RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Let a pen cost ₹$p$ and a pencil ₹$c$; form two equations.

1. $6p+5c=259$ … (i)
2. $19(p-9)+2(c+4)=90\Rightarrow 19p+2c=253$ … (ii)
3. (i)$\times 2$: $12p+10c=518$; (ii)$\times 5$: $95p+10c=1265$ → $83p=747$, $p=9$
4. $5c=259-54=205\Rightarrow c=41$
5. $13p+2c=117+82=199$

Trick: Eliminate the variable whose coefficients are easy to equalise ($5c$ and $2c$ → $10c$).

Answer: (B) ₹$199$

Concept: A root of an equation satisfies it: substitute the common root.

1. Check: ${11}^2-9(11)-22=121-99-22=0$ ✓
2. Put $x=11$ in $2x^2-3x+c=0$: $2(121)-33+c=0$
3. $242-33+c=0\Rightarrow c=-209$

Trick: Only the equation containing c needs the substitution; the second one only confirms 11 is a root.

Answer: (D) $-209$

Concept: Solve brackets from inside out: ( ) → { } → [ ].

1. $(27÷9)=3$
2. $\{63-3\}=60$, then $60÷6=10$
3. $[7+10]=17$
4. $49+17=66$

Trick: Innermost bracket first, then division before addition.

Answer: (D) $66$

Concept: $\text{Sum}=\text{Average}\times \text{Count}$

1. First two numbers: $77\times 2=154$; third $=129$
2. Remaining two $=655-154-129=372$
3. Their average $=\frac{372}{2}=186$

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

Answer: (B) $186$

Concept: Mode is the value that occurs most often.

1. Count: 9 → 4 times; 2 → 3 times; 4 → 3 times; 5 → 3 times
2. 1, 6, 8 → once each
3. Highest frequency belongs to 9.

Trick: Tally only the repeated values – the one with the longest tally is the mode.

Answer: (D) $9$

Concept: Convert all fractions to the common denominator 16.

1. $\frac{2}{8}=\frac{4}{16}$, so $\frac{4}{16}+\frac{4}{16}+\frac{1}{16}+\frac{4}{16}=\frac{13}{16}$
2. $1=\frac{16}{16}$
3. $\frac{13}{16}-\frac{16}{16}=-\frac{3}{16}$

Trick: $3\times \frac{2}{8}=\frac{3}{4}$; $\frac{3}{4}+\frac{1}{16}$ is less than 1, so the answer must be negative.

Answer: (C) $-\frac{3}{16}$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

1. Mode 165 lies at the end of 160–165, so take modal class 160–165: $l=160$, $h=5$, $f_1=37$, $f_0=26$, $f_2=x$
2. $165=160+\frac{37-26}{74-26-x}\times 5$
3. $5=\frac{55}{48-x}\Rightarrow 48-x=11$
4. $x=37$

Trick: Plug the options in: only $x=37$ makes $\frac{11}{48-x}\times 5=5$.

Answer: (C) $37$

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. Interior sum $={360}^{\circ }+100\%\text{ of }{360}^{\circ }={720}^{\circ }$
2. $(n-2)\times 180=720\Rightarrow n=6$
3. Each interior angle $=\frac{720}{6}={120}^{\circ }$

Trick: Interior sum $=k\times {360}^{\circ }$ → $n=2k+2$. Here $k=2$, so $n=6$.

Answer: (D) $120$

Concept: Successive discounts multiply the remaining fractions: $\text{Remaining}=(1-d_1)(1-d_2)(1-d_3)$.

1. $0.94\times 0.72=0.6768$
2. $0.6768\times 0.85=0.57528$
3. Single discount $=1-0.57528=0.42472=42.47\%$

Trick: Two at a time: $6\%,28\%\to 6+28-\frac{6\times 28}{100}=32.32\%$; then with $15\%$: $32.32+15-\frac{32.32\times 15}{100}=42.47\%$.

Answer: (D) $42.47\%$

Concept: Convert each ratio to a decimal (or compare by cross-multiplication).

1. $\frac{16}{57}\approx 0.281$, $\frac{17}{53}\approx 0.321$
2. $\frac{24}{54}=\frac{4}{9}\approx 0.444$
3. $\frac{28}{60}=\frac{7}{15}\approx 0.467$
4. Check: $28\times 54=1512>24\times 60=1440$, so $28:60$ is the greatest.

Trick: Simplify first: $\frac{4}{9}$ vs $\frac{7}{15}$ → cross-multiply $60$ vs $63$.

Answer: (D) $28:60$

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

1. "All bikes are roads" puts Bikes inside Roads – it never makes all roads bikes, so (I) does not follow.
2. "Some roads are TVs" means only a part of Roads meets TVs; roads outside TVs are possible, so (II) does not follow.
3. Both conclusions are "All" type, but the statements give no such link.

Trick: "All A are B" never gives "All B are A" – the reverse of an All statement is only "Some B are A".

Answer: (C) Neither conclusion (I) nor (II) follows

Concept: Find the shift of the first letters and of the second letters separately.

1. OL-QN-SU: O→Q→S is $+2,+2$; L→N→U is $+2,+7$.
2. LI-NK-PR: L→N→P is $+2,+2$; I→K→R is $+2,+7$. Same rule.
3. RO-TQ-VX: R→T→V is $+2,+2$; O→Q→X is $+2,+7$ ✔. In (C) Q→Y is $+8$, in (A)/(D) O→T or O→R breaks $+2$.

Trick: Check only the last letter first: second-letter jump must be $+7$ (Q→X).

Answer: (B) RO-TQ-VX

Concept: Interchange the signs first, then solve using BODMAS.

1. Swap ÷ ↔ – and × ↔ +: $89-4984÷89+9\times 2$
2. $4984÷89=56$ and $9\times 2=18$
3. $89-56+18=51$

Trick: Rewrite the whole expression once with new signs before calculating – avoids half-swapped mistakes.

Answer: (B) $51$

Concept: Interchange the signs, then apply BODMAS.

1. Swap + ↔ − and × ↔ ÷: $405\times 2-369÷9+32$
2. $405\times 2=810$, $369÷9=41$
3. $810-41+32=801$

Trick: Do division/multiplication first; only then add and subtract left to right.

Answer: (B) $801$

Concept: Compare letter positions of the pair: each letter moves by a different step.

1. J→K ($+1$), K→M ($+2$), L→O ($+3$); PQR→QSU also $+1,+2,+3$.
2. X $+1$ = Y; Y $+2$ = A (Z, A – wrapping after Z); Z $+3$ = C (A, B, C).
3. So XYZ → YAC.

Trick: After Z, count again from A (circular alphabet).

Answer: (D) YAC

Concept: Identify the highest and lowest numbers, then pick the required digits.

1. Highest number = 884 → second digit = 8.
2. Lowest number = 316 → third digit = 6.
3. $8+6=14$

Trick: Compare only first digits to find highest/lowest quickly: 8 (884) and 3 (316).

Answer: (A) $14$

Concept: Facing the centre: right = anticlockwise, left = clockwise.

1. Fix F. C is immediate left of F, G is immediate right of F ⇒ G, F, C are together.
2. D is third to the right of C. B is not next to C and A is second to the right of B ⇒ only B with A two places right of it fits, so E takes the last seat.
3. Clockwise order: F, C, A, E, B, D, G.
4. Going left (clockwise) from G: F, C, A, then E ⇒ three persons between.

Trick: Fix the pair with "immediate" clues first, then place third/second-to clues.

Answer: (A) Three

Concept: Apply the rule to each digit separately.

1. 3762184: odd digits $+1$ → 3→4, 7→8, 1→2.
2. Even digits $-2$ → 6→4, 2→0, 8→6, 4→2.
3. New number 4840262: highest 8, lowest 0 ⇒ $8+0=8$.

Trick: Write the new digit under each old digit – no mix-ups.

Answer: (D) $8$

Concept: Track the letters and the numbers separately.

1. First letters: G, H, I, J → K ($+1$). Second: O, P, Q, R → S. Third: Q, R, S, T → U.
2. Numbers: 35, 32, 29, 26 → $26-3=23$.
3. Next term = KSU 23.

Trick: Number pattern ($-3$) alone removes the 22 options; letters $+1$ give KSU.

Answer: (A) KSU 23

Concept: Start with the fixed clues ("no one above", "immediately below"), then use gap clues.

1. No one above I ⇒ I on floor 6. Two between I and H ⇒ H on 3.
2. K odd, not 1, not 3 ⇒ K on 5; X immediately below ⇒ X on 4.
3. Two between K and G ⇒ G on 2; J on the remaining floor 1.
4. Between J (1) and X (4): G and H ⇒ two persons.

Trick: A "top" clue (no one above) is the best anchor – place it first.

Answer: (A) Two