RRB NTPC Graduate CBT 1 2024 PYQs

Concept: B’s one-day work $=\frac{1}{18}-\frac{1}{24}$.

1. $\frac{1}{18}-\frac{1}{24}=\frac{4-3}{72}=\frac{1}{72}$ → B alone takes $72$ days.
2. One-fourth work $=\frac{72}{4}=18$ days.

Trick: Total work $=\text{LCM}(18,24)=72$ units; A+B $=4$, A $=3$, so B $=1$ unit/day.

Answer: (A) 18

Concept: Downstream speed $=b+s$, upstream speed $=b-s$; time $=\frac{\text{distance}}{\text{speed}}$.

1. Downstream $=17+16=33\text{km/h}$; upstream $=17-16=1\text{km/h}$.
2. Time $=\frac{26.4}{33}+\frac{26.4}{1}=0.8+26.4=27.2$ hours.

Trick: When $b-s$ is tiny, the upstream trip dominates – the answer must exceed $26.4$.

Answer: (D) 27.2

Concept: Use ${\csc }^{2}A-1={\cot }^{2}A$ and ${\sec }^{2}A-1={\tan }^{2}A$.

1. $3\cot A=2\Rightarrow \cot A=\frac{2}{3}$, $\tan A=\frac{3}{2}$.
2. $\sqrt{\frac{3{\cot }^{2}A}{{\tan }^{2}A}}=\sqrt{3{\cot }^{4}A}=\sqrt{3}{\cot }^{2}A$.
3. $=\sqrt{3}\times \frac{4}{9}=\frac{4\sqrt{3}}{9}$.

Trick: $\frac{{\cot }^{2}A}{{\tan }^{2}A}={\cot }^{4}A$ – the root then just gives ${\cot }^{2}A$.

Answer: (C) $\frac{4\sqrt{3}}{9}$

Concept: Effective price $=MP\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $0.92\times 0.83\times 0.83=0.92\times 0.6889$.
2. $=0.633788\approx 63.38\%$ of marked price.

Trick: Discounts cannot simply be added ($42\%$ would give $58\%$) – always multiply the remaining fractions.

Answer: (C) $63.38\%$

Concept: Let a chair cost $c$ and a table $t$; solve the two equations.

1. $15c+3t=7800$ … (i), $14c+6t=9200$ … (ii).
2. $2\times$(i) $-$ (ii): $30c-14c=15600-9200\Rightarrow 16c=6400\Rightarrow c=400$.
3. $3t=7800-6000=1800\Rightarrow t=600$.
4. $17c+18t=6800+10800=17600$.

Trick: Doubling (i) makes the table terms equal ($6t$), so subtract directly.

Answer: (C) ₹17,600

Concept: Expenditure $=(100-\text{saving}\%)$ of income.

1. Expenditure $=80\%$ of $7915$.
2. $=\frac{4}{5}\times 7915=4\times 1583=6332$.

Trick: $80\%=\frac{4}{5}$: divide by 5, multiply by 4.

Answer: (B) ₹6,332

Concept: Speed in still water $=\frac{\text{downstream}+\text{upstream}}{2}$.

1. Downstream speed $=\frac{91}{13}=7\text{km/h}$.
2. Upstream speed $=\frac{39}{13}=3\text{km/h}$.
3. Still-water speed $=\frac{7+3}{2}=5\text{km/h}$.

Trick: Stream speed $=\frac{7-3}{2}=2$ – useful if asked next.

Answer: (A) 5

Concept: Integer roots need the discriminant $D=b^2-4ac$ to be a perfect square.

1. $D=(a-1{)}^2-4(4a+7)=a^2-18a-27=(a-9{)}^2-108$.
2. Let $D=k^2$: $(a-9-k)(a-9+k)=108$; both factors even → $(2,54)$ or $(6,18)$.
3. $(6,18)$: $a-9=12\Rightarrow a=21$; $(2,54)$: $a=37$. Least odd positive $a=21$.
4. Check: $x^2-20x+91=(x-7)(x-13)$ ✔.

Trick: Option test: put each option in $(a-9{)}^2-108$; only $a=21$ gives $144-108=36$, a perfect square.

Answer: (C) 21

Concept: Rotating a rectangle about its length makes length $=h$ and breadth $=r$; $\frac{V}{CSA}=\frac{\pi r^{2}h}{2\pi rh}=\frac{r}{2}$.

1. $\frac{r}{2}=\frac{4851}{462}=10.5\Rightarrow r=21\text{cm}$.
2. $2\times \frac{22}{7}\times 21\times h=462\Rightarrow 132h=462\Rightarrow h=3.5\text{cm}$.
3. Perimeter $=2(21+3.5)=49\text{cm}$.

Trick: Divide volume by CSA to get $\frac{r}{2}$ at once.

Answer: (B) 49

Concept: The perpendicular from the centre bisects a chord: $r^2=(\text{half chord}{)}^2+(\text{distance}{)}^2$.

1. Let AB be at distance $d$ from O; CD (same side) is at $d+4$.
2. $r^2=7^2+d^2=3^2+(d+4{)}^2\Rightarrow 49=25+8d\Rightarrow d=3$.
3. $r^2=49+9=58\Rightarrow r=\sqrt{58}$; diameter $=2\sqrt{58}$.

Trick: Longer chord is nearer the centre, so on the same side the shorter chord is at $d+4$.

Answer: (B) $2\sqrt{58}$

Concept: Use alphabet positions (A = 1 … Z = 26) and find the change in each letter and in the number.

1. KX 3 → OV −9: $K(11)+4=O(15)$, $X(24)-2=V(22)$, $3-12=-9$.
2. Check IR 11 → MP −1: $I+4=M$, $R-2=P$, $11-12=-1$ ✓.
3. JG −2: $J(10)+4=N(14)$, $G(7)-2=E(5)$, $-2-12=-14$ → NE −14.

Trick: Pattern = (+4, −2, −12). Test it on the second pair before applying.

Answer: (A) NE −14

Concept: Find how the 2nd and 3rd numbers are made from the 1st number.

1. (36, 27, 41): $36-9=27$ and $36+5=41$.
2. (45, 36, 50): $45-9=36$ and $45+5=50$ ✓.
3. Check (53, 44, 58): $53-9=44$, $53+5=58$ ✓. Others fail, e.g. $74+5=79\ne 89$.

Trick: Rule: (x, x − 9, x + 5). Equivalently 3rd − 2nd = 14 in every set.

Answer: (D) (53, 44, 58)

Concept: Facing the centre: left = clockwise, right = anticlockwise. 7 seats in all.

1. Fix S. Two people between S and T from the left of S → T is 3rd to the left of S; Y is 4th to the left of S.
2. R is 4th to the right of T → R is the immediate right of S.
3. The empty seats are on the left of S (1st, 2nd) and right of Y. V cannot sit next to R or T → V is 1st left of S; U cannot be next to V → U is between Y and R; Q is between V and T.
4. From Q, count three to the right: V, S, R.

Trick: In a 7-seat circle, "4th to the right" = "3rd to the left" — use whichever is shorter.

Answer: (D) R

Concept: When codes are jumbled, match common letters with common digits.

1. LOST → 6832 and SALT → 2136.
2. Common letters: L, S, T; common digits: 6, 3, 2.
3. The only letter left in SALT is A, and the only digit left in 2136 is 1 → A = 1.

Trick: Remove common letters and common digits — the leftovers match each other.

Answer: (C) 1

Concept: Look at the differences between consecutive terms.

1. Differences: $-1,+4,-9,+16,+25?$ → they are squares with alternating signs: $-1^2,+2^2,-3^2,+4^2$.
2. Next difference $=-5^2=-25$.
3. $1524-25=1499$.

Trick: Alternating up–down series → check $\pm n^2$ differences.

Answer: (B) 1499

Concept: Draw the minimum possible Venn diagram: "All A are B" puts circle A fully inside circle B.

1. Pulse ⊂ Rice ⊂ Wheat ⊂ Corn (nested circles).
2. (I) Pulse lies inside Corn → "All pulse are corn" follows.
3. (II) Rice lies inside Corn → some corn are surely rice → follows.

Trick: Chain of "All" statements: All first are last; and "Some last are any middle" is always true.

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

Concept: Change each digit according to whether it is even or odd, then read the new number.

1. 413257: 4 (even) → $4+3=7$; 1 → 3; 3 → 5; 2 → 5; 5 → 7; 7 (odd) → $7+2=9$.
2. New number: 735579.
3. First + last $=7+9=16$.

Trick: Only first and last digits matter: $4+3=7$ and $7+2=9$ — skip the middle.

Answer: (D) 16

Concept: Track the position on a coordinate grid (East = +x, North = +y); a right turn from East faces South.

1. A = (0, 0). 9 km East → (9, 0).
2. Right → South 16 km → (9, −16). Right → West 28 km → (−19, −16).
3. Right → North 21 km → (−19, 5). Right → East 19 km → P = (0, 5).
4. P is 5 km directly North of A → he must drive 5 km South.

Trick: Net E–W: $9-28+19=0$; net N–S: $-16+21=+5$ (North) → go back 5 km South.

Answer: (A) 5 km to the South

Concept: Check each letter position and the number separately.

1. 1st letters: H, I, J, K → L (+1).
2. 2nd letters: K, L, M, N → O; 3rd letters: P, Q, R, S → T.
3. Numbers: 22, 31, 40, 49 (+9) → 58.

Trick: All three letters move +1, number moves +9 → LOT58.

Answer: (B) LOT58

Concept: Place the fixed person first, then use the sum and "immediately below" conditions.

1. C is on floor 6.
2. B is on an odd floor except 3 → 1 or 5. If B = 1, D = 6 (taken by C) ✗. So B = 5, D = 2.
3. Free floors: 1, 3, 4. E immediately below F → E = 3, F = 4; A = 1.
4. Above E (floor 3): F, B, C = 3 people.

Trick: Use "sum = 7" pairs (1+6, 2+5, 3+4) to fix positions quickly.

Answer: (C) 3