Solution:

For the cylindrical part:

Radius $r = 0.5/2 = 0.25$ cm
Height $h_c = 10$ cm
Volume $V_c = \pi r^2 h_c = \pi (0.25)^2 (10) = 0.625\pi$ cm$^3$.

For the conical part:

Radius $r = 0.25$ cm (same as cylinder)
Height $h_{cone} = 1$ cm
Volume $V_{cone} = \frac{1}{3}\pi r^2 h_{cone} = \frac{1}{3}\pi (0.25)^2 (1) = \frac{0.0625}{3}\pi$ cm$^3$.

Total volume of the pencil $= V_c + V_{cone} = 0.625\pi + \frac{0.0625}{3}\pi = \pi \left(0.625 + \frac{0.0625}{3}\right)$

$= \pi \left(\frac{1.875 + 0.0625}{3}\right) = \pi \left(\frac{1.9375}{3}\right) \approx 2.03$ cm$^3$.

Class 10: Statistics and Probability

Statistics

Mean ($\bar{x}$):
- Direct method: $\bar{x} = \frac{\sum f_ix_i}{\sum f_i}$
- Assumed Mean method: $\bar{x} = A + \frac{\sum f_id_i}{\sum f_i}$, where $d_i = x_i - A$
- Step Deviation method: $\bar{x} = A + \left(\frac{\sum f_iu_i}{\sum f_i}\right)h$, where $u_i = \frac{x_i - A}{h}$
Median: The middle-most value of the data when arranged in ascending or descending order.
For grouped data: Median $= L + \left(\frac{n/2 - cf}{f}\right)h$, where $L$ is lower limit of median class, $n$ is sum of frequencies, $cf$ is cumulative frequency of class preceding median class, $f$ is frequency of median class, $h$ is class size.
Mode: The value that appears most frequently in a data set.
For grouped data: Mode $= L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right)h$, where $L$ is lower limit of modal class, $f_1$ is frequency of modal class, $f_0$ is frequency of class preceding modal class, $f_2$ is frequency of class succeeding modal class, $h$ is class size.
Empirical Relationship: $3 \text{ Median } = \text{ Mode } + 2 \text{ Mean}$

Probability

Probability of an event E: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$
$0 \le P(E) \le 1$
$P(E) + P(\text{not } E) = 1$
An event that is impossible has probability 0.
An event that is sure or certain has probability 1.

Solved Example: Probability

Problem: A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

Solution:

Total number of balls in the bag $= 3 + 5 = 8$.

(i) Let E be the event that the ball drawn is red.

Number of red balls = 3.

$P(E) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{3}{8}$.

(ii) Let F be the event that the ball drawn is not red (i.e., black).

Number of black balls = 5.

$P(F) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{5}{8}$.

Alternatively, $P(\text{not red}) = 1 - P(\text{red}) = 1 - \frac{3}{8} = \frac{5}{8}$.

Class 11: Sets, Relations and Functions

Sets

Set: A well-defined collection of objects.
Representation: Roster form (list elements) or Set-builder form (describe property).
Types of Sets: Empty set ($\emptyset$), Finite set, Infinite set, Singleton set.
Subset: $A \subseteq B$ if every element of A is in B.
Power Set: The collection of all subsets of a set A, denoted by $P(A)$. If $|A|=n$, then $|P(A)|=2^n$.
Universal Set ($U$): The basic set for a given context.
Operations on Sets:
- Union ($A \cup B$): Elements in A or B or both.
- Intersection ($A \cap B$): Elements common to A and B.
- Difference ($A - B$): Elements in A but not in B.
- Complement ($A'$ or $A^c$): Elements in U but not in A. $A' = U - A$.
De Morg