1 Mark
11
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. $80 \text{ cm}^2$
Concept:
1
When two identical cubes are joined end-to-end, their lengths add up to form a cuboid ($l = 2a$), but their breadth and height remain exactly the same as the original cube ($b = a, h = a$). Calculate the side of the cube from its volume, form the new cuboid dimensions, and apply the Total Surface Area (TSA) formula: $2(lb + bh + hl)$.
Hint:
1
Volume = $a^3 = 64$. So, side $a = 4 \text{ cm}$.
2
New Cuboid: length = $4+4=8$, breadth = $4$, height = $4$.
Mistakes to Avoid:
!
Adding Surface Areas Directly: TSA of one cube = $6a^2 = 96$. Multiplying by two cubes = 192 (Option A). This is wrong because joining them perfectly hides two faces (one from each cube) where they touch! The actual area is $192 - 2(16) = 160$.
Official CBSE Solution:
Volume of cube = $64 \implies a^3 = 64 \implies a = 4\text{ cm}$.
Dimensions of resulting cuboid:
length ($l$) = $8\text{ cm}$, breadth ($b$) = $4\text{ cm}$, height ($h$) = $4\text{ cm}$.
Total surface area of cuboid:
$= 2(lb + bh + hl)$
$= 2(8 \times 4 + 4 \times 4 + 8 \times 4)$
$= 2(32 + 16 + 32) = 2(80) = 160\text{ cm}^2$
Ans: Option (B) $160\text{ cm}^2$
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
12
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
Concept:
1
Use the weighted mean formula: $\text{Combined Mean} = \frac{(n_1 \times \text{mean}_1) + (n_2 \times \text{mean}_2)}{n_1 + n_2}$. Assign variables for the number of men ($x$) and women ($y$), solve for their ratio, and convert to a percentage.
Hint:
1
$\frac{38x + 30y}{x + y} = 35$.
2
Cross multiply: $38x + 30y = 35x + 35y$. Solve for the ratio $\frac{x}{y}$.
Mistakes to Avoid:
!
Finding the wrong percentage: Ensure you calculate the percentage of *women* ($\frac{y}{x+y} \times 100$), not men.
Official CBSE Solution:
Let number of men be $x$ and women be $y$.
$\frac{38x + 30y}{x + y} = 35$
$38x + 30y = 35x + 35y$
$3x = 5y \implies \frac{x}{y} = \frac{5}{3}$
The ratio of Men to Women is $5 : 3$. Total parts = 8.
% age of women = $\frac{3}{8} \times 100 = 37.5$
Ans: Option (D) 37.5
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
13
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. $\frac{1}{6}$
D. $\frac{1}{36}$
Concept:
1
Total outcomes for two dice is $6 \times 6 = 36$. For *each* die to have a number less than 4, it must roll a 1, 2, or 3. Count all combinations of $(1,2,3)$ and divide by 36.
Hint:
1
Valid outcomes for Die 1: {1, 2, 3}. Valid for Die 2: {1, 2, 3}.
2
Total valid pairs = $3 \times 3 = 9$. Probability = $9/36$.
Mistakes to Avoid:
!
Including 4: "Less than 4" strictly means $1, 2, 3$. If you include 4, you get $16/36$ which is incorrect.
Official CBSE Solution:
Total outcomes = 36
Favourable outcomes = (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)
Total Favourable = 9
$P(E) = \frac{9}{36} = \frac{1}{4}$
Ans: Option (A) $\frac{1}{4}$
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
14
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
Concept:
1
A median connects a vertex to the midpoint of the opposite side. First, find coordinates of M (midpoint of QR). Then use the Distance Formula between P and M.
Hint:
1
Midpoint M of Q(6, 4) and R(0, 0): $(\frac{6+0}{2}, \frac{4+0}{2}) = (3, 2)$.
2
Distance PM from (5, -6) to (3, 2): $\sqrt{(5-3)^2 + (-6-2)^2}$.
Mistakes to Avoid:
!
Wrong coordinates: Be careful with signs. $(-6-2)^2$ is $(-8)^2 = 64$. Subtracting a negative creates a positive, but here both are negative.
Official CBSE Solution:
Coordinates of midpoint of QR i.e., M are:
$(\frac{6+0}{2}, \frac{4+0}{2}) = (3, 2)$
Median $PM$ length:
$= \sqrt{(5-3)^2 + (-6-2)^2}$
$= \sqrt{2^2 + (-8)^2}$
$= \sqrt{4 + 64} = \sqrt{68}$
$= \sqrt{4 \times 17} = 2\sqrt{17}$ units
Ans: Option (A) $2\sqrt{17}$ units
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
15
The arc of a circle is of length $6\pi$ cm and the sector it bound has an area of $24\pi \text{ cm}^2$. The radius of the circle is:
(Note: The CBSE Question Paper erroneously prints "6 cm" without the $\pi$, but their official marking scheme strictly solves the question using $6\pi$. The correct intended value is $6\pi$.)
(Note: The CBSE Question Paper erroneously prints "6 cm" without the $\pi$, but their official marking scheme strictly solves the question using $6\pi$. The correct intended value is $6\pi$.)
A. 4 cm
B. 8 cm
C. 16 cm
D. 18 cm
Concept:
1
A highly useful shortcut formula relates the Area ($A$) of a sector, its arc length ($l$), and the radius ($r$): $A = \frac{1}{2} \times l \times r$.
Hint:
1
Substitute the values: $24\pi = \frac{1}{2} \times 6\pi \times r$.
2
Cancel $\pi$ from both sides and solve for $r$.
Mistakes to Avoid:
!
Trusting the typo: If you use exactly "6 cm" without the $\pi$, you get $r = 8\pi$, which is not in the options. Always trust the mathematical relationships over a misprinted question paper.
Official CBSE Solution:
Using the Sector Area Formula:
$A = \frac{1}{2} \times l \times r$
$24\pi = \frac{1}{2} \times 6\pi \times r$
$24\pi = 3\pi \times r$
Divide by $3\pi$:
$r = \frac{24\pi}{3\pi} = 8\text{ cm}$
Ans: Option (B) 8 cm
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
16
The mean and median of the data are 35.5 and 32 respectively. The value of mode for this data is:
A. 23
B. 26
C. 25
D. 30
Concept:
1
Use the empirical relationship between the three measures of central tendency: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$.
Hint:
1
Substitute values: $\text{Mode} = 3(32) - 2(35.5)$.
Mistakes to Avoid:
!
Swapping Mean and Median: A common error is writing $3(\text{Mean}) - 2(\text{Median})$. Always remember: Mode = $3\text{Median} - 2\text{Mean}$.
Official CBSE Solution:
$\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$
$\text{Mode} = 3(32) - 2(35.5)$
$= 96 - 71$
$= 25$
Ans: Option (C) 25
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
17
A flying kite is tied to a point on the ground with the help of a string. The string makes an angle $\theta$ with the ground level such that $\tan\theta = \frac{12}{5}$. If the length of the string is 52 m, then the height of the kite above the ground is:
A. 40 m
B. 45.5 m
C. 48 m
D. 50 m
Concept:
1
Since the string length (hypotenuse) and height (Opposite) are involved, we need $\sin\theta$. Given $\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{12}{5}$, we can find the hypotenuse of the ratio triangle using Pythagoras ($12^2 + 5^2 = 13^2$). Thus, $\sin\theta = \frac{12}{13}$. Use this to find the actual height.
Hint:
1
$\sin\theta = \frac{h}{52}$.
2
Equate ratios: $\frac{h}{52} = \frac{12}{13}$.
Mistakes to Avoid:
!
Using tan directly: $\frac{h}{52} \neq \frac{12}{5}$. 52 is the hypotenuse, not the base. Tangent involves base, so you MUST convert tan to sin first.
Official CBSE Solution:
Given $\tan\theta = \frac{12}{5}$.
Using ratio hypotenuse = $\sqrt{12^2 + 5^2} = 13$.
Therefore, $\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13}$.
In the kite triangle:
$\sin\theta = \frac{\text{height}}{\text{string length}}$
$\frac{h}{52} = \frac{12}{13}$
$h = \frac{12 \times 52}{13} = 12 \times 4 = 48\text{ m}$.
Ans: Option (C) 48 m
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
18
2 cards of diamonds and 4 cards of spades are missing from a pack of 52 cards. A card is drawn at random from this pack. The probability of getting a card of heart is:
A. $\frac{13}{52}$
B. $\frac{13}{46}$
C. $\frac{11}{52}$
D. $\frac{11}{46}$
Concept:
1
Calculate the new total number of cards in the deck by subtracting the missing cards. Then, determine how many heart cards are currently in the deck. $P = \frac{\text{Favourable}}{\text{Total Remaining}}$.
Hint:
1
Total remaining cards = $52 - (2 + 4) = 46$.
2
Were any hearts removed? No, so there are still 13 hearts.
Mistakes to Avoid:
!
Using 52 as denominator: The pack is no longer complete. You must adjust the total denominator to reflect the missing cards.
Official CBSE Solution:
Total missing cards = 2 + 4 = 6.
Total remaining cards = $52 - 6 = 46$.
Number of heart cards remaining = 13 (none were lost).
P(a heart card) = $\frac{13}{46}$
Ans: Option (B) $\frac{13}{46}$
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
19
Assertion (A): If three vertices of a parallelogram taken in order are (-1, -6), (2, -5) and (7, 2), then its fourth vertex is (4,1).
Reason (R): Diagonals of a parallelogram bisect each other.
Reason (R): Diagonals of a parallelogram bisect each other.
A. Both A and R are true and R is the correct explanation of A.
B. Both A and R are true but R is not the correct explanation.
C. A is true but R is false.
D. A is false but R is true.
Concept:
1
In a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal AC is exactly the same point as the midpoint of diagonal BD.
Hint:
1
Calculate midpoint of AC: $(\frac{-1+7}{2}, \frac{-6+2}{2}) = (3, -2)$.
2
Calculate midpoint of BD using given fourth vertex (4,1): $(\frac{2+4}{2}, \frac{-5+1}{2}) = (3, -2)$. Since they match, A is true.
Mistakes to Avoid:
!
Assuming distances: Checking side lengths (AB = CD) takes much longer and doesn't directly utilize the property stated in the Reason. Midpoint formula is the fastest test.
Official CBSE Solution:
Let A(-1, -6), B(2, -5), C(7, 2), and D(4, 1) be the vertices.
Midpoint of AC = $\left(\frac{-1+7}{2}, \frac{-6+2}{2}\right) = (3, -2)$.
Midpoint of BD = $\left(\frac{2+4}{2}, \frac{-5+1}{2}\right) = (3, -2)$.
Since the midpoints are exactly the same, the diagonals bisect each other, making ABCD a parallelogram.
Assertion (A) is true. Reason (R) is true and is the exact property used to prove A.
Ans: Option (A)
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
1 Mark
20
Assertion (A): If zeroes of the polynomial $(2k-1)x^2+4x-3$ are reciprocal of each other, then $k=-1$.
Reason (R): If $a=c$, then zeroes of the polynomial $ax^2+bx+c, a \neq 0$ are reciprocal of each other.
Reason (R): If $a=c$, then zeroes of the polynomial $ax^2+bx+c, a \neq 0$ are reciprocal of each other.
A. Both A and R are true and R is the correct explanation of A.
B. Both A and R are true but R is not the correct explanation.
C. A is true but R is false.
D. A is false but R is true.
Concept:
1
If roots are reciprocals ($\alpha$ and $1/\alpha$), their product is exactly 1. Since Product of zeroes = $c/a$, setting $c/a = 1$ leads directly to $c = a$.
Hint:
1
Here, $a = (2k-1)$ and $c = -3$.
2
Set $a = c$: $2k - 1 = -3$. Solve for $k$.
Mistakes to Avoid:
!
Confusing sum and product: Don't use $-b/a$ (sum of roots). The reciprocal condition specifically triggers the product rule ($c/a = 1$).
Official CBSE Solution:
Reason (R): If zeroes are reciprocal, product $\alpha \times \frac{1}{\alpha} = 1$. Since Product = $c/a$, we get $c/a = 1 \implies c = a$. Reason R is True.
Assertion (A): Applying $c = a$ to the polynomial $(2k-1)x^2+4x-3$:
$a = 2k - 1$
$c = -3$
$2k - 1 = -3 \implies 2k = -2 \implies k = -1$.
Assertion A is True, and R perfectly explains it.
Ans: Option (A)
π― CBSE Marking Scheme:
• Correct Option selection: 1 Mark
• Correct Option selection: 1 Mark
