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RIMC Mathematics Mock Test 2 Overview (200 Marks)
The Rashtriya Indian Military College (RIMC) Entrance Examination 2026 Mathematics paper carries 200 maximum marks and tests candidates through a rigorous 90-minute descriptive pen-paper test. This Mock Test 2 is specifically crafted with fresh problems covering Number Systems, Fractions, Ratio & Proportion, Profit & Loss, Simple Interest, Speed & Distance, Mensuration, Algebra, and Geometry.
Formula → Substitution → Calculation → Final Unit. Click the ▶ View Working & Solution toggle button under each question to verify your working!SECTION A – Basic Mathematics
Questions 1–10 | 4 Marks Each | Total: 40 MarksQ1. Number System (4 Marks)
Find the HCF and LCM of: 18, 30 and 42
▶ View Working & Solution
HCF = 6, LCM = 630
1. Prime Factorisation:
• 18 = 2 × 3²
• 30 = 2 × 3 × 5
• 42 = 2 × 3 × 7
2. HCF = 2 × 3 = 6
3. LCM = 2¹ × 3² × 5¹ × 7¹ = 2 × 9 × 5 × 7 = 630
Q2. Fractions (4 Marks)
Simplify and give your answer in the simplest form:
$\frac{7}{8} + \frac{5}{12} - \frac{1}{6}$
▶ View Working & Solution
Final Answer: 9/8 (or 1 1/8)
1. LCM of denominators (8, 12, 6) = 24
2. Express with common denominator:
= (7×3 + 5×2 - 1×4) / 24
= (21 + 10 - 4) / 24
= 27 / 24 = 9/8 = 1⅛
Q3. Decimal Numbers (4 Marks)
A shopkeeper has 12.5 kg rice, 8.75 kg wheat, and 6.25 kg flour. Find the total weight.
▶ View Working & Solution
Total Weight = 27.5 kg
12.50 kg + 8.75 kg + 6.25 kg = 27.50 kg
Q4. Ratio (4 Marks)
The ratio of red balls to blue balls is 3 : 5. If there are 45 blue balls, find:
(i) Number of red balls
(ii) Total number of balls
▶ View Working & Solution
(i) Red balls = 27 | (ii) Total balls = 72
1. Let 5 units = 45 blue balls → 1 unit = 45/5 = 9
2. Red balls = 3 × 9 = 27
3. Total balls = 27 + 45 = 72
Q5. Percentage (4 Marks)
A student scored 72 marks out of 90. Find his percentage.
▶ View Working & Solution
Percentage = 80%
Percentage = (72 / 90) × 100 = (4 / 5) × 100 = 80%
Q6. Average (4 Marks)
Find the average of: 16, 22, 28, 34, 40
▶ View Working & Solution
Average = 28
Sum = 16 + 22 + 28 + 34 + 40 = 140
Average = 140 / 5 = 28
Q7. Simplification (4 Marks)
Simplify using BODMAS:
$72 \div 8 + 6 \times 4 - 7$
▶ View Working & Solution
Result = 26
1. Division: 72 ÷ 8 = 9
2. Multiplication: 6 × 4 = 24
3. Expression: 9 + 24 - 7 = 33 - 7 = 26
Q8. Unitary Method (4 Marks)
If 6 pens cost ₹90, find the cost of 14 pens.
▶ View Working & Solution
Cost = ₹210
1. Cost of 1 pen = 90 / 6 = ₹15
2. Cost of 14 pens = 14 × 15 = ₹210
Q9. Profit and Loss (4 Marks)
A book was bought for ₹450 and sold for ₹540. Find:
(i) Profit
(ii) Profit percentage
▶ View Working & Solution
(i) Profit = ₹90 | (ii) Profit % = 20%
1. Profit = SP - CP = 540 - 450 = ₹90
2. Profit % = (90 / 450) × 100 = 20%
Q10. Simple Interest (4 Marks)
Find the simple interest on ₹4,000 at 5% per annum for 3 years. Also find the total amount.
▶ View Working & Solution
SI = ₹600 | Amount = ₹4,600
1. SI = (P × R × T) / 100 = (4000 × 5 × 3) / 100 = ₹600
2. Amount = P + SI = 4000 + 600 = ₹4,600
SECTION B – Intermediate Mathematics
Questions 11–20 | 6 Marks Each | Total: 60 MarksQ11. Fraction Application (6 Marks)
A tank is 3/8 full. Another 1/4 of the tank's capacity is added. What fraction of the tank is now filled?
▶ View Working & Solution
Fraction Filled = 5/8
Total fraction = 3/8 + 1/4 = 3/8 + 2/8 = 5/8 of the tank.
Q12. Percentage (6 Marks)
A school has 1,200 students. If 45% are girls, find:
(i) Number of girls
(ii) Number of boys
▶ View Working & Solution
(i) Girls = 540 | (ii) Boys = 660
1. Girls = 45% of 1200 = (45/100) × 1200 = 540
2. Boys = 1200 - 540 = 660
Q13. Ratio and Proportion (6 Marks)
The ratio of the ages of two brothers is 4 : 7. Their total age is 33 years. Find their individual ages.
▶ View Working & Solution
Ages = 12 years and 21 years
1. Total parts = 4 + 7 = 11 parts = 33 years → 1 part = 3 years
2. Younger brother = 4 × 3 = 12 years
3. Elder brother = 7 × 3 = 21 years
Q14. Speed, Time and Distance (6 Marks)
A cyclist travels at 18 km/h. How far will the cyclist travel in:
(i) 2 hours 30 minutes?
(ii) 45 minutes?
▶ View Working & Solution
(i) Distance = 45 km | (ii) Distance = 13.5 km
1. Time (i) = 2.5 hours → Distance = 18 × 2.5 = 45 km
2. Time (ii) = 45/60 = 0.75 hours → Distance = 18 × 0.75 = 13.5 km
Q15. Time and Work (6 Marks)
A worker can complete a job in 15 days. Assuming equal work per day:
(i) What fraction of work is completed in 5 days?
(ii) What fraction remains?
▶ View Working & Solution
(i) Completed = 1/3 | (ii) Remaining = 2/3
1. Work in 1 day = 1/15
2. Work in 5 days = 5/15 = 1/3
3. Remaining work = 1 - 1/3 = 2/3
Q16. Algebra (6 Marks)
Solve and verify:
$5x - 9 = 36$
▶ View Working & Solution
x = 9 (Verified)
1. 5x = 36 + 9 = 45 → x = 45 / 5 = 9
2. Verification: LHS = 5(9) - 9 = 45 - 9 = 36 = RHS. Verified!
Q17. Algebraic Expression (6 Marks)
Simplify: $4(3x - 2) + 3(x + 5)$. Then find its value when $x = 2$.
▶ View Working & Solution
Simplified = 15x + 7 | Value at x=2 is 37
1. Expand: 12x - 8 + 3x + 15 = 15x + 7
2. For x = 2: 15(2) + 7 = 30 + 7 = 37
Q18. Geometry (6 Marks)
Two angles of a triangle are 55° and 65°. Find:
(i) The third angle
(ii) Whether the triangle is acute-angled, right-angled or obtuse-angled.
▶ View Working & Solution
(i) Third angle = 60° | (ii) Acute-angled triangle
1. Sum of angles = 180° → Third angle = 180° - (55° + 65°) = 180° - 120° = 60°
2. Since all angles (55°, 65°, 60°) are < 90°, it is an acute-angled triangle.
Q19. Rectangle (6 Marks)
A rectangular garden is 32 m long and 18 m wide. Find:
(i) Perimeter
(ii) Area
▶ View Working & Solution
(i) Perimeter = 100 m | (ii) Area = 576 sq m
1. Perimeter = 2(L + W) = 2(32 + 18) = 2(50) = 100 m
2. Area = L × W = 32 × 18 = 576 sq m
Q20. Square (6 Marks)
A square has a perimeter of 96 cm. Find:
(i) Length of each side
(ii) Area of the square
▶ View Working & Solution
(i) Side = 24 cm | (ii) Area = 576 sq cm
1. Side = Perimeter / 4 = 96 / 4 = 24 cm
2. Area = Side² = 24² = 576 sq cm
SECTION C – Advanced Application
Questions 21–25 | 10 Marks Each | Total: 50 MarksQ21. LCM Application (10 Marks)
Three school bells ring at intervals of 8, 12, and 18 minutes. They ring together at 9:00 AM. After how many minutes will they ring together again, and at what time?
▶ View Working & Solution
(i) 72 minutes | (ii) Next Ring Time = 10:12 AM
1. LCM of (8, 12, 18):
8 = 2³, 12 = 2² × 3, 18 = 2 × 3²
LCM = 2³ × 3² = 8 × 9 = 72 minutes (1 hour 12 mins)
2. Next time = 9:00 AM + 1 hr 12 mins = 10:12 AM
Q22. Discount and Profit (10 Marks)
A shopkeeper marks a school bag at ₹2,000 and gives a 15% discount. If he originally bought it for ₹1,500, find:
(i) Discount amount
(ii) Selling price
(iii) Profit
(iv) Profit percentage
▶ View Working & Solution
(i) Discount = ₹300 | (ii) SP = ₹1,700 | (iii) Profit = ₹200 | (iv) Profit % = 13.33%
1. Discount = 15% of 2000 = ₹300
2. SP = 2000 - 300 = ₹1,700
3. Profit = 1700 - 1500 = ₹200
4. Profit % = (200 / 1500) × 100 = 40/3 % = 13.33% (or 13 1/3%)
Q23. Simple Interest (10 Marks)
A sum of ₹7,500 is deposited at 8% p.a. for 2 years. Find:
(i) Simple Interest
(ii) Total Amount
▶ View Working & Solution
(i) SI = ₹1,200 | (ii) Total Amount = ₹8,700
1. SI = (7500 × 8 × 2) / 100 = ₹1,200
2. Total Amount = 7500 + 1200 = ₹8,700
Q24. Mensuration (10 Marks)
A playground is 50 m long and 30 m wide. A path 2 m wide is constructed inside the boundary. Find:
(i) Length & Width of inner rectangle
(ii) Area of playground
(iii) Area of inner rectangle
(iv) Area of path
▶ View Working & Solution
(i) Inner 46m × 26m | (ii) Playground Area = 1500 m² | (iii) Inner Area = 1196 m² | (iv) Path Area = 304 m²
1. Inner Length = 50 - 2(2) = 46 m, Inner Width = 30 - 2(2) = 26 m
2. Playground Area = 50 × 30 = 1500 sq m
3. Inner Area = 46 × 26 = 1196 sq m
4. Path Area = 1500 - 1196 = 304 sq m
Q25. Speed and Distance (10 Marks)
A bus travels first 120 km at 40 km/h and next 180 km at 60 km/h. Find:
(i) Time taken for 1st part
(ii) Time taken for 2nd part
(iii) Total time & Total distance
(iv) Average speed
▶ View Working & Solution
(i) T1 = 3h | (ii) T2 = 3h | (iii) Total Time = 6h, Total Dist = 300km | (iv) Avg Speed = 50 km/h
1. T1 = 120 / 40 = 3 hours
2. T2 = 180 / 60 = 3 hours
3. Total Time = 3 + 3 = 6 hours, Total Distance = 120 + 180 = 300 km
4. Avg Speed = Total Distance / Total Time = 300 / 6 = 50 km/h
SECTION D – RIMC Challenge
Questions 26–30 | 10 Marks Each | Total: 50 MarksQ26. Fraction Challenge (10 Marks)
A student spends 2/5 of his ₹900 on books and 1/3 of the remaining money on stationery. Find:
(i) Money spent on books
(ii) Money remaining after books
(iii) Money spent on stationery
(iv) Final money left
▶ View Working & Solution
(i) Books = ₹360 | (ii) Rem = ₹540 | (iii) Stationery = ₹180 | (iv) Final Left = ₹360
1. Books = (2/5) × 900 = ₹360
2. Remaining = 900 - 360 = ₹540
3. Stationery = (1/3) × 540 = ₹180
4. Final Left = 540 - 180 = ₹360
Q27. Consecutive Numbers (10 Marks)
The sum of four consecutive natural numbers is 106. Find all four numbers.
▶ View Working & Solution
Numbers are 25, 26, 27, and 28
1. Let numbers be x, x+1, x+2, x+3
2. x + (x+1) + (x+2) + (x+3) = 106 → 4x + 6 = 106
3. 4x = 100 → x = 25
4. The four numbers are 25, 26, 27, 28.
Q28. Geometry Challenge (10 Marks)
A field measures 40 m × 25 m. A square pond of side 10 m is constructed inside. Find:
(i) Area of field & pond
(ii) Remaining area
(iii) Perimeter of field
▶ View Working & Solution
(i) Field = 1000m², Pond = 100m² | (ii) Remaining = 900m² | (iii) Field Perimeter = 130m
1. Field Area = 40 × 25 = 1000 sq m
2. Pond Area = 10 × 10 = 100 sq m
3. Remaining Area = 1000 - 100 = 900 sq m
4. Field Perimeter = 2(40 + 25) = 2(65) = 130 m
Q29. Average Challenge (10 Marks)
The average of six numbers is 24. Five numbers are 18, 21, 25, 27, and 31. Find the sixth number.
▶ View Working & Solution
Sixth Number = 22
1. Total sum of 6 numbers = 6 × 24 = 144
2. Sum of 5 numbers = 18 + 21 + 25 + 27 + 31 = 122
3. Sixth number = 144 - 122 = 22
Q30. Mixed Challenge (10 Marks)
A school purchased 25 notebooks at ₹32 each and 15 pens at ₹12 each. Total discount given = ₹100. Remaining paid equally by 5 departments. Find:
(i) Cost of notebooks & pens
(ii) Total cost before discount
(iii) Amount paid after discount
(iv) Amount paid by each department
▶ View Working & Solution
(i) Notebooks = ₹800, Pens = ₹180 | (ii) Before Disc = ₹980 | (iii) After Disc = ₹880 | (iv) Per Dept = ₹176
1. Notebooks = 25 × 32 = ₹800, Pens = 15 × 12 = ₹180
2. Total before discount = 800 + 180 = ₹980
3. Amount after discount = 980 - 100 = ₹880
4. Amount per department = 880 / 5 = ₹176
RIMC Mathematics Marks Distribution & Weightage
| Section | Topic & Difficulty Level | Questions | Marks per Question | Total Marks |
|---|---|---|---|---|
| Section A | Basic Arithmetic, Ratios, Simplification & SI | 1–10 | 4 Marks | 40 Marks |
| Section B | Fractions, Percentages, Geometry & Algebra | 1–20 | 6 Marks | 60 Marks |
| Section C | Advanced Application & Mensuration | 21–25 | 10 Marks | 50 Marks |
| Section D | RIMC High-Level Challenge Problems | 26–30 | 10 Marks | 50 Marks |
| Total | Full Mathematics Descriptive Paper | 30 | — | 200 Marks |
Suggested 90-Minute Time Management Strategy
0–20 Mins
20 MinsComplete Section A Basic Questions (Q1–Q10).
20–45 Mins
25 MinsSolve Section B Intermediate Problems (Q11–Q20).
45–70 Mins
25 MinsSolve Section C Advanced Applications (Q21–Q25).
70–90 Mins
20 MinsSolve Section D RIMC Challenge Problems & Recheck calculations.