10th Class Board Examination - Mathematics (2-Mark Questions)

1. Find the LCM and HCF of 12, 15, and 21 by applying the prime factorization method.

2. Find a quadratic polynomial where the sum and product of its zeroes are $-3$ and $2$, respectively.

3. Given that $\sin A = \frac{3}{4}$, calculate the values of $\cos A$ and $\tan A$.

4. Find the coordinates of the point which divides the line segment joining $(-1, 7)$ and $(4, -3)$ in the ratio $2:3$.

5. A standard die is thrown once. Find the probability of getting:

(i) A prime number.

(ii) A number lying strictly between 2 and 6.

Answer Key

Answer 1:

First, find the prime factorization of each number:

$12 = 2^2 \times 3$

$15 = 3 \times 5$

$21 = 3 \times 7$

  • HCF is the product of the smallest power of each common prime factor: $\text{HCF} = 3$

  • LCM is the product of the greatest power of each prime factor involved: $\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420$

Answer 2:

Let the zeroes of the polynomial be $\alpha$ and $\beta$.

We are given the sum of the zeroes $(\alpha + \beta) = -3$ and the product of the zeroes $(\alpha\beta) = 2$.

The general form of a quadratic polynomial is $k(x^2 - (\alpha + \beta)x + \alpha\beta)$.

Substituting the given values, the polynomial is $x^2 - (-3)x + 2$, which simplifies to:

$x^2 + 3x + 2$

Answer 3:

We know that $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{4}$.

Let Opposite $= 3k$ and Hypotenuse $= 4k$.

Using the Pythagorean theorem to find the adjacent side (Base):

$\text{Base} = \sqrt{(4k)^2 - (3k)^2} = \sqrt{16k^2 - 9k^2} = \sqrt{7k^2} = k\sqrt{7}$

  • $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{7}}{4}$

  • $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{\sqrt{7}}$

Answer 4:

Using the section formula: $x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$ and $y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Given ratio $m_1:m_2 = 2:3$. Points are $(x_1, y_1) = (-1, 7)$ and $(x_2, y_2) = (4, -3)$.

  • $x = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1$

  • $y = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3$

    The coordinates of the point are $(1, 3)$.

Answer 5:

The total possible outcomes when throwing a die are 6 $\{1, 2, 3, 4, 5, 6\}$.

(i) The prime numbers on a die are 3 $\{2, 3, 5\}$.

Probability $= \frac{3}{6} = \textbf{\frac{1}{2}}$

(ii) The numbers lying strictly between 2 and 6 are 3 $\{3, 4, 5\}$.

Probability $= \frac{3}{6} = \textbf{\frac{1}{2}}$

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