RRB NTPC Graduate Level (18th March 2026 Shift 2) — RRB NTPC 2025

• Given: Let the number be $x$. $20\%$ of $x = 20\%$ of $620 + 170$.

• Basic Method (Solution):
Step 1: Calculate $20\%$ of $620$:
$$20\% \text{ of } 620 = \frac{20}{100} \times 620 = 124$$
Step 2: Formulate the equation:
$$0.20x = 124 + 170$$
$$0.20x = 294$$
Step 3: Solve for $x$:
$$x = \frac{294}{0.20} = \frac{2940}{2} = 1470$$

• Shortcut Method: $20\%$ of $(x - 620) = 170 \implies \frac{1}{5}(x - 620) = 170 \implies x - 620 = 850 \implies x = 850 + 620 = 1470$.

• Final Answer: 1470

• Given conditions: 7 people (P, Q, R, S, T, U, V) have exams from Monday to Sunday.

• Deduction process:
1. P is on Wednesday (Day 3).
2. Only 3 people after R $\implies$ R is on Thursday (Day 4).
3. S is after Q but before P $\implies$ Q must be Monday (Day 1) and S must be Tuesday (Day 2).
4. Remaining days are Friday (Day 5), Saturday (Day 6), Sunday (Day 7). Remaining people: T, U, V.
5. U is before V, but U is NOT on Friday. Thus, U must be on Saturday (Day 6) and V must be on Sunday (Day 7).
6. The only remaining slot for T is Friday (Day 5).

• Full Schedule: Mon: Q, Tue: S, Wed: P, Thu: R, Fri: T, Sat: U, Sun: V.

• The people between T (Friday) and Q (Monday) are S, P, and R $\implies$ 3 people.

• Exam Trick: Fill fixed positions immediately in a 1-7 numbered list. P=3, R=4. Q and S fitting before P strictly means Q=1, S=2.

• Final Answer: Three.

• Given conditions: The original number is 4568923.

• Deduction process:
1. Original order: 4, 5, 6, 8, 9, 2, 3
2. Ascending order: 2, 3, 4, 5, 6, 8, 9
3. Compare the positions:
- 1st: 4 vs 2 (Changed)
- 2nd: 5 vs 3 (Changed)
- 3rd: 6 vs 4 (Changed)
- 4th: 8 vs 5 (Changed)
- 5th: 9 vs 6 (Changed)
- 6th: 2 vs 8 (Changed)
- 7th: 3 vs 9 (Changed)

• Final Answer: None of the digits retained their original position.

• Given: Total numbers $n = 80$, Initial Mean $= 49$. Incorrect values $= 69, 74$. Correct values $= 92, 83$.

• Basic Method (Solution):
Step 1: Calculate the incorrect total sum $= 80 \times 49 = 3920$.
Step 2: Remove the incorrect numbers: $3920 - (69 + 74) = 3920 - 143 = 3777$.
Step 3: Add the correct numbers: $3777 + (92 + 83) = 3777 + 175 = 3952$.
Step 4: Find the new correct mean $= \frac{3952}{80} = 49.4$.

• Shortcut Method: New Mean $=$ Old Mean $+$ $\frac{\text{Correct values} - \text{Incorrect values}}{\text{Total numbers}}$
Change in sum $= (92 + 83) - (69 + 74) = 175 - 143 = +32$.
Change in mean $= \frac{+32}{80} = +0.4$.
Correct Mean $= 49 + 0.4 = 49.4$.

• Final Answer: 49.4

• Key Fact: Unveiled by Indian scientists at leading institutions in 2025, the OncolMark AI-powered cancer diagnostic model represents a major breakthrough in precision oncology, utilizing deep learning algorithms to analyze cancer patterns at the molecular level for more accurate classification and targeted therapy planning.

• Supporting Detail: This model bridges the gap between conventional biopsy imaging and genomic sequencing.

• Why others wrong: GeneMap, BioScan, and Tumour Track are fabricated or general names that do not refer to the specific 2025 Indian scientific milestone asked in the question.

• Additional Info: The integration of AI in molecular oncology significantly reduces the time taken to profile tumors compared to traditional manual pathological analysis.

• Key Fact: Hibernate mode uses the absolute least amount of electricity while saving the current system state. It saves all open documents and running applications from RAM to the hard drive (in a hibernation file) and then completely powers off the computer.

• Supporting Detail: When turned back on, it restores the session exactly as it was left.

• Related Concept: Sleep (or Standby) mode keeps the session in RAM and continues to draw a small amount of power to keep the memory refreshed.

• Why others wrong: Sleep and Standby consume low power but still draw electricity for RAM. Active mode consumes full power.

• Exam Trick: Zero power + Saved session = Hibernate. Low power + Saved session in RAM = Sleep.

• Additional Info: Hibernation was originally designed primarily for laptops to prevent data loss when battery levels drop critically low.

• Given: Total frequency $\sum f_i = 64$ and Mean $\bar{X} = 55$.

• Basic Method (Solution):
Step 1: Total frequency equation:
$$4 + 19 + x + 5 + y = 64 \implies 28 + x + y = 64 \implies x + y = 36 \quad \text{--- (Eq 1)}$$
Step 2: Find Class Marks ($m_i$) and calculate $\sum f_i m_i$:
- For $0-20$: $m_1 = 10 \implies f_1m_1 = 10 \times 4 = 40$
- For $20-40$: $m_2 = 30 \implies f_2m_2 = 30 \times 19 = 570$
- For $40-60$: $m_3 = 50 \implies f_3m_3 = 50x$
- For $60-80$: $m_4 = 70 \implies f_4m_4 = 70 \times 5 = 350$
- For $80-100$: $m_5 = 90 \implies f_5m_5 = 90y$
Sum $\sum f_i m_i = 40 + 570 + 50x + 350 + 90y = 960 + 50x + 90y$.
Step 3: Use the Mean formula:
$$\text{Mean} = \frac{\sum f_i m_i}{\sum f_i} \implies 55 = \frac{960 + 50x + 90y}{64}$$
$$3520 = 960 + 50x + 90y \implies 50x + 90y = 2560 \implies 5x + 9y = 256 \quad \text{--- (Eq 2)}$$
Step 4: Solve Eq 1 and Eq 2 simultaneously. Multiply Eq 1 by 5:
$$5x + 5y = 180$$
Subtract from Eq 2:
$$(5x + 9y) - (5x + 5y) = 256 - 180 \implies 4y = 76 \implies y = 19$$
Substitute $y = 19$ into Eq 1: $x + 19 = 36 \implies x = 17$.
Step 5: Calculate $(7x + 5y)$:
$$7(17) + 5(19) = 119 + 95 = 214$$

• Shortcut Method: Use step-deviation method with assumed mean $A = 50$, $h=20$, but solving the 2 linear equations derived from standard mean formula is equally direct.

• Final Answer: 214

• Given: Group A ($n_1 = 48, A_1 = 477$), Group B ($n_2 = 46, A_2 = 236$), Group C ($n_3 = 36, A_3 = 348$).

• Basic Method (Solution):
Step 1: Calculate total amount spent by each group.
Sum A $= 48 \times 477 = 22,896$
Sum B $= 46 \times 236 = 10,856$
Sum C $= 36 \times 348 = 12,528$
Step 2: Total amount spent by all members.
Total Sum $= 22,896 + 10,856 + 12,528 = 46,280$
Step 3: Calculate the total number of members.
Total Members $= 48 + 46 + 36 = 130$
Step 4: Find the overall average.
$$\text{Overall Average} = \frac{46,280}{130} = 356$$

• Shortcut Method: Use the weighted average formula directly $\frac{n_1A_1 + n_2A_2 + n_3A_3}{n_1 + n_2 + n_3}$. To simplify, take a common factor from members if possible. Here members are $48, 46, 36$. Ratio is $24 : 23 : 18$. The sum of weights is $65$.
$$\text{Average} = \frac{24 \times 477 + 23 \times 236 + 18 \times 348}{65} = \frac{11448 + 5428 + 6264}{65} = \frac{23140}{65} = 356$$

• Final Answer: 356

• Key Fact: A Minister of State (MoS) with Independent Charge holds full responsibility and heads a specific smaller ministry or department directly without being attached to or reporting to a Cabinet Minister.

• Supporting Detail: In contrast, a regular Minister of State (without independent charge) assists a Cabinet Minister in managing a larger ministry.

• Related Concept: Cabinet Ministers form the core decision-making body of the government and attend all cabinet meetings. An MoS with Independent Charge is not a member of the Cabinet and attends Cabinet meetings only when matters related to their specific ministry are on the agenda.

• Why others wrong: Options A, C, and D are factually incorrect because an MoS with independent charge does exercise significant powers over their assigned portfolio and holds very specific responsibilities.

• Exam Trick: 'Independent Charge' literally means 'Independently heading a department without a Cabinet boss'.

• Final Answer: A Minister of State with independent charge may head a ministry without being under a Cabinet Minister.

• Given: Total Amount $= ₹1407$. Interest rate $R = 10\%$. Time for Veer $= 14\text{ years}$, Time for Nitin $= 15\text{ years}$. Final amounts are equal.

• Basic Method (Solution):
Step 1: Let the share of Veer be $V$ and Nitin be $N$. Then $V + N = 1407$.
Step 2: According to the formula for compound interest Amount $= P\left(1 + \frac{R}{100}\right)^T$:
$$V \left(1 + \frac{10}{100}\right)^{14} = N \left(1 + \frac{10}{100}\right)^{15}$$
Step 3: Divide both sides by $(1.1)^{14}$:
$$V = N \left(1 + \frac{10}{100}\right)^1 = N \left(\frac{11}{10}\right) \implies \frac{V}{N} = \frac{11}{10}$$
Step 4: The ratio of their shares is $V : N = 11 : 10$. Total parts $= 11 + 10 = 21$.
Step 5: Nitin's share $N = \frac{10}{21} \times 1407 = 10 \times 67 = 670$.

• Shortcut Method: Whenever amounts are equal at CI and there is a time gap of $t$ years, the ratio of their principals is $\left(1 + \frac{R}{100}\right)^t$. Here the time gap is $15 - 14 = 1\text{ year}$. Ratio $= \frac{110}{100} = \frac{11}{10}$. Since Nitin gets more time to accumulate interest, his initial principal must be smaller. So Nitin gets the 10-part share: $\frac{10}{21} \times 1407 = 670$.

• Final Answer: 670