RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Divisibility by 7 (also 11, 13): make 3-digit groups from the right and take the alternating sum.

1. The number is odd and ends in 1 → not divisible by 8, 12 or 15
2. Groups from the right: 814 | 242 | 541 → $541-242+814=1113$
3. $1113=7\times 159$ → divisible by 7; indeed $814242541=7\times 116320363$

Trick: Last digit 1 kills every even option and 15 instantly; only 7 remains.

Answer: (A) 7

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; doubling speed halves the time.

1. Let Anmol's time be $T$; then $T-7$ = Nikhil's time
2. With double speed Anmol takes $\frac{T}{2}=(T-7)-8$
3. $\frac{T}{2}=15\Rightarrow T=30$ h
4. Speed $=\frac{30}{30}=1\text{km/hr}$

Trick: The gap $7+8=15$ h equals half of Anmol's time.

Answer: (B) 1 km/hr

Concept: Successive % changes multiply: $P\times \left(1\pm \frac{r}{100}\right)$ for each year.

1. $12500\times 1.52=19000$
2. $19000\times 1.30=24700$
3. $24700\times 0.85=20995$

Trick: Last digit check: only an answer ending in 5 can come from multiplying by 0.85 here.

Answer: (A) 20,995

Concept: Consecutive numbers lie close to $\sqrt{702}\approx 26.5$.

1. $26\times 27=702$
2. Greater number $=27$

Trick: Unit digit: $6\times 7=42$ ends in 2, matching 702.

Answer: (B) 27

Concept: Make decimal places equal, take LCM of the whole numbers, then put the decimal back.

1. $12.8=12.800$ and $0.004$ → 12800 and 4 (thousandths)
2. LCM(12800, 4) $=12800$
3. Answer $=12.800=12.8$

Trick: 12.8 is itself a multiple of 0.004 ($12.8÷0.004=3200$), so LCM = 12.8.

Answer: (D) 12.8

Concept: Discount% $=\frac{\text{free items}}{\text{total items received}}\times 100$.

1. Customer gets $30+10=40$ articles, pays for 30
2. Discount $=\frac{10}{40}\times 100=25\%$

Trick: Divide by items received (40), not items bought (30) – 33% is the trap.

Answer: (B) 25%

Concept: Expand; the $x^2$ terms cancel, leaving a linear equation.

1. $10x^2-35-10x^2-16x+4=12$
2. $-16x-31=12$
3. $-16x=43\Rightarrow x=-\frac{43}{16}$

Trick: Watch the sign: $-2\times (-2)=+4$.

Answer: (A) $-\frac{43}{16}$

Concept: If $3A=6B=2C$, then $A:B:C=\frac{1}{3}:\frac{1}{6}:\frac{1}{2}$.

1. Multiply by 6: $A:B:C=2:1:3$
2. Total parts $=6$
3. $A=\frac{2}{6}\times 12300=4100$

Trick: Shares are inversely proportional to the multipliers 3, 6, 2.

Answer: (A) ₹4,100

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (left to right).

1. ${50}^{-6}÷{50}^{18}={50}^{-24}$
2. ${50}^{-24}\times {50}^{-5}={50}^{-29}$

Trick: Exponent $=-6-18-5=-29$.

Answer: (B) ${50}^{-29}$

Concept: Median = middle value after arranging in order; for $n$ odd it is the $\frac{n+1}{2}$th value.

1. Order: 8, 13, 14, 15, 23, 26, 35, 36, 50
2. $n=9$ → $\frac{9+1}{2}=5$th value
3. 5th value $=23$

Trick: Always sort first – 26 (the 5th value as printed) is the trap.

Answer: (B) 23

Concept: Place the towns on a grid from the given relations.

1. H is east of R → H on R's right
2. D is south of H → D is below H
3. So D is to the east and to the south of R → South-East (E is not needed)

Trick: East + South together = South-East.

Answer: (A) Southeast

Concept: Check every symbol: the element just before must be a number and the one just after must be a symbol.

1. 5 * \ → * qualifies
2. 4 $ & → $ qualifies
3. Others fail: 9 & 6, 6 # 6 (followed by numbers); 3 < has nothing after it

Trick: Scan for the pattern number-symbol-symbol only.

Answer: (B) Two

Concept: Facing the centre: right = anticlockwise, left = clockwise.

1. Q is immediately right of D; going left (clockwise) from Q, three people sit before S
2. Three people between D and C; one such seat is S's, so C sits next to S
3. A is immediately right of R → R and A take the two seats left after D, i.e. order to the right: D, Q, B, C, S, R, A
4. From R going right: R → A → D → one person (A) between

Trick: Always draw the circle and mark right/left as anticlockwise/clockwise first.

Answer: (A) One

Concept: Step 1: swap the first two letters. Step 2: move the last letter to the front.

1. JUNK → UJNK → KUJN; LACK → ALCK → KALC
2. FROM → RFOM → MRFO ✓
3. BUNS → UBNS should give SUBN, not SUNB; MOST → OMST should give TOMS

Trick: Check both steps on every option; most options break at step 2.

Answer: (C) FROM - RFOM - MRFO

Concept: Replace the symbols, then apply BODMAS.

1. $51+8-8÷80\times 4$ becomes $51-8\times 8+80÷4$
2. $=51-64+20$
3. $=7$

Trick: Do the symbol swap on paper first; never compute with the printed signs.

Answer: (D) 7

Concept: Find the position gaps between consecutive letters.

1. OJR: $-5,+8$; JEM: $-5,+8$; LGO: $-5,+8$
2. GBI: G → B is $-5$ but B → I is $+7$
3. So GBI is odd

Trick: Write letter positions (O = 15, J = 10, R = 18) to see gaps quickly.

Answer: (C) GBI

Concept: Differences are perfect squares.

1. $90+0^2=90$, $90+1^2=91$, $91+2^2=95$, $95+3^2=104$
2. $104+4^2=120$

Trick: Differences 0, 1, 4, 9, 16 → squares of 0, 1, 2, 3, 4.

Answer: (B) 120

Concept: Letters common to both words share the common digits.

1. EARS = 4259, SARI = 9725
2. Common letters S, A, R ↔ common digits 2, 5, 9
3. Left: E = 4 and I = 7

Trick: Cancel common letters with common digits; what remains pairs up.

Answer: (D) 7

Concept: Number the positions 1 (bottom) to 7 (top), fix the certain boxes first, then check whether the clues give one unique stack.

1. Three boxes below F $\Rightarrow$ F = 4. Two boxes between F and A $\Rightarrow$ A = 1 or 7.
2. D is immediately above B, and E is below C which is below G.
3. Several stacks satisfy every clue (bottom to top): A, B, D, F, E, C, G gives 1 box between B and F; B, D, E, F, C, G, A gives 2; A, E, C, F, B, D, G gives 0.
4. Options 1 and 2 are both possible, so the answer is not unique as printed.

Note: In "how many" arrangement questions, try to build a second valid stack; if it gives a different count, the answer is not unique. 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: Check the gaps between consecutive letters.

1. LNH: $+2,-6$; GIC: $+2,-6$; IKE: $+2,-6$
2. EGB: E → G $+2$, G → B $-5$
3. So EGB is odd

Trick: Use positions: E = 5, G = 7, B = 2.

Answer: (C) EGB