RRB NTPC Mathematics Practice Questions – May 26, 2025
1. Time & Work
Question: P and Q can complete a job in 12 and 16 days respectively. They work together for 4 days. What portion of the work is left?
Solution:
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LCM of 12 and 16 = 48 units (Total work)
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P’s 1-day work = 48/12 = 4 units
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Q’s 1-day work = 48/16 = 3 units
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Combined 1-day work = 4 + 3 = 7 units
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Work done in 4 days = 7 × 4 = 28 units
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Remaining work = 48 – 28 = 20 units
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Fraction left = 20/48 = 5/12
Answer: 5/12 by RRB NTPC Mathematics Practice Questions
2. Mensuration
Question: A cone of height 32 cm and base radius 8 cm is melted and recast into a sphere. Find the radius of the sphere.
Solution:
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Volume of cone = (1/3)πr²h = (1/3)π×8²×32 = (1/3)π×64×32
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Volume of sphere = (4/3)πR³
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Equating volumes: (1/3)π×64×32 = (4/3)πR³
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Simplify: 64×32 = 4R³ ⇒ R³ = (64×32)/4 = 512
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R = ∛512 = 8 cm
Answer: 8 cm
3. Profit & Loss
Question: An article sold for Rs. 3360 yields a 12% profit. What should be the selling price for an 8% profit?
Solution:
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Cost Price (CP) = 3360 / (1 + 12/100) = 3360 / 1.12 = Rs. 3000
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New Selling Price (SP) = 3000 × (1 + 8/100) = 3000 × 1.08 = Rs. 3240
Answer: Rs. 3240
4. LCM & HCF
Question: Find the HCF of the numbers: 2×3²×5², 5×3×2², and 5²×3×2².
Solution:
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Prime factors:
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First: 2×3²×5²
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Second: 2²×3×5
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Third: 2²×3×5²
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Common factors: 2×3×5 = 30
Answer: 30
5. Efficiency Ratio
Question: P and Q can complete a task in 15 and 20 days respectively. What is the ratio of their efficiencies?
Solution:
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Efficiency ∝ 1/Time
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Ratio = 1/15 : 1/20 = 20:15 = 4:3
Answer: 4:3
6. Profit Percentage Comparison
Question: Which transaction offers the highest profit percentage?
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Option A: CP = Rs. 40, Profit = Rs. 15
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Option B: CP = Rs. 60, Profit = Rs. 24
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Option C: CP = Rs. 36, Profit = Rs. 18
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Option D: CP = Rs. 50, Profit = Rs. 24
Solution:
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Option A: (15/40)×100 = 37.5%
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Option B: (24/60)×100 = 40%
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Option C: (18/36)×100 = 50%
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Option D: (24/50)×100 = 48%
Answer: Option C (50%)
7. Trigonometry
Question: Evaluate: tan10° × tan15° × tan80° × tan75°
Solution:
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tan80° = cot10°, tan75° = cot15°
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Expression becomes: tan10°×tan15°×cot10°×cot15° = (tan10°×cot10°)×(tan15°×cot15°) = 1×1 = 1
Answer: 1
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