RRB NTPC Graduate CBT 1 2025 PYQs

Concept: $a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3$.

1. Take $a=4.23$, $b=0.23$: $0.69=3b$ and $12.69=3a$.
2. Numerator $=a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3=4^3=64$.
3. Denominator $=0.4\times 0.4\times 4=0.64$ ⇒ $\frac{64}{0.64}=100$.

Trick: Spot $0.69=3\times 0.23$ and $12.69=3\times 4.23$ — the sign pattern $+,-,+,-$ screams $(a-b{)}^3$.

Answer: (C) 100

Concept: When D is added, A, B, C keep their shares; the new total is 140% of the old total.

1. Old total $=3+5+7=15$ parts.
2. New total $=15\times 1.4=21$ parts.
3. $x=21-15=6$.

Trick: $40\%$ of 15 $=6$ — the increase itself is D's share.

Answer: (D) 6

Concept: Proportion of boys $=\frac{\text{boys}}{\text{boys}+\text{girls}}$; compare these fractions.

1. First $\frac{4}{9}\approx 0.444$; Second $\frac{3}{5}=0.6$; Third $\frac{6}{11}\approx 0.545$; Fourth $\frac{5}{11}\approx 0.455$.
2. Highest is $0.6$ — the second class.

Trick: Only $3:2$ and $6:5$ have more boys than girls; $3:2=1.5$ boys per girl beats $6:5=1.2$.

Answer: (A) Second class (3 : 2)

Concept: $\sqrt[3]{2197}=13$ (since ${13}^3=2197$) and $\sqrt[4]{10000}=10$ (since ${10}^4=10000$).

1. Expression $=\frac{13}{10}\times \frac{5}{39}\times 48$.
2. $\frac{13}{39}=\frac{1}{3}$, $\frac{5}{10}=\frac{1}{2}$ ⇒ $\frac{1}{6}\times 48=8$.

Trick: Cancel before multiplying: 13 with 39, 5 with 10 — the answer drops out as $\frac{48}{6}$.

Answer: (A) 8

Concept: Overall % $=\frac{\text{total SP}-\text{total CP}}{\text{total CP}}\times 100$.

1. CP each $=\frac{54000}{45}=1200$.
2. 30 at 25% profit: $30\times 1500=45000$; 15 at 20% loss: $15\times 960=14400$.
3. Total SP $=59400$; profit $=5400$ ⇒ $\frac{5400}{54000}\times 100=10\%$ profit.

Trick: Weighted: $\frac{30\times 25-15\times 20}{45}=\frac{750-300}{45}=10\%$ — CP per item is not even needed.

Answer: (B) 10% profit

Concept: In a right triangle, $\tan A=\cot C$ when $A+C={90}^{\circ }$ (complementary angles).

1. $BC=\sqrt{{13}^2-{12}^2}=\sqrt{25}=5$ cm.
2. $\tan A=\frac{BC}{AB}=\frac{5}{12}$; $\cot C=\frac{BC}{AB}=\frac{5}{12}$ (adjacent of C is BC, opposite is AB).
3. $\tan A-\cot C=0$.

Trick: $\angle A+\angle C={90}^{\circ }$ ⇒ $\cot C=\cot ({90}^{\circ }-A)=\tan A$ — the difference is 0 without finding BC.

Answer: (C) 0

Concept: If he uses $w$ g instead of 1000 g and sells at $80\%$ of the price of 1000 g, then $\frac{\text{SP}}{\text{CP}}=\frac{0.8\times 1000}{w}$.

1. Gain 40% ⇒ $\frac{800}{w}=1.4$.
2. $w=\frac{800}{1.4}=\frac{4000}{7}\approx 571.4$ g.

Trick: $\frac{\text{true weight}}{\text{false weight}}=\frac{100+\text{gain}}{100-\text{loss}}=\frac{140}{80}$ ⇒ false weight $=1000\times \frac{80}{140}=571.4$ g.

Answer: (C) 571.4 gm

Concept: A number is divisible by 3 if its digit sum is divisible by 3.

1. Digit sum of 445A0208 $=4+4+5+A+0+2+0+8=23+A$.
2. $23+A$ divisible by 3 ⇒ $A=1,4,7$.
3. Maximum $A=7$ ($23+7=30$).

Trick: $23\equiv 2\left(\mathrm{mod}3\right)$, so $A\equiv 1\left(\mathrm{mod}3\right)$: 1, 4, 7 — pick the largest digit.

Answer: (D) 7

Concept: Add the per-hour rates (filling positive, emptying negative).

1. Net rate $=\frac{1}{4}+\frac{1}{32}-\frac{1}{32}=\frac{1}{4}$ cistern per hour.
2. Full cistern in 4 hours ⇒ one-fourth in $\frac{1}{4}\times 4=1$ hour.

Trick: The 32-hour filler and 32-hour emptier cancel exactly — only the 4-hour tap matters.

Answer: (B) 1

Concept: For 2 years, $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $P\times {\left(\frac{4}{100}\right)}^2=53$ ⇒ $P\times \frac{16}{10000}=53$.
2. $P=\frac{53\times 10000}{16}=33125$.

Trick: Difference for 2 years = SI on the first year's interest; $P=\frac{\text{diff}\times {100}^2}{R^2}$.

Answer: (B) 33125

Concept: When people face the centre, "left" means clockwise and "right" means anticlockwise. Fix one person and place the rest.

1. Fix F. H is 3rd to the left of F. Only F sits between G and I, so G and I are F's two neighbours.
2. I is 2nd to the left of V. If I were on the clockwise side of F, V would land on G's seat – impossible. So I is on the anticlockwise side, G on the clockwise side, and V sits next to H (clockwise).
3. Two seats remain, one next to H. W is not H's neighbour, so X sits between G and H and W between V and I.
4. Clockwise order: F, G, X, H, V, W, I. Counting from the right of X (anticlockwise): G, F, then I → 2 people.

Trick: Whenever a clue gives two possibilities, test both against the next clue – one usually breaks immediately.

Answer: (A) Two

Concept: When two words share exactly one letter, the digit common to both codes is that letter's code.

1. GLOW → 4628, WEAR → 3574. The only common letter is W.
2. The only common digit is 4. Hence W = 4.

Trick: Match common letters with common digits – no need to decode the whole word.

Answer: (C) 4

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

1. Letters: HGH, FEF, DCD, BAB – each cluster moves back 2 letters (H→F→D→B). After B comes Z (wrapping), so the cluster is ZYZ.
2. Numbers: 85, 74, 63, 52 – each decreases by 11, so next is 41.
3. Required term: ZYZ 41.

Trick: Letter clusters of the form XYX with X going back by 2: H, F, D, B, Z, X …

Answer: (B) ZYZ 41

Concept: Find one rule that turns # into RWU and KPN into % – the same rule on both sides.

1. Option (C): SXV → RWU: S−1=R, X−1=W, V−1=U ✓.
2. KPN → JOM: K−1=J, P−1=O, N−1=M ✓. Same rule (each letter −1) on both sides.
3. Other options break the rule, e.g. (A) TXY→RWU would need −2, −1, −4.

Trick: Check the second pair (KPN : %) first – it is fully given and quickly eliminates options.

Answer: (C) # = SXV, % = JOM

Concept: Write the alphabet positions and compare the gaps between consecutive letters.

1. MTP: M(13)→T(20) = +7, T→P(16) = −4. GNJ: G(7)→N(14) = +7, N→J(10) = −4. JQM: J(10)→Q(17) = +7, Q→M(13) = −4.
2. PWR: P(16)→W(23) = +7, but W→R(18) = −5. Pattern breaks, so PWR is the odd one.

Trick: Pattern +7, −4 in three clusters; the fourth has −5.

Answer: (A) PWR

Concept: Replace the letters by the given signs and apply BODMAS.

1. 22 A 2 B 1 C 3 D 2 → $22÷2\times 1+3-2$.
2. $22÷2=11$; $11\times 1=11$; $11+3-2=12$.

Trick: Division and multiplication first (left to right), then addition/subtraction.

Answer: (C) 12

Concept: Find the letter-wise shift from the first cluster to the second and apply it to the options.

1. AOE → BQH: A+1=B, O+2=Q, E+3=H. OST → PUW: O+1=P, S+2=U, T+3=W. Rule: +1, +2, +3.
2. SHF → TJI: S+1=T, H+2=J, F+3=I ✓.
3. Others: DAF→FCH is +2,+2,+2; ADP→BGQ is +1,+3,+1; LAJ→MDN is +1,+3,+4.

Trick: Check the first letter of each option (+1) – it already removes DAF:FCH; then test the middle letter.

Answer: (C) SHF:TJI

Concept: Letters common to both words get the digits common to both codes; the leftover letter gets the leftover digit.

1. LITE → 8196, FELT → 6481. Common letters: L, T, E. Common digits: 8, 1, 6.
2. The letter unique to FELT is F; the digit unique to 6481 is 4. So F = 4.

Trick: Find the uncommon letter and the uncommon digit – they match.

Answer: (A) 4

Concept: Draw the Venn diagram of the statements; a conclusion follows only if it is true in every possible diagram.

1. All coppers ⊂ irons ⊂ silvers ⊂ golds – four nested circles.
2. (I) Some golds are irons: irons lie entirely inside golds, so the common part is definite → follows.
3. (II) All silvers are coppers: the reverse of the given chain – silvers may lie outside coppers → does not follow.

Trick: "All A are B" always gives "Some B are A", never "All B are A".

Answer: (A) Only conclusion (I) follows

Concept: Place the fixed person first, then use "immediately after" and "not on" clues to fill the week.

1. T = Tuesday. Two people between R and T → R = Friday.
2. P immediately after S, P not Sunday: the only free consecutive pair is Wed–Thu, so S = Wed, P = Thu.
3. Mon, Sat, Sun remain for Q, U, V. Q not Monday and Q before V → Q = Sat, V = Sun, U = Mon.
4. Between U (Mon) and P (Thu): T and S → 2 people.

Trick: Consecutive-pair clues (S then P) are the strongest – find free adjacent slots first.

Answer: (C) Two