16 Surface Areas & Volumes
The total surface area of a solid hemisphere of diameter '2d' is:
Concept:
1
Total Surface Area of Hemisphere: A solid hemisphere has two distinct surface areas: the curved bowl part ($2\pi r^2$) and the flat circular top ($\pi r^2$). Added together, the Total Surface Area (TSA) is exactly $3\pi r^2$.
Hint:
1
First, find the radius ($r$) in terms of $d$. Since the diameter is given as $2d$, the radius must be $d$. Simply plug $r = d$ into the TSA formula $3\pi r^2$.
Mistakes to Avoid:
!
Forgetting the Base: A very common error is calculating only the Curved Surface Area ($2\pi r^2$) and completely forgetting to add the flat circular top. "Solid" implies the top is covered!
👉 Scroll horizontally to read the full handwritten solution.
The diameter of the solid hemisphere is given as $2d$.
Therefore, the radius ($r$) of the hemisphere is:
$r = \frac{\text{Diameter}}{2} = \frac{2d}{2} = d$
The Total Surface Area (TSA) of a solid hemisphere consists of two parts:
1. The Curved Surface Area (CSA) $= 2\pi r^2$
2. The Area of the flat circular base $= \pi r^2$
$\text{TSA} = \text{CSA} + \text{Base Area} = 2\pi r^2 + \pi r^2 = 3\pi r^2$
Substituting the value of the radius $r = d$ into the formula:
$\text{TSA} = 3\pi (d)^2 = 3\pi d^2$ [1 M]
Correct Option: (a) $3\pi d^2$
17 Statistics
If the mean and mode of a data are 12 and 21 respectively, then its median is:
Concept:
1
Empirical Formula: For moderately skewed distributions, there is a fixed relationship between the three measures of central tendency: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$.
Hint:
1
Substitute the given values directly into the formula: $21 = 3(\text{Median}) - 2(12)$. Now, use basic algebra to solve for the Median!
Mistakes to Avoid:
!
Swapping the Coefficients: The most common mistake here is messing up the 3 and the 2. Students often mistakenly write $\text{Mean} = 3(\text{Median}) - 2(\text{Mode})$. Always remember it starts with "Mode = ".
👉 Scroll horizontally to read the full handwritten solution.
The empirical relationship between Mean, Median, and Mode is:
$\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$
We are given:
Mean = 12
Mode = 21
Substitute the given values into the empirical formula:
$21 = 3(\text{Median}) - 2(12)$
$21 = 3(\text{Median}) - 24$
Rearranging the equation to solve for the Median:
$3(\text{Median}) = 21 + 24$
$3(\text{Median}) = 45$
$\text{Median} = \frac{45}{3} = 15$ [1 M]
Correct Option: (c) 15
18 Probability
A die is thrown once. Probability of getting a number other than 3 is:
Concept:
1
Complementary Events: The probability of an event NOT happening is exactly 1 minus the probability of it happening: $P(\text{not } E) = 1 - P(E)$. You can also simply count the favorable outcomes.
Hint:
1
The sample space of a die is $\{1, 2, 3, 4, 5, 6\}$. How many of those numbers are "not 3"? Put that number over the total number of sides (6).
Mistakes to Avoid:
!
Misreading the Question: A classic error is reading this hastily as "Probability of getting a 3" and immediately selecting $\frac{1}{6}$. Always read probability scenarios twice!
👉 Scroll horizontally to read the full handwritten solution.
When a fair six-sided die is thrown once, the sample space of possible outcomes is:
$S = \{1, 2, 3, 4, 5, 6\}$
Total number of possible outcomes = 6.
The event is "getting a number other than 3".
The favorable outcomes for this event are:
$E = \{1, 2, 4, 5, 6\}$
Number of favorable outcomes = 5.
The probability $P(E)$ is given by:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{5}{6}$ [1 M]
Alternative (Complementary Event):
$P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}$
Correct Option: (c) $\frac{5}{6}$
19 Probability (Assertion-Reason)
Assertion (A): The probability that a leap year has 53 Mondays is $\frac{2}{7}$.
Reason (R): The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$.
Reason (R): The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$.
Concept:
1
Calendar Probability: Every year has 52 full weeks (which accounts for 52 Mondays, Tuesdays, etc.). A leap year (366 days) has exactly 2 extra days left over, while a non-leap year (365 days) has exactly 1 extra day left over. These extra days determine the probability of having a 53rd specific weekday.
Hint:
1
For the non-leap year (Reason R), there is 1 extra day left over. That single day could be any of the 7 days of the week. So, what is the chance that it is a Monday? Is it really $5/7$?
Mistakes to Avoid:
!
Guessing Non-Leap Probabilities: Students sometimes guess that non-leap years have a $5/7$ probability because a standard work week has 5 days. This makes no mathematical sense. The extra day can be any of the 7 days equally, so the chance is exactly $1/7$.
👉 Scroll horizontally to read the full handwritten solution.
Analyzing Assertion (A):
A leap year has 366 days. $366 \div 7 = 52$ weeks and exactly 2 extra days.
The 52 full weeks account for 52 Mondays.
The 2 extra days can be any consecutive pair: (Sun, Mon), (Mon, Tue),
(Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), or (Sat, Sun).
Out of these 7 combinations, exactly 2 contain a Monday.
Thus, the probability of 53 Mondays in a leap year is $\frac{2}{7}$.
Assertion (A) is TRUE. [½ M]
Analyzing Reason (R):
A non-leap year has 365 days. $365 \div 7 = 52$ weeks and exactly 1 extra day.
The 1 extra day can be any of the 7 days of the week.
There is exactly 1 chance out of 7 that this extra day is a Monday.
Thus, the probability is $\frac{1}{7}$, not $\frac{5}{7}$.
Reason (R) is FALSE. [½ M]
Conclusion: Assertion (A) is true, but Reason (R) is false.
Correct Option: (c)
20 Polynomials (Assertion-Reason)
Assertion (A): The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
Reason (R): A quadratic polynomial can have at most two zeroes.
Concept:
1
Fundamental Theorem of Algebra: A polynomial of degree $n$ can have at most $n$ real zeroes. Therefore, a quadratic (degree 2) has a maximum of 2 zeroes. However, to prove a specific quadratic has exactly two real zeroes, you must verify its discriminant ($b^2 - 4ac > 0$) or factorize it successfully.
Hint:
1
Factorize the polynomial $y^2 + 4y + 3$ using middle-term splitting. Since it factors perfectly, Assertion (A) is true. But does the general rule in Reason (R) actually explain why this specific polynomial has exactly two zeroes?
Mistakes to Avoid:
!
"At Most" Misinterpretation: Choosing Option A is a common trap. Reason (R) provides an upper limit ("at most two"). It does not guarantee that $p(y)$ has exactly two (it could have 0 or 1). The actual explanation for A being true is that its specific discriminant $D > 0$.
👉 Scroll horizontally to read the full handwritten solution.
Analyzing Assertion (A):
Let's find the zeroes of $p(y) = y^2 + 4y + 3$ by factorizing:
$p(y) = y^2 + 3y + y + 3$
$p(y) = y(y + 3) + 1(y + 3)$
$p(y) = (y + 1)(y + 3)$
Setting $p(y) = 0$, we get $y = -1$ and $y = -3$.
Since it has exactly two distinct real zeroes, Assertion (A) is TRUE. [½ M]
Analyzing Reason (R):
A quadratic polynomial has a degree of 2, so by the fundamental
theorem of algebra, it can have at most two zeroes.
Reason (R) is TRUE. [½ M]
Conclusion:
Both statements are true. However, Reason (R) states a general rule
that a quadratic can have *at most* two zeroes. It does not exactly
explain why this *specific* polynomial $p(y)$ has exactly two
(which requires calculating its discriminant $D > 0$).
Therefore, (R) is not the correct mathematical explanation for (A).
Correct Option: (b)
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