RBSE Class 5th 2026 HALF-YEARLY-MATHEMATICS-011225 Previous Year Papers
HALF-YEARLY-MATHEMATICS-011225 from the 2026 exam year is part of the Class 5th previous year papers archive on RBSE Solution. Many learners start here after finishing the textbook to see how questions were actually framed on the Rajasthan Board of Secondary Education (RBSE) paper.
Treat this question paper as a mock under gentle timing first, then as a marking exercise the second time. The introduction on this page is written only for this subject-and-year pair, not copied from other pages.
Bookmark the link if you coach juniors — the layout stays stable for search engines and classroom sharing.
Paper details
Quick reference for this previous year papers page — confirm board, class, and year & subject before you study.
| Board | RBSE |
|---|---|
| Class | Class 5th |
| Exam year | 2026 |
| Subject | HALF-YEARLY-MATHEMATICS-011225 |
| Resource type | Previous Year Papers |
| Category | RBSE Previous Year Question Papers |
| Website | RBSE Solution |
The table summarises this Previous Year Papers resource. Confirm RBSE, Class 5th, year 2026, and subject HALF-YEARLY-MATHEMATICS-011225 before studying.
RBSE Solution organises previous year papers so each URL carries chapter-specific guidance — better for students and for search engines than one generic paragraph for the whole class.
Turning 2026 papers into insight
One HALF-YEARLY-MATHEMATICS-011225 paper reveals style; several from the same year reveal pattern. After this page, open sibling subjects listed below to see whether marks cluster in certain units.
Keep rough work dated in your notebook. Examiners in Class 5th expect clear numbering even in practice sessions.
RBSE Class 5th 2026 HALF-YEARLY-MATHEMATICS-011225
Scroll through the Previous Year Papers pages for HALF-YEARLY-MATHEMATICS-011225 (2026).
Rajasthan Board Class 5th HALF-YEARLY-MATHEMATICS-011225 2026 solved Previous Year Question Papers
Class: 5th | 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. Write the expanded form given below as a number. (5 Marks)
- 40000 + 5000 + 700 + 70 + 2
- 75000 + 2000 + 800 + 100 + 4
- 6000 + 8000 + 000 + 20 + 3
Solution:
- 40000 + 5000 + 700 + 70 + 2 = 45772
- 75000 + 2000 + 800 + 100 + 4 = 77904
- 6000 + 8000 + 000 + 20 + 3 = 14023
Q.2 Solve the following additions and subtractions. (5 Marks)
- 6075 + 3246
- 8808 − 5303
- 3758 + 2637
- 7594 − 5302
- 4632 + 4598
Solution:
- 6075 + 3246 = 9321
- 8808 − 5303 = 3505
- 3758 + 2637 = 6395
- 7594 − 5302 = 2292
- 4632 + 4598 = 9230
Q.3 The sum of two numbers is 6732. One number is 3846. Find the other number. (2 Marks)
Solution:
Other number = Sum − Given number
= 6732 − 3846 = 2886
Thus, the other number is 2886.
Q.4 Find the sum of the largest and smallest 4-digit numbers. (2 Marks)
Solution:
Largest 4-digit number = 9999
Smallest 4-digit number = 1000
Sum = 9999 + 1000 = 10999
Q.5 If the cost of one chair is Rs 678, find the cost of 296 chairs. (3 Marks)
Solution:
Cost of 1 chair = Rs 678
Cost of 296 chairs = 678 × 296
= 678 × (300 − 4)
= (678 × 300) − (678 × 4)
= 203400 − 2712
= Rs 200688
Thus, the cost of 296 chairs is Rs 200688.
Q.6 Write the deviation on the basis of 10. (2 Marks)
- Deviation of 4
- Deviation of 3
- Deviation of 8
- Deviation of 9
Solution:
Deviation = Number − Base (Base = 10)
- Deviation of 4 = 4 − 10 = −6
- Deviation of 3 = 3 − 10 = −7
- Deviation of 8 = 8 − 10 = −2
- Deviation of 9 = 9 − 10 = −1
Q.7 Subtract with the help of the formula ‘Ek Nyunen Purven’ and ‘Param Mitra’ number. (3 Marks)
- 753 − 584
- 832 − 765
Solution:
(i) 753 − 584
Using Ek Nyunen Purven and Param Mitra:
- Step 1: 3 − 4 → 3 is less than 4. Param Mitra of 4 is 6. So, 3 + 6 = 9. Write 9. Ek Nyunen Purven: 5 becomes 4.
- Step 2: 4 − 8 → 4 is less than 8. Param Mitra of 8 is 2. So, 4 + 2 = 6. Write 6. Ek Nyunen Purven: 7 becomes 6.
- Step 3: 6 − 5 = 1. Write 1.
Answer = 169
(ii) 832 − 765
- Step 1: 2 − 5 → 2 is less than 5. Param Mitra of 5 is 5. So, 2 + 5 = 7. Write 7. Ek Nyunen Purven: 3 becomes 2.
- Step 2: 2 − 6 → 2 is less than 6. Param Mitra of 6 is 4. So, 2 + 4 = 6. Write 6. Ek Nyunen Purven: 8 becomes 7.
- Step 3: 7 − 7 = 0. Write 0.
Answer = 67
Q.8 Multiply using the Nikhilam formula. (4 Marks)
- 74 × 3
- 9 × 42
Solution:
(i) 74 × 3
Using Nikhilam (Base 10):
- 74 is 64 above 10 (Deviation = +64)
- 3 is −7 below 10 (Deviation = −7)
- Left part: 74 − 7 = 67 (or 3 + 64 = 67)
- Right part: 64 × (−7) = −448
- Adjust: 67 × 10 = 670; 670 − 448 = 222
Answer = 222
(ii) 9 × 42
Using Nikhilam (Base 10):
- 9 is −1 below 10 (Deviation = −1)
- 42 is 32 above 10 (Deviation = +32)
- Left part: 9 + 32 = 41 (or 42 − 1 = 41)
- Right part: (−1) × 32 = −32
- Adjust: 41 × 10 = 410; 410 − 32 = 378
Answer = 378
Q.0 Find the greatest factor of 5, 27, and 36.
Solution:
Factors of 5: 1, 5
Factors of 27: 1, 3, 9, 27
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Common factors: 1
Greatest common factor = 1
Q.1 Write the given decimal numbers in words. (3 Marks)
- 42.356
- 7.02
- 0.409
Answer:
(i) 42.356 = Forty-two point three five six
(ii) 7.02 = Seven point zero two
(iii) 0.409 = Zero point four zero nine
Q.2 Write the following improper fractions in the form of mixed fractions. (3 Marks)
- (i) 7/2
- (ii) 11/3
- (iii) 9/4
Answer:
(i) 7/2 = 3 ½
(ii) 11/3 = 3 ⅔
(iii) 9/4 = 2 ¼
Q.3 Write four equivalent fractions equal to ½. (4 Marks)
Answer:
Four equivalent fractions of ½ are:
2/4, 3/6, 4/8, 5/10
Q.4 Shade the equivalent fractions. (4 Marks)
(i) ½ and 2/4
(ii) ⅓ and 2/6
Answer: Shade half of the first shape for ½ and two-quarters of the second shape for 2/4. Similarly, shade one-third of the first shape for ⅓ and two-sixths of the second shape for 2/6.
Q.5 Identify the pattern and fill in the blanks. (5 Marks)
- 7, 12, 17, 22, ____, ____, ____
- 2, 6, 18, 54, ____, ____, ____
Answer:
(i) Pattern: Add 5 each time. So, 7, 12, 17, 22, 27, 32, 37
(ii) Pattern: Multiply by 3 each time. So, 2, 6, 18, 54, 162, 486, 1458