RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Correct total = wrong total − wrong value + correct value.

1. Wrong total $=10\times 55=550$
2. Correct total $=550-48+84=586$
3. Correct average $=\frac{586}{10}=58.6$

Trick: Change in average $=\frac{84-48}{10}=3.6$, so $55+3.6=58.6$.

Answer: (D) 58.6

Concept: Take total work = LCM of days; work backwards from the end.

1. Total work $=\text{LCM}(16,10)=80$ units; Ramesh $=5$, Suresh $=8$ units/day.
2. Last 4 days Suresh alone: $4\times 8=32$ units.
3. Remaining $80-32=48$ units done together at 13 units/day: $\frac{48}{13}$ days.
4. Total $=\frac{48}{13}+4=\frac{100}{13}=7\frac{9}{13}$ days.

Trick: "Leaves x days before completion" → subtract the other person's x-day work first.

Answer: (B) $7\frac{9}{13}$ days

Concept: Sum = average × count.

1. First two numbers: $2\times 77=154$
2. Remaining two $=655-154-101=400$
3. Average $=\frac{400}{2}=200$

Trick: Subtract known parts from the total, then divide by how many are left.

Answer: (C) 200

Concept: Exterior angle bisectors at B and C meet at the excentre P, and $\angle P={90}^{\circ }-\frac{\angle A}{2}$.

1. $\angle CBD={180}^{\circ }-B$, $\angle BCE={180}^{\circ }-C$; halves are ${90}^{\circ }-\frac{B}{2}$ and ${90}^{\circ }-\frac{C}{2}$.
2. In $△BPC$: $\angle P={180}^{\circ }-({90}^{\circ }-\frac{B}{2})-({90}^{\circ }-\frac{C}{2})=\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$
3. $\angle P={90}^{\circ }-{42}^{\circ }={48}^{\circ }$

Trick: Internal bisectors: ${90}^{\circ }+\frac{A}{2}$; external bisectors: ${90}^{\circ }-\frac{A}{2}$.

Answer: (C) ${48}^{\circ }$

Concept: For $n$ odd, median = ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value after arranging in order.

1. Ascending order: 24, 26, 38, 39, 43, 49, 60, 61, 62, 72, 95
2. $n=11$, so median is the $6^{\text{th}}$ value.
3. Median $=49$

Trick: Always sort first – the 6th number in the given (unsorted) list is not the median.

Answer: (D) 49

Concept: In $a:b::c:d$, product of extremes = product of means.

1. $7:14::8.7:x\Rightarrow x=\frac{14\times 8.7}{7}=17.4$
2. $5:35::7:y\Rightarrow y=\frac{35\times 7}{5}=49$
3. $x:y=17.4:49=174:490=87:245$

Trick: $x=2\times 8.7$ and $y=7\times 7$ – the ratios are just "double" and "seven times".

Answer: (C) $87:245$

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

1. Upstream speed $=\frac{176}{16}=11\text{km/h}$
2. Downstream speed $=\frac{85}{5}=17\text{km/h}$
3. Boat speed $=\frac{17+11}{2}=14\text{km/h}$ (stream $=3\text{km/h}$)

Trick: Boat = average of the two speeds; stream = half their difference.

Answer: (C) $14\text{km/h}$

Concept: Successive gains multiply: selling price $=\text{CP}\times \frac{110}{100}\times \frac{115}{100}\times \frac{125}{100}$.

1. Combined factor $=1.10\times 1.15\times 1.25=1.58125$.
2. Cost of production $=\frac{31900}{1.58125}\approx 20173.91$, i.e. about ₹20,173.91, not a whole number of rupees in the options.
3. Check each option: $20000\times 1.58125=31625$, $23200\times 1.58125=36685$, $25000\times 1.58125=39531.25$, $22000\times 1.58125=34787.5$; none gives ₹31,900.
4. So no option is correct as printed.

Note: Multiply the option by the combined factor to check it quickly. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

Concept: $\text{CSA}=2\pi rh$, $\text{TSA}=2\pi r(h+r)$

1. $2\pi rh=\frac{1}{2}\times 2\pi r(h+r)$
2. $2h=h+r\Rightarrow r=h=12\text{cm}$
3. Diameter $=2r=24\text{cm}$

Trick: CSA = half of TSA means CSA = area of the two ends, i.e. $2\pi rh=2\pi r^2\Rightarrow h=r$.

Answer: (A) 24

Concept: BODMAS – solve innermost brackets first, then multiplication, then addition, then division.

1. $(5-7)=-2$
2. $9\times (-2)=-18$
3. $\{21-18\}=3$
4. $81÷3=27$

Trick: Watch the sign: the small bracket is negative, so the curly bracket becomes $21-18$.

Answer: (A) 27

Concept: After changing signs, solve using BODMAS.

1. Replace: ‘+’ → ‘−’, ‘×’ ↔ ‘÷’.
2. New expression: $20\times 5-8÷2-1$
3. $=100-4-1=95$

Trick: Rewrite the whole expression first, then apply BODMAS – never solve while changing signs.

Answer: (A) $95$

Concept: Find the highest and lowest of the given numbers, then use their first digits.

1. Numbers: 411, 559, 916, 247, 769.
2. Highest = 916 → first digit 9; Lowest = 247 → first digit 2.
3. $9-2=7$

Trick: Compare only the hundreds digit first – 9 (916) is highest, 2 (247) is lowest.

Answer: (B) $7$

Concept: Find the letter shifts from # to EPV and apply the same to RFS.

1. Try option (D): D → E (+1), N → P (+2), U → V (+1) gives EPV.
2. Apply +1, +2, +1 to RFS: R → S, F → H, S → T = SHT.
3. So # = DNU and % = SHT.

Trick: Check options quickly: only DNU gives EPV with a neat +1, +2, +1 pattern.

Answer: (D) # = DNU, % = SHT

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

1. MICE → 6398 and PEAR → 3215.
2. Common letter: E. Common digit: 3.
3. So E = 3.

Trick: Common letter ↔ common digit (order of digits does not matter).

Answer: (C) $3$

Concept: Check each number: the element before it must be a number and the element after it must be a symbol.

1. 8 9 # → 9 (preceded by 8, followed by #) ✔
2. 1 9 © → 9 ✔; 3 5 © → 5 ✔
3. Other numbers fail (e.g. 5 after ©, 4 followed by 4).
4. Count = 3.

Trick: Scan for the pattern “number–number–symbol”; the middle number is counted.

Answer: (A) $3$

Concept: Fix the people with the most definite clues first (all face north, so left = left end side).

1. Five between G and D → they sit at the two ends. F is immediate left of D, so D = 7 (right end), F = 6, G = 1.
2. One person between F and M → M = 4.
3. N, O, E fill 2, 3, 5 with N left of O left of E → N = 2, O = 3, E = 5.
4. Third from left = O.

Trick: “Only five between” in a row of 7 always means the two ends.

Answer: (D) O

Concept: Interchange the signs, then solve by BODMAS.

1. ‘+’ ↔ ‘−’ and ‘×’ ↔ ‘÷’.
2. New expression: $215\times 4-822÷3+45$
3. $=860-274+45=631$

Trick: Do division/multiplication first: $822÷3=274$, $215\times 4=860$.

Answer: (B) $631$

Concept: Look at the differences between consecutive terms.

1. Differences: $3,9,15,21,27$ (increase by 6).
2. Next difference $=27+6=33$
3. $96+33=129$

Trick: Second difference constant (6) → next first difference is 33.

Answer: (D) $129$

Concept: Use letter positions: in each pair the 2nd letter = 1st + 3, and each pair starts 2 letters before the previous one.

1. QT-OR-MP: Q→T (+3), Q→O→M (−2 each).
2. XA-VY-TW: X→A (+3, wrapping), X→V→T (−2).
3. TW-RU-PS: T→W, R→U, P→S (+3) and T→R→P (−2) ✔
4. JM-HK-FJ fails (F→J is +4).

Trick: Check “+3 inside, −2 across” on each option.

Answer: (C) TW-RU-PS

Concept: Track the net east–west and north–south displacement.

1. South 37, right (west) 18, right (north) 58, left (west) 61, left (south) 21.
2. North–south: $-37+58-21=0$ → Z is level with Y.
3. East–west: $18+61=79$ km west of Y.
4. So he must drive 79 km towards the east.

Trick: When the vertical moves cancel, the answer is just the sum of the horizontal moves.

Answer: (D) 79 km towards the east