RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: New average $=\frac{\text{old total}+\text{new score}}{\text{new count}}$.

1. Runs in 9 matches $=9\times 55=495$.
2. Total after 10th match $=495+85=580$.
3. New average $=\frac{580}{10}=58$.

Trick: Excess over old average $=85-55=30$, spread over 10 matches $=+3$ → $55+3=58$.

Answer: (B) 58

Concept: Profit $=$ cost $\times$ profit%, computed separately for each lot.

1. 3 sets: $3\times 375\times \frac{16}{100}=180$.
2. 4 sets: $4\times 375\times \frac{24}{100}=360$.
3. Total profit $=180+360=540$.

Trick: Profit per set: $16\%$ of 375 $=60$, $24\%$ of 375 $=90$ → $3(60)+4(90)=540$.

Answer: (C) ₹540

Concept: Form two equations in adult price $a$ and child price $c$ and eliminate.

1. $3a+4c=380$ and $2a+3c=270$.
2. Multiply by 2 and 3: $6a+8c=760$, $6a+9c=810$ → $c=50$.
3. $3a+200=380\Rightarrow a=60$.

Trick: Subtract the equations: $a+c=110$; then $2(a+c)+c=270$ gives $c=50$ directly.

Answer: (C) ₹60

Concept: Direct proportion – packets per hour stay constant.

1. Rate $=\frac{1520}{5}=304$ packets/hour.
2. In 17 hours: $304\times 17=5168$.

Trick: $304\times 17=304\times 10+304\times 7=3040+2128=5168$.

Answer: (C) 5,168

Concept: Sum of interior angles $=(n-2)\times {180}^{\circ }$; diagonals $=\frac{n(n-3)}{2}$.

1. $(n-2)\times 180=1800\Rightarrow n-2=10\Rightarrow n=12$.
2. Diagonals $=\frac{12\times 9}{2}=54$.

Yaad rakhein: Each vertex joins $n-3$ others by a diagonal; divide by 2 to avoid double counting.

Answer: (D) 54

Concept: Profit is shared in the ratio of investments (equal time).

1. Ratio $60000:80000:120000=3:4:6$ (total 13 parts).
2. One part $=\frac{78000}{13}=6000$.
3. C $-$ B $=(6-4)\times 6000=12000$.

Trick: Only the difference of parts matters: $2$ parts $\times$ ₹6,000.

Answer: (B) ₹12,000

Concept: Area $=\frac{1}{2}{d}_1{d}_2$; half-diagonals and a side form a right triangle.

1. Let diagonals be $d$ and $d+10$: $\frac{1}{2}d(d+10)=600\Rightarrow d^2+10d-1200=0\Rightarrow d=30$.
2. Diagonals 30 cm and 40 cm; half-diagonals 15 and 20.
3. Side $=\sqrt{{15}^2+{20}^2}=25$ cm.
4. Perimeter $=4\times 25=100$ cm.

Trick: Half-diagonals 15, 20 → the 3-4-5 triple scaled by 5 → side 25.

Answer: (C) 100 cm

Concept: Selling price $=MP\times (1-d_1)(1-d_2)$.

1. After 18%: $1000\times 0.82=820$.
2. After 15%: $820\times 0.85=697$.

Trick: Single equivalent discount $=18+15-\frac{18\times 15}{100}=30.3\%$ → pay $69.7\%$ of 1000.

Answer: (D) 697

Concept: Compare the effective price paid per toy (marked price $=100$) under each scheme.

1. (A) $100\times 0.57\times 0.81=46.17$ per toy.
2. (B) $100\times 0.59\times 0.79\times 0.51\approx 23.77$ per toy.
3. (C) First 5 toys at 35% off → $65$ per toy.
4. (D) 9 toys for the price of 5 → $\frac{500}{9}\approx 55.56$ per toy.
5. Highest effective price $=65$ → scheme (C) is least beneficial.

Trick: Convert every offer to "price per toy"; the largest number is the worst deal.

Answer: (D) C

Concept: Take tank $=\text{LCM}(8,4)=8$ units; P fills 1 unit/h, Q fills 2 units/h.

1. One cycle (Q then P) of 2 hours fills $2+1=3$ units.
2. After 2 cycles (4 h): 6 units; remaining 2 units.
3. 5th hour is Q's turn: it fills exactly 2 units.
4. Total time $=5$ hours.

Trick: In alternate-pipe problems, stop cycling when the remainder is less than one cycle and check whose turn it is.

Answer: (A) 5 hours

Concept: Check each position of the clusters separately for a constant jump in the alphabet (A=1 … Z=26, cyclic).

1. 1st letters: G → Z → S → L : each $-7$ (G(7)−7 = 0 → Z(26)). Next: L(12)−7 = 5 → E.
2. 2nd letters: J → C → V → O : each $-7$. Next: O(15)−7 = 8 → H.
3. 3rd letters: E → X → Q → J : each $-7$. Next: J(10)−7 = 3 → C.

Trick: Subtracting 7 is the same as adding 19; when you go below A, add 26 and continue.

Answer: (A) EHC

Concept: When two coded sentences share exactly one word, the code common to both sentences is the code of that word.

1. Sentence 1: steak tooth badge → qj hc sv.
2. Sentence 2: badge poet lemon → tz qj wv.
3. Only "badge" is common and only "qj" is common → badge = qj.

Trick: Match the common word with the common code; order of codes in a sentence does not matter.

Answer: (C) qj

Concept: Decode each symbol step by step and build the family tree, then read the relation from P's side.

1. A @ D : A is the son of D. D £ J : D is the father of J → A and J are brothers.
2. J ≠ L : J is the brother of L → L is also a child of D, hence A's sister.
3. L ≤ P : L is the wife of P → A is the brother of P's wife.

Trick: Children of the same father are siblings; the sister's husband sees you as "wife's brother" (saala).

Answer: (B) Wife's brother

Concept: If a person is $m^{th}$ from the top and $n^{th}$ from the bottom, total students $=m+n-1$ (the person is counted twice).

1. $m=27,n=64$.
2. Total $=27+64-1=90$.

Trick: Top rank + bottom rank − 1 = total (subtract 1 for the double-counted person).

Answer: (C) 90

Concept: Number the seats 1–7 from left to right (facing north, left = west). Place the people with fixed positions first.

1. Six people right of E → E at 1. Five people left of Q → Q at 6. Three between M and Q → M at 2.
2. Seats 3, 4, 5, 7 remain for B, L, G, F. B is between L and G, so B at 4 with L, G at 3 and 5; F takes 7.
3. Three between F(7) and L → L at 3, so G at 5. Order: E M L B G Q F.
4. Between G(5) and F(7) sits only Q → one person.

Trick: "Only n people sit to the right/left" pins a person to an exact seat – start from such clues.

Answer: (A) One

Concept: Total students $=(\text{rank from top})+(\text{rank from bottom})-1$.

1. Rank from top $=44$, from bottom $=25$.
2. Total $=44+25-1=68$.

Trick: Add the two ranks and subtract 1 (Ayush himself is counted in both).

Answer: (A) 68

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

1. Letters: N(14) → I(9) → D(4) → Y(25) : each $-5$ (D−5 = −1 → wrap to 25 = Y). Next: Y(25)−5 = 20 → T.
2. Numbers: 32 → 25 → 18 → 11 : each $-7$. Next: $11-7=4$.
3. So the next term is T 4.

Trick: Going backwards past A, add 26: $4-5=-1\Rightarrow -1+26=25$ (Y).

Answer: (A) T 4

Concept: Compare corresponding letters of the two clusters in each pair and find the jump.

1. FE−IH : F→I (+3), E→H (+3).
2. LK−ON : L→O (+3), K→N (+3). CB−FE : C→F (+3), B→E (+3).
3. KJ−MO : K→M (+2), J→O (+5) – breaks the pattern.

Trick: In three pairs the second cluster is simply "each letter + 3"; the odd pair has unequal jumps.

Answer: (D) KJ−MO

Concept: Vowel → +2 in the alphabet, consonant → −2. Convert every letter of REFERTILIZES and count.

1. R→P, E→G, F→D, E→G, R→P, T→R, I→K, L→J, I→K, Z→X, E→G, S→Q.
2. New group: P G D G P R K J K X G Q.
3. Counts: G = 3, P = 2, K = 2, R = 1, X = 1. Among the options only K appears exactly twice.

Trick: Convert letter by letter in order and tally only the letters given in the options.

Answer: (B) K

Concept: Find the rule for letters and number separately from both given pairs, then apply it.

1. KY 1 → MU 17: K→M (+2), Y→U (−4), $1\to 17$ (+16).
2. FI 8 → HE 24: F→H (+2), I→E (−4), $8\to 24$ (+16). Rule confirmed.
3. RF 3: R+2 = T, F−4 = B, $3+16=19$ → TB 19.

Trick: Verify the rule on both examples before applying; here (+2, −4, +16).

Answer: (C) TB 19