RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Multiply each part of the ratio by its own percentage factor.

1. A: $4\times \frac{120}{100}=4.8$; B: $5\times \frac{90}{100}=4.5$.
2. New ratio $4.8:4.5=48:45=16:15$.

Trick: $4\times 12:5\times 9=48:45$ – drop the 100s, they cancel.

Answer: (A) 16 : 15

Concept: Downstream speed = speed in still water + speed of stream; time $=\frac{\text{distance}}{\text{speed}}$.

1. Downstream speed $=12+3=15\text{km/h}$.
2. Time $=\frac{60}{15}=4$ hours.

Trick: Down = add, Up = subtract. Downstream is always the faster, shorter trip.

Answer: (A) 4

Concept: Opposite angles of a cyclic quadrilateral add to ${180}^{\circ }$; co-interior angles between parallel sides also add to ${180}^{\circ }$.

1. PS ∥ QR, PQ is a transversal: $\angle P+\angle Q={180}^{\circ }\Rightarrow \angle Q={95}^{\circ }$.
2. Cyclic: $\angle P+\angle R={180}^{\circ }\Rightarrow \angle R={95}^{\circ }$.
3. Cyclic: $\angle Q+\angle S={180}^{\circ }\Rightarrow \angle S={85}^{\circ }$.

Trick: A cyclic trapezium is always isosceles – base angles equal in pairs: $\angle P=\angle S$, $\angle Q=\angle R$.

Answer: (D) $\angle Q={95}^{\circ },\angle R={95}^{\circ },\angle S={85}^{\circ }$

Concept: Round-trip time with boat speed $b$ and stream $s$: $\frac{d}{b+s}+\frac{d}{b-s}=\frac{2bd}{b^2-s^2}$.

1. With $b=7$, $d=40$: time $=\frac{560}{49-s^2}$ hours.
2. 560 minutes $=\frac{28}{3}$ h. Condition: $\frac{560}{49-(s+2{)}^2}-\frac{560}{49-s^2}=\frac{28}{3}$.
3. Try $s=3$: $\frac{560}{24}-\frac{560}{40}=\frac{70}{3}-14=\frac{28}{3}$. ✓

Trick: Options are small – plug them in. $49-s^2$ should divide 560 nicely: $s=3$ gives 40 and $s=5$ gives 24.

Answer: (D) 3

Concept: Multiply numerator and denominator by 10 to remove decimals, then divide.

1. $\frac{1.2}{0.3}=\frac{12}{3}=4$.
2. $\frac{2.4}{0.6}=\frac{24}{6}=4$.
3. Sum $=4+4=8$.

Trick: $0.3\times 4=1.2$ and $0.6\times 4=2.4$ – both quotients are 4 by inspection.

Answer: (D) 8

Concept: Profit% $=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$; compare on the same unit (one dozen).

1. CP of a dozen = ₹400. SP of a dozen $=58\times 12=$ ₹696.
2. Profit per dozen $=696-400=296$.
3. Profit% $=\frac{296}{400}\times 100=74\%$.

Trick: The '52 dozens' is a distraction – percentage profit is the same for one dozen.

Answer: (B) 74%

Concept: Profit is shared in the ratio of investments (same time). Let the total profit be $12k$ (sum of ratio parts).

1. A $=3k$, B $=5k$, C $=4k$.
2. After B gives ₹15,000 to A: $3k+15000=4k\Rightarrow k=15000$.
3. Total profit $=12k=12\times 15000=$ ₹1,80,000.

Trick: The transfer equals the gap between A and C, i.e. one ratio part – so one part = ₹15,000 and total = 12 parts.

Answer: (D) 1,80,000

Concept: Take CP = 100; marked price and selling price follow from the given percentages.

1. MP $=100+100\%=200$. Required SP for 2% gain $=102$.
2. Discount $=200-102=98$.
3. Discount% $=\frac{98}{200}\times 100=49\%$.

Trick: $\text{MP}\times (1-d)=\text{CP}\times (1+g)$: $2(1-d)=1.02\Rightarrow d=0.49$.

Answer: (A) 49%

Concept: Total salary ratio = (number ratio) × (salary ratio) for each group.

1. Trained : Untrained total $=5\times 4:3\times 1=20:3$.
2. Untrained share $=\frac{3}{23}\times 4,60,000$.
3. $4,60,000÷23=20,000$; $\times 3=$ ₹60,000.

Trick: 4,60,000 is exactly $23\times 20,000$ – the clean division confirms the 20:3 split.

Answer: (C) ₹60,000

Concept: Take successive differences and look for a simple pattern in them.

1. 23 → 26 → 32 → 41 → 53: differences $3,6,9,12$ (multiples of 3).
2. Next difference $=15$, so $53+15=68$.

Trick: Differences growing by a constant means the next term = last term + (last difference + that constant).

Answer: (B) 68

Concept: If the bisectors of $\angle A$ and $\angle B$ of quadrilateral ABCD meet at O, then $\angle AOB=\frac{1}{2}(\angle C+\angle D)$.

1. In $△AOB$: $\angle AOB={180}^{\circ }-\frac{\angle A+\angle B}{2}$.
2. Since $\angle A+\angle B={360}^{\circ }-(\angle C+\angle D)$, we get $\angle AOB=\frac{\angle C+\angle D}{2}$.
3. $\angle C+\angle D=2\times {41}^{\circ }={82}^{\circ }$.

Trick: Angle between two adjacent angle bisectors of a quadrilateral = half the sum of the other two angles.

Answer: (C) ${82}^{\circ }$

Concept: Write the alphabetical positions and find the gap pattern between letters.

1. QMO: Q(17) − 4 = M(13), M(13) + 2 = O(15).
2. SOQ: S(19) − 4 = O(15), O(15) + 2 = Q(17). OKM: O(15) − 4 = K(11), K(11) + 2 = M(13).
3. MIL: M(13) − 4 = I(9) ✓ but I(9) + 2 = K(11), not L(12). Pattern breaks.

Trick: Pattern is (−4, +2) in each cluster; MIL has (−4, +3).

Answer: (A) MIL

Concept: Treat letters and numbers as two separate series.

1. Letters: M→K→I→G (each −2), P→N→L→J (−2), L→J→H→F (−2). Next: G−2 = E, J−2 = H, F−2 = D → EHD.
2. Numbers: 79, 87, 95, 103 (+8 each). Next: 103 + 8 = 111.

Trick: Only option with EHD is (D); number 111 confirms it.

Answer: (D) EHD 111

Concept: Find one rule connecting the three numbers that works for both given sets.

1. (9, 17, 7): $9+7+1=17$ ✓. (6, 12, 5): $6+5+1=12$ ✓. Rule: first + third + 1 = middle.
2. (4, 9, 4): $4+4+1=9$ ✓.
3. (6, 12, 4): 11 ✗; (2, 5, 3): 6 ✗; (9, 17, 5): 15 ✗.

Trick: Middle − (first + third) = 1 in every valid set.

Answer: (B) (4, 9, 4)

Concept: Common letters in two words get common digits; the leftover letter gets the leftover digit.

1. LITE → 3 4 9 2 and TIME → 3 9 5 4.
2. Common letters I, T, E ↔ common digits 3, 4, 9.
3. Leftover in TIME is M ↔ leftover digit 5.

Trick: Cancel the common letters and digits; what remains is the code.

Answer: (D) 5

Concept: Number the seats 1–7 from the left (all face north, so right = higher number) and place the fixed person first.

1. Q is at seat 2. Two people between Q and R → R at seat 5.
2. One person between R and S → S at 3 (seat 7 would leave no room for C on S's right). C immediately right of S → C at 4.
3. Seats 1, 6, 7 are left for P, O, B. O immediately left of B → O at 6, B at 7, so P at 1.
4. Between P (1) and O (6): Q, S, C, R = four people.

Trick: Final order P Q S C R O B.

Answer: (D) Four

Concept: Replace the letters by the given signs, then apply BODMAS.

1. 19 D 54 A 6 C 4 B 2 → $19-54÷6+4\times 2$.
2. $54÷6=9$, $4\times 2=8$.
3. $19-9+8=18$.

Trick: Do ÷ and × first, then go left to right.

Answer: (C) 18

Concept: Remove every symbol and count from the right end of what remains.

1. Digits only (left to right): 8 6 2 1 5 9 7 5 3 1 4 – eleven digits.
2. From the right: 4 (1st), 1 (2nd), 3 (3rd), 5 (4th), 7 (5th), 9 (6th).

Trick: 6th from right of 11 items = 6th from left: 8, 6, 2, 1, 5, 9.

Answer: (B) 9

Concept: Draw the family tree generation by generation.

1. A is the wife of D → A and D are a couple.
2. D's children are X (daughter) and M (son) → they are A's children too.
3. M is married to Y → Y is the wife of A's son.

Trick: Son's wife = daughter-in-law (bahu).

Answer: (B) Son's wife

Concept: Draw the Venn diagram with the least possible overlap and test each conclusion.

1. Gorillas lie inside Monkeys; Baboons partly overlap Monkeys; Baboons lie inside Chimpanzees.
2. (I) 'All chimpanzees are monkeys' – only baboons are forced inside Chimpanzees; other chimpanzees can be outside Monkeys. Does not follow.
3. (II) 'Some chimpanzees are gorillas' – Gorillas and Chimpanzees need not touch at all. Does not follow.

Trick: 'All A are B' never gives 'All B are A'; and a 'Some' link in the chain breaks any definite conclusion.

Answer: (A) Neither conclusion (I) nor (II) follows.