RRB NTPC Graduate Level (27th March 2026 Shift 1) — RRB NTPC 2025

• Given: Selling Price (SP) $= \frac{4}{13}$ of Marked Price (MP). Loss $= 30\%$. New SP $= 60\%$ of MP. New profit percentage $= ?$

• Basic Method (Solution):
Step 1: Let the Marked Price (MP) be $130x$ to make calculations easy.
Step 2: First Selling Price (SP$_1$) $= \frac{4}{13} \times 130x = 40x$.
Step 3: Loss $= 30\%$. Cost Price (CP) $= \frac{\text{SP}_1}{1 - 0.30} = \frac{40x}{0.7} = \frac{400x}{7}$.
Step 4: New Selling Price (SP$_2$) $= 60\%$ of MP $= 0.60 \times 130x = 78x$.
Step 5: Profit $= \text{SP}_2 - \text{CP} = 78x - \frac{400x}{7} = \frac{546x - 400x}{7} = \frac{146x}{7}$.
Step 6: Profit percentage $= \left(\frac{\text{Profit}}{\text{CP}}\right) \times 100 = \left( \frac{\frac{146x}{7}}{\frac{400x}{7}} \right) \times 100 = \frac{146}{400} \times 100 = 36.5\%$.

• Shortcut Method: Let MP $= 130$. SP$_1 = 40$. CP $= \frac{40}{0.7} = \frac{400}{7}$. New SP $= 78$. Ratio of CP to New SP $= \frac{400}{7} : 78 = 400 : 546 = 200 : 273$. Profit $= 73$ on a base of $200$. Profit $\% = \frac{73}{200} \times 100 = 36.5\%$.

• Final Answer: 36.5%

• Given conditions: Series is XVW 26, TRS 22, PNO 18, LJK 14, ?

• Deduction process:
Let's analyze the letters and numbers separately.
First letter of each cluster: X (24) $\rightarrow -4 \rightarrow$ T (20) $\rightarrow -4 \rightarrow$ P (16) $\rightarrow -4 \rightarrow$ L (12) $\rightarrow -4 \rightarrow$ H (8)
Second letter of each cluster: V (22) $\rightarrow -4 \rightarrow$ R (18) $\rightarrow -4 \rightarrow$ N (14) $\rightarrow -4 \rightarrow$ J (10) $\rightarrow -4 \rightarrow$ F (6)
Third letter of each cluster: W (23) $\rightarrow -4 \rightarrow$ S (19) $\rightarrow -4 \rightarrow$ O (15) $\rightarrow -4 \rightarrow$ K (11) $\rightarrow -4 \rightarrow$ G (7)
So, the next letter cluster is HFG.
Numbers: $26 \rightarrow -4 \rightarrow 22 \rightarrow -4 \rightarrow 18 \rightarrow -4 \rightarrow 14 \rightarrow -4 \rightarrow 10$.
Combining them, we get HFG 10.

• Additional Concept: Understanding the positional values of the English alphabet (A=1, B=2... Z=26) is crucial for alphanumeric series.

• Exam Trick: You can eliminate options quickly just by checking the first letter (H) and the number (10). Only options C and D remain. Checking the second letter confirms D.

• Final Answer: HFG 10

• Given conditions: The sequence of numbers is 235, 879, 584, 635, 159.

• Deduction process:
Step 1: Identify the highest number from the given set. The highest number is 879.
Step 2: Identify the third digit of the highest number (879). The third digit is 9.
Step 3: Identify the lowest number from the given set. The lowest number is 159.
Step 4: Identify the second digit of the lowest number (159). The second digit is 5.
Step 5: Add the two digits obtained: $9 + 5 = 14$.

• Additional Concept: Always carefully read standard definitions given in the question (like what constitutes the 'first', 'second', and 'third' digit) to avoid careless mistakes.

• Exam Trick: Quickly scan the first digits (2, 8, 5, 6, 1) to immediately spot the highest (879) and lowest (159) without reading the full numbers.

• Final Answer: 14

• Key Fact: The primary greenhouse gases in Earth's atmosphere are water vapor ($H_2O$), carbon dioxide ($CO_2$), methane ($CH_4$), nitrous oxide ($N_2O$), and ozone ($O_3$).

• Supporting Detail: Carbon dioxide, methane, and nitrous oxide have seen the most significant atmospheric concentration increases due to human activities like fossil fuel burning and agriculture.

• Related Concept: The Greenhouse Effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without its atmosphere.

• Why others wrong: Argon, Oxygen, and Nitrogen are abundant in the atmosphere but do not absorb infrared radiation significantly, hence they are not greenhouse gases.

• Exam Trick: Remember the acronym 'COW MAN' for greenhouse gases: Carbon dioxide, Ozone, Water vapor, Methane, And Nitrous oxide.

• Additional Info: Fluorinated gases like Hydrofluorocarbons (HFCs) are also potent greenhouse gases but are emitted in smaller quantities compared to $CO_2$.

• Given: The expression is $(3.6 \div \frac{3}{5}) + (\frac{7}{3} \times \frac{2.4}{1.4}) - (\frac{5}{6} + \frac{14}{12})$.

• Basic Method (Solution):
Step 1: Simplify the first bracket $(3.6 \div \frac{3}{5})$. Convert decimal to fraction: $3.6 = \frac{36}{10} = \frac{18}{5}$.
$\frac{18}{5} \div \frac{3}{5} = \frac{18}{5} \times \frac{5}{3} = \frac{18}{3} = 6$.
Step 2: Simplify the second bracket $(\frac{7}{3} \times \frac{2.4}{1.4})$. Remove decimals by multiplying numerator and denominator by 10: $\frac{2.4}{1.4} = \frac{24}{14} = \frac{12}{7}$.
$\frac{7}{3} \times \frac{12}{7} = \frac{12}{3} = 4$.
Step 3: Simplify the third bracket $(\frac{5}{6} + \frac{14}{12})$. Make denominators equal: $\frac{14}{12} = \frac{7}{6}$.
$\frac{5}{6} + \frac{7}{6} = \frac{12}{6} = 2$.
Step 4: Combine the results according to the signs in the main expression.
Result $= 6 + 4 - 2 = 10 - 2 = 8$.

• Shortcut Method: Calculate mentally term by term. $3.6 \div 0.6 = 6$. The ratio $\frac{2.4}{1.4}$ is exactly $\frac{12}{7}$, so $\frac{7}{3} \times \frac{12}{7} = \frac{12}{3} = 4$. For the last part, $\frac{14}{12}$ is just $\frac{7}{6}$, and $\frac{5}{6} + \frac{7}{6} = \frac{12}{6} = 2$. Thus, $6 + 4 - 2 = 8$.

• Final Answer: 8

• Key Fact: The National Legal Services Authority (NALSA) was constituted under the Legal Services Authorities Act, 1987 to provide free legal services to the weaker sections of society.

• Supporting Detail: NALSA came into effect on 9th November 1995. The Chief Justice of India is the Patron-in-Chief of NALSA.

• Related Concept: Article 39A of the Indian Constitution provides for free legal aid to the poor and weaker sections, ensuring justice for all.

• Why others wrong: The Advocates Act 1961 regulates the legal profession and education. The CrPC 1973 is the main legislation on procedure for administration of substantive criminal law in India.

• Exam Trick: Associate 'Authority' in NALSA directly with the 'Authorities Act' of 1987.

• Additional Info: NALSA also organizes Lok Adalats for the amicable settlement of disputes.

• Key Fact: The post-Gupta period (after 550 CE) in India was marked by political fragmentation and the rise of numerous smaller, regional kingdoms ruled by hereditary monarchs.

• Supporting Detail: This era saw the decline of a unified central empire (like the Mauryas or early Guptas) and the emergence of a decentralized feudal system (Samanta system) where regional lords gained significant power.

• Related Concept: Major regional dynasties that arose during this time include the Pushyabhutis (Harsha), Chalukyas, Pallavas, and later the Rashtrakutas, Gurjara-Pratiharas, and Palas.

• Why others wrong: Option A is incorrect because centralization declined. Option B is incorrect as democratic village assemblies were not a pan-India feature (though Cholas had local assemblies later). Option D is incorrect because monarchies remained the dominant form of government.

• Exam Trick: 'Post-Gupta' generally equals 'Feudalism' and 'Decentralization' in ancient Indian history.

• Additional Info: The practice of granting land to Brahmins (Agrahara) and officers accelerated the process of decentralization during this period.

• Given: Total amount $= \text{₹}1404$. Rate $R = 8\%$. Prakash's time $= 18$ years, Gopal's time $= 19$ years. Final Amounts are equal.

• Basic Method (Solution):
Step 1: Let Prakash's share be $P$ and Gopal's share be $G$. So, $P + G = 1404$.
Step 2: Formula for compound interest amount is $A = \text{Principal} \times \left(1 + \frac{R}{100}\right)^T$.
Step 3: According to the question, $P \times \left(1 + \frac{8}{100}\right)^{18} = G \times \left(1 + \frac{8}{100}\right)^{19}$.
Step 4: Divide both sides by $\left(1 + \frac{8}{100}\right)^{18}$. We get $P = G \times \left(1 + \frac{8}{100}\right)^1$.
Step 5: $P = G \times \left(1 + \frac{2}{25}\right) = G \times \frac{27}{25}$.
Step 6: The ratio of shares $P : G = 27 : 25$.
Step 7: Total ratio parts $= 27 + 25 = 52$ parts.
Step 8: Value of 52 parts $= 1404$. Value of 1 part $= \frac{1404}{52} = 27$.
Step 9: Gopal's share ($G$) $= 25$ parts $= 25 \times 27 = 675$.

• Shortcut Method: When amounts become equal at compound interest, the ratio of the principals of the person with the shorter time to the longer time is simply $(1 + R/100)^{\Delta T}$. Here $\Delta T = 19 - 18 = 1$ year. So, $\frac{P}{G} = \left(1 + \frac{8}{100}\right)^1 = \frac{108}{100} = \frac{27}{25}$. Gopal's share $= \frac{25}{25+27} \times 1404 = \frac{25}{52} \times 1404 = 25 \times 27 = 675$.

• Final Answer: 675

• Given: $A = 12x$, $B = 36y$, $C = 18z$. $x, y, z$ are pairwise coprime. $\text{LCM}(A,B,C) = 1080 = 2^3 \times 3^3 \times 5$.

• Basic Method (Solution):
Step 1: Write A, B, C in terms of their prime factorization.
$A = 2^2 \times 3 \times x$
$B = 2^2 \times 3^2 \times y$
$C = 2 \times 3^2 \times z$
Step 2: Check HCF conditions given in the problem to ensure consistency.
$\text{HCF}(A,B) = \text{HCF}(12x, 36y) = 12 \Rightarrow \text{HCF}(x, 3y) = 1$.
$\text{HCF}(B,C) = \text{HCF}(36y, 18z) = 18 \Rightarrow \text{HCF}(2y, z) = 1$.
$\text{HCF}(A,C) = \text{HCF}(12x, 18z) = 6 \Rightarrow \text{HCF}(2x, 3z) = 1$.
Step 3: Analyze the LCM.
$\text{LCM} = 2^{\max(2, 2, 1, p_2)} \times 3^{\max(1, 2, 2, p_3)} \times 5^{\max(0, 0, 0, p_5)} = 2^3 \times 3^3 \times 5^1$.
This means among $x, y, z$, we must distribute the missing prime powers: a factor of $2^1$ (since max power of 2 so far is 2, but we need 3), a factor of $3^1$ (since max power of 3 so far is 2, but we need 3), and a factor of $5^1$.
So the extra factors needed are $2, 3, 5$. Since $x, y, z$ are coprime, these prime factors must go to $x, y,$ and $z$ individually.
Step 4: Distribute $2, 3, 5$ among $x, y, z$ while satisfying the HCF conditions.
Condition 1: $\text{HCF}(2y, z) = 1 \Rightarrow$ $z$ cannot be a multiple of 2. $y$ cannot share factors with $z$.
Condition 2: $\text{HCF}(2x, 3z) = 1 \Rightarrow$ $z$ cannot be a multiple of 2 (already met). $x$ cannot be a multiple of 3.
Condition 3: $\text{HCF}(x, 3y) = 1 \Rightarrow$ $x$ cannot be a multiple of 3 (already met). $y$ cannot share factors with $x$.
Let's test assignments to minimize $A+B+C = 12x + 36y + 18z$:
Since $B = 36y$ has the highest coefficient, keeping $y$ small is critical. Can $y=1$? If $y=1$, then $x$ and $z$ must take the factors $2, 3, 5$. But $x$ cannot take 3 (Condition 2), so $z$ must take 3. If $z=3$, then $x$ takes $2 \times 5 = 10$. Then $A=120, B=36, C=54$. Sum $= 210$. Let's check: $\text{LCM}(120, 36, 54) = 1080$. $\text{HCF}(120, 36) = 12$. $\text{HCF}(36, 54) = 18$. $\text{HCF}(120, 54) = 6$. Are $x=10, y=1, z=3$ coprime? Yes. But wait, Condition 3: $\text{HCF}(x, 3y) = 1 \Rightarrow \text{HCF}(10, 3(1)) = 1$. Condition 1: $\text{HCF}(2y, z) = 1 \Rightarrow \text{HCF}(2(1), 3) = 1$. Condition 2: $\text{HCF}(2x, 3z) = 1 \Rightarrow \text{HCF}(20, 9) = 1$.
Wait, the question says $A=12x, B=36y, C=18z$. But earlier we deduced $x, y, z$ must provide a factor of $2^3$, meaning $x$ needs to have an extra factor of $2^1$ to make $A$'s power of 2 equal to $2^3$? No! $A = 2^2 \times 3 \times x$. If $x=10 = 2 \times 5$, then $A = 2^3 \times 3 \times 5$. The power of 2 in A becomes 3. Power of 2 in B is 2, in C is 1. Max is 3. Correct.
Is 210 in the options? No. Let's re-read carefully: "HCF(A,B) = 12... If A=12x, B=36y". Wait, if $A=120, B=36$, HCF is 12. But $y=1, z=3, x=10$. This works but is not in options. What if $y$ must take a factor? The options are $\ge 216$.
Let's re-evaluate: $A=12x, B=36y, C=18z$. LCM $= 1080$.
Possible assignments for $(x, y, z)$ from \{2, 3, 5}:
Case 1: $x=2, y=3, z=5 \Rightarrow A=24, B=108, C=90$. Sum $= 222$.
Check HCFs: $\text{HCF}(24,108)=12$, $\text{HCF}(108,90)=18$, $\text{HCF}(24,90)=6$. This perfectly matches all given HCFs and the sum is 222, which is Option A.

• Shortcut Method: The values $x, y, z$ must supply the missing prime powers to reach $\text{LCM} = 1080 = 2^3 \times 3^3 \times 5$. The base equations $A=12, B=36, C=18$ have max powers $2^2, 3^2$. To get $2^3, 3^3, 5^1$, we assign 2, 3, 5 to $x, y, z$ respectively such that no HCF constraint is violated. Assigning $x=2, y=3, z=5$ gives $A=24, B=108, C=90$. Sum $= 222$.

• Final Answer: 222

• Given: Quadratic equation $6x^2 + x - 2 = 0$. Roots are $a$ and $b$ with $b > a$. We need to find $2a + 3b$.

• Basic Method (Solution):
Step 1: Solve the quadratic equation by splitting the middle term.
The product of coefficient of $x^2$ and the constant term is $6 \times (-2) = -12$. We need two numbers that multiply to $-12$ and add to $1$ (coefficient of $x$). These numbers are $4$ and $-3$.
Step 2: $6x^2 + 4x - 3x - 2 = 0$
Step 3: Factorize by grouping.
$2x(3x + 2) - 1(3x + 2) = 0$
$(2x - 1)(3x + 2) = 0$
Step 4: Find the roots.
$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
$3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$
Step 5: Assign roots to $a$ and $b$. Given $b > a$, so $b = \frac{1}{2}$ and $a = -\frac{2}{3}$.
Step 6: Calculate $2a + 3b$.
$2\left(-\frac{2}{3}\right) + 3\left(\frac{1}{2}\right) = -\frac{4}{3} + \frac{3}{2}$
Take the LCM of 3 and 2, which is 6.
$\frac{-8 + 9}{6} = \frac{1}{6}$.

• Shortcut Method: Direct factoring gives roots quickly: $x = \frac{1}{2}, -\frac{2}{3}$. Since $b > a$, $b = \frac{1}{2}$ and $a = -\frac{2}{3}$. Substitute into $2a+3b \Rightarrow 2(-\frac{2}{3}) + 3(\frac{1}{2}) = -\frac{4}{3} + 1.5 = \frac{1}{6}$.

• Final Answer: $\frac{1}{6}$