2 Marks
21
(A) A hall has a length of 9.75 m, breadth of 6.75 m and height of 5.25 m. What is the length of the longest unmarked ruler that can exactly measure the dimensions of the hall?
Concept:
1
When asked for the "longest" measuring unit that divides multiple dimensions exactly, you must calculate the Highest Common Factor (HCF). Because decimals are difficult to factorize, first convert all measurements from meters to centimeters to work with whole integers.
Hint:
1
Convert to cm: L = 975 cm, B = 675 cm, H = 525 cm.
2
Find the prime factorization of 975, 675, and 525.
3
Find the HCF (lowest power of common prime factors).
Mistakes to Avoid:
!
Finding LCM: Calculating the LCM will give you a massive number (the smallest hall size that could be built using these blocks). We want a small ruler that fits inside, which requires HCF.
Official CBSE Solution:
Convert all dimensions to cm:
Length = 9.75 m = 975 cm
Breadth = 6.75 m = 675 cm
Height = 5.25 m = 525 cm
Length of longest unmarked ruler = HCF (975, 675, 525).
Prime Factorizations:
$975 = 3 \times 5^2 \times 13$
$675 = 3^3 \times 5^2$
$525 = 3 \times 5^2 \times 7$
$\text{HCF}(975, 675, 525) = 3 \times 5^2 = 3 \times 25 = 75$
Ans: The length of the longest unmarked ruler is 75 cm or 0.75 m.
🎯 CBSE Marking Scheme (2 Marks):
• Prime factorisation / setup for HCF: ½ Mark
• Finding correct HCF (75): 1 Mark
• Final answer with units: ½ Mark
• Prime factorisation / setup for HCF: ½ Mark
• Finding correct HCF (75): 1 Mark
• Final answer with units: ½ Mark
2 Marks
21
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.?
Concept:
1
To find when independent cyclic events synchronize (happen at the same time), calculate the Least Common Multiple (LCM) of their intervals. Then, determine how many times that common interval fits into the total duration.
Hint:
1
Find LCM of 5, 6, and 8.
2
Convert the LCM from minutes to hours.
3
How many hours pass between 10:00 AM and 7:00 PM?
Mistakes to Avoid:
!
Counting the 10:00 AM toll: The question explicitly asks how many *more* times they toll together *till* the park is closed. Do not count the initial toll at 10:00 AM in your final answer.
Official CBSE Solution:
Find the LCM to determine the synchronised tolling interval:
LCM of 5, 6, and 8 is 120.
i.e., all the bells toll together after an interval of 120 minutes or 2 hours.
Determine the tolling schedule:
Starting at 10:00 a.m.
They will toll at: 12:00 noon, 2:00 p.m., 4:00 p.m., and 6:00 p.m.
Total number of times = 4.
Ans: The bells will toll together 4 more times till 7:00 p.m.
🎯 CBSE Marking Scheme (2 Marks):
• Finding LCM = 120 / stating 2 hours: 1 Mark
• Correctly identifying the 4 times: 1 Mark
• Finding LCM = 120 / stating 2 hours: 1 Mark
• Correctly identifying the 4 times: 1 Mark
2 Marks
22
(A) ABCD is a trapezium in which AB is parallel to DC and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.
Concept:
1
Because AB is parallel to DC, the diagonals act as transversals. This creates Alternate Interior Angles that are equal. Prove that $\triangle AOB \sim \triangle COD$ using the AA similarity criterion. Once proven similar, the ratio of their corresponding sides must be equal.
Hint:
1
Identify alternate interior angles: $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$.
2
Write the similarity statement: $\triangle AOB \sim \triangle COD$.
3
Write the proportional sides: $\frac{AO}{CO} = \frac{BO}{DO}$, and then cross-multiply/rearrange to match the "To Prove".
Mistakes to Avoid:
!
Wrong correspondence: Be careful writing the similarity. It is $\triangle AOB \sim \triangle COD$, NOT $\triangle AOB \sim \triangle DOC$. A corresponds to C, and B corresponds to D. Matching the correct vertices is essential for the correct side ratio.
Official CBSE Solution:
In $\triangle AOB$ and $\triangle COD$:
$\angle OAB = \angle OCD$ (Alternate interior angles, since AB || DC)
$\angle OBA = \angle ODC$ (Alternate interior angles, since AB || DC)
$\angle AOB = \angle COD$ (Vertically opposite angles - optional)
Therefore, by AA similarity criterion:
$\triangle AOB \sim \triangle COD$
Since corresponding sides of similar triangles are proportional:
$\frac{AO}{CO} = \frac{BO}{DO}$
Rearranging the terms (cross-exchange CO and BO):
$\frac{AO}{BO} = \frac{CO}{DO}$
Ans: Hence Proved.
🎯 CBSE Marking Scheme (2 Marks):
• Stating correct equal angles with reason: ½ Mark
• Concluding $\triangle AOB \sim \triangle COD$ (AA criterion): 1 Mark
• Writing correct ratio and rearranging: ½ Mark
• Stating correct equal angles with reason: ½ Mark
• Concluding $\triangle AOB \sim \triangle COD$ (AA criterion): 1 Mark
• Writing correct ratio and rearranging: ½ Mark
2 Marks
22
OR (B) In the given figure, if $\angle PQR = \angle QSP$, $PQ = 6\text{ cm}$ and $PS = 3\text{ cm}$, then find the length of PR.
Concept:
1
Identify two triangles that share an angle and have the given equal angles. Prove $\triangle PRQ \sim \triangle PQS$ using the AA similarity criterion. Once proven similar, set up the ratio of corresponding sides to solve for PR.
Hint:
1
Look at the big triangle $\triangle PRQ$ and the small left triangle $\triangle PQS$.
2
$\angle PQR = \angle QSP$ (given). $\angle P$ is shared by both! So they are similar.
3
Ratio: $\frac{PR}{PQ} = \frac{PQ}{PS}$.
Mistakes to Avoid:
!
Wrong side ratios: When $\triangle PRQ \sim \triangle PQS$, the side opposite the marked angle in the big triangle is PR, and the side opposite the marked angle in the small triangle is PQ. Therefore, PR pairs with PQ.
Official CBSE Solution:
In $\triangle PRQ$ and $\triangle PQS$:
$\angle PQR = \angle QSP$ (Given)
$\angle QPR = \angle QPS$ (Common Angle P)
Therefore, by AA similarity criterion:
$\triangle PRQ \sim \triangle PQS$
Using corresponding proportional sides:
$\frac{PQ}{PS} = \frac{PR}{PQ}$
$\frac{6}{3} = \frac{PR}{6}$
$2 = \frac{PR}{6} \implies PR = 12\text{ cm}$
Ans: Length of PR = 12 cm
🎯 CBSE Marking Scheme (2 Marks):
• Proving $\triangle PRQ \sim \triangle PQS$ (AA criterion): 1 Mark
• Setting up ratio $\frac{6}{3} = \frac{PR}{6}$: ½ Mark
• Final calculation ($PR=12$): ½ Mark
• Proving $\triangle PRQ \sim \triangle PQS$ (AA criterion): 1 Mark
• Setting up ratio $\frac{6}{3} = \frac{PR}{6}$: ½ Mark
• Final calculation ($PR=12$): ½ Mark
2 Marks
23
If $\tan\theta = \frac{1}{\sqrt{5}}$, then find the value of $\frac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta}$.
Concept:
1
You can solve this using Pythagorean identities. Since $\tan\theta = \frac{1}{\sqrt{5}}$, you know $\cot\theta = \sqrt{5}$. Use the identities $\sec^2\theta = 1 + \tan^2\theta$ and $\csc^2\theta = 1 + \cot^2\theta$ to find the individual terms, then substitute them into the expression.
Hint:
1
$\sec^2\theta = 1 + (\frac{1}{\sqrt{5}})^2 = 1 + \frac{1}{5} = \frac{6}{5}$.
2
$\csc^2\theta = 1 + (\sqrt{5})^2 = 1 + 5 = 6$.
3
Substitute $\frac{6}{5}$ and $6$ into the fraction.
Mistakes to Avoid:
!
Finding hypotenuse unnecessarily: While drawing a right triangle and finding the hypotenuse ($\sqrt{1^2 + (\sqrt{5})^2} = \sqrt{6}$) works perfectly, using trigonometric identities is faster and less prone to radical arithmetic errors. Both methods are valid.
Official CBSE Solution:
Given $\tan\theta = \frac{1}{\sqrt{5}} \implies \cot\theta = \sqrt{5}$.
Using Trigonometric Identities:
$\sec^2\theta = 1 + \tan^2\theta = 1 + \left(\frac{1}{\sqrt{5}}\right)^2 = 1 + \frac{1}{5} = \frac{6}{5}$
$\csc^2\theta = 1 + \cot^2\theta = 1 + (\sqrt{5})^2 = 1 + 5 = 6$
Substitute values into the expression:
$= \frac{6 - \frac{6}{5}}{6 + \frac{6}{5}}$
Simplify the numerator and denominator:
$= \frac{\frac{30 - 6}{5}}{\frac{30 + 6}{5}} = \frac{\frac{24}{5}}{\frac{36}{5}}$
$= \frac{24}{36} = \frac{2}{3}$
Ans: Value is 2/3
🎯 CBSE Marking Scheme (2 Marks):
• Finding $\sec^2\theta = 6/5$: ½ Mark
• Finding $\csc^2\theta = 6$: ½ Mark
• Substitution into expression: ½ Mark
• Final evaluation ($2/3$): ½ Mark
• Finding $\sec^2\theta = 6/5$: ½ Mark
• Finding $\csc^2\theta = 6$: ½ Mark
• Substitution into expression: ½ Mark
• Final evaluation ($2/3$): ½ Mark
2 Marks
24
Find the area of a sector of a circle with diameter 28 cm, if the length of the corresponding arc is 22 cm.
Concept:
1
While the standard sector area formula is $\frac{\theta}{360} \times \pi r^2$, you don't know the angle $\theta$. Instead, use the direct relationship between Area ($A$), arc length ($l$), and radius ($r$): $A = \frac{1}{2} \times l \times r$.
Hint:
1
Diameter = 28, so Radius ($r$) = 14 cm.
2
Substitute into $A = \frac{1}{2} \times 22 \times 14$.
Mistakes to Avoid:
!
Using diameter as radius: A classic trap. The question gives $d = 28$. If you calculate $\frac{1}{2} \times 22 \times 28 = 308$, it is incorrect. Always halve the diameter first.
Official CBSE Solution:
Given Diameter = 28 cm.
Radius ($r$) = $\frac{28}{2} = 14\text{ cm}$.
Given Arc Length ($l$) = 22 cm.
Area of sector formula:
$A = \frac{1}{2} \times l \times r$
$A = \frac{1}{2} \times 22 \times 14$
$A = 11 \times 14 = 154\text{ cm}^2$
Ans: Area is 154 cm²
🎯 CBSE Marking Scheme (2 Marks):
• Finding radius = 14 cm: ½ Mark
• Stating correct formula ($1/2 \cdot l \cdot r$): ½ Mark
• Final calculation (154 cm²): 1 Mark
• Finding radius = 14 cm: ½ Mark
• Stating correct formula ($1/2 \cdot l \cdot r$): ½ Mark
• Final calculation (154 cm²): 1 Mark
2 Marks
25
Two dice are thrown simultaneously. Find the probability that the product of the numbers appearing on them is a prime number.
Concept:
1
A prime number has exactly two distinct factors: 1 and itself. Since rolling two dice means multiplying two numbers from 1 to 6, the product can ONLY be prime if one of the dice rolls a 1, and the other die rolls a prime number (2, 3, or 5).
Hint:
1
List the valid combinations: (1, 2), (1, 3), (1, 5).
2
Don't forget the reverse rolls! (2, 1), (3, 1), (5, 1).
3
Total valid = 6. Total possible = 36.
Mistakes to Avoid:
!
Assuming 1 is prime: 1 is NOT a prime number. Do not include (1, 1) in your list of favourable outcomes.
Official CBSE Solution:
Total number of outcomes = $6 \times 6 = 36$.
For the product to be prime, one die must be 1, and the other must be a prime number (2, 3, or 5).
Favourable outcomes = {(1,2), (1,3), (1,5), (2,1), (3,1), (5,1)}
Total favourable outcomes = 6.
Probability (Prime Product):
$P = \frac{6}{36} = \frac{1}{6}$
Ans: 1/6
🎯 CBSE Marking Scheme (2 Marks):
• Stating total outcomes = 36: ½ Mark
• Listing correct favourable outcomes {(1,2),(1,3),(1,5),(2,1),(3,1),(5,1)}: 1 Mark
• Final probability ($1/6$): ½ Mark
• Stating total outcomes = 36: ½ Mark
• Listing correct favourable outcomes {(1,2),(1,3),(1,5),(2,1),(3,1),(5,1)}: 1 Mark
• Final probability ($1/6$): ½ Mark
