Word Problems and Applications

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 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).

Set the given height equal to 15: \(20t-5t^2=15\u0005. Rearranging and dividing by \(-5\u0005 gives \(t^2-4t+3=0\u0005. Factoring produces \((t-1)(t-3)=0\u0005, so the possible times are \(t=1\u0005 s and \(t=3\u0005 s. Both are physically meaningful because the ball reaches the same height while rising and later while falling.

Therefore option A is correct. Substitution confirms that at either time the height is 15 m: at 1 s, \(20-5=15\u0005, and at 3 s, \(60-45=15\u0005. A single time would miss the two stages of the motion.

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}).

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}).

The direct answer is option B, ₹10 per pen. Let the actual price of one pen be \(x\) rupees. Then the number of pens bought is \(800/x\). If the price decreases by ₹2, the new price is \(x-2\), and the number bought is \(800/(x-2)\). The new number is 20 more, so \(\frac{800}{x-2}-\frac{800}{x}=20\). Combining the fractions gives \(\frac{1600}{x(x-2)}=20\), so \(x(x-2)=80\), or \(x^2-2x-80=0\). Factoring gives \((x-10)(x+8)=0\). Price must be positive and greater than ₹2, so \(x=10\). Check: at ₹10, 80 pens are bought; at ₹8, 100 pens are bought, exactly 20 more. Option A, ₹8, is the reduced price, not the actual price. Option B passes the check. Option C, ₹12, gives 66⅔ pens and does not produce the stated difference. Option D, ₹16, also fails. Memory cue: identify the original price first; the cheaper price is \(x-2\), not \(x\).

Let the smaller positive odd integer be x. Consecutive odd integers differ by 2, so the other integer is x+2. Their squared sum gives x²+(x+2)²=394. Expanding and simplifying produces 2x²+4x+4=394, hence x²+2x−195=0. Factoring gives (x+15)(x−13)=0, so x=13 or x=−15. Since the integers are positive, x=13 is the valid value.

Thus the smaller integer is 13, and option B is correct. Checking confirms the result: 13²+15²=169+225=394. The negative algebraic root is rejected because the question asks for positive integers. The supplied explanation reaches the correct answer, although it skips the intermediate division by 2 when displaying the simplified quadratic equation.

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 unknown number be x. The number that is 5 greater than it is x + 5. Their product is given as 204, so the condition is x(x + 5) = 204. This creates a quadratic equation, and both algebraic roots should be considered before selecting the intended positive number.

Expanding gives x² + 5x − 204 = 0. Factoring, (x − 12)(x + 17) = 0, so x = 12 or x = −17. The positive number asked for in the choices is 12, and 12 × 17 = 204. Thus option A is correct. The negative root also satisfies the equation algebraically but is not listed as the intended answer.

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.

The direct answer is C, 12. Let one positive number be x; the other is 21−x because their sum is 21. The product condition is x(21−x)=108, giving x²−21x+108=0. This factors as (x−9)(x−12)=0, so the two numbers are 9 and 12. The larger is 12. Option A, 9, is the smaller number, although it is part of the correct pair. Option B, 10, would require the other number to be 11, whose product is 110. Option C is correct because 9+12=21 and 9×12=108. Option D, 13, would pair with 8 and give product 104. Memory cue: when the sum is fixed, use x and sum−x, then choose the larger root.

The cost price is (1200-200=1000), and profit percentage is (\frac{200}{1000}\times100=20). This does not match (\frac{1000}{20}=50), so the statement is inconsistent.

Let the height be \(x\text{ cm}\); then the base is \((x+6)\text{ cm}\). Using the triangle-area formula, \(\frac{1}{2}x(x+6)=140\), which gives \(x^2+6x-280=0\). Factoring, \((x-14)(x+20)=0\), so \(x=14\) or \(x=-20\). Since a length cannot be negative, the height is \(14\text{ cm}\). Exam tip: after solving a quadratic word problem, reject any root that is not physically meaningful.

Let the fraction be (\frac{x}{x+4}), then (\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}). This gives (x=5), so the fraction is (\frac{5}{9}).