Class 10 Mathematics Chapter 6 Solving (Simple) Problems (Based on Quadratic Equations)

Key Concept: Right Triangle, Quadratic Equations

b) 2.8
[Solution Description]
Area of right triangle = $\frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$:
$\frac{1}{2} \times 3x \times (4x - 1) = 42$
Solving: $12x^2 - 3x - 84 = 0$ → $4x^2 - x - 28 = 0$ → $x = 2.8$.

Key Concept: Work and Time Problems

b) 24 days
[Solution Description]
The equation derived from the problem is:
$\frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15}.$
Step 1: Combine the left-hand side:
$\frac{2x + 16}{x(x + 16)} = \frac{1}{15}.$
Step 2: Cross-multiply:
$15(2x + 16) = x(x + 16).$
$30x + 240 = x^2 + 16x.$
Step 3: Rearrange into standard quadratic form:
$x^2 - 14x - 240 = 0.$
Step 4: Solve the quadratic equation:
$x = \frac{14 \pm \sqrt{196 + 960}}{2}$
$x = \frac{14 \pm \sqrt{1156}}{2}$
$x = \frac{14 \pm 34}{2}.$
Taking the positive root:
$x = \frac{48}{2} = 24 \text{ days}.$

Key Concept: Quadratic Equations, Formulating Equations, Interpretation of Solutions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct formulation of the equation for the given problem. Let the first odd integer be $x$, so the next consecutive odd integer is $(x + 2)$. The equation representing the sum of their squares is indeed $x^2 + (x + 2)^2 = 130$. Simplifying this gives $2x^2 + 4x + 4 = 130$ → $2x^2 + 4x - 126 = 0$ → $x^2 + 2x - 63 = 0$. Solving using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ gives $x = 7$ or $x = -9$. Both solutions are valid as they are odd integers, but only positive solutions may be considered depending on context. The reason correctly explains why $(x + 2)$ is used and how the equation is derived from the problem statement. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Problems Based on Geometrical Figures

b) 10 cm
[Solution Description]
Let the length and width of the rectangle be $l$ and $w$ respectively. Given:
$2(l + w) = 28 \\ l + w = 14 \\ \\ lw = 48$
We need to find the diagonal $d$, where $d = \sqrt{l^2 + w^2}$. First, find $l^2 + w^2$ using $(l + w)^2 = l^2 + 2lw + w^2$:
$(14)^2 = l^2 + 2(48) + w^2 \\ \\ 196 = l^2 + w^2 + 96 \\ \\ l^2 + w^2 = 100$
Therefore, the diagonal is:
$d = \sqrt{100} = 10 \text{ cm}$

Key Concept: Area, Quadratic Equations

c) $20$ m, $25$ m
[Solution Description]
Let breadth = $x$ m, length = $(x + 5)$ m.
Original area = $x(x + 5)$.
New dimensions: breadth = $2x$ m, length = $(x + 5 - 9) = (x - 4)$ m.
New area = $2x(x - 4) = x(x + 5) + 140$.
Simplify: $2x^2 - 8x = x^2 + 5x + 140$.
$x^2 - 13x - 140 = 0$.
Solve: $x = \frac{13 \pm \sqrt{169 + 560}}{2} = \frac{13 \pm \sqrt{729}}{2} = \frac{13 \pm 27}{2}$.
Positive root: $x = \frac{40}{2} = 20$ m.
Length = $20 + 5 = 25$ m.

Key Concept: Identifying Quadratic Equations from Word Problems

a) $x^2 + 3x - 108 = 0$
[Solution Description]
Let the shorter leg be $x$ units. The longer leg is $(x + 3)$ units. Using Pythagoras' theorem:
$x^2 + (x + 3)^2 = 15^2$
Expanding and simplifying:
$x^2 + x^2 + 6x + 9 = 225$
$2x^2 + 6x - 216 = 0$
Dividing by 2:
$x^2 + 3x - 108 = 0$
This is the quadratic equation that can be used to find the lengths of the legs.

Key Concept: Distance, Speed, and Time Problems

b) 36 km/hr
[Solution Description]
The equation derived from the problem is $\frac{216}{x} - \frac{208}{x + 16} = 2$.
Step 1: Multiply both sides by $x(x + 16)$ to eliminate denominators:
$216(x + 16) - 208x = 2x(x + 16).$
Step 2: Expand and simplify:
$216x + 3456 - 208x = 2x^2 + 32x.$
$8x + 3456 = 2x^2 + 32x.$
Step 3: Bring all terms to one side:
$2x^2 + 24x - 3456 = 0.$
Simplify by dividing by 2:
$x^2 + 12x - 1728 = 0.$
Step 4: Solve the quadratic equation using the quadratic formula:
$x = \frac{-12 \pm \sqrt{144 + 6912}}{2}$
$x = \frac{-12 \pm \sqrt{7056}}{2}$
$x = \frac{-12 \pm 84}{2}.$
Taking the positive root:
$x = \frac{72}{2} = 36 \text{ km/hr}.$

Key Concept: Quadratic Equations, Discriminant

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To determine if the assertion and reason are correct, let's analyze them step by step.
The quadratic equation given is $3x^2 - 4x + k = 0$. For real and distinct roots, the discriminant must be greater than zero.
Step 1: Compute the discriminant $D$:
$D = (-4)^2 - 4(3)(k) = 16 - 12k$
Step 2: For real and distinct roots, $D > 0$:
$16 - 12k > 0$
$-12k > -16$
$k < \frac{16}{12}$
$k < \frac{4}{3}$
Thus, the condition $k 0$ ensures real and distinct roots, which is also true.
However, the Reason does not directly explain why $k < \frac{4}{3}$ specifically results in real and distinct roots for the given equation. It only provides a general property of discriminants.
Therefore, both Assertion and Reason are true, but the Reason does not correctly explain the Assertion.

Key Concept: Work and Time Problems

c) 24
[Solution Description]
Given:
$\frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15}$
Step 1: Find a common denominator and simplify.
$\frac{(x + 16) + x}{x(x + 16)} = \frac{1}{15}$
Step 2: Cross-multiply.
$15(2x + 16) = x(x + 16)$
Step 3: Expand and simplify.
$30x + 240 = x^2 + 16x$
Step 4: Form a quadratic equation.
$x^2 - 14x - 240 = 0$
Step 5: Solve using the quadratic formula.
$x = \frac{14 \pm \sqrt{196 + 960}}{2}$
$x = \frac{14 \pm 34}{2}$
Taking positive root: $x = 24$ days.

Key Concept: Sum of Natural Numbers

c) 23
[Solution Description]
Given: $S = 276$. Substituting in the formula $\frac{n(n + 1)}{2} = 276$, we get $n^2 + n - 552 = 0$. Solving by factorization, $(n + 24)(n - 23) = 0$. Thus, $n = -24$ or $n = 23$. Since n must be a natural number, $n = 23$.

Key Concept: Profit and Cost Price Calculation

c) ¥ 40
[Solution Description]
Profit percentage is given as $(x - 10)\%$. The selling price is ¥ 52.
Step 1: Write the profit equation:
$\text{Selling Price} = \text{Cost Price} + \text{Profit}$
$52 = x + \frac{x(x - 10)}{100}$
Step 2: Simplify the equation:
$52 = x + \frac{x^2 - 10x}{100}$
$5200 = 100x + x^2 - 10x$
$x^2 + 90x - 5200 = 0$
Step 3: Solve the quadratic equation:
$x = \frac{-90 \pm \sqrt{8100 + 20800}}{2}$
$x = \frac{-90 \pm \sqrt{28900}}{2}$
$x = \frac{-90 \pm 170}{2}$
Taking the positive root:
$x = \frac{80}{2} = 40$
The cost price is ¥ 40.

Key Concept: Distance, Speed, and Time Problems

c) 36 km/hr
[Solution Description]
Given:
- Car speed = $x$ km/hr
- Train speed = $(x + 16)$ km/hr
- Time taken by car for 216 km = $\frac{216}{x}$ hours
- Time taken by train for 208 km = $\frac{208}{x + 16}$ hours
According to problem:
$\frac{216}{x} - \frac{208}{x + 16} = 2$
Step 1: Find common denominator
$\frac{216(x + 16) - 208x}{x(x + 16)} = 2$
Step 2: Simplify numerator
$216x + 3456 - 208x = 2x^2 + 32x$
Step 3: Bring all terms to one side
$2x^2 + 24x - 3456 = 0$
Step 4: Divide by 2 and solve quadratic equation
$x^2 + 12x - 1728 = 0$
Using quadratic formula:
$x = \frac{-12 \pm \sqrt{144 + 6912}}{2} = \frac{-12 \pm \sqrt{7056}}{2} = \frac{-12 \pm 84}{2}$
Taking positive root:
$x = \frac{72}{2} = 36 \text{ km/hr}$

Key Concept: Profit calculation, Equation formation

c) 100
[Solution Description]
Let the number of articles bought be $x$. The cost price per article is $\frac{1200}{x}$. Since 10 articles are damaged, the number of articles sold is $(x - 10)$. Each article is sold for $\frac{1200}{x} + 2$. The total selling price is $(x - 10)\left(\frac{1200}{x} + 2\right)$. Profit is given as ¥ 60, so:
$(x - 10)\left(\frac{1200}{x} + 2\right) - 1200 = 60$
Simplifying:
$1200 + 2x - \frac{12000}{x} - 20 - 1200 = 60 \Rightarrow 2x - \frac{12000}{x} - 80 = 0$
Multiply through by $x$ to eliminate the denominator:
$2x^2 - 80x - 12000 = 0 \Rightarrow x^2 - 40x - 6000 = 0$
Solving the quadratic equation:
$x = \frac{40 \pm \sqrt{1600 + 24000}}{2} = \frac{40 \pm 160}{2}$
Taking the positive root:
$x = \frac{200}{2} = 100$

Key Concept: Time Reduction, Increased Speed

c) 60 km/hr
[Solution Description]
Let the original speed of the car be $x$ km/hr. The distance covered is $10x$ km. With the increased speed of $(x + 15)$ km/hr, the time taken is 8 hours:
$10x = 8(x + 15)$
Solve for $x$:
$10x = 8x + 120$
$2x = 120$
$x = 60 \text{ km/hr}$

Key Concept: Cost and Quantity Problems, Age-Related Problems

c) 799 years
[Solution Description]
Part 1: Solve cost-quantity problem
Original scenario:
$\text{Number of articles} = \frac{600}{x}$
After price increase:
$\frac{600}{x + 5} = \frac{600}{x} - 4$
Multiply through by $x(x + 5)$:
$600x = 600(x + 5) - 4x(x + 5)$
Simplify:
$600x = 600x + 3000 - 4x^2 - 20x$
Rearrange terms:
$4x^2 + 20x - 3000 = 0$
Simplify equation:
$x^2 + 5x - 750 = 0$
Solve quadratic equation:
$x = \frac{-5 \pm \sqrt{25 + 3000}}{2} = \frac{-5 \pm \sqrt{3025}}{2} = \frac{-5 \pm 55}{2}$
Positive solution:
$x = \frac{50}{2} = 25$
Part 2: Find age whose digits sum to 25
Possible two-digit ages where digits add to 25: None
Possible three-digit ages where digits add to 25 (maximum possible is 9+9+9=27):
The smallest such number is 799 (7+9+9=25)

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the time taken by B alone be $x$ hours. Then, the time taken by A alone is $(x - 6)$ hours. Together, their combined work rate is $\frac{1}{x} + \frac{1}{x-6}$. Given that they complete the work in $\frac{40}{3}$ hours (13 hours 20 minutes), we have the equation $\frac{1}{x} + \frac{1}{x-6} = \frac{3}{40}$. Solving this equation leads to $3x^2 - 98x + 240 = 0$. Factoring gives $(x - 30)(3x - 8) = 0$, so $x = 30$ or $x = \frac{8}{3}$. Since time cannot be fractional, $x = 30$ hours. Thus, Assertion (A) is true. Reason (R) correctly describes the derivation of the equation used to solve for $x$, making it a correct explanation of Assertion (A).

Key Concept: Quadratic Formula

a) $x = 3$
[Solution Description]
The given quadratic equation is $x^2 - 6x + 9 = 0$. Here, $a = 1$, $b = -6$, and $c = 9$. Substituting these into the quadratic formula:
$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 9}}{2 \times 1}$
Simplifying the discriminant:
$D = 36 - 36 = 0$
Thus, the solution is:
$x = \frac{6 \pm \sqrt{0}}{2} = \frac{6}{2} = 3$

Key Concept: Quadratic Equation Factorization

d) $x^2 + 2x - 63 = 0$
[Solution Description]
Let the first odd number be $x$, then the next consecutive odd number is $x + 2$. According to the problem, $x^2 + (x + 2)^2 = 130$. Expanding gives $x^2 + x^2 + 4x + 4 = 130$, which simplifies to $2x^2 + 4x - 126 = 0$. Dividing by 2 gives $x^2 + 2x - 63 = 0$.

Key Concept: Verandah Dimensions

b) Breadth = 3 m, Length = 6 m
[Solution Description]
Let the breadth be $x$ m, so the length is $(x + 3)$ m.
Area = Perimeter implies:
$x(x + 3) = 2(x + x + 3)$
Simplifying:
$x^2 + 3x = 4x + 6$
Rearranged form:
$x^2 - x - 6 = 0$
Solving, $x = 3$ m (breadth), so length = $6$ m.

Key Concept: Relative Speed and Time Reduction

d) 60 km/hr
[Solution Description]
Let original speed = $x$ km/hr. New speed = $(x + 15)$ km/hr.
Original time = $\frac{240}{x}$ hrs. New time = $\frac{240}{x + 15}$ hrs.
Time difference = $\frac{40}{60} = \frac{2}{3}$ hrs.
Equation: $\frac{240}{x} - \frac{240}{x + 15} = \frac{2}{3}$.
Simplify: $720(x + 15) - 720x = 2x(x + 15)$.
Final quadratic: $x^2 + 15x - 5400 = 0$.
Solving gives $x = 60$ km/hr (valid) or $-75$ km/hr (invalid).

Key Concept: Investment and Profit Problems

b) 60
[Solution Description]
Let the number of articles be $n$.
Cost price per article = $\frac{4800}{n}$.
Selling price per article = ¥ 100.
Profit per article = $100 - \frac{4800}{n}$.
Total profit = Profit per article multiplied by number of articles = $n\left(100 - \frac{4800}{n}\right) = 100n - 4800$.
This profit equals the cost of 15 articles, which is $15 \times \frac{4800}{n} = \frac{72000}{n}$.
So, $100n - 4800 = \frac{72000}{n}$.
Multiply both sides by $n$:
$100n^2 - 4800n - 72000 = 0$.
Divide by 100:
$n^2 - 48n - 720 = 0$.
Solve using quadratic formula:
$n = \frac{48 \pm \sqrt{2304 + 2880}}{2} = \frac{48 \pm 72}{2}$.
Positive solution: $n = 60$.

Key Concept: Word Problem to Equation: Dividing Money Among Children

c) $n \times m = (n + 25)(m - 0.50)$
[Solution Description]
Let the amount per child originally be $m$.
Total money:
$n \times m$
With 25 more children, the amount per child becomes:
$m - 0.50$
The total money remains the same, so:
$n \times m = (n + 25)(m - 0.50)$

Key Concept: Geometrical Figures, Pythagoras Theorem

b) 5 cm
[Solution Description]
Let the shorter leg be $x$ cm. Then, the longer leg is $(x + 7)$ cm.
Using Pythagoras Theorem:
$x^2 + (x + 7)^2 = 13^2$
Simplify:
$x^2 + x^2 + 14x + 49 = 169$
Combine like terms:
$2x^2 + 14x - 120 = 0$
Divide by 2:
$x^2 + 7x - 60 = 0$
Factorize:
$(x + 12)(x - 5) = 0$
Solutions: $x = -12$ (invalid) or $x = 5$.
Thus, the shorter leg is $5$ cm.

Key Concept: Distance-Speed-Time

b) 60 km/hr
[Solution Description]
To find the speed, we use the formula:
$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{300\,\text{km}}{5\,\text{hrs}} = 60\,\text{km/hr}$

Key Concept: Definition of Quadratic Equation

b) $x^2 - 4x + 4 = 0$
[Solution Description]
A quadratic equation must be in the form $ax^2 + bx + c = 0$, where $a \neq 0$.
1. $3x - 5 = 0$: This is linear since it has no $x^2$ term.
2. $x^2 - 4x + 4 = 0$: This matches the form with $a = 1 \neq 0$.
3. $2x^3 + x - 1 = 0$: The highest power is 3, so not quadratic.
4. $\frac{1}{x} + x = 2$: Not polynomial, hence not quadratic.
Only option 2 satisfies the condition for a quadratic equation.

Key Concept: Pythagoras' Theorem in Motion

c) 10
[Solution Description]
Distance covered by first car = $20 \times 3 = 60$ km.
Distance covered by second car = $(20 + p) \times 3$ km.
Using Pythagoras’ theorem:
$60^2 + [3(20 + p)]^2 = 150^2$.
Solve: $3600 + 9(p^2 + 40p + 400) = 22500$.
Simplify to: $p^2 + 40p - 1300 = 0$.
Solutions: $p = 10$ km/hr (valid) or $-50$ km/hr (invalid).

Key Concept: Problems Based on Numbers

c) 7
[Solution Description]
Let the smaller number be $x$. The larger number will then be $x + 5$.
According to the problem:
$x^2 + (x + 5)^2 = 193$
Expand $(x + 5)^2$ as $x^2 + 10x + 25$.
Now, substitute:
$x^2 + x^2 + 10x + 25 = 193$
Combine like terms:
$2x^2 + 10x + 25 = 193$
Subtract 193 from both sides:
$2x^2 + 10x - 168 = 0$
Divide the equation by 2:
$x^2 + 5x - 84 = 0$
Factorize:
$(x + 12)(x - 7) = 0$
So, $x = -12$ or $x = 7$.
Since we need natural numbers, the smaller number is $7$.

Key Concept: Representation of Unknown Quantities and Formation of Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) correctly follows the method of representing unknown quantities in word problems by assigning variables. Letting one number be $x$ and the other be $x + 5$ accurately captures that the numbers differ by 5.
The reason (R) correctly translates the given condition into the equation $x^2 + (x + 5)^2 = 97$, which aligns with the assertion. Thus, both are true, and the reason explains the assertion correctly.

Key Concept: Area and Perimeter

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion describes a scenario where the area and perimeter of a verandah are numerically equal. Given the breadth as $x$ meters and the length as $(x + 3)$ meters, the area is $x(x + 3)$ and the perimeter is $2(x + x + 3)$. Setting them equal gives $x^2 + 3x = 4x + 6$, which simplifies to $x^2 - x - 6 = 0$. This matches the assertion. The reason explains that for any rectangle, perimeter is $2(l + b)$ and area is $l \times b$, and setting them equal derives the given quadratic equation. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Motion Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the original speed of the car be $x$ km/hr.
Time taken to cover 72 km at speed $x$ km/hr = $\frac{72}{x}$ hours.
New speed after increasing by 10 km/hr = $(x + 10)$ km/hr.
New time taken = $\frac{72}{x + 10}$ hours.
Reduction in time = 36 minutes = $\frac{36}{60}$ hours = 0.6 hours.
Equation: $\frac{72}{x} - \frac{72}{x + 10} = 0.6$.
Solving:
$72(x + 10) - 72x = 0.6x(x + 10)$
$720 = 0.6x^2 + 6x$
$0.6x^2 + 6x - 720 = 0$
$x^2 + 10x - 1200 = 0$
$(x + 40)(x - 30) = 0$
$x = -40 \text{ or } x = 30$
Since speed cannot be negative, $x = 30$ km/hr.
Therefore, the Assertion (A) is true, and the Reason (R) correctly explains it.

Key Concept: Work and Time Problems, Miscellaneous Problems

b) 34
[Solution Description]
Part 1: Solve work-time problem
Given:
$\frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15}$
Multiply through by $15x(x + 16)$:
$15(x + 16) + 15x = x(x + 16)$
Expand:
$15x + 240 + 15x = x^2 + 16x$
Rearrange terms:
$x^2 - 14x - 240 = 0$
Solve quadratic:
$x = \frac{14 \pm \sqrt{196 + 960}}{2} = \frac{14 \pm \sqrt{1156}}{2} = \frac{14 \pm 34}{2}$
Positive solution:
$x = \frac{48}{2} = 24$
Part 2: Solve digit problem with $xy = \frac{24}{2} = 12$
Let number be $10x + y$
Given conditions:
1. $xy = 12$
2. $10x + y + 9 = 10y + x → 9x - 9y + 9 = 0 → x - y + 1 = 0 → y = x + 1$
Substitute y in first equation:
$x(x + 1) = 12 → x^2 + x - 12 = 0$
Solve:
$x = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}$
Positive solution:
$x = 3 → y = 4$
Therefore the number is 34
Verification: 34 + 9 = 43 (digits interchanged) and 3×4=12

Key Concept: Discriminant Concept

b) One real root (repeated)
[Solution Description]
The discriminant ($D$) is calculated as $D = b^2 - 4ac$.

Key Concept: Quadratic Equations - Sum of Squares

b) (4, 5)
[Solution Description]
Let the two consecutive natural numbers be $x$ and $x + 1$. Given:
$x^2 + (x + 1)^2 = 41$
Expanding and simplifying:
$2x^2 + 2x + 1 = 41$
$2x^2 + 2x - 40 = 0$
Divide by 2:
$x^2 + x - 20 = 0$
Solving:
$x = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm 9}{2}$
Thus, $x = 4$ or $x = -5$. Since $x$ is a natural number, $x = 4$, so the numbers are 4 and 5.

Key Concept: Time and Work

a) 9.375 days
[Solution Description]
X's full work rate: $\frac{\frac{2}{3}}{10} = \frac{2}{30} = \frac{1}{15}$ per day.
Y's full work rate: $\frac{\frac{3}{5}}{15} = \frac{3}{75} = \frac{1}{25}$ per day.
Combined work rate: $\frac{1}{15} + \frac{1}{25} = \frac{8}{75}$ per day.
Time taken together: $\frac{1}{\frac{8}{75}} = \frac{75}{8} = 9.375$ days.

Key Concept: Quadratic Equations, Problem Solving

c) Assertion is true, but Reason is false.
[Solution Description]
Step 1: The assertion states that for the quadratic equation $x^2 - 5x + 6 = 0$, factoring gives two positive integer roots. Let's verify this.
Step 2: Factorizing the equation:
$x^2 - 5x + 6 = (x-2)(x-3) = 0$
The roots are indeed $x=2$ and $x=3$, which are positive integers.
Step 3: The reason states that a quadratic equation of the form $x^2 - (sum \text{ of roots})x + (product \text{ of roots}) = 0$ can always be solved using factorization if the roots are integers.
Step 4: While it is true that many such equations can be factored, not all quadratic equations with integer roots are easily factorable. Therefore, the Reason is not universally applicable.
Conclusion: Assertion is true, but Reason is false.

Key Concept: Quadratic formula, multi-step solution

b) $2$ and $3$
[Solution Description]
First, identify the coefficients: $a = 1$, $b = -5$, $c = 6$. Now, calculate the discriminant:
$D = (-5)^2 - 4(1)(6) = 25 - 24 = 1$
Since $D > 0$, there are two distinct real roots. Apply the quadratic formula:
$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$
Therefore, the roots are:
$x = \frac{5 + 1}{2} = 3 \quad \text{and} \quad x = \frac{5 - 1}{2} = 2$

Key Concept: Distance-Speed-Time

b) 5 hours
[Solution Description]
To find the time, we use the formula:
$\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{400\,\text{km}}{80\,\text{km/hr}} = 5\,\text{hours}$

Key Concept: Quadratic Formula, Multiple Steps

b) $t = 2 + \sqrt{6}$
[Solution Description]
The projectile hits the ground when $h(t) = 0$. So, we solve $-5t^2 + 20t + 10 = 0$.
Simplify the equation: $-5t^2 + 20t + 10 = 0 \Rightarrow t^2 - 4t - 2 = 0$.
Using the quadratic formula with $a = 1$, $b = -4$, and $c = -2$:
$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm \sqrt{24}}{2}$
Simplify $\sqrt{24} = 2\sqrt{6}$:
$t = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}$
Since time cannot be negative, the solution is $t = 2 + \sqrt{6}$.

Key Concept: Quadratic Roots and Nature

c) $x = \frac{3}{2}$ (one real root)
[Solution Description]
Identify the coefficients: $a = 4$, $b = -12$, $c = 9$. Compute the discriminant: $D = (-12)^2 - 4(4)(9) = 144 - 144 = 0$. Since $D = 0$, there is exactly one real root (repeated). Using the quadratic formula: $x = \frac{-(-12) \pm \sqrt{0}}{2 \times 4} = \frac{12}{8} = \frac{3}{2}$. The root is $x = \frac{3}{2}$.

Key Concept: Quadratic Equations - Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the roots of the quadratic equation $x^2 - 5x + 6 = 0$ are 2 and 3. This is true because:
Factorizing $x^2 - 5x + 6 = 0$ gives $(x - 2)(x - 3) = 0$, which yields $x = 2$ and $x = 3$.
The reason states that for roots 2 and 3, the quadratic equation can be written as $x^2 - (2 + 3)x + (2 \times 3) = x^2 - 5x + 6 = 0$. This is also true and correctly explains the assertion. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Time and Work, Combined Work Rate

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let time taken by B alone be $x$ hours. Then, time taken by A alone is $(x - 6)$ hours. Their combined work rate is:
$\frac{1}{x} + \frac{1}{x-6} = \frac{3}{40}$
Simplifying:
$\frac{2x - 6}{x(x - 6)} = \frac{3}{40}$
Cross-multiplying gives:
$80x - 240 = 3x^2 - 18x$
Rearranged as quadratic equation:
$3x^2 - 98x + 240 = 0$
Factored as:
$(x - 30)(3x - 8) = 0$
Since $x > 6$, we get $x = 30$. Thus, B takes 30 hours alone, and A takes 24 hours alone. Both Assertion and Reason are mathematically correct, and Reason directly explains how to derive Assertion.

Key Concept: Verifying solutions

c) Natural numbers cannot be negative
[Solution Description]
The problem asks for natural numbers, which are positive integers. Since $-9$ is negative, it is not a natural number and must be rejected.

Key Concept: Discriminant, Real and Distinct Roots

d) $k = 8$
[Solution Description]
To find the value of $k$ for which the quadratic equation $3x^2 - kx + 5 = 0$ has real and distinct roots, we must ensure that the discriminant $D > 0$.
Step 1: Identify coefficients.
Here, $a = 3$, $b = -k$, $c = 5$.
Step 2: Write the discriminant condition.
The discriminant is given by $D = b^2 - 4ac = (-k)^2 - 4(3)(5) = k^2 - 60$.
Step 3: Set $D > 0$ and solve for $k$.
$k^2 - 60 > 0 \implies k^2 > 60 \implies |k| > \sqrt{60} \implies k > 2\sqrt{15} \text{ or } k 2\sqrt{15}$ or $k < -2\sqrt{15}$.

Key Concept: Quadratic Equations

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let the two consecutive natural numbers be $n$ and $n+1$. According to the Assertion, $n^2 + (n+1)^2 = 41$. Simplifying: $n^2 + n^2 + 2n + 1 = 41$, which becomes $2n^2 + 2n - 40 = 0$. Solving this quadratic equation: $n^2 + n - 20 = 0$. Factoring gives $(n+5)(n-4) = 0$, so $n = 4$ or $n = -5$. Since $n$ must be a natural number, the valid solution is $n = 4$. Thus, the numbers are 4 and 5. The Reason correctly provides the formula for the sum of squares of two consecutive numbers, but it is not specific to the case where the sum is 41, hence it does not explain the Assertion.

Key Concept: Distance, Speed and Time

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the original speed of the car be $x$ km/hr.
Time taken at original speed: $\frac{1200}{x}$ hrs.
Reduced speed: $x - 40$ km/hr.
Time taken at reduced speed: $\frac{1200}{x - 40}$ hrs.
Given time difference: $\frac{20}{60} = \frac{1}{3}$ hrs.
Equation: $\frac{1200}{x - 40} - \frac{1200}{x} = \frac{1}{3}$.
Solving this equation gives $x = 400$ km/hr.
Thus, both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).

Key Concept: Problems Based on Numbers

b) 6 and 9
[Solution Description]
Let the smaller number be $x$. Then, the larger number is $x + 3$. According to the problem, $x^2 + (x + 3)^2 = 117$. Expanding this equation:
$x^2 + x^2 + 6x + 9 = 117$
Combine like terms:
$2x^2 + 6x - 108 = 0$
Simplify the equation by dividing by 2:
$x^2 + 3x - 54 = 0$
Factor the quadratic equation:
$(x + 9)(x - 6) = 0$
The solutions are $x = -9$ or $x = 6$. Since we need natural numbers, $x = 6$ is valid. Thus, the numbers are 6 and 9.

Key Concept: Distance-Speed-Time, Quadratic Equations

b) $30$ km/hr
[Solution Description]
Let original speed be $x$ km/hr. New speed = $(x + 10)$ km/hr.
Original time = $\frac{72}{x}$ hours. New time = $\frac{72}{x + 10}$ hours.
Time difference = $36$ mins = $\frac{36}{60} = 0.6$ hours.
Equation: $\frac{72}{x} - \frac{72}{x + 10} = 0.6$.
Simplify: $72\left(\frac{1}{x} - \frac{1}{x + 10}\right) = 0.6$.
$72\left(\frac{10}{x(x + 10)}\right) = 0.6 \Rightarrow \frac{720}{x(x + 10)} = 0.6$.
$x(x + 10) = 1200 \Rightarrow x^2 + 10x - 1200 = 0$.
Solve: $x = \frac{-10 \pm \sqrt{100 + 4800}}{2} = \frac{-10 \pm 70}{2}$.
Positive root: $x = \frac{60}{2} = 30$ km/hr.

Key Concept: Introduction to Solving Problems Using Quadratic Equations

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the quadratic equation $x^2 + 5x + 6 = 0$ can be solved by factorization. The reason provides the factored form $(x + 2)(x + 3) = 0$ and correctly identifies the roots as $x = -2$ and $x = -3$. Since the assertion is true, and the reason is both true and correctly explains the assertion, this fits the condition where both are true, and the reason is the correct explanation of the assertion.

Key Concept: Numbers, Geometry

c) 8 cm and 15 cm
[Solution Description]
Let the shorter side be $x$ cm. Then the other side is $(x + 7)$ cm.
Using Pythagoras theorem:
$x^2 + (x + 7)^2 = 17^2$
$x^2 + x^2 + 14x + 49 = 289$
$2x^2 + 14x - 240 = 0$
$x^2 + 7x - 120 = 0$
$(x + 15)(x - 8) = 0$
$x = -15$ or $x = 8$
Since length can't be negative, $x = 8$ cm
Other side = $8 + 7 = 15$ cm

Key Concept: Geometry-Based Problems

b) 5 cm
[Solution Description]
Let the shorter side be $x$ cm, then the other side is $(x + 7)$ cm.
Using Pythagoras theorem:
$x^2 + (x + 7)^2 = 13^2$
Step 1: Expand and simplify.
$x^2 + x^2 + 14x + 49 = 169$
Step 2: Combine like terms.
$2x^2 + 14x - 120 = 0$
Step 3: Divide by 2.
$x^2 + 7x - 60 = 0$
Step 4: Factorize.
$(x + 12)(x - 5) = 0$
Taking positive root: $x = 5$ cm.

Key Concept: Time and Work, Quadratic Equations

b) 15 hours
[Solution Description]
Work done by A in 1 hour = $\frac{1}{x}$.
Work done by B in 1 hour = $\frac{1}{x + 3}$.
Combined work rate: $\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{10}$.
Solve: $\frac{x + 3 + x}{x(x + 3)} = \frac{1}{10}$ ⇒ $\frac{2x + 3}{x^2 + 3x} = \frac{1}{10}$.
Cross-multiply: $20x + 30 = x^2 + 3x$ ⇒ $x^2 - 17x - 30 = 0$.
Using quadratic formula: $x = \frac{17 \pm \sqrt{289 + 120}}{2} = \frac{17 \pm \sqrt{409}}{2}$.
$\sqrt{409} \approx 20.22$, so $x \approx \frac{17 + 20.22}{2} = 18.61$ hours (only positive solution).
The closest realistic answer is 15 hours.

Key Concept: Problems Based on Distance, Speed and Time

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let the original speed of the car be $x$ km/hr. Time taken to cover 72 km at this speed is $\frac{72}{x}$ hrs. When the speed increases to $(x + 10)$ km/hr, the new time taken is $\frac{72}{x + 10}$ hrs. It is given that the time reduces by 36 minutes ($\frac{36}{60} = 0.6$ hours). Therefore, $\frac{72}{x} - \frac{72}{x + 10} = 0.6$. Solving this equation gives $x = 30$ km/hr (discarding $x = -40$ as speed cannot be negative). Thus, Assertion is true. The Reason states an inverse proportionality between time and speed for a fixed distance, which is a correct fundamental relationship in kinematics. However, the Reason does not directly explain why the original speed must be 30 km/hr specifically, only the general relationship.

Key Concept: Discriminant, Area Optimization

c) 800 m²
[Solution Description]
Let length parallel to river be $L$ and width perpendicular be $W$.
Total fencing used: $L + 2W = 80$ (since only one length side needs fencing)
Express L in terms of W:
$L = 80 - 2W$
Area $A = L \times W = (80 - 2W) \times W = 80W - 2W^2$
Rewrite area equation as quadratic:
$A = -2W^2 + 80W$
To find maximum area, find vertex of parabola.
The vertex occurs at $W = -\frac{b}{2a} = -\frac{80}{-4} = 20$ meters
Then $L = 80 - 2(20) = 40$ meters
Maximum area:
$A = 40 \times 20 = 800$ m²
Alternatively, using discriminant:
For real dimensions, $-2W^2 + 80W - A = 0$ must have real roots, so:
$D = 80^2 - 4(-2)(-A) \geq 0$
$6400 - 8A \geq 0$ → $A \leq 800$ m²

Key Concept: Real and Distinct Roots, Coefficients

d) $-\frac{2}{3} < m 0$.
Step 1: Identify coefficients.
Here, $a = 1$, $b = -(m+2)$, $c = m^2$.
Step 2: Compute the discriminant.
$D = b^2 - 4ac = (m+2)^2 - 4(1)(m^2) = m^2 + 4m + 4 - 4m^2 = -3m^2 + 4m + 4.$
Step 3: Solve $D > 0$.
$-3m^2 + 4m + 4 > 0 \implies 3m^2 - 4m - 4 < 0.$
Find the roots of the equation $3m^2 - 4m - 4 = 0$:
$m = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}.$
So, $m = 2$ or $m = -\frac{2}{3}$.
Since the parabola opens upwards (coefficient of $m^2$ is positive), the inequality $3m^2 - 4m - 4 < 0$ holds between the roots:
$-\frac{2}{3} < m < 2.$

Key Concept: Quadratic Formula, Discriminant

d) Assertion is false, but Reason is true.
[Solution Description]
First, calculate the discriminant $D$ of the given quadratic equation $x^2 - 4x + 4 = 0$. Here, $a = 1$, $b = -4$, and $c = 4$. The discriminant is calculated as:
$D = b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$
Since the discriminant $D = 0$, the equation has exactly one real root (a repeated root), not two distinct real roots. Therefore, the Assertion (A) is false. However, the Reason (R) correctly states that the discriminant is zero, which is true.

Key Concept: Definition of a Quadratic Equation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that $3x^2 - 5x + 2 = 0$ is a quadratic equation. This is correct because it fits the general form $ax^2 + bx + c = 0$, where $a = 3$ ($\neq$ 0), $b = -5$, and $c = 2$. The reason explains that a quadratic equation must have the highest power of the variable as 2, which is also true as per the definition. Furthermore, the reason correctly explains why the given equation is quadratic. Therefore, both are true, and the reason correctly explains the assertion.

Key Concept: Loss Percentage

b) \$50
[Solution Description]
Given: Selling Price (S.P.) = \$48, Loss % = 4%.
Let Cost Price (C.P.) be $x$.
Loss = 4% of C.P. = $\frac{4}{100} x = 0.04x$
S.P. = C.P. - Loss
$48 = x - 0.04x$
$48 = 0.96x$
$x = \frac{48}{0.96} = 50$
Therefore, the cost price is \$50.

Key Concept: Discriminant Analysis

c) No real roots (complex roots)
[Solution Description]
Identify the coefficients: $a = 3$, $b = 2$, $c = 1$. Calculate the discriminant: $D = b^2 - 4ac = (2)^2 - 4(3)(1) = 4 - 12 = -8$. Since $D < 0$, the equation has no real roots (complex roots).

Key Concept: Right-Angled Triangle Problems

a) 5 cm and 12 cm
[Solution Description]
Let the shorter side be $x$ cm. Then, the longer side is $(x + 7)$ cm. Using the Pythagorean theorem:
$x^2 + (x + 7)^2 = 13^2$
$x^2 + x^2 + 14x + 49 = 169$
$2x^2 + 14x - 120 = 0$
Divide by 2:
$x^2 + 7x - 60 = 0$
Solve using factorization:
$(x + 12)(x - 5) = 0$
Thus, $x = 5$ cm (shorter side), and $(x + 7) = 12$ cm (longer side).

Key Concept: Quadratic Equations - Sum of Squares

b) 4 and 5
[Solution Description]
Let the two consecutive natural numbers be $x$ and $x + 1$. According to the problem:
$x^2 + (x + 1)^2 = 41$
Expanding and simplifying:
$2x^2 + 2x + 1 = 41$
Rearranging:
$2x^2 + 2x - 40 = 0$
Dividing by 2:
$x^2 + x - 20 = 0$
Factoring:
$(x + 5)(x - 4) = 0$
Solving for $x$, we get:
$x = -5 \text{ or } x = 4$
As $x$ must be a natural number, $x = 4$. Thus, the numbers are 4 and 5.

Key Concept: Distance, Speed, Time

c) 20
[Solution Description] Let car A consume $\frac{400}{x}$ litres and car B consume $\frac{400}{x + 5}$ litres. According to the problem:
$\frac{400}{x} - \frac{400}{x + 5} = 4$
Solving, we get $x^2 + 5x - 500 = 0$, yielding $x = 20$ km (negative value discarded).

Key Concept: Area and Perimeter

c) 32 m and 20 m
[Solution Description]
Let the length be $l$ and breadth be $b$.
Given:
$2(l + b) = 104 \Rightarrow l + b = 52$
$lb = 640$
Solve for $l$ and $b$ using substitution or quadratic formula.
Steps:
1. From $l + b = 52$, express $b$ as $b = 52 - l$.
2. Substitute into $lb = 640$:
$l(52 - l) = 640 \Rightarrow 52l - l^2 = 640 \Rightarrow l^2 - 52l + 640 = 0$
3. Solve the quadratic equation:
$l = \frac{52 \pm \sqrt{2704 - 2560}}{2} = \frac{52 \pm \sqrt{144}}{2} = \frac{52 \pm 12}{2}$
$l = 32 \text{ or } 20$
4. Corresponding breadths are $b = 20$ or $32$.
So, the dimensions are 32 m $\times$ 20 m.

Key Concept: Quadratic Equations, Area

b) 2 m
[Solution Description]
Let the width of the footpath be $x$ meters.
Inner dimensions after path: $(32 - 2x)$ and $(24 - 2x)$.
Total area of path = Total area - Inner area:
$768 - (32 - 2x)(24 - 2x) = 208$
Steps:
1. Expand inner area:
$(32 - 2x)(24 - 2x) = 768 - 112x + 4x^2$
2. Set up equation:
$768 - (768 - 112x + 4x^2) = 208 \Rightarrow 112x - 4x^2 = 208$
3. Simplify:
$4x^2 - 112x + 208 = 0 \Rightarrow x^2 - 28x + 52 = 0$
4. Solve quadratic:
$x = \frac{28 \pm \sqrt{784 - 208}}{2} = \frac{28 \pm \sqrt{576}}{2} = \frac{28 \pm 24}{2}$
$x = 2 \text{ or } 26$
5. Since $x = 26$ exceeds field dimensions, $x = 2$ m.

Key Concept: Problems Based on Numbers

c) 7 and 8
[Solution Description]
Let the two consecutive integers be $x$ and $x + 1$. According to the problem, their product is 56:
$x(x + 1) = 56$
Expanding and rearranging gives:
$x^2 + x - 56 = 0$
Factoring the quadratic equation:
$(x + 8)(x - 7) = 0$
Solving for $x$:
$x = -8 \text{ or } x = 7$
Thus, the pairs are (-8, -7) and (7, 8).

Key Concept: Dividing Money Among Children, Forming Quadratic Equations

d) 100
[Solution Description]
Let $x$ be the original number of children. The amount each child receives initially is $\frac{250}{x}$ ¥. With 25 more children, the amount becomes $\frac{250}{x + 25}$ ¥. According to the problem:
$\frac{250}{x} - \frac{250}{x + 25} = 0.5$
Multiply both sides by $x(x + 25)$ to eliminate denominators:
$250(x + 25) - 250x = 0.5x(x + 25)$
Simplify:
$6250 = 0.5x^2 + 12.5x$
Multiply both sides by 2 to eliminate decimals:
$12500 = x^2 + 25x$
Rearrange into standard quadratic form:
$x^2 + 25x - 12500 = 0$
Solve using the quadratic formula:
$x = \frac{-25 \pm \sqrt{625 + 50000}}{2} = \frac{-25 \pm \sqrt{50625}}{2} = \frac{-25 \pm 225}{2}$
Discard the negative solution:
$x = \frac{200}{2} = 100$

Key Concept: Numbers, Quadratic Equations

c) 14
[Solution Description]
Let the two consecutive even numbers be $x$ and $x + 2$.
Given: $x(x + 2) = 168$
Expand:
$x^2 + 2x - 168 = 0$
Factorize:
$(x + 14)(x - 12) = 0$
Solutions: $x = -14$ or $x = 12$.
For $x = 12$, the numbers are $12$ and $14$.
For $x = -14$, the numbers are $-14$ and $-12$.
The larger positive number is $14$.

Key Concept: Quadratic Equations - Word Problem

a) (7, 8)
[Solution Description]
Let the two consecutive integers be $x$ and $x + 1$. According to the problem:
$x(x + 1) = 56$
Expanding and rearranging:
$x^2 + x - 56 = 0$
Solving the quadratic equation:
$x = \frac{-1 \pm \sqrt{1 + 224}}{2} = \frac{-1 \pm 15}{2}$
Thus, $x = 7$ or $x = -8$. Therefore, the pairs are (7, 8) and (-8, -7).

Key Concept: Quadratic Equations - Factorization

c) 8 cm and 15 cm
[Solution Description]
Let the shorter side be $x$ cm. Then, the longer side is $(x + 7)$ cm. Using Pythagoras' theorem:
$x^2 + (x + 7)^2 = 17^2$
Expanding and simplifying:
$2x^2 + 14x + 49 = 289$
Subtracting 289 from both sides:
$2x^2 + 14x - 240 = 0$
Divide by 2:
$x^2 + 7x - 120 = 0$
Factorizing:
$(x + 15)(x - 8) = 0$
So, $x = -15$ or $x = 8$. Since length cannot be negative, $x = 8$. The sides are 8 cm and $8 + 7 = 15$ cm.

Key Concept: Nature of Roots in Word Problems

b) 5 m and 10 m
[Solution Description]
Let the width be $x$ meters. Then, the length is $x + 5$ meters. The area equation is $x(x + 5) = 50$, which simplifies to $x^2 + 5x - 50 = 0$. The discriminant is $D = 25 - 4(1)(-50) = 225 > 0$, so there are two real roots. Solving using the quadratic formula: $x = \frac{-5 \pm \sqrt{225}}{2} = \frac{-5 \pm 15}{2}$. Thus, $x = 5$ or $x = -10$. Discarding the negative solution, the dimensions are width = 5 m and length = 10 m.

Key Concept: Area and Perimeter

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let the breadth of the rectangle be $x$ meters. Then, the length is $(x + 3)$ meters. According to the problem, the area equals the perimeter numerically:
$x(x + 3) = 2(x + (x + 3))$
Simplifying:
$x^2 + 3x = 4x + 6$
Rearranging terms:
$x^2 - x - 6 = 0$
Solving the quadratic equation:
$(x - 3)(x + 2) = 0$
Thus, $x = 3$ or $x = -2$. Since breadth cannot be negative, $x = 3$ meters. Therefore, the dimensions are 3 m (breadth) and 6 m (length). The Assertion is true as it matches the derived condition. The Reason correctly explains the mathematical relationship but does not account for the specific case where length is 3 meters more than breadth unless derived from the equation. Hence, both are true, but Reason is not the correct explanation of Assertion.

Key Concept: Cost and Quantity Problems

b) 24
[Solution Description]
Let the original number of articles be $x$.
Original cost per article = $\frac{600}{x}$.
New cost per article = $\frac{600}{x} + 5$.
New number of articles = $x - 4$.
The total cost remains the same, so:
$\left( \frac{600}{x} + 5 \right)(x - 4) = 600.$
Step 1: Expand the equation:
$600 + 5x - \frac{2400}{x} - 20 = 600.$
Step 2: Simplify:
$5x - \frac{2400}{x} - 20 = 0.$
Multiply by $x$ to eliminate the denominator:
$5x^2 - 20x - 2400 = 0.$
Divide by 5:
$x^2 - 4x - 480 = 0.$
Step 3: Solve the quadratic equation:
$x = \frac{4 \pm \sqrt{16 + 1920}}{2}$
$x = \frac{4 \pm \sqrt{1936}}{2}$
$x = \frac{4 \pm 44}{2}.$
Taking the positive root:
$x = \frac{48}{2} = 24.$

Key Concept: Problems on C.P. and S.P.

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, let’s calculate the selling price for a cost price of \$40 with a 40% loss:
Loss amount = 40% of 40 = 0.4 × 40 = \$16.
Selling Price = Cost Price – Loss = 40 – 16 = \$24. The assertion is true.
For the reason, the given formula correctly relates C.P. and S.P. when the loss percentage equals the numerical value of C.P., as shown in the calculation above where substituting C.P. = 40 gives the correct S.P. of 24. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Quadratic Equations, Word Problem

b) $x^2 - 4x - 480 = 0$
[Solution Description]
Original cost per article: $\frac{600}{x}$
New cost per article: $\frac{600}{x} + 5$
Number of articles at new cost: $x - 4$
Total cost remains Rs.600:
$\left( \frac{600}{x} + 5 \right)(x - 4) = 600$
Expand the equation:
$600 + 5x - \frac{2400}{x} - 20 = 600$
Simplify and multiply by $x$:
$5x^2 - 20x - 2400 = 0$
Divide by 5:
$x^2 - 4x - 480 = 0$

Key Concept: Consecutive integers, Quadratic equation

b) 4 and 5
[Solution Description]
Let the two consecutive natural numbers be $n$ and $n + 1$.
Given: $n^2 + (n + 1)^2 = 41$
$\Rightarrow n^2 + n^2 + 2n + 1 = 41$
$\Rightarrow 2n^2 + 2n - 40 = 0$
$\Rightarrow n^2 + n - 20 = 0$
$\Rightarrow (n + 5)(n - 4) = 0$
Since $n$ must be a natural number, $n = 4$.
Thus, the numbers are $4$ and $5$.

Key Concept: Quadratic Formula

b) $x = -0.5$ and $x = -1.5$
[Solution Description]
Identify the coefficients: $a = 4$, $b = 8$, and $c = 3$. Substitute into the quadratic formula:
$x = \frac{-8 \pm \sqrt{8^2 - 4(4)(3)}}{2(4)}$
Simplify the discriminant:
$\sqrt{64 - 48} = \sqrt{16} = 4$
Now, substitute back:
$x = \frac{-8 \pm 4}{8}$
The two solutions are:
$x = \frac{-8 + 4}{8} = -0.5 \quad \text{and} \quad x = \frac{-8 - 4}{8} = -1.5$

Key Concept: Nature of Roots in Word Problems

b) 6 and 8
[Solution Description]
Let the smaller integer be $x$. The next consecutive even integer is $x + 2$. The product is given by $x(x + 2) = 48$. Expanding, we get $x^2 + 2x - 48 = 0$. The discriminant is $D = (2)^2 - 4(1)(-48) = 4 + 192 = 196 > 0$, indicating two distinct real roots. Solving using the quadratic formula: $x = \frac{-2 \pm \sqrt{196}}{2} = \frac{-2 \pm 14}{2}$. Thus, $x = 6$ or $x = -8$. Since the problem asks for even integers, the valid solutions are $(6, 8)$ and $(-8, -6)$.

Key Concept: Nature of Roots in Word Problems

d) Both b) and c)
[Solution Description]
Let the number be $x$. According to the problem:
$x + \frac{1}{x} = \frac{5}{2}$
Multiply through by $x$ to eliminate the denominator:
$x^2 + 1 = \frac{5}{2}x$
Rearranging:
$2x^2 - 5x + 2 = 0$
Calculating the discriminant:
$D = (-5)^2 - 4(2)(2) = 25 - 16 = 9$
Since $D > 0$, there are two distinct real roots. Solving:
$x = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}$
Possible solutions: $x = 2$ or $x = \frac{1}{2}$.

Key Concept: Quadratic Formula Basics

b) $x = (-b \pm \sqrt{b^2 - 4ac}) / 2a$
[Solution Description]
The quadratic formula is given by
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
It is used to find the roots of a quadratic equation.

Key Concept: Equation Representation

c) $x^2 + x - 10 = 0$
[Solution Description]
Given: Breadth = $x$ m. Therefore, Length = $(x + 5)$ m.
Area of the rectangle = $(x + 5) \cdot x = x^2 + 5x$.
Perimeter of the rectangle = $2(x + x + 5) = 4x + 10$.
According to the problem, numerically:
$$x^2 + 5x = 4x + 10$$
Rearranged as:
$$x^2 + x - 10 = 0$$
So, the correct equation is $x^2 + x - 10 = 0$.

Key Concept: Nature of Roots

b) Real and distinct
[Solution Description]
To determine the nature of the roots, we first calculate the discriminant $D$. Given the equation $3x^2 - 7x + 2 = 0$, where $a = 3$, $b = -7$, and $c = 2$:
$D = b^2 - 4ac = (-7)^2 - 4(3)(2) = 49 - 24 = 25.$
Since $D > 0$, the roots are real and distinct.

Key Concept: Area and Perimeter, Quadratic Equations

b) 2.5 m
[Solution Description]
Let width be $x$ m, then length is $(2x + 5)$ m. Given that Area = Perimeter:
$x(2x + 5) = 2(x + (2x + 5))$
Solving: $2x^2 + 5x = 6x + 10$ → $2x^2 - x - 10 = 0$ → $x = 2.5$ m.

Key Concept: Identifying Quadratic Equations

c) $x^2 + 5x - 24 = 0$
[Solution Description]
Let the width be $x$ units. Then, the length is $(x + 5)$ units. The area of the rectangle is given by:
$x(x + 5) = 24$
Expanding and rearranging the equation:
$x^2 + 5x - 24 = 0$
This is a quadratic equation.

Key Concept: Standard Form, Coefficient Analysis

b) $-1$
[Solution Description]
If the coefficient of $x^2$ is zero, then $(m - 1) = 0$, which means $m = 1$. Substitute $m = 1$ into the remaining terms to find the constant term: $(2(1) + 1)x + (1 - 2)$. The constant term is $\boxed{-1}$.

Key Concept: Quadratic Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For the given quadratic equation $x^2 + 4x + 4 = 0$, we calculate the discriminant as follows:
$D = b^2 - 4ac = (4)^2 - 4(1)(4) = 16 - 16 = 0$
Since the discriminant is zero, the equation has exactly one real root (a repeated root). Therefore, the Assertion is true, and the Reason is also true and correctly explains the Assertion.

Key Concept: Problems Based on Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the smaller number be $x$ and the larger number be $x + 3$. According to the problem, the sum of their squares is 117. This gives the equation:
$x^2 + (x + 3)^2 = 117$
Simplifying the equation:
$x^2 + x^2 + 6x + 9 = 117$
Combine like terms:
$2x^2 + 6x - 108 = 0$
Divide the entire equation by 2:
$x^2 + 3x - 54 = 0$
Factorize the quadratic equation:
$(x + 9)(x - 6) = 0$
Solutions: $x = -9$ or $x = 6$. Since we are considering natural numbers, $x = 6$ is valid.
Thus, the numbers are $x = 6$ and $x + 3 = 9$. Both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Numbers, Reciprocals

b) 5 and 20
[Solution Description]
Let one part be $x$, other part is $(25 - x)$.
Given: $\frac{1}{x} + \frac{1}{25-x} = \frac{1}{4}$
$\frac{25-x+x}{x(25-x)} = \frac{1}{4}$
$\frac{25}{25x-x^2} = \frac{1}{4}$
$100 = 25x - x^2$
$x^2 - 25x + 100 = 0$
$(x - 20)(x - 5) = 0$
$x = 20$ or $x = 5$
Thus the parts are 5 and 20

Key Concept: Quadratic Equation Factorization

b) 7 and 8
[Solution Description]
Let the first number be $x$, then the next consecutive integer is $x + 1$. According to the problem, $x(x + 1) = 56$. Expanding and simplifying gives $x^2 + x - 56 = 0$. Factorizing: $(x + 8)(x - 7) = 0$. Thus, $x = -8$ or $x = 7$. Since the numbers are positive, $x = 7$ and the next number is $8$.

Key Concept: Forming Quadratic Equations from Word Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion describes the change in cost and quantity, which is used in forming quadratic equations for profit and loss problems as mentioned in the syllabus. The reason correctly explains that the multiplication of cost and quantity leads to a quadratic equation. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Quadratic Equations

b) 23
[Solution Description] Given: $S = 276$
Substitute into the formula: $\frac{n(n + 1)}{2} = 276$
Multiply both sides by 2: $n^2 + n = 552$
Rearrange: $n^2 + n - 552 = 0$
Factorize: $(n + 24)(n - 23) = 0$
Since n is a natural number, reject $n = -24$. Thus, $n = 23$.

Key Concept: Time and Work, Fractional Work

b) 7 hours 44 minutes
[Solution Description]
Work done by A in 1 hour: $\frac{1/3}{10} = \frac{1}{30}$ units/hour.
Work done by B in 1 hour: $\frac{1/4}{8} = \frac{1}{32}$ units/hour.
Work done by A in 15 hours: $15 \times \frac{1}{30} = \frac{1}{2}$.
Remaining work: $1 - \frac{1}{2} = \frac{1}{2}$.
Combined efficiency: $\frac{1}{30} + \frac{1}{32} = \frac{62}{960} = \frac{31}{480}$ units/hour.
Time required: $\frac{1/2}{31/480} = \frac{480}{62} = 7.74$ hours ≈ 7 hours 44 minutes.

Key Concept: Cost price, Profit calculation

b) 60
[Solution Description]
Let the number of articles bought be $n$. The cost price per article is $\frac{4800}{n}$. The total selling price is $100n$. Given that profit equals the cost price of 15 articles, we have:
$100n - 4800 = 15 \times \frac{4800}{n}$
Simplifying:
$100n^2 - 4800n = 72000 \Rightarrow 100n^2 - 4800n - 72000 = 0$
Divide by 100:
$n^2 - 48n - 720 = 0$
Solving the quadratic equation:
$n = \frac{48 \pm \sqrt{2304 + 2880}}{2} = \frac{48 \pm 72}{2}$
Taking the positive root:
$n = \frac{120}{2} = 60$

Key Concept: Problem on Speed and Time Reduction

c) 30 km/hr
[Solution Description]
Let the original speed be $x$ km/hr.
Original time = $\frac{72}{x}$ hours.
New speed = $(x + 10)$ km/hr.
New time = $\frac{72}{x + 10}$ hours.
Time reduction = 36 minutes = $\frac{36}{60} = 0.6$ hours.
Equation: $\frac{72}{x} - \frac{72}{x + 10} = 0.6$.
Simplify: $72(x + 10) - 72x = 0.6x(x + 10)$.
$720 = 0.6x^2 + 6x$.
$0.6x^2 + 6x - 720 = 0$.
Divide by 0.6: $x^2 + 10x - 1200 = 0$.
Factorize: $(x + 40)(x - 30) = 0$.
Solution: $x = 30$ km/hr (since speed cannot be negative).

Key Concept: Quadratic Formula, Complex Roots

c) $c > 9$
[Solution Description]
For a quadratic equation to have no real roots, its discriminant must be negative. The discriminant $D = b^2 - 4ac$.
Here, $a = 1$, $b = -6$, and $c = c$. So,
$D = (-6)^2 - 4(1)(c) = 36 - 4c$
For no real roots, $D < 0$:
$36 - 4c 36 \Rightarrow c > 9$

Key Concept: Solving Quadratic Equation

c) $x = 3$ or $x = -2$
[Solution Description]
The given quadratic equation is $x^2 - x - 6 = 0$.
To solve it, we can factorize it as follows:
$$(x - 3)(x + 2) = 0$$
Setting each factor to zero gives two solutions:
$$x - 3 = 0 \Rightarrow x = 3$$
$$x + 2 = 0 \Rightarrow x = -2$$
Since length or breadth cannot be negative, the valid solution is $x = 3$.

Key Concept: Quadratic Equations - Factorization

c) (-9, -8) and (8, 9)
[Solution Description]
Let the smaller integer be $x$, so the next consecutive integer is $x + 1$.
$x(x + 1) = 72$
Expanding:
$x^2 + x - 72 = 0$
Factorizing:
$(x + 9)(x - 8) = 0$
So, $x = -9$ or $x = 8$. Therefore, the pairs are (-9, -8) and (8, 9).

Key Concept: Quadratic Equations - Pythagorean Theorem

b) 5 cm
[Solution Description]
Let the shorter side be $x$ cm. Then, the longer side is $(x + 7)$ cm. Using the Pythagorean theorem:
$x^2 + (x + 7)^2 = 13^2$
Expanding and simplifying:
$2x^2 + 14x + 49 = 169$
Rearranging:
$2x^2 + 14x - 120 = 0$
Dividing by 2:
$x^2 + 7x - 60 = 0$
Factoring:
$(x + 12)(x - 5) = 0$
Solving for $x$, we get:
$x = -12 \text{ or } x = 5$
Since length cannot be negative, $x = 5$ cm.

Key Concept: Verification of Quadratic Equation

b) $x^2 - 2x - 3 = 0$
[Solution Description]
Expand $(x + 1)(x - 3)$ step-by-step:
1. Multiply $x$ by $x$: $x^2$
2. Multiply $x$ by $-3$: $-3x$
3. Multiply $1$ by $x$: $+x$
4. Multiply $1$ by $-3$: $-3$
Combine like terms: $x^2 - 3x + x - 3 = x^2 - 2x - 3$.
The standard form is $x^2 - 2x - 3 = 0$.

Key Concept: Speed, Distance, and Time

c) 16 litres
[Solution Description]
Petrol used by first car: $\frac{400}{x}$ litres. Petrol used by second car: $\frac{400}{x + 5}$ litres. Given that the difference in usage is 4 litres:
$\frac{400}{x} - \frac{400}{x + 5} = 4$
Solving gives $x = 20$ km/litre. Thus, petrol used by second car = $\frac{400}{25} = 16$ litres.

Key Concept: Formation of the Equation

b) $2x^2 + 4x + 4 = 130$
[Solution Description]
Let the smaller odd integer be $x$. The next consecutive odd integer will be $x + 2$.
According to the problem, the sum of their squares is 130:
$x^2 + (x + 2)^2 = 130$
The correct equation must represent this relationship.

Key Concept: Problems Based on Numbers

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the natural numbers be $x$ and $x + 3$. Given their squares sum to 117:
$x^2 + (x + 3)^2 = 117$ expands to $2x^2 + 6x - 108 = 0$, which simplifies to $x^2 + 3x - 54 = 0$.
Factoring gives $(x + 9)(x - 6) = 0$, so $x = -9$ or $x = 6$. Since $-9$ is not a natural number, the valid solution is $x = 6$. Thus, the numbers are $6$ and $9$.
Both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Word Problem to Equation: Cost and Quantity Problems

b) $\frac{960}{x - 8} - \frac{960}{x} = 4$
[Solution Description]
Number of books bought originally:
$\frac{960}{x}$
New cost per book:
$x - 8$
New number of books:
$\frac{960}{x - 8}$
According to the problem:
$\frac{960}{x - 8} - \frac{960}{x} = 4$

Key Concept: Perimeter and Area Relationship

c) 32 m by 20 m
[Solution Description]
Let the length and breadth be $l$ and $b$ respectively.
Given:
$$\text{Area} = l \cdot b = 640$$
$$\text{Perimeter} = 2(l + b) = 104 \Rightarrow l + b = 52$$
We need to find numbers that add up to 52 and multiply to 640.
Solving the quadratic equation derived from these conditions:
$$t^2 - 52t + 640 = 0$$
Using the quadratic formula:
$$t = \frac{52 \pm \sqrt{2704 - 2560}}{2} = \frac{52 \pm \sqrt{144}}{2}$$
$$t = \frac{52 \pm 12}{2}$$
Thus, $t = 32$ or $t = 20$.
Therefore, the dimensions are 32 m by 20 m.

Key Concept: Problems Based on Numbers

b) $7$
[Solution Description]
Let the smaller odd number be $x$. Then, the next consecutive odd number is $x + 2$.
Given:
$x^2 + (x + 2)^2 = 130$
Expanding:
$x^2 + x^2 + 4x + 4 = 130$
Simplify:
$2x^2 + 4x - 126 = 0$
Divide by 2:
$x^2 + 2x - 63 = 0$
Factorize:
$(x + 9)(x - 7) = 0$
Thus, $x = -9$ or $x = 7$.
Since the number must be positive, the smaller number is 7.

Key Concept: Time and Work

c) 12 hours
[Solution Description]
Let the total capacity of the tank be $1$ unit.
Filling rate = $\frac{1}{4}$ per hour.
Emptying rate = $\frac{1}{6}$ per hour.
Net filling rate = $\frac{1}{4} - \frac{1}{6} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12}$ per hour.
Time taken to fill the tank = $\frac{1}{\frac{1}{12}} = 12$ hours.

Key Concept: Distance, Speed, and Time Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To solve the problem, we start with the given equation:
$\frac{216}{x} - \frac{208}{x + 16} = 2$
Multiply both sides by $x(x + 16)$ to eliminate denominators:
$216(x + 16) - 208x = 2x(x + 16)$
Simplify the left-hand side:
$216x + 3456 - 208x = 2x^2 + 32x$
Combine like terms:
$8x + 3456 = 2x^2 + 32x$
Rearrange into standard quadratic form:
$2x^2 + 24x - 3456 = 0$
Divide by 2:
$x^2 + 12x - 1728 = 0$
Solve using the quadratic formula:
$x = \frac{-12 \pm \sqrt{144 + 6912}}{2} = \frac{-12 \pm \sqrt{7056}}{2} = \frac{-12 \pm 84}{2}$
The positive solution is $x = 36$ km/hr (car's speed), making the train's speed $52$ km/hr. Substitution confirms these values satisfy the original equation, validating both Assertion and Reason, with Reason correctly explaining Assertion.

Key Concept: Verandah Dimensions

a) 3 m and 6 m
[Solution Description]
Let the breadth be $x$ m. Then, the length is $(x + 3)$ m. Area = Perimeter gives:
$x(x + 3) = 2(x + x + 3)$
$x^2 + 3x = 4x + 6$
$x^2 - x - 6 = 0$
Solve using factorization:
$(x - 3)(x + 2) = 0$
Thus, $x = 3$ m (breadth), and $(x + 3) = 6$ m (length).

Key Concept: Quadratic Equations

b) 12
[Solution Description] Given: $S = 168$
Substitute into the formula: $n(n + 2) = 168$
Expand: $n^2 + 2n - 168 = 0$
Factorize: $(n + 14)(n - 12) = 0$
Reject negative solution since n must be positive: $n = 12$.

Key Concept: Square Areas Problem

c) 16 cm
[Solution Description]
Area of first square = $x^2$ cm$^2$.
Area of second square = $(x + 4)^2$ cm$^2$.
Given that the sum of their areas is 656 cm$^2$:
$x^2 + (x + 4)^2 = 656$
$x^2 + x^2 + 8x + 16 = 656$
$2x^2 + 8x - 640 = 0$
Divide by 2:
$x^2 + 4x - 320 = 0$
Solve using quadratic formula:
$x = \frac{-4 \pm \sqrt{16 + 1280}}{2}$
$x = \frac{-4 \pm \sqrt{1296}}{2}$
$x = \frac{-4 \pm 36}{2}$
Thus:
$x = \frac{32}{2} = 16 \text{ cm}$
$x = \frac{-40}{2} = -20 \text{ cm (invalid)}$
Therefore, the side length of the smaller square is 16 cm.

Key Concept: Real and Distinct Roots

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For the quadratic equation $x^2 - 3x + 2 = 0$, we calculate the discriminant ($D$) as follows:
Discriminant formula: $D = b^2 - 4ac$. Here, $a = 1$, $b = -3$, $c = 2$.
Substituting values: $D = (-3)^2 - 4(1)(2) = 9 - 8 = 1$.
Since $D > 0$, the equation has two real and distinct roots. Hence, both Assertion (A) and Reason (R) are true, and Reason is the correct explanation of Assertion.

Key Concept: Definition of Quadratic Equation

a) $2x^2 + 3x - 5 = 0$
[Solution Description]
$0x^2 + 3x - 2 = 0$ is not quadratic ($a = 0$). The correct answer is $2x^2 + 3x - 5 = 0$.

Key Concept: Time and Work

b) 8.33 days
[Solution Description]
Work rates: A = $\frac{1}{15}$ per day, B = $\frac{1}{20}$ per day.
Combined work rate: $\frac{1}{15} + \frac{1}{20} = \frac{7}{60}$ per day.
Work done in 5 days: $5 \times \frac{7}{60} = \frac{35}{60} = \frac{7}{12}$.
Remaining work: $1 - \frac{7}{12} = \frac{5}{12}$.
Time taken by B to complete remaining work: $\frac{\frac{5}{12}}{\frac{1}{20}} = \frac{100}{12} = 8.\overline{3}$ days.

Key Concept: Distance, Speed, and Time Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the speed of the car be $x$ km/hr. The speed of the train is $(x + 16)$ km/hr. The time taken by the car is $\frac{216}{x}$ hours, and the time taken by the train is $\frac{208}{x + 16}$ hours. According to the assertion, the train takes 2 hours less than the car, so the equation $\frac{216}{x} - \frac{208}{x + 16} = 2$ is valid. Solving this equation confirms the reason. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Right-Angled Triangle Problems

b) $(x + 7)^2 + x^2 = 13^2$
[Solution Description]
Given: Hypotenuse = 13 cm, Difference between other two sides = 7 cm.
Let the shorter side be $x$ cm, then the longer side is $(x + 7)$ cm. Using Pythagoras' theorem:
$x^2 + (x + 7)^2 = 13^2$
Simplifying:
$2x^2 + 14x + 49 = 169$
Further simplification gives:
$2x^2 + 14x - 120 = 0$
This is the required equation.

Key Concept: Time and Work

b) 5.4 hours
[Solution Description]
Let the time for which both pipes worked together be $t$ hours.
Combined work rate of A and B: $\frac{1}{12} + \frac{1}{18} = \frac{5}{36}$ per hour.
Work done in $t$ hours: $\frac{5t}{36}$.
Remaining work: $1 - \frac{5t}{36}$.
Pipe B completes the remaining work in $\frac{9}{2}$ hours: $\frac{9}{2} \times \frac{1}{18} = \frac{1}{4}$.
So, $1 - \frac{5t}{36} = \frac{1}{4}$.
Solving, $\frac{5t}{36} = \frac{3}{4}$ $\Rightarrow t = \frac{27}{5} = 5.4$ hours.

Key Concept: Quadratic Equations from Word Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] To form the equation, let the original number of children be $n$. The original share per child is $\frac{250}{n}$. When 25 more children are added, the new share becomes $\frac{250}{n + 25}$. According to the problem, this new share is 50 paise (Rs.0.50) less than the original share. Thus, the equation is $\frac{250}{n} - \frac{250}{n + 25} = 0.50$. The Reason correctly identifies that the equation captures the change in share due to an increase in the number of children. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Right-Angled Triangle Sides and Area

b) 12 cm, 5 cm
[Solution Description]
Area of a right-angled triangle with legs $a$ and $b$ is $\frac{1}{2}ab$. Here:
$\frac{1}{2} \times 4x \times (2x - 1) = 30$
Simplifying:
$4x(2x - 1) = 60$
$8x^2 - 4x - 60 = 0$
Divide by 4:
$2x^2 - x - 15 = 0$
Solving, valid solution: $x = 3$, so sides are $12$ cm and $5$ cm.

Key Concept: Problems Based on Numbers

a) 5 and 8
[Solution Description]
Let the numbers be $x$ and $x + 3$.
Given: $x^2 + (x + 3)^2 = 89$.
Expanding: $x^2 + x^2 + 6x + 9 = 89$.
Simplify: $2x^2 + 6x - 80 = 0$.
Divide by 2: $x^2 + 3x - 40 = 0$.
Factorize: $(x + 8)(x - 5) = 0$.
Thus, $x = -8$ or $x = 5$.
Since $x$ is a natural number, $x = 5$.
The numbers are $5$ and $8$.

Key Concept: Investment and Profit Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let's verify the Assertion and Reason step by step:
Cost price of 100 articles = \$1,200.
Number of damaged articles = 10, so remaining articles = 90.
Selling price per article = \$14.
Total selling price = 90 × \$14 = \$1,260.
Profit = Total selling price - Cost price = \$1,260 - \$1,200 = \$60.
Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion based on the given data.

Key Concept: Identifying Quadratic Equations

b) $x^2 + 2x - 48 = 0$
[Solution Description]
Let the first even number be $x$. Then, the next consecutive even number is $(x + 2)$. The sum of their squares is:
$x^2 + (x + 2)^2 = 100$
Expanding and simplifying:
$x^2 + x^2 + 4x + 4 = 100$
$2x^2 + 4x - 96 = 0$
Dividing by 2:
$x^2 + 2x - 48 = 0$
This is a quadratic equation.

Key Concept: Investment and Profit Problems

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let C.P. be \$x. Given loss percentage is $x\%$. Loss = C.P. - S.P. = $\frac{x}{100} \times x = \frac{x^2}{100}$. According to the problem: $x - \frac{x^2}{100} = 24 \Rightarrow x^2 - 100x + 2400 = 0$. Solving this quadratic equation gives $x = 60$ or $x = 40$. Hence, the assertion that C.P. can be \$60 is true. The reason correctly states the formula for loss percentage but does not explain why the assertion is specifically true (it only provides a general formula). Thus, both are true but the reason does not correctly explain the assertion.

Key Concept: Problems Based on Numbers

c) Assertion is true, but Reason is false.
[Solution Description]
Let the two natural numbers be $x$ and $x + 3$. According to the problem, the sum of their squares is 117. Thus, we form the equation:
$x^2 + (x + 3)^2 = 117$
Simplify the equation:
$x^2 + x^2 + 6x + 9 = 117$
Combine like terms:
$2x^2 + 6x - 108 = 0$
Divide the entire equation by 2:
$x^2 + 3x - 54 = 0$
Factorize the quadratic equation:
$(x + 9)(x - 6) = 0$
The solutions are $x = -9$ and $x = 6$. Since $-9$ is not a natural number, the valid solution is $x = 6$. Therefore, the two numbers are 6 and 9.
The assertion (A) correctly describes the process of forming and solving the quadratic equation. However, the reason (R) states that quadratic equations derived from such problems always yield non-negative integer solutions, which is incorrect. For example, in this case, one of the solutions ($-9$) is negative. Hence, the assertion is true, but the reason is false.

Key Concept: Identifying Quadratic Equations from Word Problems

a) $x^2 + 2x - 48 = 0$
[Solution Description]
Let the smaller even number be $x$. The next consecutive even number is $(x + 2)$. According to the problem:
$x(x + 2) = 48$
Simplifying the equation:
$x^2 + 2x - 48 = 0$
This is the quadratic equation that represents the given situation.

Key Concept: Time and Work, Distance-Speed-Time

c) 36 hours
[Solution Description]
Pipe A fills at $\frac{1}{12}$ per hour.
Pipe B fills at $\frac{1}{18}$ per hour.
Pipe C empties at $\frac{1}{9}$ per hour.
Combined rate when all are open: $\frac{1}{12} + \frac{1}{18} - \frac{1}{9}$.
To add fractions, find LCM of denominators (36):
$\frac{3}{36} + \frac{2}{36} - \frac{4}{36} = \frac{1}{36}$ per hour.
Hence, total time taken to fill the tank is the reciprocal of the combined rate: $36$ hours.