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

Key Concept: Real and Distinct Roots

a) 4 and 3
[Solution Description]
Given the quadratic equation: $x^2 - 7x + 12 = 0$.
Here, $a = 1$, $b = -7$, $c = 12$.
The discriminant is calculated as:
$D = b^2 - 4ac = (-7)^2 - 4(1)(12) = 49 - 48 = 1$.
Since $D > 0$, there are two distinct real roots.
Using the quadratic formula:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{1}}{2}$.
Thus, the roots are:
$x = \frac{7 + 1}{2} = 4$ and $x = \frac{7 - 1}{2} = 3$.

Key Concept: Speed, Distance, and Time

c) 30 km/hr
[Solution Description]
Let the original speed be $x$ km/hr. The original time taken is $\frac{72}{x}$ hours. New speed = $(x + 10)$ km/hr, and new time taken = $\frac{72}{x + 10}$ hours. Given, time reduction = 36 minutes = $\frac{3}{5}$ hours. Therefore:
$\frac{72}{x} - \frac{72}{x + 10} = \frac{3}{5}$
Solving the equation gives $x = 30$ 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 and Work

b) 7.2 days
[Solution Description]
Let the total work be $1$ unit.
A's work rate is $\frac{1}{12}$ per day.
B's work rate is $\frac{1}{18}$ per day.
Combined work rate = $\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}$ per day.
Time taken together = $\frac{1}{\frac{5}{36}} = \frac{36}{5} = 7.2$ days or $7\frac{1}{5}$ days.

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: Distance-Speed-Time

c) 15 km/hr
[Solution Description]
To find the average speed, we use the formula:
$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{45\,\text{km}}{3\,\text{hrs}} = 15\,\text{km/hr}$

Key Concept: Discriminant Concept

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

Key Concept: Wage calculation, Equation formation

c) 45
[Solution Description]
Let the original weekly wage be $x$ and the original number of workers be $y$. The initial total wage bill is $xy = 3150$. After increasing the wage by ¥ 5 and reducing the number of workers by 5, the new wage bill is $(x + 5)(y - 5) = 3250$. From the first equation, $y = \frac{3150}{x}$. Substituting into the second equation:
$(x + 5)\left(\frac{3150}{x} - 5\right) = 3250$
Expanding:
$3150 - 5x + \frac{15750}{x} - 25 = 3250 \Rightarrow -5x + \frac{15750}{x} + 3125 = 3250$
Simplifying:
$-5x + \frac{15750}{x} = 125 \Rightarrow -5x^2 + 15750 = 125x \Rightarrow 5x^2 + 125x - 15750 = 0$
Divide by 5:
$x^2 + 25x - 3150 = 0$
Solving the quadratic equation:
$x = \frac{-25 \pm \sqrt{625 + 12600}}{2} = \frac{-25 \pm 115}{2}$
Taking the positive root:
$x = \frac{90}{2} = 45$

Key Concept: Quadratic Equation in Profit/Loss

b) \$31.11
[Solution Description]
Given: Selling Price (S.P.) = \$28, Loss % = 10%.
Let Cost Price (C.P.) be $x$.
Loss = 10% of C.P. = $\frac{10}{100} x = 0.1x$
S.P. = C.P. - Loss
$28 = x - 0.1x$
$28 = 0.9x$
$x = \frac{28}{0.9} \approx 31.11$
Therefore, the cost price is approximately \$31.11.

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: Quadratic Equations - Factorization

c) $\frac{1}{3}$ and 3
[Solution Description]
Let the number be $x$. According to the problem:
$x + \frac{1}{x} = \frac{10}{3}$
Multiply through by $3x$ to eliminate denominators:
$3x^2 + 3 = 10x$
Bring all terms to one side:
$3x^2 - 10x + 3 = 0$
Factorizing:
$(3x - 1)(x - 3) = 0$
So, $x = \frac{1}{3}$ or $x = 3$.

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: Cost Price and Selling Price

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the selling price when the cost price is \$40 and the loss is 20%, we calculate the loss amount as 20% of \$40 which is \$8. The selling price is then obtained by subtracting the loss from the cost price: \$40 - \$8 = \$32. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Standard Form of Quadratic Equation

b) $ax^2 + bx + c = 0$
[Solution Description]
The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a \neq 0$. This is the universally accepted form.

Key Concept: Problems Based on Geometrical Figures

a) 16 cm and 23 cm
[Solution Description]
Let the shorter side be $x$ cm.
Then, the longer side is $(x + 7)$ cm.
Using Pythagoras' theorem:
$x^2 + (x + 7)^2 = 25^2$.
Expand: $x^2 + x^2 + 14x + 49 = 625$.
Simplify: $2x^2 + 14x - 576 = 0$.
Divide by 2: $x^2 + 7x - 288 = 0$.
Factorize: $(x + 18)(x - 16) = 0$.
Thus, $x = -18$ or $x = 16$.
Since length cannot be negative, $x = 16$.
The sides are $16$ cm and $23$ cm.

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: 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: 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: 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 $x$. Then the larger number is $(x + 3)$. According to the problem statement:
$x^2 + (x + 3)^2 = 117$
Expanding the equation:
$x^2 + x^2 + 6x + 9 = 117$
Simplifying:
$2x^2 + 6x - 108 = 0$
Further simplifying by dividing by 2:
$x^2 + 3x - 54 = 0$
Solving using factorization:
$(x + 9)(x - 6) = 0$
Therefore, $x = 6$ or $x = -9$. Since $x$ must be a natural number, $x = 6$, and the numbers are 6 and 9. Thus, the Assertion is true. The Reason correctly identifies the formation of the quadratic equation, which leads to the solution. Hence, both Assertion and Reason are true, and Reason explains the Assertion.

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: 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: 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.

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: Time and Work, Reciprocals

c) 20 hours and 30 hours
[Solution Description]
Let faster pipe take $x$ hours, slower pipe takes $(x + 10)$ hours.
Their combined work rate:
$\frac{1}{x} + \frac{1}{x+10} = \frac{1}{12}$
$\frac{x+10+x}{x(x+10)} = \frac{1}{12}$
$\frac{2x+10}{x^2+10x} = \frac{1}{12}$
$24x + 120 = x^2 + 10x$
$x^2 - 14x - 120 = 0$
$(x - 20)(x + 6) = 0$
$x = 20$ or $x = -6$
Taking positive value: Faster pipe = 20 hrs, Slower pipe = 30 hrs

Key Concept: Quadratic Equation, Discriminant

d) Assertion is false, but Reason is true.
[Solution Description]
First, we calculate the discriminant $D$ for the given quadratic equation $3x^2 - 6x + 3 = 0$. Here, $a = 3$, $b = -6$, and $c = 3$.
The formula for discriminant is:
$D = b^2 - 4ac$
Substituting the values:
$D = (-6)^2 - 4(3)(3) = 36 - 36 = 0$
Since $D = 0$, the equation has exactly one real root (a repeated root), not two distinct real roots. Therefore, Assertion (A) is false.
Reason (R) states that the discriminant is positive when $b^2 - 4ac > 0$, which is a correct statement as per the definition of discriminant. Hence, Reason (R) is true.
Thus, Assertion is false, but Reason is true.

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: Quadratic Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] To solve this, first find the discriminant $D$ of the quadratic equation $x^2 - 5x + 6 = 0$. The discriminant formula is $D = b^2 - 4ac$, where $a = 1$, $b = -5$, and $c = 6$. Substituting the values: $D = (-5)^2 - 4(1)(6) = 25 - 24 = 1$. Since $D > 0$, the assertion is true as there are two real roots. The reason correctly states that the discriminant is positive ($1 > 0$) and explains why the assertion is correct.

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: Quadratic Formula

c) 0
[Solution Description]
The discriminant ($D$) of a quadratic equation $ax^2 + bx + c = 0$ is given by:
$D = b^2 - 4ac$
For the equation $x^2 - 6x + 9 = 0$, the coefficients are $a = 1$, $b = -6$, and $c = 9$. Substitute these values:
$D = (-6)^2 - 4(1)(9) = 36 - 36 = 0$

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: Quadratic Equation in Rectangle Dimensions

c) 5.73 m
[Solution Description]
Let the breadth be $x$ m. Then, the length is $(2x + 5)$ m.
Given that the area equals the perimeter plus 50:
$x(2x + 5) = 2(x + 2x + 5) + 50$
$2x^2 + 5x = 2(3x + 5) + 50$
$2x^2 + 5x = 6x + 10 + 50$
$2x^2 + 5x - 6x - 60 = 0$
$2x^2 - x - 60 = 0$
Solving the quadratic equation using the quadratic formula:
$x = \frac{1 \pm \sqrt{1 + 480}}{4}$
$x = \frac{1 \pm \sqrt{481}}{4}$
Since $\sqrt{481} \approx 21.93$, so:
$x = \frac{1 + 21.93}{4} \approx 5.73 \text{ m}$
$x = \frac{1 - 21.93}{4} \approx -5.23 \text{ m (invalid)}$
So, the breadth is approximately 5.73 m.

Key Concept: Distance, Speed, and Time, Forming Quadratic Equations

c) 36 km/hr
[Solution Description]
Time taken by car: $\frac{216}{x}$ hours. Time taken by train: $\frac{208}{x + 16}$ hours. The difference in time is 2 hours:
$\frac{216}{x} - \frac{208}{x + 16} = 2$
Multiply both sides by $x(x + 16)$ to eliminate denominators:
$216(x + 16) - 208x = 2x(x + 16)$
Expand and simplify:
$216x + 3456 - 208x = 2x^2 + 32x$
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:
$x = \frac{-12 \pm \sqrt{144 + 6912}}{2} = \frac{-12 \pm \sqrt{7056}}{2} = \frac{-12 \pm 84}{2}$
Discard the negative solution:
$x = \frac{72}{2} = 36$

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: Problem on Petrol Consumption by Two Cars

b) 20 km/litre
[Solution Description]
Petrol used by Car A = $\frac{400}{x}$ litres.
Petrol used by Car B = $\frac{400}{x + 5}$ litres.
Given difference: $\frac{400}{x} - \frac{400}{x + 5} = 4$.
Simplify: $400(x + 5) - 400x = 4x(x + 5)$.
$2000 = 4x^2 + 20x$.
Divide by 4: $x^2 + 5x - 500 = 0$.
Factorize: $(x + 25)(x - 20) = 0$.
Solution: $x = 20$ km/litre (positive value).

Key Concept: Investment and Profit Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the number of articles bought be $x$.
Cost price per article: $\frac{1200}{x}$.
Number of articles sold: $x - 10$.
Selling price per article: $\frac{1200}{x} + 2$.
Total selling price: $(x - 10) \left( \frac{1200}{x} + 2 \right)$.
Given, profit is \$60:
$(x - 10) \left( \frac{1200}{x} + 2 \right) - 1200 = 60$.
Simplifying:
$1200 + 2x - \frac{12000}{x} - 20 - 1200 = 60$ which reduces to $2x - \frac{12000}{x} - 20 = 60$.
Further simplification gives $2x^2 - 80x - 12000 = 0$.
Solving the quadratic equation gives $x = 100$ or $x = -60$.
The solution is $x = 100$ since the number of articles cannot be negative.
Both Assertion and Reason are true, and Reason correctly explains Assertion.

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: Relative Speed, Two Objects Moving Apart

c) 70 km/hr
[Solution Description]
Let the speed of the slower car be $x$ km/hr. The speed of the faster car is $(x + 20)$ km/hr. Since they are moving in opposite directions, their relative speed is $(x + x + 20) = (2x + 20)$ km/hr. In 5 hours, they cover 600 km:
$5(2x + 20) = 600$
Solve for $x$:
$10x + 100 = 600$
$10x = 500$
$x = 50 \text{ km/hr}$
The speed of the faster car is $50 + 20 = 70$ km/hr.

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: 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: 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: 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: 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: Problems Based on Geometrical Figures

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

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: Problems Involving Speed, Time, and Distance

b) $50$ km/hr
[Solution Description]
Original time taken: $\frac{300}{x}$ hours.
New speed: $x + 10$ km/hr.
New time taken: $\frac{300}{x + 10}$ hours.
According to the problem:
$\frac{300}{x} - \frac{300}{x + 10} = 1$
Solve for $x$:
Multiply through by $x(x + 10)$:
$300(x + 10) - 300x = x(x + 10)$
Simplify:
$300x + 3000 - 300x = x^2 + 10x$
$x^2 + 10x - 3000 = 0$
Factorize:
$(x + 60)(x - 50) = 0$
Thus, $x = -60$ or $x = 50$.
Since speed cannot be negative, $x = 50$ km/hr.

Key Concept: Word Problem to Equation: Forming Quadratic Equations from Word Problems

c) $x(p + 0.25) = (x + 3)p$
[Solution Description]
Let the market price per orange be $p$.
Total cost paid by Mr. Mehra:
$x \times (p + 0.25)$
If he ate 3 oranges, the total cost would have been:
$(x + 3) \times p$
Equating both expressions:
$x(p + 0.25) = (x + 3)p$
Simplifying gives:
$0.25x = 3p$

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: Multi-Vehicle Efficiency Comparison

c) 2.4
[Solution Description]
Fuel used by Car X = $\frac{360}{12} = 30$ litres.
Fuel used by Car Y = $\frac{360}{12 + k}$ litres.
Given difference: $30 - \frac{360}{12 + k} = 5$.
Simplify: $\frac{360}{12 + k} = 25 \implies 12 + k = 14.4 \implies k = 2.4$.

Key Concept: Problems Based on Work and Time

b) 24
[Solution Description]
The combined work rate of A and B is $\frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15}$. Multiply through by $15x(x + 16)$ to eliminate denominators:
$15(x + 16) + 15x = x(x + 16)$
Simplify:
$15x + 240 + 15x = x^2 + 16x$
Combine like terms:
$30x + 240 = x^2 + 16x$
Rearrange into standard quadratic form:
$x^2 - 14x - 240 = 0$
Factor the equation:
$(x - 24)(x + 10) = 0$
The positive solution is $x = 24$ days.

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: Numbers, Sum of Reciprocals

b) 7
[Solution Description]
Let the two consecutive odd integers be $x$ and $x + 2$. Given:
$\frac{1}{x} + \frac{1}{x+2} = \frac{16}{63}$
Taking LCM:
$\frac{(x+2) + x}{x(x+2)} = \frac{16}{63}$
Simplify numerator:
$\frac{2x + 2}{x^2 + 2x} = \frac{16}{63}$
Cross-multiply:
$63(2x + 2) = 16(x^2 + 2x)$
Expand:
$126x + 126 = 16x^2 + 32x$
Rearrange to form quadratic equation:
$16x^2 - 94x - 126 = 0$
Divide by 2:
$8x^2 - 47x - 63 = 0$
Solve using quadratic formula:
$x = \frac{47 \pm \sqrt{2209 + 2016}}{16} = \frac{47 \pm \sqrt{4225}}{16} = \frac{47 \pm 65}{16}$
So, $x = \frac{112}{16} = 7$ or $x = \frac{-18}{16} = -1.125$ (discarded as not integer).
Thus, the numbers are $7$ and $9$.

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: 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: Nature of Roots in Word Problems

a) 5 units and 12 units
[Solution Description]
Let the shorter leg be $x$ units. Then, the longer leg is $x + 7$ units. The area of the triangle is given by:
$\frac{1}{2} \times x \times (x + 7) = 30$
Simplifying:
$x(x + 7) = 60$
$x^2 + 7x - 60 = 0$
Calculating the discriminant:
$D = (7)^2 - 4(1)(-60) = 49 + 240 = 289$
Since $D > 0$, there are two distinct real roots. Solving:
$x = \frac{-7 \pm \sqrt{289}}{2} = \frac{-7 \pm 17}{2}$
Possible solutions: $x = 5$ or $x = -12$. Since length cannot be negative, $x = 5$ units. Thus, the longer leg is $5 + 7 = 12$ units.

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: 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: 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: Geometry, Quadratic Equations

b) 15 cm
[Solution Description]
Let the shorter leg be $x$ cm. Then the longer leg is $x + 5$ cm.
By Pythagoras' theorem: $x^2 + (x + 5)^2 = 25^2$.
Expand: $x^2 + x^2 + 10x + 25 = 625$ ⇒ $2x^2 + 10x - 600 = 0$.
Divide by 2: $x^2 + 5x - 300 = 0$.
Factorize: $(x + 20)(x - 15) = 0$ ⇒ $x = 15$ (since length cannot be negative).

Key Concept: Representing unknown quantities

c) Represent the unknown quantity by a variable
[Solution Description]
The first step is to represent the unknown quantity by a variable, say $x$, as mentioned in the syllabus.

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: 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: 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: Percentage Profit/Loss and Cost Price

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the cost price be $\yen$ $x$. The loss is given as $x%$ of C.P., which is $\frac{x^2}{100}$. Since S.P. = C.P. - Loss, we have:
$x - \frac{x^2}{100} = 24$
Multiplying through by 100 gives:
$100x - x^2 = 2400$
Rearranging terms:
$x^2 - 100x + 2400 = 0$
Solving this quadratic equation yields $x = 60$ or $x = 40$. Thus, both Assertion and Reason are true, and Reason correctly explains how to derive the Assertion.

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: 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 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: 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: Basic Concepts

c) \$5
[Solution Description]
Profit is calculated as:
$\text{Profit} = \text{S.P.} - \text{C.P.} = 15 - 10 = \$5$

Key Concept: Problems Based on Numbers

c) 8
[Solution Description]
Let the smaller number be $x$. Then, the next consecutive natural number is $x + 1$.
According to the problem:
$x^2 + (x + 1)^2 = 145$
Expand and simplify:
$x^2 + x^2 + 2x + 1 = 145 \\ \\ 2x^2 + 2x + 1 = 145 \\ \\ 2x^2 + 2x - 144 = 0 \\ \\ x^2 + x - 72 = 0$
Factorize the quadratic equation:
$(x + 9)(x - 8) = 0$
So, $x = 8$ or $x = -9$. Since $x$ is a natural number, the solution is $x = 8$.

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: Problems Based on Geometrical Figures

b) 2 meters
[Solution Description]
Let the breadth be $x$ meters. Then the length is $x + 4$ meters.
Area of rectangle: $x(x + 4)$
Perimeter of rectangle: $2(x + x + 4) = 2(2x + 4) = 4x + 8$
According to the problem, area equals perimeter:
$x(x + 4) = 4x + 8$
Simplify:
$x^2 + 4x = 4x + 8$
Subtract $4x + 8$ from both sides:
$x^2 - 8 = 0$
Solve for $x$:
$x^2 = 8$
$x = \sqrt{8} = 2\sqrt{2}$ meters (since breadth cannot be negative).
However, none of the options match this value, indicating a possible error in framing the problem.
Re-examining the problem statement and solution steps is necessary to ensure correctness.
Here's correct calculation:
Original equation: $x(x + 4) = 4x + 8$
=> $x^2 + 4x - 4x - 8 = 0$
=> $x^2 - 8 = 0$
Thus $x = \sqrt{8} = 2.828$ meters approximately.
Since no option matches exactly, considering integer solutions may indicate a misstated problem.
However, based on the standard problems we can modify our options accordingly.

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: Time, Speed, and Distance Problems

b) 36 km/hr
[Solution Description]
Time taken by the car to travel 216 km is $\frac{216}{x}$ hours. Time taken by the train to travel 208 km is $\frac{208}{x + 16}$ hours. According to the problem:
$\frac{216}{x} - \frac{208}{x + 16} = 2$
Multiply through by $x(x + 16)$ to eliminate denominators:
$216(x + 16) - 208x = 2x(x + 16)$
Expand and simplify:
$216x + 3456 - 208x = 2x^2 + 32x$
Combine like terms:
$8x + 3456 = 2x^2 + 32x$
Rearrange into standard quadratic form:
$2x^2 + 24x - 3456 = 0$
Divide through by 2:
$x^2 + 12x - 1728 = 0$
Solve using the quadratic formula:
$x = \frac{-12 \pm \sqrt{144 + 6912}}{2} = \frac{-12 \pm \sqrt{7056}}{2} = \frac{-12 \pm 84}{2}$
Taking the positive root:
$x = \frac{72}{2} = 36$
So, the speed of the car is 36 km/hr.

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 - 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: Problems Based on Geometrical Figures

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:
$x^2 + (x + 7)^2 = 13^2$
Simplify the equation:
$x^2 + x^2 + 14x + 49 = 169$
Combine like terms:
$2x^2 + 14x - 120 = 0$
Divide by 2:
$x^2 + 7x - 60 = 0$
Factor the quadratic equation:
$(x + 12)(x - 5) = 0$
The positive solution is $x = 5$ cm. Thus, the other side is $12$ cm ($5 + 7$).

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: Problems Based on Speed, Distance, and Time, Problems Involving Cost and Quantity

d) Assertion is false, but Reason is true.
[Solution Description]
First, translate the problem into an equation. The time taken by the car is $\frac{216}{x}$ hours, and the time taken by the train is $\frac{216}{x + 16}$ hours. According to the problem, the train takes 2 hours less than the car, so:
$\frac{216}{x} - \frac{216}{x + 16} = 2$
Simplify the equation by finding a common denominator and simplifying the numerator:
$216(x + 16) - 216x = 2x(x + 16)$
Expand and simplify:
$216x + 3456 - 216x = 2x^2 + 32x$
Combine like terms:
$3456 = 2x^2 + 32x$
Divide both sides by 2:
$x^2 + 16x - 1728 = 0$
Now, solve the quadratic equation using factorization or the quadratic formula. Factoring gives:
$(x + 48)(x - 36) = 0$
Thus, $x = -48$ (invalid) or $x = 36$. Therefore, the speed of the car is 36 km/hr, and the speed of the train is $36 + 16 = 52$ km/hr, not 40 km/hr as stated in the assertion. Hence, the Assertion is false. The Reason correctly forms and simplifies the quadratic equation, but it does not match the solution, so the Reason is true.

Key Concept: Profit and Loss Problems, Forming Quadratic Equations

c) 100
[Solution Description]
Cost price per article is $\frac{1200}{x}$ ¥. Selling price per article is $\frac{1200}{x} + 2$ ¥. Total selling price for $(x - 10)$ articles is $(x - 10)\left(\frac{1200}{x} + 2\right)$ ¥. Profit is given as ¥60:
$(x - 10)\left(\frac{1200}{x} + 2\right) = 1200 + 60$
Expand the left side:
$1200 + 2x - \frac{12000}{x} - 20 = 1260$
Simplify:
$2x - \frac{12000}{x} - 80 = 0$
Multiply through by $x$ to eliminate the denominator:
$2x^2 - 80x - 12000 = 0$
Divide the equation by 2:
$x^2 - 40x - 6000 = 0$
Solve using the quadratic formula:
$x = \frac{40 \pm \sqrt{1600 + 24000}}{2} = \frac{40 \pm \sqrt{25600}}{2} = \frac{40 \pm 160}{2}$
Discard the negative solution:
$x = \frac{200}{2} = 100$

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: Distance, Speed, Time

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Let the original speed be $x$ km/hr. The original time taken is $\frac{72}{x}$ hours. After increasing the speed by 10 km/hr, the new time taken is $\frac{72}{x + 10}$ hours. The reduction in time is 36 minutes, which is $\frac{36}{60} = 0.6$ hours. Therefore, we have the equation $\frac{72}{x} - \frac{72}{x + 10} = 0.6$. Multiplying through by $x(x + 10)$ gives $72(x + 10) - 72x = 0.6x(x + 10)$. Simplifying, $720 = 0.6x^2 + 6x$. Rearranging, $0.6x^2 + 6x - 720 = 0$. Dividing by 0.6 gives $x^2 + 10x - 1200 = 0$. Solving this quadratic equation using the quadratic formula: $x = \frac{-10 \pm \sqrt{100 + 4800}}{2} = \frac{-10 \pm 70}{2}$. The positive solution is $x = 30$ km/hr. The reason correctly describes the inverse relationship between speed and time for a fixed distance, but it does not account for the specific reduction in time due to the speed increase, making it an incomplete explanation.

Key Concept: Pythagoras Theorem, Quadratic Equations

a) $x^2 + x^2 + 14x + 49 = 169$
[Solution Description]
Let the shorter side be $x$ cm. Then the longer side is $(x + 7)$ cm. Using Pythagoras theorem:
$x^2 + (x + 7)^2 = 13^2$

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, 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: Dividing Money Among Children

b) 100
[Solution Description]
Let the original number of children be $x$. The amount each child gets initially is $\frac{250}{x}$ yen. With 25 more children, the new number of children is $x + 25$, and the amount each child gets now is $\frac{250}{x + 25}$ yen. According to the problem, this new amount is 50 paise (which is ¥0.5) less than the original amount:
$\frac{250}{x} - \frac{250}{x + 25} = 0.5$
Multiply through 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 for $x$ using the quadratic formula:
$x = \frac{-25 \pm \sqrt{625 + 50000}}{2} = \frac{-25 \pm \sqrt{50625}}{2} = \frac{-25 \pm 225}{2}$
Taking the positive root:
$x = \frac{200}{2} = 100$
So, the original number of children is 100.

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: Speed, Distance, and Time

b) 75 km/hr
[Solution Description]
Let speed of ordinary train = $x$ km/hr. Express train speed = $(x + 25)$ km/hr. Time taken by ordinary train = $\frac{300}{x}$ hrs; time taken by express train = $\frac{300}{x + 25}$ hrs. Given:
$\frac{300}{x} - \frac{300}{x + 25} = 2$
Solving gives $x = 50$ km/hr (ordinary), so express train speed = $75$ km/hr.

Key Concept: Investment and Profit Problems

b) ¥ 45
[Solution Description]
Let the original number of workers be $n$.
Original weekly wage bill = $3150 = n x$.
New weekly wage = $x + 5$.
New number of workers = $n - 5$.
New weekly wage bill = $3250 = (n - 5)(x + 5)$.
From first equation, $n = \frac{3150}{x}$.
Substitute into second equation:
$\left(\frac{3150}{x} - 5\right)(x + 5) = 3250$.
Simplify:
$\frac{(3150 - 5x)(x + 5)}{x} = 3250$.
Multiply through by $x$:
$(3150 - 5x)(x + 5) = 3250x$.
Expand:
$3150x + 15750 - 5x^2 - 25x = 3250x$.
Simplify:
$-5x^2 + 3125x + 15750 = 3250x$.
Bring all terms to one side:
$-5x^2 - 125x + 15750 = 0$.
Divide by $-5$:
$x^2 + 25x - 3150 = 0$.
Solve using quadratic formula:
$x = \frac{-25 \pm \sqrt{625 + 12600}}{2} = \frac{-25 \pm \sqrt{13225}}{2} = \frac{-25 \pm 115}{2}$.
Positive solution: $x = 45$.

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: 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: Pythagoras Theorem, Area of Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the sides of a right-angled triangle are 5 cm, 12 cm, and 13 cm. Let us verify this using the Pythagoras theorem mentioned in the reason.
Step 1: According to the Pythagoras theorem, for a right-angled triangle with legs of lengths 5 cm and 12 cm, the hypotenuse $h$ should satisfy:
$h = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$
Step 2: The assertion provides the hypotenuse as 13 cm, which matches the calculation. Thus, the assertion is true.
Step 3: The reason correctly states the Pythagoras theorem, which is the fundamental property used to verify the assertion. Hence, the reason is not only true but also the correct explanation for the assertion.

Key Concept: Profit and Number of Articles

c) 60
[Solution Description]
Let the number of articles bought be $n$.
Step 1: Calculate the cost price per article:
$\text{Cost Price per Article} = \frac{4800}{n}$
Step 2: Total selling price when sold at ¥ 100 each:
$\text{Selling Price} = 100n$
Step 3: Profit is given as the cost price of 15 articles:
$\text{Profit} = 15 \times \frac{4800}{n}$
Step 4: Formulate the profit equation:
$\text{Selling Price} = \text{Cost Price} + \text{Profit}$
$100n = 4800 + \frac{72000}{n}$
Step 5: Multiply through by $n$ to eliminate the denominator:
$100n^2 = 4800n + 72000$
$100n^2 - 4800n - 72000 = 0$
$n^2 - 48n - 720 = 0$
Step 6: Solve the quadratic equation:
$n = \frac{48 \pm \sqrt{2304 + 2880}}{2}$
$n = \frac{48 \pm \sqrt{5184}}{2}$
$n = \frac{48 \pm 72}{2}$
Taking the positive root:
$n = \frac{120}{2} = 60$
The number of articles bought is 60.

Key Concept: Word Problem to Equation, Forming Quadratic Equations from Word Problems, Profit and Loss Problems

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let the original number of books be $x$. The original cost per book is $\frac{800}{x}$. If the cost per book decreases by ¥10, it becomes $\frac{800}{x} - 10$. With this reduced price, the number of books that can be bought for ¥800 becomes $x + 4$. So, the new cost per book multiplied by the new number of books should equal ¥800:
$\left(\frac{800}{x} - 10\right)(x + 4) = 800.$
Expanding and simplifying:
$800 + \frac{3200}{x} - 10x - 40 = 800.$
Subtract 800 from both sides:
$\frac{3200}{x} - 10x - 40 = 0.$
Multiply through by $x$ to eliminate the denominator:
$3200 - 10x^2 - 40x = 0.$
Rearranging terms and dividing by -10:
$x^2 + 4x - 320 = 0.$
Thus, the assertion is correct. The reason correctly explains how the condition leads to the quadratic equation.

Key Concept: Quadratic Equations, Factorization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To determine the correctness of the assertion and reason:
1. The given quadratic equation is $x^2 - 4x + 4 = 0$.
2. Calculate the discriminant ($D$) using the formula $D = b^2 - 4ac$, where $a = 1$, $b = -4$, $c = 4$.
3. Substituting the values: $D = (-4)^2 - 4(1)(4) = 16 - 16 = 0$.
4. Since $D = 0$, the equation has exactly one real root (a repeated root).
5. Thus, the Assertion (A) is true because there is only one unique root when $D = 0$.
6. The Reason (R) states that the discriminant is zero, which is also true and correctly explains why the equation has only one unique root.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

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: 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: 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: 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: 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: 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: Quadratic Equations, Pythagorean Theorem

a) $x^2 - 23x + 120 = 0$
[Solution Description]
Let the shorter side be $x$. Then, the other side is $23 - x$.
Using Pythagoras' theorem:
$x^2 + (23 - x)^2 = 17^2$
Simplify the equation:
$x^2 + 529 - 46x + x^2 = 289$
Combine like terms:
$2x^2 - 46x + 529 - 289 = 0$
Simplify further:
$2x^2 - 46x + 240 = 0$
Divide the equation by 2:
$x^2 - 23x + 120 = 0$

Key Concept: Representation of Unknown Quantities

d) $x$ and $x + 5$
[Solution Description]
Let the smaller number be $x$. Since the difference is 5, the larger number can be represented as $x + 5$.
The product of the numbers is given by:
$x(x + 5) = 126$
Simplifying:
$x^2 + 5x - 126 = 0$
The correct representation is $x$ and $x + 5$.

Key Concept: Discriminant, Consecutive Integers

d) Both b and c
[Solution Description]
Let the smaller odd integer be $x$. Then the next consecutive odd integer is $x + 2$.
Given their product is 143:
$x(x + 2) = 143$
Expand:
$x^2 + 2x - 143 = 0$
Calculate discriminant:
$D = b^2 - 4ac = (2)^2 - 4(1)(-143) = 4 + 572 = 576$
Since $D > 0$, there are two distinct real roots.
Roots are:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{576}}{2} = \frac{-2 \pm 24}{2}$
So:
$x = \frac{22}{2} = 11$ or $x = \frac{-26}{2} = -13$
Thus the pairs are (11, 13) or (-13, -11).
Checking products:
$11 \times 13 = 143$ and $-13 \times -11 = 143$

Key Concept: Distance, Speed, Time

c) 75 km/hr
[Solution Description] Let the speed of the ordinary train be $x$ km/hr and the express train be $(x + 25)$ km/hr. The time difference is given as 2 hours:
$\frac{300}{x} - \frac{300}{x + 25} = 2$
Solving, we get $x^2 + 25x - 3750 = 0$, leading to $x = 50$ km/hr (speed of ordinary train). Thus, the express train's speed is $75$ km/hr.

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: 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: 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: 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: 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: 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: 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: Quadratic Formula

b) $x = \frac{3}{2}$
[Solution Description]
The quadratic equation is $4x^2 - 12x + 9 = 0$, where $a = 4$, $b = -12$, and $c = 9$. Applying the quadratic formula:
$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \times 4 \times 9}}{2 \times 4}$
Calculating the discriminant:
$D = 144 - 144 = 0$
Thus, the solution is:
$x = \frac{12 \pm \sqrt{0}}{8} = \frac{12}{8} = \frac{3}{2}$

Key Concept: Problems Based on Geometry

b) $15$ cm
[Solution Description]
Let the shorter leg be $x$ cm. Then, the longer leg is $x + 5$ cm.
Using Pythagoras Theorem:
$x^2 + (x + 5)^2 = 25^2$
Expand:
$x^2 + x^2 + 10x + 25 = 625$
Simplify:
$2x^2 + 10x - 600 = 0$
Divide by 2:
$x^2 + 5x - 300 = 0$
Factorize:
$(x + 20)(x - 15) = 0$
Thus, $x = -20$ or $x = 15$.
Since length cannot be negative, the shorter leg is 15 cm.

Key Concept: Rectangular Area and Perimeter

b) Length = 32 m, Breadth = 20 m
[Solution Description]
Let the length be $l$ and breadth be $b$.
The perimeter is given by: $2(l + b) = 104$.
Simplifying, we get $l + b = 52$.
The area is given by: $l \times b = 640$.
From $l + b = 52$, we can express $l$ as $l = 52 - b$.
Substituting into the area equation: $(52 - b) \times b = 640$.
This simplifies to $52b - b^2 = 640$.
Rearranged, it becomes $b^2 - 52b + 640 = 0$.
Solving this quadratic equation:
Using the quadratic formula:
$b = \frac{52 \pm \sqrt{(52)^2 - 4 \times 1 \times 640}}{2}$
$b = \frac{52 \pm \sqrt{2704 - 2560}}{2}$
$b = \frac{52 \pm \sqrt{144}}{2}$
$b = \frac{52 \pm 12}{2}$
So, $b = \frac{64}{2} = 32$ or $b = \frac{40}{2} = 20$.
Therefore, the dimensions are either $32$ m and $20$ m.

Key Concept: Problems Based on Numbers

c) 4
[Solution Description]
Let the number be $x$. The reciprocal is $\frac{1}{x}$. According to the problem:
$x + \frac{1}{x} = 4.25$
Multiply through by $x$ to eliminate the fraction:
$x^2 + 1 = 4.25x$
Rearranging:
$x^2 - 4.25x + 1 = 0$
Multiply through by 4 to eliminate decimals:
$4x^2 - 17x + 4 = 0$
Factoring:
$(4x - 1)(x - 4) = 0$
Solving for $x$:
$x = \frac{1}{4} \text{ or } x = 4$
Both solutions satisfy the original equation.

Key Concept: Roots, Quadratic Formula

c) $2x^2 - 5x + 2 = 0$
[Solution Description]
Let the roots be $\alpha$ and $\frac{1}{\alpha}$. Given $\alpha + \frac{1}{\alpha} = \frac{5}{2}$, and let $\alpha = 2$. Then, $\frac{1}{\alpha} = \frac{1}{2}$. The sum of the roots is $2 + \frac{1}{2} = \frac{5}{2}$, and the product is $2 \times \frac{1}{2} = 1$. The quadratic equation is $x^2 - (\text{sum})x + \text{product} = 0$, so $x^2 - \frac{5}{2}x + 1 = 0$. Multiply by 2 to clear fractions: $2x^2 - 5x + 2 = 0$.

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: 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: Quadratic Formula Application

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

Key Concept: Quadratic formula, discriminant nature analysis

d) $2$
[Solution Description]
For no real roots, the discriminant must be negative. The discriminant ($D$) is:
$D = 4^2 - 4(m)(4) = 16 - 16m$
Set $D < 0$:
$16 - 16m < 0$
$-16m 1$
So, the equation has no real roots for $m > 1$.