SECTION - B (2 Marks Each)
Q.21
Real Numbers
[2 Marks]
(A) A hall has a length of $9.75\text{ m}$, breadth of $6.75\text{ m}$ and height of $5.25\text{ m}$. What is the length of the longest unmarked ruler that can exactly measure the dimensions of the hall?
OR
(B) There are three bells placed at different swings in a park, which toll at intervals of $5$, $6$ and $8$ minutes, respectively. They all toll together when the park opens for the visitors at 10:00 a.m. How many more times do they all toll together till the park is closed at 7:00 p.m.?
💡 Hint (A): "Longest exactly measuring" implies finding the Highest Common Factor (HCF). First, convert all measurements from meters to centimeters to avoid decimals! (Ref: www.maheshkumarsir.in)
Solution 21(A)
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Converting dimensions into centimeters:
Length ($L$) = $9.75\text{ m} = 975\text{ cm}$
Breadth ($B$) = $6.75\text{ m} = 675\text{ cm}$
Height ($H$) = $5.25\text{ m} = 525\text{ cm}$ [½ Mark]
The longest ruler that can measure all dimensions exactly is the $\text{HCF}(975, 675, 525)$.
Prime Factorization:
$975 = 3 \times 5^2 \times 13$
$675 = 3^3 \times 5^2$
$525 = 3 \times 5^2 \times 7$ [1 Mark]
$\text{HCF} = 3 \times 5^2 = 3 \times 25 = 75\text{ cm}$.
$\therefore$ Length of the longest ruler is $75\text{ cm}$. [½ Mark]
💡 Hint (B): "Toll together" implies finding the Lowest Common Multiple (LCM) of the time intervals. Find the LCM, convert it to hours, and then count how many of those chunks fit between 10:00 a.m. and 7:00 p.m.
Solution 21(B)
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The bells will toll together after an interval equal to $\text{LCM}(5, 6, 8)$.
Prime factorization:
$5 = 5^1$
$6 = 2 \times 3$
$8 = 2^3$
$\text{LCM} = 2^3 \times 3 \times 5 = 120\text{ minutes}$. [1 Mark]
$120\text{ minutes} = 2\text{ hours}$. (They toll together every 2 hours).
Total duration from 10:00 a.m. to 7:00 p.m. is $9\text{ hours}$. [½ Mark]
Number of more times they toll together $= \frac{9}{2} = 4.5$
Since they can only toll in complete intervals, we take the whole number part.
(They toll at 12:00 pm, 2:00 pm, 4:00 pm, and 6:00 pm).
$\therefore$ They will toll 4 more times. [½ Mark]
Q.22
Triangles
[2 Marks]
(A) $ABCD$ is a trapezium in which $AB$ is parallel to $DC$ and its diagonals intersect each other at the point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
OR
(B) In $\Delta PQR$, $S$ is a point on $PR$. If $\angle PQR = \angle PSQ$, $PQ = 6\text{ cm}$ and $PS = 3\text{ cm}$, then find the length of $PR$.
💡 Hint (A): Prove that $\Delta AOB$ is similar to $\Delta COD$ using the AA (Angle-Angle) similarity criterion based on vertically opposite angles and alternate interior angles from the parallel lines.
Solution 22(A)
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Given: Trapezium $ABCD$ where $AB \parallel DC$. Diagonals $AC$ and $BD$ intersect at $O$.
In $\Delta AOB$ and $\Delta COD$:
1. $\angle AOB = \angle COD$ (Vertically opposite angles) [½ Mark]
2. $\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel DC$ and $AC$ is transversal) [½ Mark]
By AA similarity criterion, $\Delta AOB \sim \Delta COD$. [½ Mark]
Since the triangles are similar, their corresponding sides are proportional:
$\frac{AO}{CO} = \frac{BO}{DO}$
Rearranging the terms (cross-multiplication):
$\frac{AO}{BO} = \frac{CO}{DO}$ (Hence Proved) [½ Mark]
💡 Hint (B): Compare the large triangle $\Delta PQR$ and the small triangle $\Delta PQS$. They share an angle, and another angle is given as equal. Prove them similar using AA similarity!
Solution 22(B)
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In $\Delta PQR$ and $\Delta PQS$:
1. $\angle P = \angle P$ (Common angle) [½ Mark]
2. $\angle PQR = \angle PSQ$ (Given) [½ Mark]
By AA similarity criterion, $\Delta PQR \sim \Delta PQS$.
Since the triangles are similar, the ratio of their corresponding sides must be equal:
$\frac{PQ}{PS} = \frac{PR}{PQ}$ [½ Mark]
Substitute the given values ($PQ = 6\text{ cm}$, $PS = 3\text{ cm}$):
$\frac{6}{3} = \frac{PR}{6}$
$2 = \frac{PR}{6}$
$PR = 2 \times 6 = 12\text{ cm}$
$\therefore \text{Length of } PR = 12\text{ cm}$ [½ Mark]
Q.23
Trigonometry
[2 Marks]
If $\tan\theta = \frac{1}{\sqrt{5}}$, then find the value of $\frac{\text{cosec}^2\theta - \sec^2\theta}{\text{cosec}^2\theta + \sec^2\theta}$.
💡 Hint: Use the powerful trigonometric identities: $\text{cosec}^2\theta = 1 + \cot^2\theta$ and $\sec^2\theta = 1 + \tan^2\theta$. You already know $\tan\theta$, so $\cot\theta$ is just its reciprocal! (Ref: www.maheshkumarsir.in)
⚠️ Common Mistake: Drawing a triangle to find the hypotenuse ($\sqrt{6}$) works, but calculating the square roots is prone to error. Using identities is much faster and cleaner.
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Given: $\tan\theta = \frac{1}{\sqrt{5}}$
Therefore, $\cot\theta = \frac{1}{\tan\theta} = \sqrt{5}$. [½ Mark]
Using Trigonometric Identities:
$\text{cosec}^2\theta = 1 + \cot^2\theta = 1 + (\sqrt{5})^2 = 1 + 5 = 6$
$\sec^2\theta = 1 + \tan^2\theta = 1 + \left(\frac{1}{\sqrt{5}}\right)^2 = 1 + \frac{1}{5} = \frac{6}{5}$ [1 Mark]
Now, substituting these values into the given expression:
Expression $= \frac{\text{cosec}^2\theta - \sec^2\theta}{\text{cosec}^2\theta + \sec^2\theta}$
$= \frac{6 - \frac{6}{5}}{6 + \frac{6}{5}}$
$= \frac{\frac{30 - 6}{5}}{\frac{30 + 6}{5}}$
$= \frac{\frac{24}{5}}{\frac{36}{5}} = \frac{24}{36}$
Simplifying the fraction: $\frac{24}{36} = \frac{2}{3}$.
$\therefore \text{The value is } \frac{2}{3}$ [½ Mark]
Q.24
Areas Related to Circles
[2 Marks]
Find the area of a sector of a circle with diameter $28\text{ cm}$, if the length of the corresponding arc is $22\text{ cm}$.
💡 Hint: Instead of finding the angle $\theta$ first, use the direct shortcut formula relating Area ($A$), arc length ($l$), and radius ($r$): $A = \frac{1}{2} \times l \times r$.
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Given:
Diameter $d = 28\text{ cm} \Rightarrow \text{Radius } (r) = \frac{28}{2} = 14\text{ cm}$. [½ Mark]
Length of arc ($l$) $= 22\text{ cm}$.
We know the direct relationship for the area of a sector:
$\text{Area } (A) = \frac{1}{2} \times l \times r$ [1 Mark]
$A = \frac{1}{2} \times 22 \times 14$
$A = 11 \times 14$
$A = 154\text{ cm}^2$.
$\therefore \text{Area of the sector is } 154\text{ cm}^2$ [½ Mark]
Q.25
Probability
[2 Marks]
Two dice are thrown simultaneously. Find the probability that the product of the numbers appearing on them is a prime number.
💡 Hint: The only way to multiply two whole positive numbers together and get a prime number is if one of the numbers is EXACTLY 1, and the other number is a prime!
⚠️ Common Mistake: Listing out pairs like (2,2) because 2 is prime. But $2 \times 2 = 4$, and 4 is a composite number! Only pairs with a '1' can work.
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When two dice are thrown, Total possible outcomes $= 6 \times 6 = 36$. [½ Mark]
A number is prime only if its factors are 1 and the number itself.
Therefore, for the product of two dice numbers to be prime, one die must show a '1' and the other must show a prime number.
Prime numbers on a die are: $\{2, 3, 5\}$. [½ Mark]
Let $E$ be the event of getting a prime product.
Favorable outcomes for $E = \{(1,2), (1,3), (1,5), (2,1), (3,1), (5,1)\}$
Number of favorable outcomes $= 6$. [½ Mark]
$P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{36} = \frac{1}{6}$
$\therefore \text{Required probability is } \frac{1}{6}$ [½ Mark]
