RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Let the numbers be $3x,2x,3x$; total cost = sum of (number × price).

1. Cost $=3x\times 280+2x\times 180+3x\times 120=840x+360x+360x=1560x$.
2. $1560x=6240\Rightarrow x=4$.
3. Articles of type Q $=2x=8$.

Trick: $1560\times 4=6240$ – spotting that $6240$ is $4\times 1560$ gives $x$ instantly.

Answer: (B) 8

Concept: Divide by $x^2$ to get $x^2+\frac{1}{x^2}$, then use ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$ and $x^3-\frac{1}{x^3}=a^3+3a$ where $a=x-\frac{1}{x}$.

1. $x^4-83x^2+1=0\Rightarrow x^2+\frac{1}{x^2}=83$.
2. ${(x-\frac{1}{x})}^2=83-2=81\Rightarrow x-\frac{1}{x}=9$ (taking the value that keeps the root real).
3. $x^3-x^{-3}=9^3+3\times 9=729+27=756$.
4. $\sqrt{756+28}=\sqrt{784}=28$.

Trick: $x^2+\frac{1}{x^2}=83\Rightarrow x-\frac{1}{x}=\sqrt{81}=9$; then $a^3+3a$ – remember ‘minus’ identity has ‘plus $3a$’.

Answer: (C) 28

Concept: For similar triangles, $\frac{A_1}{A_2}={\left(\frac{h_1}{h_2}\right)}^2$ (ratio of areas = square of ratio of corresponding altitudes/sides).

1. $\frac{h_1}{h_2}=\sqrt{\frac{841}{676}}$.
2. $\sqrt{841}=29$, $\sqrt{676}=26$.
3. Ratio $=29:26$.

Trick: Areas → take square root for any length ratio (side, altitude, median, perimeter).

Answer: (B) 29 : 26

Concept: Division and multiplication before addition and subtraction.

1. $18÷6=3$, $3\times 3=9$.
2. $16+3-9=10$.

Trick: Underline every ÷ and × pair first, solve them, then move left to right with + and −.

Answer: (C) 10

Concept: Work backwards: each selling price = cost price × multiplier.

1. R paid ₹2655 = Q’s SP $={\text{CP}}_Q\times 0.5\Rightarrow {\text{CP}}_Q=5310$.
2. Q’s CP = P’s SP $={\text{CP}}_P\times 1.77\Rightarrow {\text{CP}}_P=\frac{5310}{1.77}=3000$.

Trick: Final price $={\text{CP}}_P\times \frac{177}{100}\times \frac{50}{100}={\text{CP}}_P\times 0.885$; $\frac{2655}{0.885}=3000$.

Answer: (A) 3000

Concept: Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$.

1. $a=8\Rightarrow \text{Area}=\frac{\sqrt{3}}{4}\times 64=16\sqrt{3}$.
2. $\sqrt{3}\approx 1.732\Rightarrow 16\times 1.732=27.712$.
3. Correct to 2 decimals: $27.71{\text{cm}}^2$.

Trick: Memorise $\sqrt{3}=1.732$; $16\sqrt{3}=27.71$, $25\sqrt{3}=43.30$ appear often.

Answer: (B) 27.71 sq. cm.

Concept: Simplify the square root first, then work left to right with × and ÷.

1. $\sqrt{2025}=45$.
2. $19\times 7\times 630÷45=19\times 7\times 14=1862$.
3. $y=1862-517=1345$.

Trick: Cancel before multiplying: $\frac{630}{45}=14$, then $133\times 14=1862$.

Answer: (A) 1345

Concept: For 2 years, $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2$.

1. $P\times {\left(\frac{4}{100}\right)}^2=P\times \frac{16}{10000}=52$.
2. $P=\frac{52\times 10000}{16}=32500$.

Trick: Difference for 2 years = SI on the first year’s interest: $4\%$ of $4\%$ of $P$ = $0.16\%$ of $P$.

Answer: (B) 32500

Concept: Express CP in terms of MP from the first sale, then compare with the second SP.

1. Let MP $=100$. First SP $=30$; loss $16\%\Rightarrow \text{CP}=\frac{30}{0.84}=\frac{250}{7}$.
2. Second SP $=69$. Profit $=69-\frac{250}{7}=\frac{233}{7}$.
3. Profit$\%=\frac{233/7}{250/7}\times 100=\frac{233}{250}\times 100=93.2\%$.

Trick: New SP is $\frac{69}{30}=2.3$ times the old SP, so new multiplier $=0.84\times 2.3=1.932\Rightarrow 93.2\%$ profit.

Answer: (D) 93.2%

Concept: Area of 4 walls $=2(l+b)h$; ceiling $=l\times b$.

1. Walls $=2(20+10)\times 6=360{\text{m}}^2$.
2. Ceiling $=20\times 10=200{\text{m}}^2$.
3. Total $=560{\text{m}}^2$; cost $=560\times 15=$ ₹8400.

Trick: ‘Four walls and ceiling’ = total surface area minus floor: $2(lb+bh+hl)-lb=lb+2h(l+b)$.

Answer: (B) ₹8400

Concept: Replace each letter by its symbol and then apply BODMAS (÷ and × before + and −).

1. 13 B 2 C 6 D 8 A 4 becomes $13\times 2+6-8÷4$.
2. Division first: $8÷4=2$; multiplication: $13\times 2=26$.
3. $26+6-2=30$.

Trick: Write the decoded expression fully before calculating – most mistakes come from solving left-to-right instead of BODMAS.

Answer: (C) 30

Concept: Compare the two letters of each pair position-wise using alphabet positions.

1. ZC−AD: Z→A (+1, cyclic), C→D (+1).
2. DG−EH: D→E (+1), G→H (+1). RU−SV: R→S (+1), U→V (+1).
3. OR−OU: O→O (0), R→U (+3) – breaks the +1,+1 pattern.

Trick: Z→A counts as +1 (alphabet wraps around) – don't pick ZC−AD as odd just because it crosses Z.

Answer: (B) OR−OU

Concept: Identify the required numbers first, then pick the asked digits.

1. Numbers: 442, 405, 297, 384, 316. Highest = 442, lowest = 297.
2. Third digit of 442 = 2; second digit of 297 = 9.
3. Sum $=2+9=11$.

Trick: Underline the highest and lowest numbers as soon as you read the list – it saves re-scanning for every sub-question of this set.

Answer: (B) 11

Concept: Find one operation on the whole number that works for both given pairs.

1. $22\times 3-6=66-6=60$ ✓
2. $27\times 3-6=81-6=75$ ✓
3. $33\times 3-6=99-6=93$.

Trick: The second number is a little less than $3\times$ the first (66→60, 81→75, both −6) – test "×3 then −6" immediately.

Answer: (D) 93

Concept: Draw the Venn diagram with the minimum overlap allowed by the statements and test each conclusion.

1. All hands are legs → the "hands" circle lies inside "legs". Some hands are faces → the hands–faces overlap is non-empty, and it also lies inside legs. So some legs are faces – conclusion (I) follows.
2. Some faces are arms – the arms can overlap faces outside hands, so "some hands are arms" is not necessary. Conclusion (II) does not follow.

Trick: Two "Some" statements in a chain (Some hands are faces, Some faces are arms) never give a definite conclusion between the end terms.

Answer: (B) Only conclusion (I) follows

Concept: Treat letters and numbers as separate series.

1. Letters: DHG → CGF → BFE → AED – every letter moves back by 1. Next: A→Z, E→D, D→C = ZDC.
2. Numbers: 42, 38, 34, 30 – decreasing by 4. Next $=26$.
3. Combined: ZDC 26.

Trick: A −1 = Z (cyclic). Only option (A) starts with ZD and has 26.

Answer: (A) ZDC 26

Concept: Number the 7 seats 1–7 from left to right (all face north, so left/right are the usual).

1. Only two people left of C → C at seat 3. Three between C and F → F at seat 7.
2. B second to the right of C → B at seat 5. Two between B and D → D at seat 2 (seat 8 does not exist).
3. Seats 1, 4, 6 remain for G, E, A with G left of E left of A → G 1, E 4, A 6.
4. Left of D (seat 2): only G → 1 person.

Trick: Start with the clue that fixes a seat ("only two to the left of C") – it anchors the whole row.

Answer: (A) 1

Concept: Handle letters and numbers separately.

1. Letters: N(14)→K(11)→H(8)→E(5): each −3 → next B(2). V→S→P→M: −3 → J. A→X→U→R: −3 → O. Cluster = BJO.
2. Numbers: 7, 14, 21, 28 → +7 → 35.
3. Answer: BJO 35.

Trick: Check only the first letter and the number first: E −3 = B and 28 + 7 = 35 → option (C) is the only match.

Answer: (C) BJO 35

Concept: Number positions 1 (bottom) to 7 (top) and place the fixed blocks first.

1. Only K below R → K at 1, R at 2.
2. Only A above E → E at 6, A at 7.
3. Only S between E and T → S at 5, T at 4.
4. Remaining M at 3. Between M (3) and K (1): only R → one box.

Trick: "Only X below/above Y" fixes Y at the second position from that end – use such clues first.

Answer: (C) One

Concept: Put A at the origin, East = +x, North = +y, and track the position after each leg (every left turn rotates the direction anticlockwise).

1. East 38 → (38, 0); left = North 35 → (38, 35); left = West 39 → (−1, 35).
2. Left = South 37 → (−1, −2); left = East 1 → P = (0, −2).
3. P is directly 2 km south of A, so he must drive 2 km to the North.

Trick: Net East–West: $38-39+1=0$; net North–South: $35-37=-2$ → 2 km south of A.

Answer: (D) 2 km to the North