RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Same direction on a circular track: first meeting when the faster runner gains one full lap, i.e. time $=\frac{\text{track length}}{\text{relative speed}}$.

1. Relative speed $=7-5=2\text{m/s}$; time to gain 840 m $=\frac{840}{2}=420$ s.
2. Distance run by A $=7\times 420=2940$ m $=3\times 840+420$.
3. So A has completed 3 full rounds and is 420 m from the start (clockwise) when they meet.

Trick: B runs $5\times 420=2100=2\times 840+420$ – both are 420 m from the start, a quick consistency check.

Answer: (C) 420 m after completing 3 rounds

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$ – weight each group average by its size.

1. Total $=30\times 378+21\times 355+40\times 224=11340+7455+8960=27755$.
2. Total members $=30+21+40=91$.
3. Average $=\frac{27755}{91}=305$.

Trick: Never average the three averages ($\frac{378+355+224}{3}=319$) – group sizes differ, so weights matter.

Answer: (D) 305

Concept: Do division and multiplication before addition/subtraction, and keep track of signs.

1. $(-56)÷(-8)=+7$; $(-4)\times 7=-28$.
2. Expression $=(-9)-7+(-28)$.
3. $=-9-7-28=-44$.

Trick: Negative $÷$ negative $=$ positive; then "minus a positive" makes it $-7$. Sign slips here give the wrong options $-46$, $-47$.

Answer: (B) $-44$

Concept: First find the maximum marks from Girish's data, then convert the friend's marks to a percentage.

1. $19\%$ of maximum $=424+13=437$.
2. Maximum marks $=\frac{437}{19}\times 100=23\times 100=2300$.
3. Friend's percentage $=\frac{529}{2300}\times 100=23\%$.

Trick: $437=19\times 23$ and $529={23}^2$ – once max $=2300$, $529$ is $23\%$ by inspection.

Answer: (C) 23%

Concept: Translate the sentence into an equation and collect $A$ and $B$ terms.

1. $\frac{10A}{100}+\frac{20B}{100}=\frac{3}{4}(\frac{20A}{100}+\frac{10B}{100})$.
2. Multiply by 100 and by 4: $40A+80B=60A+30B$.
3. $50B=20A\Rightarrow \frac{A}{B}=\frac{50}{20}=\frac{5}{2}$, i.e. $A:B=5:2$.

Trick: Work with $A+2B=\frac{3}{4}(2A+B)$ (cancel the 10%) → $4A+8B=6A+3B$ → $5B=2A$.

Answer: (C) 5 : 2

Concept: Take total work $=\text{LCM}(10,15,25)=150$ units; work in units/day, then use 2-day cycles.

1. P $=15$, Q $=10$, R $=6$ units/day.
2. Odd day (P+Q) $=25$ units; even day (P+R) $=21$ units; one 2-day cycle $=46$ units.
3. 3 cycles (6 days) $=138$ units; remaining $=12$ units.
4. Day 7 is odd (25 units/day): time $=\frac{12}{25}$ day. Total $=6\frac{12}{25}$ days.

Trick: Always stop the cycle count just below the total (138 < 150), then check which worker-pair is on duty for the leftover.

Answer: (D) $6\frac{12}{25}$

Concept: Division and multiplication first (left to right), then addition and subtraction.

1. $20÷4=5$; $2\times 3=6$.
2. $16+5-6=15$.

Trick: Doing $16+20=36$ first gives wrong answers – circle the $÷$ and $\times$ parts before touching $+$/$-$.

Answer: (D) 15

Concept: Opposite directions → relative speed $=$ sum of speeds; distance covered while crossing $=$ sum of lengths.

1. Relative speed $=90+70=160\text{km/h}=160\times \frac{5}{18}=\frac{400}{9}\text{m/s}$.
2. Sum of lengths $=\frac{400}{9}\times 27=1200$ m.
3. Other train $=1200-480=720$ m.

Trick: $160\text{km/h}\times 27\text{s}$: $160\times 27=4320$, $\times \frac{5}{18}=1200$. Keep $\frac{5}{18}$ for the end to avoid fractions.

Answer: (D) 720

Concept: Look at the differences – here they are consecutive prime numbers.

1. $15-13=2,18-15=3,23-18=5,30-23=7$.
2. Differences $2,3,5,7$ → next prime $11$.
3. $?=30+11=41$.

Trick: If the differences are not squares or an AP, test primes $2,3,5,7,11,13$ – a favourite RRB pattern.

Answer: (C) 41

Concept: Add filling rates and subtract the emptying rate (work per hour).

1. Net rate $=\frac{1}{2}+\frac{1}{57}-\frac{1}{57}=\frac{1}{2}$ of the cistern per hour.
2. Full cistern takes 2 hours; half the cistern takes $\frac{1}{2}\times 2=1$ hour.

Trick: The 57-hour inlet and 57-hour outlet cancel exactly – only the 2-hour tap matters. Read numbers before calculating.

Answer: (C) 1

Concept: With equal time periods, profit is shared in the ratio of investments.

1. Profit ratio A : B $=3:2$; total parts $=5$.
2. B's share $=\frac{2}{5}\times 25000=10000$.

Trick: Each part $=25000÷5=5000$ → A gets 3 parts (₹15,000), B gets 2 parts (₹10,000). Don't pick A's share by mistake.

Answer: (C) ₹10,000

Concept: The sphere encloses the cone: the apex and the rim of the base lie on the sphere. In the cross-section, the sphere's centre O is on the axis and $OB=R$.

1. Radius $r=3$, height $h=3\sqrt{2}$. Centre O is at distance $h-R$ from the base centre M.
2. In right triangle OMB: $(h-R{)}^2+r^2=R^2\Rightarrow h^2-2hR+r^2=0$.
3. $R=\frac{h^2+r^2}{2h}=\frac{18+9}{6\sqrt{2}}=\frac{27}{6\sqrt{2}}=\frac{9}{2\sqrt{2}}$, so $R^2=\frac{81}{8}$.
4. Surface area $=4\pi R^2=4\times \frac{22}{7}\times \frac{81}{8}=\frac{891}{7}=127\frac{2}{7}{\text{cm}}^2$.

Trick: For a cone (or cylinder) inside a sphere: $R=\frac{h^2+r^2}{2h}$. Here $R\approx 3.18>3$ confirms the sphere is the outer solid.

Answer: (B) $127\frac{2}{7}$

Concept: Fix the box whose position is certain (U at the top), then place the others by counting gaps.

1. No box above U → U is at position 1 (top). Three boxes between U and P → P is 5th.
2. T is immediately above P → T is 4th. One box between R and T → R is 2nd or 6th.
3. Four boxes between R and V → R = 2nd, V = 7th (R = 6th would need V at position 1, which is U).
4. Q is above S, so Q = 3rd and S = 6th. Final stack (top to bottom): U, R, Q, T, P, S, V.
5. Between S (6th) and U (1st) lie R, Q, T, P → 4 boxes.

Trick: "Only n boxes between X and Y" means the positions differ by n + 1; start from the box with a fixed end position.

Answer: (C) 4

Concept: Count positions from the top (1 = top, 7 = bottom) and fill the fixed boxes first.

1. Two boxes below B → B is 5th. One box above G → G is 2nd.
2. One box between G and A → A is 4th (position 0 does not exist).
3. Vacant places: 1, 3, 6, 7. C immediately above F needs two consecutive places → C = 6th, F = 7th.
4. D and E take 1 and 3; E is below D → D = 1st, E = 3rd. Stack: D, G, E, A, B, C, F.
5. Below E are A, B, C, F → 4 boxes.

Trick: After fixing the definite boxes, look for the only pair of adjacent empty slots – that is where the "immediately above" pair goes.

Answer: (C) 4

Concept: Draw the Venn diagram: "All heaters are fans" puts the Heater circle inside Fans; "Some heaters are lights" makes Lights overlap Heaters; "No heater is a bulb" keeps Bulbs apart from Heaters.

1. Conclusion I: the heaters that are lights lie inside Fans too, so some fans are definitely lights → follows.
2. Conclusion II: "No heater is a bulb" can always be reversed to "No bulb is a heater" → follows.

Yaad rakhein: All A are B + Some A are C ⇒ Some B are C (definite). "No A is B" ⇔ "No B is A".

Answer: (C) Both conclusions (I) and (II) follow

Concept: For people facing the centre, "left" is clockwise and "right" is anticlockwise. Start with the pair whose relative position is fixed.

1. E is immediately right of F, and F is third to the left of D → going clockwise: D … F with two seats between, E just anticlockwise of F.
2. E is third to the right of A → A is third to the left (clockwise) of E; this places A two seats clockwise after F.
3. Only one person between G and C counting from the left of C → G is 2nd clockwise from C. The only arrangement that fits all this (clockwise): A, G, D, B, E, F, C.
4. Check: B's neighbours are D and E, not F ✓. Third to the right (anticlockwise) of G: A, C, F → F.

Trick: Facing centre → right = anticlockwise. Put the two "immediate" clues on the circle first, then test the "third to…" clues.

Answer: (A) F

Concept: Both unknowns must follow one rule; test each option's # against ZVT and apply the same rule to JFD.

1. Option (D): # = WSQ. W + 3 = Z, S + 3 = V, Q + 3 = T → each letter moves +3 to give ZVT ✓.
2. Apply +3 to JFD: J + 3 = M, F + 3 = I, D + 3 = G → MIG, which matches % in option (D).
3. Other options fail: e.g. (A) WTR → +3, +2, +2 (not uniform); (B) XSR → +2, +3, +2.

Trick: Write place values (W=23, S=19, Q=17 → Z=26, V=22, T=20); a constant gap confirms the rule instantly.

Answer: (D) # = WSQ, % = MIG

Concept: Find the shift that converts # into NXF, then apply the same shift to KUC (the alphabet wraps after Z).

1. Try # = FPX (options B, C): F + 8 = N, P + 8 = X, X + 8 → 24 + 8 = 32 → 32 − 26 = 6 = F ✓. So the rule is +8 on each letter.
2. KUC + 8: K(11) + 8 = S(19); U(21) + 8 = 29 → 3 = C; C(3) + 8 = K(11) → SCK.
3. Option (B) gives SCO (wrong third letter); option (D) EPX gives E + 8 = M ≠ N. Only (C) satisfies both.

Trick: When a sum crosses 26, subtract 26 (U + 8 = 29 → C). Check all three letters – the last letter usually separates the options.

Answer: (C) # = FPX, % = SCK

Concept: Treat the letters and the numbers as two separate series.

1. Within each cluster: O(15), Q(17), R(18) → +2, +1. Same inside TVW, YAB, DFG.
2. First letters: O → T → Y → D → ? each +5 → D(4) + 5 = I(9). Cluster: I, I+2 = K, K+1 = L → IKL.
3. Numbers: 42, 35, 28, 21 → −7 each → 14.

Trick: Check the first letters of consecutive clusters first – a constant jump (+5) plus a constant number step (−7) pins the answer.

Answer: (A) IKL 14

Concept: Use the numeric clues (sums) first – they fix floors exactly.

1. T is even but not 6 → T = 2 or 4. T + R = 3 → only T = 2, R = 1 works.
2. Q + T = 7 → Q = 5.
3. Free floors: 3, 4, 6. P immediately above U → U = 3, P = 4; S = 6.
4. Below T (floor 2) there is only R → 1 person.

Trick: "Sum = 3" with floors ≥ 1 forces the pair {1, 2} – such tiny sums are the fastest entry point.

Answer: (A) 1