RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Take capacity $=\text{LCM}(44,84,28)=924$ units. A pipe that empties the tank lowers the combined rate, so with the outlet open the tank must fill slower than with the two inlets alone.

1. Inlet 1: $\frac{924}{44}=21$ units/min; inlet 2: $\frac{924}{84}=11$ units/min; together $=32$ units/min.
2. All three pipes: $\frac{924}{28}=33$ units/min.
3. Outlet rate $=32-33=-1$ units/min. The negative value means the third pipe would be adding water, not draining it.
4. So an emptying pipe cannot give a filling time of 28 min; the time asked has no valid value and none of 916, 921, 920, 924 min is correct.
5. If the third pipe were also a filling pipe, it alone would fill the tank in $\frac{924}{1}=924$ min.

Note: Quick check: two inlets alone take $\frac{924}{32}$ min, which is more than 28 min, so the third pipe cannot be an outlet. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

Concept: Speed $=\frac{\text{Distance}}{\text{Time}}$ for the remaining part.

1. Distance covered $=\frac{2}{3}\times 222=148$ km in $\frac{3}{7}\times 14=6$ h
2. Remaining: $222-148=74$ km in $14-6=8$ h
3. Speed $=\frac{74}{8}=9.25$ km/h

Trick: Remaining distance $=\frac{1}{3}$, remaining time $=\frac{4}{7}$ of total.

Answer: (C) 9.25

Concept: A number divisible by both 27 and 19 (co-prime) is divisible by $27\times 19=513$.

1. Digit sum must be a multiple of 9: 35691 → 24 ✗, 33488 → 26 ✗, 34371 → 18 ✓, 34962 → 24 ✗
2. Check: $34371÷513=67$ exactly ✓

Trick: Use the digit-sum test for 9 to eliminate quickly, then divide only the survivor.

Answer: (C) 34371

Concept: If $a:x::x:b$, then $x^2=ab$ (x is the mean proportional).

1. $A^2=10\times 40=400$
2. $A=\sqrt{400}=20$ (positive value)

Trick: Mean proportional $=\sqrt{ab}$.

Answer: (D) 20

Concept: Follow BODMAS: bracket first, then multiplication, then addition.

1. $70+42=112$
2. $\frac{93}{28}\times 112=93\times 4=372$
3. $372+82=454$

Trick: Look for cancellation: $112=28\times 4$.

Answer: (A) 454

Concept: SP $=\text{CP}\times \frac{100\pm \%}{100}$.

1. At 25% loss, SP $=75\%$ of CP $=112$
2. Required SP $=150\%$ of CP $=112\times \frac{150}{75}=224$

Trick: Ratio method: $\frac{\text{new SP}}{\text{old SP}}=\frac{150}{75}=2$ → $112\times 2$.

Answer: (B) ₹224

Concept: Linear pair on the transversal + angle sum of triangle EGF.

1. $\angle BEF={180}^{\circ }-\angle PEB={180}^{\circ }-{42}^{\circ }={138}^{\circ }$
2. $\angle GEF=\angle BEF-\angle BEG={138}^{\circ }-{21}^{\circ }={117}^{\circ }$
3. In $△EGF$: $\angle EGF={180}^{\circ }-{117}^{\circ }-{34}^{\circ }={29}^{\circ }$

Trick: Shortcut: $\angle EGF=\angle PEB+\angle BEG-\angle GFE={42}^{\circ }+{21}^{\circ }-{34}^{\circ }={29}^{\circ }$.

Answer: (A) ${29}^{\circ }$

Concept: Expenditure $=$ Income $-$ Savings.

1. Incomes $4k$, $3k$; expenditures $4k-15000$ and $3k-9000$.
2. $\frac{4k-15000}{3k-9000}=\frac{16}{17}\Rightarrow 68k-255000=48k-144000\Rightarrow k=5550$
3. Expenditures: $22200-15000=7200$ and $16650-9000=7650$
4. Total $=7200+7650=14850$

Trick: Check: $7200:7650=16:17$ ✓.

Answer: (A) ₹14,850

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. $\text{SI}=\frac{350\times 4\times 2.5}{100}=\frac{3500}{100}=35$

Trick: $4\%\times 2.5=10\%$ of 350 $=35$.

Answer: (D) ₹35

Concept: Greatest number leaving the same remainder = HCF of the differences.

1. Differences: $2636-1272=1364$, $4209-2636=1573$
2. HCF(1364, 1573): $1573-1364=209$; $1364=6\times 209+110$; $209=110+99$; $110=99+11$; $99=9\times 11$ → $x=11$
3. $1272=11\times 115+7$ → $y=7$
4. $3x-14y=33-98=-65$

Trick: Same remainder → use differences, never the numbers themselves.

Answer: (B) $-65$

Concept: Check each condition against the letter group; apply only the one that fits.

1. TAEX: T = $, A = 2, E = @, X = 6 (from table).
2. Condition 1: first letter T is not a vowel – no. Condition 4: T not a vowel – no.
3. Condition 2: 2nd letter A is a vowel and last letter X is a consonant – applies, so A and X both become W.
4. Condition 3: 3rd letter E is a vowel – no. Code = $ W @ W.

Trick: Test conditions quickly by marking vowels (A, E, I, O, U) in the word first.

Answer: (C) $ W @ W

Concept: Arrange the digits in descending order and count from the right end.

1. Digits of 6954381: 6, 9, 5, 4, 3, 8, 1.
2. Descending: 9 8 6 5 4 3 1.
3. From right: 1st = 1, 2nd = 3, 3rd = 4.

Trick: Third from right in descending order = third smallest digit = 4.

Answer: (D) 4

Concept: Find the positional shift of each letter in the given pairs.

1. DTUH → BQWK: D−2=B, T−3=Q, U+2=W, H+3=K.
2. Check: ZNYN → XKAQ: Z−2=X, N−3=K, Y+2=A, N+3=Q ✔.
3. VHCT: V−2=T, H−3=E, C+2=E, T+3=W → TEEW.

Trick: Pattern $-2,-3,+2,+3$; count backward past A wraps to Z.

Answer: (D) TEEW

Concept: Draw the least-overlap Venn diagram satisfying all statements.

1. Some nuts are black → Nuts and Black overlap partly; All black are crows → Black lies inside Crows, which lies inside Pens.
2. I: All nuts are black – the diagram shows nuts outside Black too, so not certain ✗.
3. II: Some black are crows – since every black is a crow, some black are surely crows ✔.

Trick: "All A are B" always gives "Some A are B" (and "Some B are A").

Answer: (A) Only conclusion (II) follows.

Concept: Track the two letters and the number separately.

1. 1st letters: K, J, I, H → next G (−1).
2. 2nd letters: N, M, L, K → next J (−1).
3. Numbers: 33, 41, 49, 57 → next $57+8=65$.

Trick: Both letters move back by 1, number goes up by 8 → GJ65.

Answer: (A) GJ65

Concept: Compare the letters of each pair by their alphabet positions.

1. PQ − NO: P−2=N, Q−2=O.
2. KL − IJ: K−2=I, L−2=J; GH − EF: G−2=E, H−2=F.
3. MH − IP: M−4=I, H+8=P – also the first cluster is not consecutive letters.

Trick: Three pairs are consecutive letters shifted back by 2; MH − IP breaks it.

Answer: (D) MH − IP

Concept: Look at successive differences.

1. $301-1=300,300-3=297,297-5=292,292-7=285$
2. Subtract odd numbers 1, 3, 5, 7, 9 …
3. Next: $285-9=276$

Trick: Also $301-n^2$: $301-0,1,4,9,16,25$ → 276.

Answer: (B) 276

Concept: Find number–NUMBER–number triplets in the series.

1. Number runs in the series: "2", "6", "4 5 3", "1 1", "6 7", "7 4", "6".
2. A middle number with numbers on both sides needs a run of at least 3 numbers.
3. Only "4 5 3" qualifies → the number 5. Count = 1.

Trick: Mark consecutive-number blocks; each block of length $k\ge 3$ gives $k-2$ answers.

Answer: (D) 1

Concept: Find the rule for each part of the cluster.

1. TK 14 → YH 5: T+5=Y, K−3=H, $14-9=5$.
2. Check MN 5 → RK −4: M+5=R, N−3=K, $5-9=-4$ ✔.
3. SV 0 → S+5=X, V−3=S, $0-9=-9$ → XS −9.

Trick: Rule: $+5$, $-3$, $-9$.

Answer: (D) XS −9

Concept: Interchange the signs first, then apply BODMAS.

1. After swapping: $121\times 6-384÷3+31$
2. $=726-128+31$
3. $=629$

Trick: Rewrite the whole expression with new signs before calculating.

Answer: (C) 629