RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Take distance $MN=D$; speed $=\frac{D}{\text{journey time}}$; after both start, gap closes at the sum of speeds.

1. A takes $9$ h (speed $\frac{D}{9}$), B takes $6$ h (speed $\frac{D}{6}$).
2. By 8:35 AM, A has run $2$ h: covered $\frac{2D}{9}$, gap $=\frac{7D}{9}$.
3. Relative speed $=\frac{D}{9}+\frac{D}{6}=\frac{5D}{18}$; time $=\frac{7}{9}÷\frac{5}{18}=2.8$ h $=2$ h $48$ min.
4. Meeting time $=8:35+2:48=11:23$ AM.

Trick: Distance cancels – always use $D$ or LCM of times ($18$ units).

Answer: (D) 11:23 AM

Concept: Write both numbers in terms of one variable.

1. Let numbers be $5x$ and $x$: $5x^2=1805$ ⇒ $x^2=361$ ⇒ $x=19$.
2. Numbers $95$ and $19$; sum $=114$.

Trick: Sum $=6x$, so it must be a multiple of $6$: only $114$ and $126$ qualify; $x=19$ gives $114$.

Answer: (C) $114$

Concept: Form two equations and solve by substitution.

1. $F+5S=42$ and $2.8F+S=76$.
2. $F=42-5S$ ⇒ $2.8(42-5S)+S=76$ ⇒ $117.6-13S=76$ ⇒ $S=3.2$.
3. $F=42-16=26$.

Trick: Check options: $F=26$ ⇒ $S=3.2$, and $2.8\times 26+3.2=76$ ✓.

Answer: (A) $26$

Concept: Total work in man-days stays constant; convert women into men.

1. Work $=9\times 8=72$ man-days $=4\times 9=36$ woman-days.
2. So $1$ woman $=2$ men.
3. $2$ women $+4$ men $=8$ men ⇒ days $=\frac{72}{8}=9$.

Trick: Equate total work: $72M=36W$ ⇒ $W=2M$.

Answer: (C) $9$ days

Concept: Savings $=$ Income $-$ Expenditure; increase income and expenditure separately.

1. Savings $=18.5\%$ of $44000=8140$; expenditure $=35860$.
2. New income $=44000\times 1.17=51480$; new expenditure $=35860\times 1.4=50204$.
3. New savings $=51480-50204=1276$.
4. Change $=8140-1276=6864$ decrease.

Trick: Expenditure grows faster ($40\%$ vs $17\%$) on a big base – savings must fall, ruling out A and D.

Answer: (B) ₹6,864 decrease

Concept: Median of an even number of observations $=$ average of the two middle values (after sorting).

1. Sorted: $3,6,18,32,37,48$.
2. Middle values: $18$ and $32$.
3. Median $=\frac{18+32}{2}=25$.

Trick: Always sort first – never take the middle of the given order.

Answer: (A) $25$

Concept: $\text{HCF}\times \text{LCM}=$ product of the two numbers.

1. $4\times 2600=10400$.
2. Other number $=\frac{10400}{104}=100$.

Trick: Check: $\text{HCF}(104,100)=4$ ✓.

Answer: (D) $100$

Concept: For equal distances, average speed $=\frac{2xy}{x+y}$.

1. $\frac{2\times 15\times 30}{15+30}=\frac{900}{45}=20\text{km/h}$.

Trick: Average speed is always below the simple average $22.5$.

Answer: (C) $20$

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{70}^2}{28}=\frac{4900}{28}=175$.

Trick: $\frac{70}{28}=2.5$, so $x=70\times 2.5=175$.

Answer: (B) $175$

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

1. ${55}^{-6}÷{55}^5={55}^{-11}$.
2. ${55}^{-11}\times {55}^{-8}={55}^{-19}$.

Trick: Just add/subtract powers: $-6-5-8=-19$.

Answer: (B) ${55}^{-19}$

Concept: Look at the differences between consecutive terms.

1. Differences: $1,2,4,8$ – each doubles.
2. Next difference $=16$: $27+16=43$.

Trick: Doubling differences: $1,2,4,8,16$.

Answer: (A) $43$

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

1. BALD → 6184, LAMB → 4136.
2. Common letters B, A, L ↔ common digits 6, 1, 4.
3. Left over: M in LAMB ↔ 3 in 4136 (D ↔ 8).

Trick: Strike out common letters and digits; the leftovers pair up.

Answer: (A) 3

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

1. Fix Q. P is 3rd to the left of Q; I and R sit on either side of Q.
2. If R were on Q's left, S (two to R's right) would clash with I – so R is on Q's right, I on Q's left, and S is two places right of R.
3. H is not next to P, so H sits between R and S; J takes the last seat (next to P and I).
4. Going right from J: I, Q, then R – 2 persons between.

Trick: Fix the person mentioned most often (Q) first.

Answer: (B) Two

Concept: Write the digits in ascending order and compare position by position.

1. Original: 8 2 3 4 6 1 9.
2. Ascending: 1 2 3 4 6 8 9.
3. Same positions: 2, 3, 4, 6, 9 – five digits.

Trick: Line the two numbers one under the other and tick matches.

Answer: (B) Five

Concept: Interchange the signs first, then apply BODMAS.

1. After swapping (÷↔−, ×↔+): $45\times 5-5+6÷3\times 24+9-3+2$.
2. $45\times 5=225$; $6÷3\times 24=48$.
3. $225-5+48+9-3+2=276$.

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

Answer: (C) $276$

Concept: Find the letter-position shifts in the given pairs.

1. PGS → LBO: P−4=L, G−5=B, S−4=O.
2. HWK → DRG: H−4=D, W−5=R, K−4=G.
3. XQA: X−4=T, Q−5=L, A−4=W (wrap back from A) → TLW.

Trick: Pattern $-4,-5,-4$; for A, count back to W.

Answer: (D) XQA : TLW

Concept: Track each letter position and the number separately.

1. 1st letters: I, H, G, F, E ($-1$).
2. 2nd letters: V, Q, L, G, B ($-5$); 3rd letters: P, M, J, G, D ($-3$).
3. Numbers: $16,5,-6,-17,-28$ ($-11$).
4. Missing term: FGG $-17$.

Trick: Check the number first: only $-17$ fits $-6-11$.

Answer: (C) FGG $-17$

Concept: Draw the Venn diagram with the minimum possible overlap.

1. Games ⊂ Ludo ⊂ Drums.
2. (I) Ludo lies inside Drums, so some drums are ludo – follows.
3. (II) Games inside Ludo inside Drums – all games are drums – follows.

Trick: "All A are B" + "All B are C" ⇒ "All A are C"; and "All" always gives "Some" in reverse.

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

Concept: Replace the signs as told, then use BODMAS.

1. $26-14\times 2+56÷7$.
2. $14\times 2=28$, $56÷7=8$.
3. $26-28+8=6$.

Trick: Write the new expression fully before solving.

Answer: (A) $6$

Concept: Compare the letter positions of the two groups in each pair.

1. DH-KB: D+7=K, H−6=B.
2. MQ-TK: M+7=T, Q−6=K; GK-NE: G+7=N, K−6=E.
3. JN-QG: J+7=Q but N−6=H, not G – odd one.

Trick: Check the second letter: N(14)−6 = H(8), while G is 7.

Answer: (C) JN-QG