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.

Consecutive even numbers are (x) and (x+2). From (x^2+(x+2)^2=3044), (x=38), so the larger number is (40).

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 area of a square is (side)^2, so (x+17)^2=3249=57^2. Since a side length is positive, x+17=57; hence x=57-17=40. Option D is incorrect because 57 is the side length, not the value of x. Exam tip: When the area of a square is given, first take its positive square root to find the side.

The area of a square is the square of its side, so \((x-12)^2=1849=43^2\). Thus, \(x-12=43\) or \(x-12=-43\). Since a tile cannot have a negative side length, \(x-12=43\), giving \(x=55\). The negative algebraic root would make the side \(-43\) cm, which is not physically valid. In an exam, always apply the positive-length condition when taking a square root.

By Pythagoras, (x^2+(x+30)^2=150^2), giving (x=90). Form the equation by taking the hypotenuse as the largest side.

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.

If the base distance is (x), then (x^2+84^2=(x+42)^2), giving (x=63). The hypotenuse is always the greatest length.

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.