# Quadratic Equations - Class 10 - Mathematics

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## Extra Questions - Quadratic Equations | NCERT | Mathematics | Class 10

By increasing the list price of a book by ₹10, a person can buy 20 books less for ₹1200. The original price of the book was _____.

A) ₹0

B) ₹20

C) ₹30

D) ₹24

The problem provides the scenario where increasing the list price of a book by ₹10 results in buying 20 fewer books for ₹1200. We are asked to find the **original price** of the book.

Let's denote the original price of the book as $ x $. Hence, after the price increase, the new price of the book becomes $ x + 10. $

The number of books that can be bought initially with ₹1200 is given by $ \frac{1200}{x},$ and the number of books that can be bought after the price increase is $ \frac{1200}{x + 10}. $

According to the problem, the difference in the number of books bought before and after the price increase is 20 books, leading to the equation: $$ \frac{1200}{x} - \frac{1200}{x + 10} = 20 $$

We can simplify and solve this equation by first clearing the denominators: $$ 1200(x + 10) - 1200x = 20x(x + 10) $$ $$ 12000 = 20x^2 + 200x $$

Dividing all terms by 20 to simplify: $$ 600 = x^2 + 10x $$

Rearranging into a standard quadratic equation form: $$ x^2 + 10x - 600 = 0 $$

Solving this quadratic equation using the quadratic formula where $ a = 1 $, $ b = 10 $, and $ c = -600 $: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-10 \pm \sqrt{100 + 2400}}{2} $$ $$ x = \frac{-10 \pm \sqrt{2500}}{2} $$ $$ x = \frac{-10 \pm 50}{2} $$

We have two potential solutions for $ x $: $$ x = \frac{40}{2} = 20 \quad \text{and} \quad x = \frac{-60}{2} = -30 $$

Rejecting the negative value, $ x = -30 $, as a price cannot be negative, the original price of the book is therefore **₹20**. Hence, the correct answer is:

**B) ₹20**

The area of a rectangular plot is $528 \mathrm{~m}^{2}$. The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth of the plot.

A) $30 \mathrm{~m}$ and $10 \mathrm{~m}$

B) $35 \mathrm{~m}$ and $12 \mathrm{~m}$

C) $34 \mathrm{~m}$ and $15 \mathrm{~m}$

D) $33 \mathrm{~m}$ and $16 \mathrm{~m}$

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If $a, \beta$ are the roots of the equation $x^{2} - ax + b = 0$, and the roots of the equation $bx^{2} - 4x + 4 = 0$ are $a + \frac{\beta^{2}}{a}, \beta + \frac{a^{2}}{\beta}$, then $a + b$ can be -

A. 0

B. 1

C. 2

D. 3

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If the line $y = mx + a$ meets the parabola $y^{2} = 4ax$ in two points whose abscissas are $x_{1}$ and $x_{2}$, then $x_{1} + x_{2} = 0$ if

(A) $m = -1$ B) $m = 1$ C) $m = 2$ D) $m = \frac{-1}{2}$

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Let $f(x) = (\lambda^{2} + \lambda - 2) x^{2} + (\lambda + 2) x$ be a quadratic polynomial. The sum of all integral values of $\lambda$ for which $f(x) < 1$ for all $x \in \mathbb{R}$ is:

A) -1 B) -3 C) 0 D) -2

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For a quadratic equation of the form $a x^{2} + b x + c = 0$, the solutions are $x_{1,2} = \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a}$.

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If $x=\frac{2}{3}$ is a solution of the quadratic equation $7x^{2}+mx-3=0$, find the value of $m$.

(A) $m=\frac{1}{6}$

(B) $m=\frac{-1}{3}$

(C) $m=\frac{-1}{6}$

(D) $m=\frac{1}{3}$

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Solve for $x$:

$$ x^2 - 11x + 28 = 0 $$