RRB NTPC Graduate Level (25th March 2026 Shift 1) — RRB NTPC 2025

• Key Fact: Globalisation in India, initiated through the 1991 economic reforms, aimed to integrate the Indian economy with the world. A major part of this was liberalizing (relaxing rules for) Foreign Institutional Investments (FIIs), not making them more restrictive.

• Supporting Detail: The 1991 LPG (Liberalisation, Privatisation, Globalisation) reforms included FDI liberalisation, the devaluation of the rupee to boost exports, and partial convertibility of the rupee on the current account to facilitate international trade.

• Why others wrong: Options A, C, and D are actual, prominent impacts or steps taken during the 1991 economic reforms in India.

• Exam Trick: Globalisation = Opening up markets = Less restriction (liberalisation). Anything saying 'more restrictive' contradicts the core premise of globalisation.

• Given conditions: Set 1: 50 - 10 - 5 - 15. Set 2: 200 - 40 - 20 - 30.

• Deduction process:
Let's analyze the mathematical operations between terms in Set 1 ($50 \rightarrow 10 \rightarrow 5 \rightarrow 15$).
1st to 2nd: $50 \rightarrow 10$. Logic: $\div 5$.
2nd to 3rd: $10 \rightarrow 5$. Logic: $\div 2$.
3rd to 4th: $5 \rightarrow 15$. Logic: $+ 10$.
The operation pattern is: $\div 5$, $\div 2$, $+ 10$.
Let's verify this pattern on Set 2 ($200 \rightarrow 40 \rightarrow 20 \rightarrow 30$).
$200 \div 5 = 40$. (Correct).
$40 \div 2 = 20$. (Correct).
$20 + 10 = 30$. (Correct). The pattern is definitively $\div 5, \div 2, + 10$.
Now apply this pattern to the options:
Option A: $160 \div 5 = 32 \neq 40$. (Incorrect).
Option B: $120 \div 5 = 24 \neq 60$. (Incorrect).
Option C: $100 \div 5 = 20$. $20 \div 2 = 10 \neq 40$. (Incorrect).
Option D: $400 \div 5 = 80$. $80 \div 2 = 40$. $40 + 10 = 50$. (Matches perfectly!).

• Exam Trick: You can often just look at the last transition. $5 \rightarrow 15$ and $20 \rightarrow 30$ is clearly a 'add 10' step. Check the options for the final step: $20+10 \neq 30$, $30+10 \neq 40$, $40+10 \neq 60$, but $40+10 = 50$. Option D is the only one that works for the last step. Solved in 5 seconds.

• Final Answer: 400 - 80 - 40 - 50

• Given conditions: 7 people facing centre. Right = Anti-Clockwise (ACW), Left = Clockwise (CW). Find 4th to the left of X.

• Deduction process:
1) Let's assign positions 1 to 7 ACW. Fix G at Position 1.
2) 'I sits 3rd to the right of G': Right is ACW. From G(1), 3rd position ACW is 1 + 3 = 4. So I is at Pos 4.
3) 'Only three people sit between G and V': In a 7-person circle, a gap of 3 people on one side means they are separated by 2 people on the other side.
From G(1), going ACW, the 3 people are Pos 2, 3, 4. So V would be at Pos 5.
Going CW from G(1), the 3 people are Pos 7, 6, 5. So V would be at Pos 4. But I is already at 4.
Therefore, V MUST be at Position 5.
4) 'Only two people sit between I and F': I is at Pos 4.
Gap of 2 ACW: skip 5, 6. F would be at Pos 7.
Gap of 2 CW: skip 3, 2. F would be at Pos 1. But G is already at 1.
Therefore, F MUST be at Position 7.
Layout so far (ACW): 1-G, 2-empty, 3-empty, 4-I, 5-V, 6-empty, 7-F.
5) 'H sits immediate left of W': Left is CW. This means moving clockwise from W, you hit H. So in anti-clockwise (Right) order, the sequence is H $\rightarrow$ W.
The empty spots are 2, 3, and 6. For H $\rightarrow$ W to fit consecutively, they must be at 2 and 3.
So H = 2, W = 3.
6) The only person left is X, who takes the last spot: X = 6.
Final Arrangement (Pos 1 to 7 ACW): 1-G, 2-H, 3-W, 4-I, 5-V, 6-X, 7-F.
7) Question asks: 'Who sits 4th to the left of X?'
Left is CW. Start at X(6) and move 4 positions CW: X(6) $\rightarrow$ V(5) $\rightarrow$ I(4) $\rightarrow$ W(3) $\rightarrow$ H(2).
The person is H.

• Exam Trick: For center-facing, 'Left' is Clockwise. "H sits immediate left of W" means W is on the right (ACW) of H. The biggest mistake is placing W at 2 and H at 3.

• Final Answer: H

• Key Fact: Francisco de Almeida, the first Governor and Viceroy of the Portuguese State of India, implemented the "Blue Water Policy" (the Cartaze system).

• Supporting Detail: The policy aimed to establish Portuguese mastery over the Indian Ocean rather than building terrestrial fortresses. He firmly believed that whoever controlled the sea would ultimately control the trade and wealth of India.

• Why others wrong: Alfonso de Albuquerque (the second governor) reversed this policy, shifting focus to building fortresses and capturing key terrestrial strongholds (like Goa). Vasco da Gama was the explorer who found the sea route to India. Cabral discovered Brazil and commanded the second India armada.

• Exam Trick: Blue Water = Ocean mastery = First Governor = Francisco de Almeida. (Albuquerque = Real founder/Terrestrial forts).

• Key Fact: During the post-reform period of 1991–92 to 1998–99, the 'Basic and Capital Goods' sector experienced a significant decline in its average annual growth rate according to the Index of Industrial Production (IIP).

• Supporting Detail: This structural decline was largely due to sudden exposure to cheap foreign imports following aggressive trade liberalisation and tariff reductions, coupled with a lack of domestic demand and the structural adjustment challenges faced by heavy industries during the initial phase of India's economic reforms.

• Why others wrong: Manufacturing, electricity, and mining did not show as stark and clear a structural decline in growth during this specific transition window compared to the heavy capital goods sector, which bore the immediate brunt of tariff reductions.

• Exam Trick: 1991 Reforms $\rightarrow$ Import Tariff cuts $ ightarrow$ Cheaper imported machines flooded the market $ ightarrow$ Decline in domestic 'Capital Goods' sector growth.

• Given: Discrete frequency distribution data. Marks ($x_i$) = {11, 34, 51, 62, 98}. No. of Students ($f_i$) = {57, 23, 95, 14, 96}. Find Mean.

• Basic Method (Solution):
Step 1: The formula for mean is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
Step 2: Calculate the total number of students ($\sum f_i$).
$\sum f_i = 57 + 23 + 95 + 14 + 96 = 285$.
Step 3: Calculate the product $f_i x_i$ for each term:
$11 \times 57 = 627$
$34 \times 23 = 782$
$51 \times 95 = 4845$
$62 \times 14 = 868$
$98 \times 96 = (100 - 2) \times 96 = 9600 - 192 = 9408$
Step 4: Sum up the products ($\sum f_i x_i$).
$627 + 782 + 4845 + 868 + 9408 = 16530$.
Step 5: Calculate the Mean.
$Mean = \frac{16530}{285}$.
Divide numerator and denominator by 5: $\frac{3306}{57}$.
Divide by 3: $\frac{1102}{19}$.
$1102 \div 19 = 58$.

• Shortcut Method: Use an Assumed Mean ($A = 50$).
Deviations ($d_i = x_i - A$): -39, -16, +1, +12, +48.
Calculate $f_i d_i$: $57(-39) = -2223$; $23(-16) = -368$; $95(1) = 95$; $14(12) = 168$; $96(48) = 4608$.
Sum of $f_i d_i = -2223 - 368 + 95 + 168 + 4608 = -2591 + 4871 = 2280$.
Mean = $A + \frac{\sum f_i d_i}{\sum f_i} = 50 + \frac{2280}{285} = 50 + 8 = 58$. This avoids multiplying 98 by 96 entirely.

• Final Answer: 58

• Key Fact: The ICAR-Central Marine Fisheries Research Institute (CMFRI), headquartered in Kochi, hosted the crucial national stakeholder consultation on the High Seas Treaty (BBNJ - Biodiversity Beyond National Jurisdiction) in December 2025.

• Supporting Detail: The consultation aimed to gather inputs from marine scientists, policymakers, and the fishing industry to shape India's stance and regulatory frameworks regarding the conservation of marine biodiversity in international waters.

• Why others wrong: While MoEFCC and ICAR are overarching government bodies involved in policy formulation, the actual physical hosting, organization, and technical facilitation of this specific marine consultation in Kochi was done by CMFRI.

• Exam Trick: Marine research + Kochi location = CMFRI.

• Key Fact: In the National Institutional Ranking Framework (NIRF) 2025, the Indian Institute of Technology (IIT) Madras continued its dominant streak by securing the top 'Overall' position in India.

• Supporting Detail: IIT Madras has consistently held the No. 1 spot in the "Overall" category for several consecutive years, owing to its consistent excellence in teaching, intensive research, and superior graduation outcomes compared to other institutions.

• Why others wrong: IISc Bengaluru typically tops the "University" and "Research" categories. AIIMS Delhi tops the "Medical" category. IIT Bombay is usually in the top 5 but behind IIT Madras in the overall composite score.

• Exam Trick: NIRF 'Overall' Category Winner = IIT Madras (it has held this spot consistently for years).

• Given: A sequence: 11, 13, 15, ..., 71. This is an Arithmetic Progression (AP) with first term $a = 11$, common difference $d = 2$, and last term $l = 71$. Standard deviation = $m$. Find integer part of $m$.

• Basic Method (Solution):
Step 1: Find the number of terms ($n$).
Using the formula $l = a + (n-1)d$:
$71 = 11 + (n-1)2 \implies 60 = 2(n-1) \implies n-1 = 30 \implies n = 31$.
Step 2: The standard deviation (SD) of an arithmetic progression with common difference $d$ and $n$ terms is given by the formula: $SD = |d| imes \sqrt{\frac{n^2 - 1}{12}}$.
Step 3: Substitute $d = 2$ and $n = 31$ into the formula.
$m = 2 \times \sqrt{\frac{31^2 - 1}{12}} = 2 \times \sqrt{\frac{961 - 1}{12}} = 2 \times \sqrt{\frac{960}{12}}$.
Step 4: Simplify the term inside the square root.
$m = 2 \times \sqrt{80} = 2 \times \sqrt{16 \times 5} = 2 \times 4\sqrt{5} = 8\sqrt{5}$.
Step 5: Find the approximate decimal value to get the integral part.
We know $\sqrt{5} \approx 2.236$.
$m = 8 \times 2.236 = 17.888$.
Step 6: The integral part (the integer before the decimal) is 17.

• Shortcut Method: $SD = 2 \times \sqrt{(31^2-1)/12} = 2 \times \sqrt{960/12} = 2 \sqrt{80} = \sqrt{320}$. We know $17^2 = 289$ and $18^2 = 324$. Since 320 is firmly between 289 and 324, $\sqrt{320}$ is "17 point something". Therefore, the integer part is exactly 17.

• Final Answer: 17

• Given: Length of train 1 ($L_1$) = 120 m. Speed of train 1 ($V_1$) = 60 km/hr. Speed of train 2 ($V_2$) = 24 km/hr. Direction = Same. Time to cross = 30 seconds. Find Length of train 2 ($L_2$).

• Basic Method (Solution):
Step 1: Calculate the relative speed. Since they are travelling in the same direction, relative speed is the difference between their individual speeds.
Relative Speed = $V_1 - V_2 = 60 - 24 = 36 \text{ km/hr}$.
Step 2: Convert relative speed from km/hr to m/s by multiplying by $\frac{5}{18}$.
Relative Speed in m/s $= 36 \times \frac{5}{18} = 2 \times 5 = 10 \text{ m/s}$.
Step 3: The total distance covered when two trains cross each other is the sum of their lengths ($L_1 + L_2$).
Distance = Speed $\times$ Time.
$L_1 + L_2 = 10 \text{ m/s} \times 30 \text{ seconds}$.
$120 + L_2 = 300$.
Step 4: Solve for $L_2$.
$L_2 = 300 - 120 = 180 \text{ m}$.

• Shortcut Method: Relative speed = $36 \text{ km/h} = 10 \text{ m/s}$. In 30 sec, total distance covered = $10 \times 30 = 300$ m. This is the total length of both trains. Second train length = $300 - 120 = 180$ m. Calculation can be done in mind in 10 seconds.

• Final Answer: 180