योग्यं पर्यायं चिनुत। (3 तः 2)
त्वं द्रष्टुं \(\underline{\hspace{1cm}}\) । (शक्नोषि/शक्नोति)
योग्यं पर्यायं चिनुत। (3 तः 2)
त्वं द्रष्टुं \(\underline{\hspace{1cm}}\) । (शक्नोषि/शक्नोति)
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Solution and Explanation
Top Questions on संस्कृत व्याकरण
पदे शुद्धं पूर्णं च लिखत। (3 तः 2)
अभ्यासनाम् ............................ मतिदातृत्तम्॥
पदे शुद्धं पूर्णं च लिखत। (3 तः 2)
यत्र ............................ ह्रदयते॥
पदे शुद्धं पूर्णं च लिखत। (3 तः 2)
वैद्यराज ............................ धनानि च॥
मञ़्जुषातः नामानि सर्वनामानि च पृथक्कुरुत। (5 तः 4)
मञ़्जुषातः क्रियापदानि धातुसाधित-विशेषणानि च पृथक्कुरुत। (5 तः 4)
Questions Asked in Maharashtra Class X Board exam
- Appreciation of the poem:
Read the following poem and write an appreciation of it with the help of the points given below:
Stopping by Woods on a Snowy Evening
Whose woods these are I think I know.
His house is in the village, though;
He will not see me stopping here
To watch his woods fill up with snow
My little horse must think it queer
To stop without a farmhouse near
Between the woods and frozen lake
The darkest evening of the year.
He gives his harness bells a shake
To ask if there is some mistake.
The only other sound’s the sweep
Of easy wind and downy flake.
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
Robert Frost
- Maharashtra Class X Board - 2025
- Reading Comprehension
- Pick out the infinitive from the following sentence :
He was asking to go for the concert.- Maharashtra Class X Board - 2025
- Grammar and Language Usage
- Out of the following which is a Pythagorean triplet ?
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
- Maharashtra Class X Board - 2025
- Properties of Triangles
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.