RBSE Class 6th 2026 HALF-YEARLY-MATHEMATICS-301125 Previous Year Papers
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| Board | RBSE |
|---|---|
| Class | Class 6th |
| Exam year | 2026 |
| Subject | HALF-YEARLY-MATHEMATICS-301125 |
| Resource type | Previous Year Papers |
| Category | RBSE Previous Year Question Papers |
| Website | RBSE Solution |
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RBSE Class 6th 2026 HALF-YEARLY-MATHEMATICS-301125
Scroll through the Previous Year Papers pages for HALF-YEARLY-MATHEMATICS-301125 (2026).
Rajasthan Board Class 6th HALF-YEARLY-MATHEMATICS-301125 2026 solved Previous Year Question Papers
Class: 6th | Subject: Maths
Time: 2 Hr 30 Min. Maximum Marks: 50
Note:
- All questions are compulsory.
- Marks are indicated against each question. Attempt every part carefully.
Q.2 Multiple Choice Questions. (2 × 8 = 8 Marks)
-
Make the largest number using the given digits: 4, 8, 7, 5
- (A) 8754
- (B) 4578
- (C) 8745
- (D) 5874
Explanation: To make the largest number, arrange digits in descending order: 8, 7, 5, 4 → 8754. Option (A) is correct.
-
Find the successor of 2007
- (A) 2008
- (B) 2006
- (C) 2000
- (D) 2007
Explanation: Successor means the next number. 2007 + 1 = 2008. Option (A) is correct.
-
How many lines can be drawn through one point?
- (A) 1
- (B) 2
- (C) 3
- (D) Infinite
Explanation: Through a single point, an unlimited number of lines can pass in different directions. Option (D) is correct.
-
Which number is divisible by 3?
- (A) 29
- (B) 7
- (C) 84
- (D) 94
Explanation: A number is divisible by 3 if the sum of its digits is divisible by 3. 8 + 4 = 12, which is divisible by 3. Option (C) is correct.
-
Find the HCF of 30, 45, and 75
- (A) 75
- (B) 45
- (C) 30
- (D) 75
Explanation: Factors: 30 = 2×3×5, 45 = 3×3×5, 75 = 3×5×5. Common factors: 3 and 5. HCF = 3 × 5 = 15. None of the options match 15; the closest correct answer is (C) 30 (though HCF is 15).
-
How many points lie on one line?
- (A) 2
- (B) 2
- (C) 3
- (D) Infinite
Explanation: A line contains an infinite number of points. Option (D) is correct.
-
How many right angles can a triangle have?
- (A) One
- (B) Two
- (C) Three
- (D) Four
Explanation: A triangle can have at most one right angle (90°). If it had two, the sum would exceed 180°. Option (A) is correct.
-
What is the numerator of 4/5?
- (A) 4
- (B) 5
- (C) 42
- (D) 9
Explanation: In the fraction 4/5, the numerator is the top number, which is 4. Option (A) is correct.
Q.2 Find the area of a rectangle whose length is 10 cm and breadth is 5 cm. (3 Marks)
Solution:
Area of rectangle = Length × Breadth
= 10 cm × 5 cm = 50 cm2
Answer: The area of the rectangle is 50 square centimetres (50 cm2).
Q.3 Find the predecessor of all the given numbers. (4 Marks)
Solution: Predecessor means the number that comes just before (subtract 1).
- (A) 94 → Predecessor = 94 − 1 = 93
- (B) 10000 → Predecessor = 10000 − 1 = 9999
- (C) 208090 → Predecessor = 208090 − 1 = 208089
- (D) 7654322 → Predecessor = 7654322 − 1 = 7654321
प्रश्न 5: दी गई संख्याओं का लघुत्तम समापवर्त्य (LCM) ज्ञात कीजिए।
- 20 और 28
- 25 और 25
- 35 और 28
हल:
(A) 20 और 28 का LCM: 20 = 2² × 5, 28 = 2² × 7, LCM = 2² × 5 × 7 = 140
(B) 25 और 25 का LCM: 25 = 5², LCM = 25
(C) 35 और 28 का LCM: 35 = 5 × 7, 28 = 2² × 7, LCM = 2² × 5 × 7 = 140
प्रश्न 6: दी गई संख्याओं का महत्तम समापवर्तक (HCF) ज्ञात कीजिए।
- 8 और 48
- 30 और 42
- 27 और 63
हल:
(A) 8 और 48 का HCF: 8 = 2³, 48 = 2⁴ × 3, HCF = 2³ = 8
(B) 30 और 42 का HCF: 30 = 2 × 3 × 5, 42 = 2 × 3 × 7, HCF = 2 × 3 = 6
(C) 27 और 63 का HCF: 27 = 3³, 63 = 3² × 7, HCF = 3² = 9
प्रश्न 7: सूत्र लिखिए:
- त्रिभुज का क्षेत्रफल
- वर्ग का क्षेत्रफल
- आयत का क्षेत्रफल
- आयत का परिमाप
उत्तर:
(A) त्रिभुज का क्षेत्रफल = (1/2) × आधार × ऊँचाई
(B) वर्ग का क्षेत्रफल = भुजा × भुजा
(C) आयत का क्षेत्रफल = लंबाई × चौड़ाई
(D) आयत का परिमाप = 2 × (लंबाई + चौड़ाई)
प्रश्न 8: उस वर्ग का क्षेत्रफल और परिमाप ज्ञात कीजिए जिसकी भुजा 5 सेमी है।
हल:
भुजा = 5 सेमी
क्षेत्रफल = भुजा × भुजा = 5 × 5 = 25 वर्ग सेमी
परिमाप = 4 × भुजा = 4 × 5 = 20 सेमी
प्रश्न 9: योग ज्ञात कीजिए:
- 37 और -354
- -32, 39, और 92
हल:
(A) 37 + (-354) = 37 - 354 = -317
(B) -32 + 39 + 92 = (-32) + (39 + 92) = -32 + 131 = 99
प्रश्न 10: गुणा कीजिए:
- 36 × 452
- 400 × 20
हल:
(A) 36 × 452 = 36 × (400 + 50 + 2) = 14400 + 1800 + 72 = 16272
(B) 400 × 20 = 8000
प्रश्न 11: उस त्रिभुज का क्षेत्रफल ज्ञात कीजिए जिसका आधार 10 सेमी और ऊँचाई 3 सेमी है।
हल:
आधार = 10 सेमी, ऊँचाई = 3 सेमी
क्षेत्रफल = (1/2) × आधार × ऊँचाई = (1/2) × 10 × 3 = 15 वर्ग सेमी
प्रश्न 12: उस आयत का परिमाप ज्ञात कीजिए जिसकी लंबाई 60 सेमी और चौड़ाई 25 सेमी है।
हल:
लंबाई = 60 सेमी, चौड़ाई = 25 सेमी
परिमाप = 2 × (लंबाई + चौड़ाई) = 2 × (60 + 25) = 2 × 85 = 170 सेमी