Word Problems and Applications

Let breadth be (x), then (x(x+4)=240), so (x=12) and length is (16). The perimeter is (2(12+16)=56\text{ m}), so choose (56\text{ m}).

Let the actual price be (x), then (\frac{800}{x-2}-\frac{800}{x}=20). This gives (x^2-2x-80=0), so (x=10).

Let the smaller odd integer be (x), then (x^2+(x+2)^2=394). This gives (x^2+2x-195=0), so (x=13).

Let the number be \(x\). Then \(x^2=9x+22\), so \(x^2-9x-22=0\). Factoring gives \((x-11)(x+2)=0\), hence \(x=11\) or \(x=-2\). Since the number is positive, \(-2\) is rejected, making 11 the correct answer. Exam tip: Translate the wording into an equation first, then apply the condition that the required number must be positive.

After moving the foot \(2\text{ m}\) farther, the new horizontal distance from the wall is \(6+2=8\text{ m}\). The ladder, wall and ground form a right-angled triangle. By the Pythagorean theorem, \(h^2+8^2=10^2\), so \(h^2=36\) and \(h=6\text{ m}\), since height is positive. The option \(8\text{ m}\) is the new horizontal distance, not the height. Exam tip: Treat the ladder as the hypotenuse and recalculate the perpendicular height after updating the horizontal distance.

Let correct answers be (x), so wrong answers are (30-x), and (4x-(30-x)=85). This is a linear check inside applications, giving (x=23).

Let the photo breadth be (x), so outer dimensions are ((x+2)) and ((x+9)). From ((x+2)(x+9)=180), we get (x=9).

Let the number be (x), then (x(x+5)=204). This gives (x^2+5x-204=0), so the positive number is (12).

Let the onward speed be (x), then (\frac{150}{x-25}-\frac{150}{x}=1). This gives (x^2-25x-3750=0), so (x=75).

Let breadth be (x), then (x^2+(x+7)^2=17^2). This gives (x=8), length (15), and area (120\text{ cm}^2).

Let the first friend's time be (x), so the second's is (x+10), and (\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}). This gives (x^2-14x-120=0), so (x=20).

Let the side of the original square be x centimetres. Cutting away a strip 3 centimetres wide from one side leaves a rectangle whose dimensions are x and x-3. Its area is therefore x(x-3), and the given information produces x(x-3)=70. Since a side length must be positive and greater than 3, only a physically meaningful positive solution can be used.

Expanding gives x^2-3x-70=0. Factoring yields (x-10)(x+7)=0, so x=10 or x=-7. A length cannot be negative, so x=-7 is rejected. With x=10, the remaining rectangle has dimensions 10 cm and 7 cm, and its area is 70 square centimetres. Therefore the original square side was 10 cm, option B.

Let breadth be (x) and length (x+12), then path area is (x(x+12)-(x-4)(x+8)). The given values should give (8x+32); for (24\text{ m}), the path area is (224\text{ m}^2).

From (1) to (9), (9) digits are used, and from (10) to (99), (180) digits are used. The remaining (83) digits do not make complete three-digit pages, so this data is inconsistent for a whole page count.

Let the smaller side be \(x\) cm. Then the larger side is \((x+9)\) cm. Using the area, \(x(x+9)=252\), so \(x^2+9x-252=0\). Factoring gives \((x-12)(x+21)=0\), yielding \(x=12\) or \(x=-21\). Since a length cannot be negative, the smaller side is \(12\text{ cm}\). Exam tip: Always reject a negative root when solving for a physical length.

Let the number be (x), then (x^2-14x=-45), i.e. (x^2-14x+45=0). The roots are (5) and (9), so the larger number is (9).

Let the number of rows be \(x\). Then the total number of students is \(x(x+3)=180\), so \(x^2+3x-180=0\). Factoring gives \((x+15)(x-12)=0\), hence \(x=12\) or \(x=-15\). Since the number of rows cannot be negative, the valid answer is 12. Exam tip: For arrangement problems, write total objects = number of rows × objects in each row.

Let pipe (A)'s time be (x), then (\frac{1}{x}+\frac{1}{x+4}=\frac{1}{4}). This gives (x^2-4x-16=0), so the actual time is (2+2\sqrt{5}\text{ h}), not an integer option.

Let the numbers be (x) and (21-x), then (x(21-x)=108). The roots are (9) and (12), so the larger number is (12).

Let breadth be (x), then length is (3x-2), and (x(3x-2)=161). This gives (3x^2-2x-161=0), so (x=7).