RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: ${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$.

1. $P=\sec \theta +\tan \theta \Rightarrow \frac{1}{P}=\sec \theta -\tan \theta$.
2. $\frac{P^2+1}{2P}=\frac{1}{2}(P+\frac{1}{P})$.
3. $=\frac{1}{2}(2\sec \theta )=\sec \theta$.

Trick: Put $\theta =0^{\circ }$: $P=1$, value $=\frac{2}{2}=1=\sec 0^{\circ }$ – but $\tan 0^{\circ }=0$, so try $\theta ={60}^{\circ }$ to separate: matches $\sec \theta$.

Answer: (D) $\sec \theta$

Concept: Consecutive numbers $n(n+1)$ – their product lies just above $n^2$.

1. $\sqrt{272}\approx 16.5$, so try $16$ and $17$.
2. $16\times 17=272$ ✓
3. Greater number $=17$.

Trick: Unit digit 2 ($6\times 7$) confirms $16\times 17$.

Answer: (D) $17$

Concept: Add the hourly filling rates: $\frac{1}{17}+\frac{1}{27}+\frac{1}{12}$.

1. LCM $(17,27,12)=1836$ (tank capacity units).
2. Rates: $108+68+153=329$ units/hour.
3. Time $=\frac{1836}{329}=5\frac{191}{329}$ hours.

Trick: Time must be less than the fastest pipe alone ($12$ h) – only $5\frac{191}{329}$ fits.

Answer: (C) $5\frac{191}{329}$ hours

Concept: Expand brackets, then combine like terms.

1. $x(3x-6)=3x^2-6x$
2. $4(x^2-5)=4x^2-20$
3. Total $=7x^2-6x-20+15=7x^2-6x-5$

Trick: Constant term: $-20+15=-5$ rules out options with $+5$.

Answer: (D) $7x^2-6x-5$

Concept: Find each pipe's full time, then the net rate.

1. P: $\frac{5}{9}$ in $25$ h → full in $45$ h; Q: $\frac{3}{7}$ in $27$ h → full in $63$ h.
2. In 3 h: $\frac{3}{45}+\frac{3}{63}=\frac{1}{15}+\frac{1}{21}=\frac{4}{35}$ of tank.
3. R empties $\frac{4}{35}$ in $4$ h → empties full tank in $35$ h.
4. Net: $\frac{1}{45}+\frac{1}{63}-\frac{1}{35}=\frac{7+5-9}{315}=\frac{1}{105}$ → $105$ h.

Trick: Full time $=\text{time}÷\text{fraction}$: $25÷\frac{5}{9}=45$.

Answer: (C) $105$ hours

Concept: If $4A=7B=2C=k$, then $A:B:C=\frac{1}{4}:\frac{1}{7}:\frac{1}{2}$.

1. Multiply by LCM $28$: $A:B:C=7:4:14$.
2. Total units $=25$; 1 unit $=\frac{13700}{25}=548$.
3. A $=7\times 548=3836$.

Trick: Shares are inversely proportional to the multipliers $4,7,2$.

Answer: (B) ₹3,836

Concept: Walk one way + drive one way = given time; driving both ways saves 6 h.

1. $W+D=7\frac{3}{4}$ h.
2. $2D=7\frac{3}{4}-6=1\frac{3}{4}$ h → $D=\frac{7}{8}$ h.
3. $W=7\frac{3}{4}-\frac{7}{8}=6\frac{7}{8}$ h.
4. Walking both ways $=2W=13\frac{3}{4}$ h $=13$ h $45$ min.

Trick: Walk both ways $=$ total $+$ time gained $=7\text{ h }45\text{ min}+6\text{ h}=13\text{ h }45\text{ min}$.

Answer: (A) 13 hours 45 minutes

Concept: Make the decimal places equal, find LCM of the integers, then restore the decimal.

1. $2.8=2.800$ and $0.217$ → work with $2800$ and $217$ (thousandths).
2. $2800=2^4\times 5^2\times 7$; $217=7\times 31$.
3. LCM $=2800\times 31=86800$ → $86.800=86.8$.

Trick: LCM of decimals is a multiple of the larger number: $86.8÷2.8=31$ ✓, $86.8÷0.217=400$ ✓.

Answer: (A) $86.8$

Concept: $(a+b{)}^2=a^2+2ab+b^2$.

1. ${34}^2=(30+4{)}^2$
2. $=900+240+16=1156$

Trick: Square ending in 4 ends in 6; near ${35}^2=1225$: $1225-35-34=1156$.

Answer: (C) $1156$

Concept: Compute both quantities and subtract.

1. $65\%$ of $65=0.65\times 65=42.25$
2. $\frac{3}{5}$ of $45=27$
3. Difference $=42.25-27=15.25$

Trick: $65\%$ of $65=\frac{{65}^2}{100}=\frac{4225}{100}$.

Answer: (B) $15.25$

Concept: Find the shift from FJEC to LPKI and apply it to LPKI.

1. F→L, J→P, E→K, C→I: each $+6$.
2. Check: IMHF $+6$ → OSNL ✓.
3. LPKI $+6$: L→R, P→V, K→Q, I→O = RVQO.

Trick: Same $+6$ to every letter; keep the order of letters.

Answer: (C) RVQO

Concept: Common letters share common digits.

1. TEAR = 7165, SEAT = 5678.
2. Common letters T, E, A ↔ common digits 5, 6, 7.
3. TEAR's extra letter R ↔ extra digit 1; SEAT's extra letter S ↔ extra digit 8.

Trick: Cancel the common part; the leftover letter takes the leftover digit.

Answer: (A) $8$

Concept: Track the position with coordinates (East $+x$, North $+y$).

1. A(0,0) → 15 km E → (15,0) → left 6 km N → (15,6).
2. Left 12 km W → (3,6) → left 8 km S → (3,−2).
3. Right turn while facing S = West: 3 km → P(0,−2).
4. P is 2 km South of A, so drive 2 km towards North.

Trick: Net E–W: $15-12-3=0$; net N–S: $6-8=-2$ → P is 2 km south of A.

Answer: (A) 2 km towards North

Concept: Fix definite positions first (positions 1 = top to 7 = bottom).

1. Z at 1 (nothing above); three between Z and W → W at 5; R immediately above W → R at 4.
2. One between V and R → V at 2 or 6; four between V and X → V = 2, X = 7.
3. Remaining 3, 6: Q above Y → Q = 3, Y = 6.
4. Second below Q (3) is position 5 = W.

Trick: Start with 'no box above Z' – it anchors everything.

Answer: (D) W

Concept: Check each number: the element before and after it must both be numbers.

1. Series: 4 7 + $ £ 3 & 8 5 @ 1 Ω # * % 6 2 9.
2. Pairs of adjacent numbers: (4,7), (8,5), (6,2,9).
3. Only 2 has a number on both sides (6 before, 9 after). 7 and 5 are followed by symbols.

Trick: Look for three numbers in a row – the middle one counts.

Answer: (C) $1$

Concept: Total $=$ rank from top $+$ rank from bottom $-1$.

1. Total $=39+11-1$
2. $=49$

Yaad rakhein: Subtract 1 because Mayank is counted twice.

Answer: (A) $49$

Concept: Look at the differences.

1. Differences: $+5,+10,+5,+10,+5,\dots$
2. $585+10=595$
3. Check: $595+5=600$ ✓

Trick: Alternating $+5$, $+10$ pattern.

Answer: (B) $595$

Concept: Use the 'only' clues to fix top and bottom (1 = top).

1. Only S above X → S = 1, X = 2.
2. Two boxes between X and Z → Z = 5.
3. Only P below R → P = 7, R = 6.
4. Q not immediately above Z → Q ≠ 4, so Q = 3, Y = 4. Between Y (4) and P (7): Z, R = two.

Trick: 'Only S above X' means X is second from top.

Answer: (A) Two

Concept: Find the positional gaps between consecutive letters.

1. EHC: E $+3$ H, H $-5$ C.
2. YBW: Y $+3$ B, B $-5$ W; SVQ: S $+3$ V, V $-5$ Q.
3. NQK: N $+3$ Q, but Q $-6$ K → odd.

Trick: Write letter positions: N(14) Q(17) K(11) – the second gap is 6, not 5.

Answer: (B) NQK

Concept: Compare first letters of the two clusters and second letters separately.

1. LQ-NJ: L $+2$ N, Q $-7$ J.
2. JO-LH: J $+2$ L, O $-7$ H; HM-JF: H $+2$ J, M $-7$ F.
3. NS-PK: N $+2$ P, but S $-8$ K → odd.

Trick: In the first cluster the gap is 5 in all (L–Q, N–S, J–O, H–M), so the second cluster decides.

Answer: (B) NS-PK