RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Successive discounts multiply the remaining fractions of price.

1. Remaining after 29%, 18%, 9%: $0.71\times 0.82\times 0.91$
2. $0.71\times 0.82=0.5822$; $0.5822\times 0.91=0.529802$
3. Single discount $=(1-0.529802)\times 100=47.02\%$

Trick: Take two at a time: $a+b-\frac{ab}{100}$; $29,18\to 41.78$; $41.78,9\to 47.02\%$.

Answer: (A) $47.02\%$

Concept: Take CP = 100; after swapping, the old SP becomes the new CP.

1. Profit 79% → SP = 179
2. Swapped: CP = 179, SP = 100, loss = 79
3. $\text{Loss}\%=\frac{79}{179}\times 100=44.13\%$

Trick: Swap rule: $\text{Loss}\%=\frac{P}{100+P}\times 100$.

Answer: (D) $44.13\%$

Concept: Split the middle term: find two numbers with product $5\times (-17)=-85$ and sum $-12$.

1. The numbers are $-17$ and $5$.
2. $5x^2-17x+5x-17=x(5x-17)+1(5x-17)=(x+1)(5x-17)$
3. $x=-1$ or $x=\frac{17}{5}$

Trick: If $a-b+c=0$ (here $5+12-17=0$), one root is $-1$ and the other is $-\frac{c}{a}=\frac{17}{5}$.

Answer: (B) $-1,\frac{17}{5}$

Concept: Centroid $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

1. $\frac{x+3-6}{3}=3.5\Rightarrow x-3=10.5\Rightarrow x=13.5$
2. $\frac{5-4+y}{3}=4.5\Rightarrow 1+y=13.5\Rightarrow y=12.5$
3. $(x+3,y-4)=(16.5,8.5)$

Trick: Sum of x-coordinates $=3\times 3.5=10.5$; sum of y-coordinates $=3\times 4.5=13.5$.

Answer: (A) $(16.5,8.5)$

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100-\text{Loss}\%}$; new $\text{SP}=\text{CP}\times \frac{100+\text{Gain}\%}{100}$.

1. $\text{CP}=\frac{120000\times 100}{60}=200000$
2. Required $\text{SP}=200000\times \frac{155}{100}=310000$

Trick: Direct: $120000\times \frac{155}{60}=310000$.

Answer: (C) ₹$3,10,000$

Concept: When interest is compounded every 10 months, convert both rate and time to 10-month periods.

1. Rate per period $=12\times \frac{10}{12}=10\%$
2. Time $=2\frac{1}{2}$ years $=30$ months $=3$ periods
3. $x\times (1.1{)}^3=83853\Rightarrow x\times 1.331=83853$
4. $x=\frac{83853}{1.331}=63000$

Trick: $(1.1{)}^3=1.331=\frac{1331}{1000}$; $83853÷1331=63$.

Answer: (D) $63,000$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$

1. $1000=\frac{2500\times 10\times T}{100}$
2. $1000=250T\Rightarrow T=4$

Trick: Interest per year $=10\%$ of 2500 $=250$; $1000÷250=4$.

Answer: (C) $4$ years

Concept: $\frac{x+1}{x}=1+\frac{1}{x}$, so the given condition fixes $\frac{1}{x}$ directly; also $\frac{x^2+1}{x^2}=1+\frac{1}{x^2}$.

1. $\frac{x+1}{x}=17\Rightarrow 1+\frac{1}{x}=17\Rightarrow \frac{1}{x}=16$.
2. $\frac{x^2+1}{x^2}=1+{\left(\frac{1}{x}\right)}^2=1+{16}^2=1+256=257$.
3. $257$ is not among the options (279, 288, 277, 287), so no option is correct.
4. If the condition had been $x+\frac{1}{x}=17$, then $x^2+\frac{1}{x^2}={17}^2-2=287$, option (D).

Note: Read the fraction carefully: $\frac{x+1}{x}$ is not the same as $x+\frac{1}{x}$. 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: $(a\pm b{)}^2=a^2+b^2\pm 2ab$

1. $(a+b{)}^2=90+54=144\Rightarrow a+b=\pm 12$
2. $(a-b{)}^2=90-54=36\Rightarrow a-b=\pm 6$
3. $\frac{a+b}{a-b}=\frac{12}{6}=2$ (or $-2$); the possible value in the options is 2

Trick: Also: $a=9,b=3$ fits ($81+9=90,ab=27$) → $\frac{12}{6}=2$.

Answer: (A) $2$

Concept: x must divide the LCM and must supply every prime power of the LCM that 48 and 88 lack.

1. $48=2^4\times 3$, $88=2^3\times 11$, $4752=2^4\times 3^3\times 11$
2. x must contain $3^3=27$ and have no prime other than 2, 3, 11.
3. $216=2^3\times 3^3$ ✓; 202 has 101, 307 is prime, 123 has 41 – none divide 4752.

Trick: First test: does the option divide the LCM? $4752÷216=22$ ✓.

Answer: (C) $216$

Concept: Total students = rank from top + rank from bottom − 1.

1. Students above Firoz = $38-1=37$; below him = $28-1=27$.
2. Total = $37+1+27=65$.

Trick: Total $=T+B-1=38+28-1=65$.

Answer: (A) 65

Concept: Facing the centre, 'left' means clockwise and 'right' means anticlockwise.

1. Fix S. T is third to the left of S. Only S sits between L and M, so L and M are the two neighbours of S.
2. M is second to the left of A. If M were on S's left, A would land on L's seat, so M is on S's right (anticlockwise) side, L on S's left, and A is two places to the right of M.
3. The two seats left go to B and Z. B cannot sit next to T, so B takes the seat beside A on M's side, and Z sits next to T.
4. Going right from Z: L, S, then M, so 2 people sit between them.

Trick: Draw the circle first, place S and T, then the pair L–S–M; the rest fall in place.

Answer: (B) 2

Concept: Fix the definite clues first (Q on Sunday), then the pair clues.

1. Four between P and U ⇒ positions (Mon, Sat) or (Tue, Sun). Sunday is Q, so P and U are Mon and Sat.
2. T is right after P; if P were Sat, T would be Sun (Q's day). So P = Mon, T = Tue, U = Sat.
3. V, R, S fill Wed, Thu, Fri in order V < R < S.
4. Between V (Wed) and P (Mon): only T → one person.

Trick: Clue with 'only four between' in a 7-slot row has just two placements – test both.

Answer: (A) One

Concept: Check every symbol: the element just before and just after it must both be numbers.

1. Series: 5 @ 6 # 8 $ % 1 9 ^ 7 ≺ 2 3 ⋟ 4
2. @ (5, 6) ✓, # (6, 8) ✓, $ (8, %) ✗, % ($, 1) ✗
3. ^ (9, 7) ✓, ≺ (7, 2) ✓, ⋟ (3, 4) ✓
4. Total = 5, i.e. more than three.

Trick: Two symbols together ($ %) can never qualify – skip them quickly.

Answer: (D) More than three

Concept: Look for the pattern symbol – number – symbol.

1. Series: € & % & 5 6 3 1 © 1 4 6 @ £ @ $ 8 8 £ 8 % 1
2. 5 (after &) is followed by 6 ✗; 1 after © is followed by 4 ✗; 8 after $ is followed by 8 ✗.
3. £ 8 % ✓ – the only one; the last 1 has nothing after it.
4. Count = 1.

Trick: Scan only numbers that stand alone between two symbols.

Answer: (D) 1

Concept: Write the direct codes, then apply only the conditions that are satisfied.

1. Direct codes: 3 → A, ÷ → E, 7 → S, @ → W, 4 → T.
2. Condition (i): first 3 is prime and last 4 is even → swap codes: first = T, last = A.
3. Condition (ii): third element 7 is not a symbol – no change. Condition (iii): second element ÷ is not a number – no change.
4. Code = T E S W A.

Trick: Check conditions one by one; most groups satisfy only one.

Answer: (D) TESWA

Concept: Find how KRX changes; apply the same change to get MTZ from #.

1. Positions: H(8) O(15) U(21) → +5 each → M(13) T(20) Z(26).
2. Same rule on KRX: K+5 = P, R+5 = W, X+5 = C (wrap after Z).
3. So # = HOU and % = PWC.

Trick: Check one letter first: X + 5 = C eliminates PWP straight away.

Answer: (D) # = HOU, % = PWC

Concept: Look at the differences between consecutive terms.

1. $1320-1312=8$, $1330-1320=10$, $1342-1330=12$, $1356-1342=14$
2. Differences rise by 2 → next difference $=16$.
3. $1356+16=1372$

Trick: Differences 8, 10, 12, 14, 16 – an even-number ladder.

Answer: (A) 1372

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

1. YOUR → 5683 and UNIT → 2519.
2. Common letter: U. Common digit: 5.
3. So U = 5.

Trick: Match common letters with common digits.

Answer: (A) 5

Concept: Track each move on a grid (East +x, North +y).

1. P(0, 0) → 11 km South → (0, −11).
2. Left (now East) 6 km → (6, −11); right (South) 12 km → (6, −23).
3. Left (East) 1 km → (7, −23); left (North) 23 km → Q(7, 0).
4. Q is 7 km East of P, so he must drive 7 km towards the West.

Trick: Net N–S: −11 −12 +23 = 0; net E–W: 6 + 1 = 7 East → return 7 km West.

Answer: (C) 7 km towards the West