Q.11
Surface Areas & Volumes
[1 Mark]
Two cubes each of volume $64\text{ cm}^3$ are joined end to end to form a solid. The surface area of the resultant cuboid is:
A $192\text{ cm}^2$
B $160\text{ cm}^2$
C $96\text{ cm}^2$
D $80\text{ cm}^2$
💡 Hint: Volume of a cube $= a^3$. First, find the side '$a$'. When two identical cubes are joined end-to-end, only the length doubles ($2a$), while the breadth and height remain '$a$'. (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Students often find the surface area of one cube ($6a^2$) and just multiply it by $2$ to get $192\text{ cm}^2$. But when cubes are joined, the two faces touching each other are hidden inside and must be subtracted!
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Volume of each cube $= 64\text{ cm}^3$.
Let the side of the cube be '$a$'.
$a^3 = 64 \Rightarrow a = \sqrt[3]{64} = 4\text{ cm}$. [½ Mark]
When two cubes are joined end to end, a cuboid is formed.
Dimensions of the resulting cuboid:
Length ($l$) = $4 + 4 = 8\text{ cm}$
Breadth ($b$) = $4\text{ cm}$
Height ($h$) = $4\text{ cm}$
Surface area of cuboid = $2(lb + bh + hl)$
$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$
$= 2(32 + 16 + 32)$
$= 2(80) = 160\text{ cm}^2$. [½ Mark]
$\therefore$ Correct Option is (B)
Q.12
Statistics
[1 Mark]
The mean age of a combined group of men and women is 35 years. If the mean ages of the men and women are 38 years and 30 years respectively, then the percentage of women in the group is:
A 15
B 25.5
C 35
D 37.5
💡 Hint: Use the weighted mean formula: $\text{Combined Mean} = \frac{n_1 x_1 + n_2 x_2}{n_1 + n_2}$. Let the number of men be $M$ and women be $W$. Find the ratio of $W$ to $M$ first.
⚠️ Common Mistake: Finding the ratio of Women to Men (which is $3:5$) and then stopping or calculating the wrong percentage. You need the percentage of women in the entire group, which means dividing by the total parts $(3+5=8)$.
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Let the number of men be $M$ and the number of women be $W$.
Given combined mean $= 35$.
Mean of men $= 38$, Mean of women $= 30$.
Using the weighted average formula:
$35 = \frac{38M + 30W}{M + W}$
Cross multiply:
$35(M + W) = 38M + 30W$
$35M + 35W = 38M + 30W$ [½ Mark]
Rearranging terms:
$35W - 30W = 38M - 35M$
$5W = 3M \Rightarrow \frac{W}{M} = \frac{3}{5}$
Ratio of Women to Men is $3:5$.
Total parts in the group $= 3 + 5 = 8$.
Percentage of women $= \frac{\text{Parts of Women}}{\text{Total Parts}} \times 100$
$= \frac{3}{8} \times 100 = 37.5\%$ [½ Mark]
$\therefore$ Correct Option is (D)
Q.13
Probability
[1 Mark]
Two dice are rolled simultaneously. The probability of getting number less than 4 on each die is:
A \(\frac{1}{4}\)
B \(\frac{1}{9}\)
C \(\frac{1}{36}\)
D \(\frac{1}{6}\)
💡 Hint: The favorable numbers on a single die must be strictly less than 4 (i.e., 1, 2, or 3). Calculate how many combinations exist where both Die 1 AND Die 2 show these numbers. (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Including the number "4" in your count. "Less than 4" does not mean "Less than or equal to 4".
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When two dice are rolled, total possible outcomes $= 6 \times 6 = 36$.
We need a number strictly "less than 4" on each die.
Favorable numbers on Die 1: $\{1, 2, 3\}$ $\rightarrow$ ($3$ options)
Favorable numbers on Die 2: $\{1, 2, 3\}$ $\rightarrow$ ($3$ options)
Total favorable outcomes $= 3 \times 3 = 9$.
(These are: (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)) [½ Mark]
Probability $P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$
$P(E) = \frac{9}{36} = \frac{1}{4}$ [½ Mark]
$\therefore$ Correct Option is (A)
Q.14
Coordinate Geometry
[1 Mark]
$PM$ is a median of $\Delta PQR$ with vertices $P(5, -6)$, $Q(6, 4)$ and $R(0, 0)$. The length of $PM$ is:
A $2\sqrt{17}$ units
B $2\sqrt{15}$ units
C $\sqrt{101}$ units
D 10 units
💡 Hint: The median $PM$ drops from vertex $P$ to the exact midpoint of the opposite side $QR$. First, use the Midpoint formula on $Q$ and $R$ to find point $M$. Then, use the Distance formula between $P$ and $M$.
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$PM$ is the median from vertex $P$ to side $QR$.
Therefore, $M$ is the midpoint of $QR$.
Given $Q(6, 4)$ and $R(0, 0)$.
Coordinates of $M = \left( \frac{6+0}{2}, \frac{4+0}{2} \right) = (3, 2)$. [½ Mark]
Now, we need the length of $PM$, where $P(5, -6)$ and $M(3, 2)$.
Using the Distance Formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
$PM = \sqrt{(3 - 5)^2 + (2 - (-6))^2}$
$PM = \sqrt{(-2)^2 + (2 + 6)^2}$
$PM = \sqrt{4 + (8)^2}$
$PM = \sqrt{4 + 64} = \sqrt{68}$
Simplify the surd: $\sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$ units. [½ Mark]
$\therefore$ Correct Option is (A)
Q.15
Areas Related to Circles
[1 Mark]
The arc of a circle is of length $6\pi\text{ cm}$ and the sector it bounds has an area of $24\pi\text{ cm}^2$. The radius of the circle is:
A 4 cm
B 8 cm
C 16 cm
D 18 cm
💡 Hint: There is a powerful shortcut formula relating Sector Area ($A$), Arc Length ($l$), and radius ($r$): $A = \frac{1}{2} \times l \times r$. Use this to skip finding the angle $\theta$! (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Trying to calculate the central angle $\theta$ first. While it works, it takes way too long for a 1-mark MCQ and increases the chance of calculation errors.
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Let arc length be $l$ and area of the sector be $A$.
Given: $l = 6\pi\text{ cm}$ and $A = 24\pi\text{ cm}^2$. [½ Mark]
We know the direct relation:
$\text{Area of Sector } (A) = \frac{1}{2} \times l \times r$
Substitute the given values:
$24\pi = \frac{1}{2} \times (6\pi) \times r$
$24\pi = 3\pi \times r$
$r = \frac{24\pi}{3\pi}$
$r = 8\text{ cm}$ [½ Mark]
(Alternative: $\frac{A}{l} = \frac{\frac{\theta}{360}\pi r^2}{\frac{\theta}{360}2\pi r} = \frac{r}{2}$. So $\frac{24\pi}{6\pi} = \frac{r}{2} \Rightarrow 4 = \frac{r}{2} \Rightarrow r=8$.)
$\therefore$ Correct Option is (B)
