Solution of 10 Sure 2-Mark Questions for Class 10

This lesson works through ten sure-shot two-mark questions taken from previous CBSE Class 10 board papers, covering trigonometry, coordinate geometry, probability, arithmetic progressions, quadratic polynomials, and prime factorisation.

A quick correction to an earlier video first: in that question the final steps should read $-4k - 3k = -2$, so $-7k = -2$ and $k = \tfrac{2}{7}$, giving the ratio $k : 1 = 2 : 7$.

Finding two angles from sine and cosine

Given $\sin(A - B) = \tfrac{1}{2}$ and $\cos(A + B) = \tfrac{1}{2}$, with $0^\circ < A < 90^\circ$ and $A > B$, find $A$ and $B$.

Since $\sin 30^\circ = \tfrac{1}{2}$ and $\cos 60^\circ = \tfrac{1}{2}$:

$A - B = 30^\circ \quad (1), \qquad A + B = 60^\circ \quad (2).$

Adding $(1)$ and $(2)$ gives $2A = 90^\circ$, so $A = 45^\circ$. Substituting back into $(1)$, $45^\circ - B = 30^\circ$, so $B = 15^\circ$.

$A = 45^\circ, \qquad B = 15^\circ.$

Ratio in which a point divides a segment

In what ratio does the point $P(-4, 6)$ divide the segment joining $A(-6, 10)$ and $B(3, -8)$?

Let the ratio be $k : 1$, so $m_1 = k$ and $m_2 = 1$. By the section formula, the x-coordinate of $P$ is

$-4 = \frac{k(3) + 1(-6)}{k + 1} = \frac{3k - 6}{k + 1}.$

Cross-multiplying, $-4(k + 1) = 3k - 6$, so $-4k - 4 = 3k - 6$. Collecting the $k$ terms:

$-7k = -2, \qquad k = \frac{2}{7}.$

So $P$ divides $AB$ in the ratio $2 : 7$.

Two birthday probability cases

Two friends, Savitha and Hamida, are considered (ignoring leap years).

Different birthdays. The first person can have any of $365$ days; the second must avoid that one day, leaving $364$ favourable outcomes:

$P(\text{different}) = \frac{364}{365}.$

Same birthday. Only one day works for the second person:

$P(\text{same}) = \frac{1}{365}.$

Sum of multiples of 9 between two numbers

Find the sum of all integers between $92$ and $786$ that are multiples of $9$.

First term. Take the number just after the lower limit, $93$. Dividing $93$ by $9$ leaves a remainder of $3$, so add $9 - 3 = 6$ to get the first multiple:

$a_1 = 93 + 6 = 99.$

Last term. Take the number just before the upper limit, $785$. Dividing $785$ by $9$ leaves a remainder of $2$, so subtract it:

$a_n = 785 - 2 = 783.$

Number of terms. With common difference $d = 9$:

$n = \frac{a_n - a_1}{d} + 1 = \frac{783 - 99}{9} + 1 = 76 + 1 = 77.$

Sum.

$S_n = \frac{n}{2}(a_1 + a_n) = \frac{77}{2}(99 + 783) = \frac{77}{2}(882) = 77 \times 441 = 33957.$

Finding k from a condition on the roots

If $\alpha$ and $\beta$ are the zeros of $f(x) = kx^2 + 4x + 4$ with $\alpha^2 + \beta^2 = 24$, find $k$.

Here $a = k$, $b = 4$, $c = 4$, so

$\alpha + \beta = -\frac{b}{a} = -\frac{4}{k}, \qquad \alpha\beta = \frac{c}{a} = \frac{4}{k}.$

Using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 24$:

$\left(-\frac{4}{k}\right)^2 - 2\left(\frac{4}{k}\right) = 24, \qquad \frac{16}{k^2} - \frac{8}{k} = 24.$

Multiplying through by $k^2$ gives $16 - 8k = 24k^2$, so

$24k^2 + 8k - 16 = 0.$

Dividing by $8$, $3k^2 + k - 2 = 0$. Splitting the middle term:

$3k^2 + 3k - 2k - 2 = 0, \qquad (k + 1)(3k - 2) = 0.$

$k = -1 \quad \text{or} \quad k = \frac{2}{3}.$

A seventh and thirteenth term of an AP

The seventh term of an AP is $32$ and the thirteenth term is $62$. Find the AP.

$a + 6d = 32 \quad (1), \qquad a + 12d = 62 \quad (2).$

Subtracting $(1)$ from $(2)$, $6d = 30$, so $d = 5$. From $(1)$, $a + 30 = 32$, so $a = 2$. The AP is

$2, \; 7, \; 12, \; 17, \dots$

Probability of drawing a blue ball

A bag has $35$ balls, of which $x$ are blue. The probability of drawing a blue ball is

$P(\text{blue}) = \frac{x}{35}.$

If $7$ more blue balls are added, there are $42$ balls in total and $x + 7$ blue ones, and this probability is double the first:

$\frac{x + 7}{42} = 2 \cdot \frac{x}{35}.$

Cross-multiplying, $35(x + 7) = 84x$, so $35x + 245 = 84x$, giving $49x = 245$ and $x = 5$.

A point from the ends of a diameter

Find the coordinates of $A$, where $AB$ is a diameter of a circle with centre $(2, -1)$ and $B = (1, 4)$.

The centre is the midpoint of the diameter. Let $A = (x, y)$:

$2 = \frac{x + 1}{2}, \qquad -1 = \frac{y + 4}{2}.$

So $x + 1 = 4$, giving $x = 3$, and $y + 4 = -2$, giving $y = -6$. Thus $A = (3, -6)$.

A trigonometric ratio from cotangent

Evaluate

$\frac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)}.$

Using $(a + b)(a - b) = a^2 - b^2$, the numerator is $1 - \sin^2\theta = \cos^2\theta$ and the denominator is $1 - \cos^2\theta = \sin^2\theta$:

$\frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta.$

With $\cot\theta = \tfrac{7}{8}$:

$\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}.$

Prime factorisation

Express $156$ as a product of prime factors. Dividing repeatedly:

$156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13.$

Key takeaways

• Reading off the standard angles for $\sin 30^\circ$ and $\cos 60^\circ$ turns a sine-and-cosine condition into two linear equations in the angles.
• For a multiple-of-$d$ range, adjust the inner endpoints by the remainder to get the first and last terms, then use the AP formulas for $n$ and $S_n$.
• A condition such as $\alpha^2 + \beta^2$ can be rewritten with $(\alpha + \beta)^2 - 2\alpha\beta$ and the sum and product of the roots.