36 Surface Areas & Volumes
A cap is cylindrical in shape and is surmounted by a hemispherical top. If the volume of the cylindrical part is equal to that of the hemispherical part, then the ratio of the height of the cylindrical part to the height of the hemispherical part is:
Concept:
1
Volume Formulas & Composite Solids: The volume of a cylinder is $V = \pi r^2 h$. The volume of a hemisphere is $V = \frac{2}{3} \pi r^3$. Furthermore, the "height" of a hemisphere is simply equal to its radius $r$.
Hint:
1
Set the two volume formulas equal to each other: $\pi r^2 h = \frac{2}{3} \pi r^3$. Cancel out common terms ($\pi$ and $r^2$) from both sides. Find the ratio of $h / r$.
Mistakes to Avoid:
!
Ratio Inversion: A common error is solving the equation properly to get $3h = 2r$, but then writing the ratio $\frac{h}{r}$ as $3/2$ instead of $2/3$. Pay close attention when rearranging terms!
👉 Scroll horizontally to read the full handwritten solution.
Let the radius of the cylindrical part be $r$.
Since the hemisphere surmounts it, its radius is also $r$.
The height of the cylindrical part is $h$.
The height of a hemisphere is equivalent to its radius, $r$.
Volume of the cylindrical part ($V_c$) = $\pi r^2 h$
Volume of the hemispherical part ($V_h$) = $\frac{2}{3} \pi r^3$
According to the problem, volumes are equal:
$\pi r^2 h = \frac{2}{3} \pi r^3$ [½ M]
Divide both sides by $\pi r^2$:
$h = \frac{2}{3} r$
We need the ratio of the height of the cylinder ($h$)
to the height of the hemisphere ($r$):
$\frac{h}{r} = \frac{2}{3}$ [½ M]
Correct Option: (A) $2 : 3$
37 Surface Areas & Volumes
The curved surface area of the largest right circular cone that can be carved out from a solid cube of edge $4 \text{ cm}$ is:
Concept:
1
Inscribed Solids: For the largest possible cone to fit perfectly inside a cube, its diameter must match the side of the cube, and its height must also match the side of the cube. The Curved Surface Area (CSA) formula for a cone is $\pi r l$, where $l$ is the slant height ($l = \sqrt{r^2 + h^2}$).
Hint:
1
Since the cube edge is 4, $h = 4$, and the diameter is 4 (so $r = 2$). Find the slant height $l = \sqrt{2^2 + 4^2} = \sqrt{20}$. Simplify $\sqrt{20}$ and plug it into $\pi r l$.
Mistakes to Avoid:
!
Radius vs Edge Confusion: A very common error is taking the radius $r$ to be $4 \text{ cm}$ (the edge of the cube) instead of the diameter. Remember, the full width of the cone spans the entire cube face, making the radius only half of it ($2 \text{ cm}$).
👉 Scroll horizontally to read the full handwritten solution.
For the largest cone carved out of a cube:
Base Diameter = Edge of cube
Height ($h$) = Edge of cube
Given the edge of the cube $a = 4 \text{ cm}$:
- Diameter = $4 \text{ cm} \implies$ Radius ($r$) = $2 \text{ cm}$.
- Height ($h$) = $4 \text{ cm}$.
Calculate the slant height ($l$) using Pythagoras theorem:
$l = \sqrt{r^2 + h^2}$
$l = \sqrt{2^2 + 4^2}$
$l = \sqrt{4 + 16} = \sqrt{20}$
$l = \sqrt{4 \times 5} = 2\sqrt{5} \text{ cm}$ [½ M]
The Curved Surface Area (CSA) of a cone is $\pi r l$:
$\text{CSA} = \pi \times 2 \times 2\sqrt{5}$
$\text{CSA} = 4\sqrt{5}\pi \text{ cm}^2$ [½ M]
Correct Option: (A) $4\sqrt{5}\pi \text{ cm}^2$
38 Probability
If the probability of a player winning a game is 0.21, then the probability of him losing the same game is:
Concept:
1
Complementary Events: The sum of the probabilities of all possible complementary events in a given scenario is always exactly $1$. Therefore, $P(\text{Winning}) + P(\text{Losing}) = 1$.
Hint:
1
Subtract the probability of winning from 1. Calculate: $1.00 - 0.21 = ?$
Mistakes to Avoid:
!
The Percentage Trap: A common mistake is selecting Option C (0.79%). While 0.79 is correct in decimal format, it translates to 79% in percentage format. 0.79% is practically zero! Always watch out for rogue percent signs in MCQ options.
👉 Scroll horizontally to read the full handwritten solution.
The sum of probabilities of complementary events is 1.
$P(\text{Winning}) + P(\text{Losing}) = 1$
We are given $P(\text{Winning}) = 0.21$.
$0.21 + P(\text{Losing}) = 1$
$P(\text{Losing}) = 1 - 0.21$
$P(\text{Losing}) = 0.79$ [1 M]
(Note: Option C is incorrect because 0.79% is 0.0079)
Correct Option: (D) 0.79
39 Statistics (Assertion-Reason)
Assertion (A): Consider the following data:
The difference of the upper limit of the median class and the lower limit of the modal class is 20.
Reason (R): The median class and modal class of grouped data are always different.
| Class | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
|---|---|---|---|---|---|---|---|
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
Reason (R): The median class and modal class of grouped data are always different.
Concept:
1
Median & Modal Class Identifiers:
• The Modal Class is simply the class interval with the absolute highest frequency.
• The Median Class is found by calculating the total frequency ($N$), finding $N/2$, and then identifying the class whose cumulative frequency is strictly greater than $N/2$.
• The Modal Class is simply the class interval with the absolute highest frequency.
• The Median Class is found by calculating the total frequency ($N$), finding $N/2$, and then identifying the class whose cumulative frequency is strictly greater than $N/2$.
Hint:
1
Calculate cumulative frequencies. Total $N = 67$. So $N/2 = 33.5$. Which class contains this cumulative value? Also, look at the frequency row—which class has the highest number? Compare both classes.
Mistakes to Avoid:
!
Trusting "Always" Statements: Reason (R) states that these classes are always different. In mathematics, words like "always" or "never" are usually red flags in true/false reasoning. As you will see in this exact problem, they can be the same!
👉 Scroll horizontally to read the full handwritten solution.
Analyzing Assertion (A):
Let's calculate cumulative frequency ($cf$) to find the median class:
65-85: $cf = 4$
85-105: $cf = 4 + 5 = 9$
105-125: $cf = 9 + 13 = 22$
125-145: $cf = 22 + 20 = 42$
Total frequency $N = 67$. Half is $\frac{N}{2} = \frac{67}{2} = 33.5$.
The $cf$ just greater than $33.5$ is $42$, which belongs to $125-145$.
Thus, the Median Class is $125-145$. Upper limit = 145.
The Modal Class is the class with highest frequency (20).
Thus, the Modal Class is also $125-145$. Lower limit = 125.
Difference = $145 - 125 = 20$. Assertion (A) is TRUE. [½ M]
Analyzing Reason (R):
The reason states they are *always* different. However, in our
data above, both classes are exactly the same ($125-145$).
Therefore, Reason (R) is FALSE. [½ M]
Correct Option: (C)
40 Probability (Assertion-Reason)
Assertion (A): The probability that a number selected at random from the numbers $1, 2, 3, \dots, 15$ is a multiple of 4, is $\frac{1}{3}$.
Reason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is $\frac{3}{4}$.
Reason (R): Two different coins are tossed simultaneously. The probability of getting at least one head is $\frac{3}{4}$.
Concept:
1
Basic Probability: Probability = $\frac{\text{Number of Favorable Outcomes}}{\text{Total Possible Outcomes}}$. Evaluate Assertion A by listing multiples of 4 up to 15. Evaluate Reason R by listing the sample space of a two-coin toss: {HH, HT, TH, TT}.
Hint:
1
Multiples of 4 between 1 and 15 are 4, 8, 12. That's 3 numbers out of 15. Simplify the fraction. Does it equal 1/3? Then, look at the coin toss sample space—how many have "at least one" H?
Mistakes to Avoid:
!
"At Least One" Misinterpretation: For Reason R, students often confuse "at least one head" with "exactly one head." "At least one" means 1 OR 2 heads. So {HH} counts! Only {TT} does not.
👉 Scroll horizontally to read the full handwritten solution.
Analyzing Assertion (A):
The sample space for selecting a number from $1$ to $15$ has 15 outcomes.
The multiples of 4 within this range are: $4, 8, \text{ and } 12$.
There are exactly 3 favorable outcomes.
$P(\text{multiple of 4}) = \frac{3}{15} = \frac{1}{5}$
The assertion claims the probability is $\frac{1}{3}$.
Therefore, Assertion (A) is FALSE. [½ M]
Analyzing Reason (R):
When two coins are tossed, the sample space is: $\{HH, HT, TH, TT\}$.
Total number of outcomes = 4.
The event "at least one head" includes $\{HH, HT, TH\}$.
There are 3 favorable outcomes.
$P(\text{at least one head}) = \frac{3}{4}$
Therefore, Reason (R) is TRUE. [½ M]
Conclusion: A is false, but R is true.
Correct Option: (D)
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