number be x; then the other number is 90 − x. Thus, \(x^2+(90-x)^2=4068\), which simplifies to \(2x^2-180x+4032=0\), or \(x^2-90x+2016=0\). Factoring gives \((x-42)(x-48)=0\), so the two numbers are 42 and 48. Therefore, the smaller number is 42. Exam tip: When the sum and the sum of squares are given, you can also use \(a^2+b^2=(a+b)^2-2ab\) to find the numbers efficiently.

If breadth is (x), then (x(x+23)=2220), giving (x=37). For a rectangle, use the product of length and breadth for area.

Let the breadth be \(x\) cm. Then the length is \(x+19\) cm. Using the area, \(x(x+19)=2772\), so \(x^2+19x-2772=0\). Factoring gives \((x-44)(x+63)=0\). Thus \(x=44\) cm, since a negative dimension is not possible, and the length is \(44+19=63\) cm. Exam tip: Reject negative roots when solving quadratic equations involving physical dimensions.

If breadth is (x), then (x^2+(x+15)^2=75^2), giving (x=45). Use Pythagoras for a rectangle with a diagonal.

Let the breadth be \(x\) cm. Then the length is \(x+27\) cm, so \(x(x+27)=2170\), or \(x^2+27x-2170=0\). The positive solution is \(x=35\), giving breadth 35 cm and length 62 cm. Hence, the perimeter is \(2(35+62)=194\) cm. Exam tip: In rectangle word problems, form the quadratic equation from the area and retain only the positive dimension.

If one side is (x), the other is (101-x), and (x(101-x)=2508). The solutions are (44) and (57), so the shorter side is (44).

The square’s area is \(56^2=3136\) square metres. Equating the two areas gives \(x(x+66)=3136\), or \(x^2+66x-3136=0\). Factoring, \((x-32)(x+98)=0\), so \(x=32\) or \(x=-98\). Since a rectangle’s breadth must be positive, \(x=32\) m is the valid answer. In word problems, always reject roots that do not have a meaningful physical interpretation.

The side length is the positive square root of the area: √3249 = 57 cm. Thus x + 17 = 57, so x = 57 − 17 = 40.

The side length is the positive square root of the area: √1849 = 43 cm. Therefore x − 12 = 43, so x = 43 + 12 = 55. The negative square root is not possible because a tile cannot have a negative side length.

The direct answer is option C: 90 cm. In a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs. The legs are \(x\) and \(x+30\), while the hypotenuse is 150, so \(x^2+(x+30)^2=150^2\). Expanding gives \(x^2+x^2+60x+900=22500\), hence \(2x^2+60x-21600=0\). Dividing by 2 gives \(x^2+30x-10800=0\). Factoring gives \((x-90)(x+120)=0\), so \(x=90\) or \(x=-120\). A length cannot be negative, so \(x=90\). The other leg is 120 cm, and 90 is the smaller leg. Option A, 75 cm, does not satisfy the equation; its other leg would be 105 and the hypotenuse would not be 150. Option B, 80 cm, similarly fails. Option C works because \(90^2+120^2=8100+14400=22500=150^2\). Option D is the larger leg, not the smaller one. Memory cue: write Pythagoras as leg squared plus leg squared equals hypotenuse squared, then reject negative length.

Let the shorter leg be \(x\) cm. Then the longer leg is \(x+37\) cm. By the Pythagorean theorem, \(x^2+(x+37)^2=185^2\), which simplifies to \(x^2+37x-16428=0\). Factoring gives \((x-111)(x+148)=0\). Therefore, \(x=111\) cm, since a length cannot be negative; \(x=-148\) is rejected. Exam tip: represent the longer side as \(x+\) the given difference before applying the Pythagorean theorem.

From the right triangle, (x^2+120^2=(x+50)^2), giving (x=119). In ladder problems, the ladder is the hypotenuse.

The direct answer is B, 63 m. Let the horizontal distance from the wall be x metres. The ladder is x+42 metres, and the vertical height is 84 m. By Pythagoras, x²+84²=(x+42)². Expanding and cancelling x² gives 7056=84x+1764, so 5292=84x and x=63. Option A, 56, gives ladder length 98 and does not satisfy the triangle equation. Option B works: 63²+84²=3969+7056=11025=105², so the ladder is 105 m. Option C, 84, would make the ladder 126 m and fails. Option D, 105, is the ladder length, not the base distance. Exam cue: the wall height and ground distance are the perpendicular sides; the ladder is the hypotenuse.

Using distance = speed × time, we get \(x(x-92)=800\). Thus, \(x^2-92x-800=0\), which factors as \((x-100)(x+8)=0\). Therefore, \(x=100\) or \(x=-8\). Since speed cannot be negative, the valid speed is 100 km per hour. In such problems, always reject a negative root when it has no physical meaning.

Using distance = speed × time, \(x(x-85)=1056\), so \(x^2-85x-1056=0\). Factoring gives \((x-96)(x+11)=0\), hence \(x=96\) or \(x=-11\). Speed cannot be negative, so \(x=96\) km/h. Therefore, the time is \(x-85=96-85=11\) hours. Exam tip: In speed–time–distance problems, first write \(D=ST\) and reject any negative value that is physically impossible.

Downstream speed is (x+5), and (8(x+5)=216), so (x=22). In boat problems, speed changes with direction.

Using distance = speed × time, we get x(x+20)=1125, or x²+20x−1125=0. Factoring gives (x−25)(x+45)=0, so x=25 or x=−45. Since time cannot be negative, x=25 hours is the valid answer. Exam tip: In speed–distance–time problems, first write distance = speed × time and reject any negative time root.

The direct answer is option B: 30 days. Let the number of days be x. Daily work is x+18 units, so total work is x(x+18)=1440. Thus x²+18x−1440=0. Factoring gives (x−30)(x+48)=0, so x=30 or x=−48. A duration cannot be negative, leaving x=30. Check: the worker does 30+18=48 units per day, and 30×48=1440 units. Option A, 24, gives 24×42=1008, so it is too small. Option B, 30, gives exactly 1440 and is correct. Option C, 36, gives 36×54=1944, too large. Option D, 48, gives 48×66=3168, also too large. The quadratic comes directly from multiplying days by daily output; no extra formula is needed. Memory cue: total production equals time multiplied by production per day, and reject the negative root because time is positive.

The total number of questions gives \(x(x+22)=1400\). Thus, \(x^2+22x-1400=0\), which factors as \((x-28)(x+50)=0\). Therefore, \(x=28\) or \(x=-50\); since the number of days cannot be negative, the valid answer is 28. Exam tip: In word problems, reject algebraic roots that are impossible in the given context.

The total number of books gives the equation \(x(x+27)=1980\), or \(x^2+27x-1980=0\). Factoring gives \((x-33)(x+60)=0\), so \(x=33\) or \(x=-60\). Since the number of shelves cannot be negative, \(x=33\) is valid. Exam tip: In word problems, reject a negative root when the variable represents a count or measurement.

The total number of chairs gives x(x + 30) = 2,275, so x² + 30x − 2,275 = 0. Factoring gives (x − 35)(x + 65) = 0, so x = 35 or x = −65. A number of rows cannot be negative, so there are 35 rows.