RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Number of identical pipes and time are inversely proportional: $N_1{T}_1=N_2{T}_2$.

1. $43\times 40=9\times T$
2. $T=\frac{1720}{9}=191.11$ minutes

Trick: More pipes → less time; just multiply and divide: $\frac{43\times 40}{9}$.

Answer: (A) $191.11$

Concept: Before taking a ratio, both quantities must be in the same unit.

1. $7\text{ days}=7\times 24=168\text{ hours}$
2. $76:168$; divide by 4: $19:42$

Trick: Always convert days to hours first: 1 day = 24 h.

Answer: (B) $19:42$

Concept: Simplify nested roots from the innermost root outward.

1. $\sqrt{900}=30$
2. $\sqrt{139+30}=\sqrt{169}=13$
3. $\sqrt{212+13}=\sqrt{225}=15$

Trick: Each step should give a perfect square: 169, then 225.

Answer: (B) 15

Concept: Monthly saving = salary − expenses; the birthday month adds no saving.

1. Monthly saving = $12000-4000-1500=6500$
2. Savings are made in only $11$ months.
3. Yearly saving = $6500\times 11=71500$

Trick: Do not take 12 months – one month's saving is fully spent.

Answer: (C) ₹71,500

Concept: In the right triangle formed, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {45}^{\circ }=\frac{h}{27.6}$
2. $\tan {45}^{\circ }=1\Rightarrow h=27.6\text{m}$

Trick: At ${45}^{\circ }$, height = distance from the foot.

Answer: (B) $27.6\text{m}$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$.

1. Here $a=1,b=3,c=k$: $D=9-4k$
2. $9-4k=9\Rightarrow k=0$

Trick: $D=b^2$ already, so $4ac$ must be $0$ → $k=0$.

Answer: (D) 0

Concept: If numbers are in ratio $a:b:c$ (co-prime parts), they are $ax,bx,cx$ with HCF $=x$ and LCM $=\text{LCM}(a,b,c)\times x$.

1. Numbers: $x,9x,7x$
2. $\text{LCM}=63x=6930$
3. $x=\frac{6930}{63}=110$ = HCF

Trick: HCF = $\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{6930}{63}$.

Answer: (B) 110

Concept: For a pure recurring decimal, $0.\bar{a}=\frac{a}{9}$.

1. $1.222\dots =1+0.\bar{2}=1+\frac{2}{9}$
2. $=\frac{9+2}{9}=\frac{11}{9}$

Trick: $1.\bar{2}=\frac{12-1}{9}=\frac{11}{9}$ (whole number − non-repeating part, over 9).

Answer: (A) $\frac{11}{9}$

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$

1. Sum $=5+8+6+7+9=35$
2. Average $=\frac{35}{5}=7\text{km}$

Trick: 5, 6, 7, 8, 9 are consecutive → average = middle value 7.

Answer: (C) $7\text{km}$

Concept: Use $(a-b)(a+b)=a^2-b^2$.

1. $891\times 909=(900-9)(900+9)$
2. $={900}^2-9^2=810000-81=809919$

Trick: Both numbers are equally far (9) from 900 → square of 900 minus square of 9. Last digits: $1\times 9=9$ ends in 9.

Answer: (A) 809919

Concept: Replace the letters by the signs and solve using BODMAS.

1. Replace: $19\times 4-7\times 8+32-11\times 2$
2. Multiplications first: $76-56+32-22$
3. Left to right: $20+32-22=30$

Trick: Do all × and ÷ first, then add all positives ($76+32=108$) and subtract all negatives ($56+22=78$): $108-78=30$.

Answer: (C) 30

Concept: A letter common to two words gets the digit common to both codes.

1. ROME → 7245 and TURN → 5689.
2. The only letter common to both words is R.
3. The only digit common to both codes is 5, so R = 5.

Trick: Match common letter with common digit – no need to decode the other letters.

Answer: (B) 5

Concept: Each letter moves +4 from one cluster to the next.

1. CX → GB → KF: C+4=G, G+4=K; X+4=B, B+4=F.
2. WR → AV → EZ: W+4=A, A+4=E; R+4=V, V+4=Z ✓
3. Others fail, e.g. FA-JE-NJ: E+4=I, not J.

Trick: Check just the second letters: X→B→F (+4); only R→V→Z also goes +4.

Answer: (B) WR-AV-EZ

Concept: A letter common to two words gets the digit common to both codes.

1. ROPE → 5391 and OILY → 2458.
2. The only common letter is O.
3. The only common digit is 5, so O = 5.

Trick: Common letter ↔ common digit.

Answer: (B) 5

Concept: Find the rule in the complete pair DAF : % from the first pair and apply.

1. Test option (C): OLQ → LIN: O−3=L, L−3=I, Q−3=N (each letter −3).
2. Same rule on DAF: D−3=A, A−3=X, F−3=C → AXC.
3. Both pairs follow −3, so # = OLQ and % = AXC.

Trick: Work backward from LIN (+3 gives OLQ) and forward from DAF (−3 gives AXC).

Answer: (C) # = OLQ, % = AXC

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise.

1. Fix T; B is 3rd to the left (clockwise) of T.
2. A and R are T's neighbours. If R were left of T, S (2nd right of R) would clash with A, so R is on T's right and A on T's left; then S is 2nd right of R.
3. Remaining two seats: the one next to B is D (C cannot sit beside B), the other is C.
4. Clockwise: T, A, D, B, S, C, R. From D going right: A, T, then R → 2 persons.

Trick: Place the fixed-distance clues (B–T, R–S) first; the neighbour clue then leaves only one arrangement.

Answer: (D) Two

Concept: Change each digit by the rule, then read the required positions.

1. 6753412: odd digits 7, 5, 3, 1 → +1 = 8, 6, 4, 2; even digits 6, 4, 2 → −2 = 4, 2, 0.
2. New number: 4 8 6 4 2 2 0
3. 2nd from left = 8, 2nd from right = 2; sum $=8+2=10$

Trick: Only transform the two digits you need: 7 → 8 and 1 → 2.

Answer: (B) 10

Concept: Compare position changes of first letters and of second letters.

1. BU – GK: B+5=G; U−10=K.
2. YR – DH: Y+5=D; R−10=H. MF – RV: M+5=R; F−10=V (6−10 → 22).
3. TN – YC: T+5=Y but N−11=C (not −10) → odd one.

Trick: First letters all move +5; check only the second letters for the break.

Answer: (C) TN – YC

Concept: Swap the signs as told, then apply BODMAS.

1. Replace: $96÷6\times 8+71-26$
2. $96÷6=16$; $16\times 8=128$
3. $128+71-26=173$

Trick: Write the converted expression fully before calculating – most errors come from half-swapped signs.

Answer: (C) 173

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

1. All cats are inside owls; some owls are rats – but the rats may lie in the part of owls outside cats.
2. So 'Some rats are cats' is not certain → (I) does not follow.
3. Bats are only kept away from rats; nothing stops bats from touching cats, so 'Some bats can be cats' is a valid possibility → (II) follows.

Trick: 'Can be' conclusions follow unless a 'No' statement blocks that exact pair.

Answer: (C) Only conclusion (II) follows