51 Probability
A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is:
Concept:
1
Mutually Exclusive Events: When drawing a single card, it cannot be both a Ten and a King at the same time. Therefore, these are mutually exclusive events. You simply add the probability of getting a Ten to the probability of getting a King.
Hint:
1
How many Tens are in a standard 52-card deck? (There are 4). How many Kings are there? (There are 4). Add them together for your total favorable outcomes, then divide by 52 and simplify the fraction.
Mistakes to Avoid:
!
Failing to Simplify: You might calculate $\frac{8}{52}$ correctly, but then look at Option D ($\frac{8}{26}$) and accidentally select it because it has an $8$ on top. Always simplify your fraction fully before comparing it to the options! ($\frac{8}{52} \div 4 = \dots$)
👉 Scroll horizontally to read the full handwritten solution.
A standard deck contains $52$ cards in total.
The event is drawing "either a ten or a king".
Number of tens in a deck = $4$
Number of kings in a deck = $4$
Total favorable outcomes = $4 + 4 = 8$.
Total possible outcomes = $52$.
Probability = $\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}}$
Probability = $\frac{8}{52}$ [½ M]
Divide numerator and denominator by $4$ to simplify:
Probability = $\frac{2}{13}$ [½ M]
Correct Option: (B) $\frac{2}{13}$
52 Surface Areas & Volumes
A camping tent in hemispherical shape of radius 1.4 m, has a door opening of area $0.50 \text{ m}^{2}$. Outer surface area of the tent is:
Concept:
1
Surface Area with Cutouts: The tent is the curved part of a hemisphere. The formula for the Curved Surface Area (CSA) of a hemisphere is $2\pi r^2$. Since there is an opening for a door, that area ($0.50 \text{ m}^2$) is literally missing material, so it must be subtracted from the total CSA.
Hint:
1
First calculate the full CSA: $2 \times \frac{22}{7} \times (1.4)^2$. Once you get that value, subtract the door area of $0.50$ to get the final outer surface area.
Mistakes to Avoid:
!
Adding Instead of Subtracting: Some students calculate the CSA ($12.32$) and then mistakenly add the door area, thinking it is an extra piece, leading to Option D ($12.86$). A door opening is a hole; you subtract holes!
👉 Scroll horizontally to read the full handwritten solution.
The tent is shaped like the curved dome of a hemisphere.
Radius ($r$) = $1.4 \text{ m}$.
First, we calculate the full Curved Surface Area (CSA) without the door:
$\text{CSA} = 2\pi r^2$
$\text{CSA} = 2 \times \frac{22}{7} \times (1.4)^2$
$\text{CSA} = 2 \times \frac{22}{7} \times 1.96$
$\text{CSA} = 44 \times 0.28 = 12.32 \text{ m}^2$ [½ M]
The door opening represents a hole in the tent material.
Outer surface area of the tent = Full CSA - Area of door
Outer surface area = $12.32 - 0.50$
Outer surface area = $11.82 \text{ m}^2$ [½ M]
Correct Option: (C) $11.82 \text{ m}^{2}$
53 Areas Related to Circles
An arc of length 2.2 cm subtends an angle $\theta$ at the centre of the circle with radius 2.8 cm. The value of $\theta$ is:
Concept:
1
Arc Length Formula (Reverse): The length of an arc $l$ is a fraction of the total circumference of the circle: $l = \frac{\theta}{360^\circ} \times 2\pi r$. By plugging in the known arc length and radius, you can isolate and solve for the central angle $\theta$.
Hint:
1
Substitute $l = 2.2$ and $r = 2.8$ into the formula. Use $\pi = \frac{22}{7}$. It helps to calculate the full circumference ($2\pi r$) first, and then divide the arc length by that circumference. Multiply by 360 to get the angle!
Mistakes to Avoid:
!
Decimal Math Errors: Calculating with decimals like $2.2$ and $2.8$ can be tricky. Try multiplying both sides of your equation by 10 early on to remove the decimals (making them 22 and 28) for a much easier cancellation!
👉 Scroll horizontally to read the full handwritten solution.
We use the formula for the length of an arc:
$l = \frac{\theta}{360^\circ} \times 2\pi r$
Substitute the given values ($l = 2.2$, $r = 2.8$, $\pi = \frac{22}{7}$):
$2.2 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 2.8$ [½ M]
Simplify the right side:
$2.2 = \frac{\theta}{360^\circ} \times 2 \times 22 \times 0.4$
$2.2 = \frac{\theta}{360^\circ} \times 44 \times 0.4$
$2.2 = \frac{\theta}{360^\circ} \times 17.6$
Isolate $\theta$:
$\theta = \frac{2.2 \times 360^\circ}{17.6}$
$\theta = \frac{1 \times 360^\circ}{8}$ (since $17.6 \div 2.2 = 8$)
$\theta = 45^\circ$ [½ M]
Correct Option: (C) $45^\circ$
54 Trigonometry
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9\csc \theta + 1}{9\csc \theta - 1}$ is:
Concept:
1
Reciprocal Trigonometric Identities: The cosecant function ($\csc \theta$) is the exact direct reciprocal of the sine function ($\sin \theta$). Therefore, $\csc \theta = \frac{1}{\sin \theta}$. No complex Pythagorean calculations are needed!
Hint:
1
If $\sin \theta = \frac{1}{9}$, then flipping the fraction gives you $\csc \theta = 9$. Now substitute the number 9 directly into the expression $\frac{9(\csc \theta) + 1}{9(\csc \theta) - 1}$.
Mistakes to Avoid:
!
The Long Route (Drawing a Triangle): Some students will draw a right triangle, label the perpendicular as 1, hypotenuse as 9, and calculate the base as $\sqrt{80}$ using Pythagoras. This is perfectly valid, but it wastes valuable exam time since the question only asks for $\csc \theta$, not $\cos$ or $\tan$!
👉 Scroll horizontally to read the full handwritten solution.
We know the reciprocal identity:
$\csc \theta = \frac{1}{\sin \theta}$
Given that $\sin \theta = \frac{1}{9}$, we simply flip the fraction:
$\csc \theta = \frac{9}{1} = 9$ [½ M]
Substitute this value into the given expression:
$\text{Expression} = \frac{9(\csc \theta) + 1}{9(\csc \theta) - 1}$
$= \frac{9(9) + 1}{9(9) - 1}$
$= \frac{81 + 1}{81 - 1}$
$= \frac{82}{80}$ [½ M]
Correct Option: (D) $\frac{82}{80}$
55 Triangles
In the given figure, $PQ \parallel YZ$ such that $XP : PY = 2 : 3$. If $PQ = 5 \text{ cm}$, then $YZ$ equals:
Concept:
1
Similar Triangles vs BPT: When a line is parallel to a side of a triangle ($PQ \parallel YZ$), it creates a smaller triangle ($\Delta XPQ$) that is exactly similar to the larger triangle ($\Delta XYZ$). The ratio of their sides is $\frac{XP}{XY} = \frac{PQ}{YZ}$. Notice that you must use the FULL side $XY$, not just the segment $PY$.
Hint:
1
You are given the pieces $XP : PY = 2 : 3$. This means $XP$ is 2 parts, and $PY$ is 3 parts. Therefore, the FULL side $XY$ is $2 + 3 = 5$ parts. Set up your ratio as $\frac{XP}{XY} = \frac{2}{5}$ and equate it to $\frac{5}{YZ}$.
Mistakes to Avoid:
!
Direct Basic Proportionality Theorem (BPT) Trap: The most common mistake is directly equating the segment ratio to the base ratio: $\frac{XP}{PY} = \frac{PQ}{YZ}$, leading to $\frac{2}{3} = \frac{5}{YZ} \implies YZ = 7.5$ (Option D). BPT only applies to the segments on the side walls, NOT the parallel bases!
👉 Scroll horizontally to read the full handwritten solution.
Since $PQ \parallel YZ$, the smaller triangle and larger triangle are similar
(by AA similarity: $\angle XPQ = \angle XYZ$ and $\angle XQP = \angle XZY$).
Therefore, $\Delta XPQ \sim \Delta XYZ$. [½ M]
The ratio of their corresponding sides is equal:
$\frac{XP}{XY} = \frac{PQ}{YZ}$
We are given the segment ratio $XP : PY = 2 : 3$.
Let $XP = 2k$ and $PY = 3k$.
The full side of the large triangle is $XY = XP + PY = 2k + 3k = 5k$.
Substitute these into our similarity proportion:
$\frac{2k}{5k} = \frac{5}{YZ}$
$\frac{2}{5} = \frac{5}{YZ}$
Cross-multiply to solve for $YZ$:
$2 \times YZ = 25$
$YZ = \frac{25}{2} = 12.5 \text{ cm}$ [½ M]
Correct Option: (A) $12.5 \text{ cm}$
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