RBSE Class 7th 2026 HALF-YEARLY-MATHEMATICS-301125 Previous Year Papers
HALF-YEARLY-MATHEMATICS-301125 from the 2026 exam year is part of the Class 7th 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.
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Paper details
Quick reference for this previous year papers page — confirm board, class, and year & subject before you study.
| Board | RBSE |
|---|---|
| Class | Class 7th |
| Exam year | 2026 |
| Subject | HALF-YEARLY-MATHEMATICS-301125 |
| 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 7th, year 2026, and subject HALF-YEARLY-MATHEMATICS-301125 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-301125 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 7th expect clear numbering even in practice sessions.
RBSE Class 7th 2026 HALF-YEARLY-MATHEMATICS-301125
Scroll through the Previous Year Papers pages for HALF-YEARLY-MATHEMATICS-301125 (2026).
Rajasthan Board Class 7th HALF-YEARLY-MATHEMATICS-301125 2026 solved Previous Year Question Papers
Class: 7
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.1 Select the correct option (2 × 8 = 8 Marks)
-
2% of Rs. 50 will be:
- (A) Rs. 1
- (B) Rs. 60
- (C) Rs. 50
- (D) Rs. 1
Explanation: 2% of Rs. 50 = (2/100) × 50 = Rs. 1.
-
Each angle of an equilateral triangle is:
- (A) 60°
- (B) 90°
- (C) 120°
- (D) 80°
Explanation: In an equilateral triangle, all angles are equal and sum to 180°, so each angle = 180°/3 = 60°.
-
The complementary angle of 50° is:
- (A) 30°
- (B) 120°
- (C) 40°
- (D) 90°
Explanation: Complementary angles sum to 90°. So, 90° - 50° = 40°.
-
The sum of two supplementary angles is:
- (A) 60°
- (B) 90°
- (C) 120°
- (D) 180°
Explanation: Supplementary angles always add up to 180°.
-
If x + 5 = 7, then the value of x will be:
- (A) 2
- (B) 3
- (C) 12
- (D) 4
Explanation: x + 5 = 7 ⇒ x = 7 - 5 = 2. So correct option is (A).
-
The measure of a right angle is:
- (A) 90°
- (B) 60°
- (C) 45°
- (D) 180°
Explanation: A right angle measures exactly 90°.
-
The solution of (-28) - (-28) will be:
- (A) -56
- (B) 56
- (C) -18
- (D) 0
Explanation: (-28) - (-28) = -28 + 28 = 0.
-
2/3 + 1/3 is equal to:
- (A) 1/3
- (B) 2/3
- (C) 1
- (D) 3/3
Explanation: 2/3 + 1/3 = (2+1)/3 = 3/3 = 1.
Q.2 Fill in the blanks (1 × 8 = 8 Marks)
-
a × 0 = 0 × a = 0
-
9 × (3) × (-1) = -27
-
Write the formula for the area of a circle: πr²
-
If x/5 = 0, then x = 0
-
The side opposite to the right angle is called hypotenuse.
-
12.5% = 1/8 (in fraction form)
-
Profit = Selling Price – Purchase Price.
-
Simple interest = (P × R × T) / 100
Question 4
Multiply and express as a mixed fraction. (2 Marks)
(Note: The original text appears to have a formatting issue. Assuming the intended question is: 5 × 6² or similar. Based on available text, the problem is unclear. Please refer to the original question paper.)
(i) 4 × 6⁵
Question 5
A car travels 76 km on 1 liter of petrol. What is the total distance the car will travel on 2 liters of petrol? (2 Marks)
Solution:
Distance traveled on 1 liter = 76 km
Distance traveled on 2 liters = 76 × 2 = 152 km
Answer: The car will travel 152 km on 2 liters of petrol.
Question 6
Find the area of a triangle whose base is 0 cm and height is 2 cm. (2 Marks)
Solution:
Area of a triangle = ½ × base × height
Base = 0 cm, Height = 2 cm
Area = ½ × 0 × 2 = 0 cm²
Answer: The area of the triangle is 0 cm².
Question 7
Solve the following equations. (5 Marks)
(Note: The original text has unclear symbols. Assuming standard equations.)
- 10P + 0 = 00 → Answer: P = 0
- 2q - 6 = 0 → Answer: q = 3
- 2q + 6 = 2 → Answer: q = -2
- 35 = 0 → Answer: No solution (equation is false)
- 2q = 6 → Answer: q = 3
Question 8
Find the unknown angle in the adjoining figure (where P || Q). (2 Marks)
(Figure not provided. Please refer to the original question paper for the diagram.)
Question 9
Find the supplement of each of the following angles. (2 Marks)
(Note: The original text has unclear angle values. Assuming standard angles.)
- Angle = 45° → Supplement = 180° - 45° = 135°
- Angle = 60° → Supplement = 180° - 60° = 120°
Question 10
Find the values of the unknown X and Y in the following figures. (2 Marks)
(Figures not provided. Please refer to the original question paper for the diagrams.)
Question 11
Consider the following data obtained from a survey conducted in a colony. (3 Marks)
| Favorite Sport | Cricket | Volleyball | Swimming | Hockey | Playing |
|---|---|---|---|---|---|
| Watching | 240 | 470 | 520 | 430 | 250 |
| Participating | 620 | 320 | 320 | 250 | 405 |
Draw a double bar graph by choosing a suitable scale.
(Graph not drawn here. Please draw on paper with scale: 1 unit = 100 people on y-axis, sports on x-axis.)
Question 12
Find out! (4 Marks)
- 15% of 250
- 75% of 1 kg
Solution:
(i) 15% of 250 = (15/100) × 250 = 37.5
(ii) 75% of 1 kg = (75/100) × 1 = 0.75 kg = 750 g
Question 13
The population of a town decreased from 25000 to 24500. Find the percentage decrease. (4 Marks)
Solution:
Original population = 25000
New population = 24500
Decrease = 25000 - 24500 = 500
Percentage decrease = (Decrease / Original) × 100 = (500 / 25000) × 100 = 2%
Answer: The population decreased by 2%.
Question 14
Find the hypotenuse of a right angle triangle whose base is 3 cm and height is 4 cm. (4 Marks)
Solution:
Using Pythagoras theorem: Hypotenuse² = Base² + Height²
Base = 3 cm, Height = 4 cm
Hypotenuse² = 3² + 4² = 9 + 16 = 25
Hypotenuse = √25 = 5 cm
Answer: The hypotenuse is 5 cm.