RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Take total work = LCM(5, 55, 66) = 330 units.

1. Efficiencies: A $=\frac{330}{5}=66$, B $=\frac{330}{55}=6$, C $=\frac{330}{66}=5$ units/day
2. 4 days together: $(66+6+5)\times 4=308$ units
3. Left: $330-308=22$ units by B + C at 11/day ⇒ $\frac{22}{11}=2$ days
4. Total $=4+2=6$ days

Trick: Units method avoids fractions – always use LCM as total work.

Answer: (D) 6

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$. Let first part $=d$, second part $=2d$.

1. Onward time $=\frac{d}{88}+\frac{2d}{16}=\frac{d}{88}+\frac{11d}{88}=\frac{12d}{88}=\frac{3d}{22}$
2. Total distance $=6d$; total time $=\frac{6d}{33}=\frac{2d}{11}=\frac{4d}{22}$
3. Return time $=\frac{4d}{22}-\frac{3d}{22}=\frac{d}{22}$
4. Return speed $=\frac{3d}{d/22}=66\text{km/h}$

Trick: Take $d=88$: onward time 12 h, total time 16 h, return time 4 h for 264 km ⇒ 66 km/h.

Answer: (C) 66

Concept: The largest circle inside a rectangle touches the two longer sides, so its diameter = the shorter side.

1. Diameter $=112\text{m}$
2. Circumference $=\pi d=\frac{22}{7}\times 112=22\times 16=352\text{m}$

Trick: Diameter = smaller side of the rectangle; the longer side (281) is irrelevant.

Answer: (A) 352

Concept: Take square roots of numerator and denominator separately.

1. $\sqrt{0.04}=0.2$ and $\sqrt{0.0004}=0.02$
2. $\frac{0.2}{0.02}=10$

Trick: $\sqrt{\frac{0.04}{0.0004}}=\sqrt{100}=10$ in one step.

Answer: (C) 10

Concept: Let pen $=p$, pencil $=c$ and solve the two equations.

1. $8p+2c=96$ and $4p+7c=72$
2. Double the second: $8p+14c=144$; subtract: $12c=48\Rightarrow c=4$
3. $8p=96-8=88\Rightarrow p=11$
4. $2p+8c=22+32=54$

Trick: Make the pen coefficients equal (8) and subtract to get pencils first.

Answer: (C) ₹54

Concept: Sum of interior angles of an $n$-sided polygon $=(n-2)\times {180}^{\circ }$.

1. $(n-2)\times 180=1800$
2. $n-2=10\Rightarrow n=12$

Trick: Divide the sum by 180 and add 2: $\frac{1800}{180}+2=12$.

Answer: (C) 12

Concept: Solve innermost brackets first, then multiplication, then addition, then division.

1. $2-6=-4$
2. $7\times (-4)=-28$
3. $30+(-28)=2$
4. $160÷2=80$

Trick: Watch the sign: $7\times (-4)$ is negative, so the curly bracket becomes just 2.

Answer: (A) 80

Concept: $\text{SP}=\text{CP}\times (1+\text{profit})$ and $\text{SP}=\text{MP}\times (1-\text{discount})$.

1. $\text{SP}=546\times 1.6=873.6$
2. $\text{MP}\times 0.78=873.6$
3. $\text{MP}=\frac{873.6}{0.78}=1120$

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{100+P}{100-D}=\frac{160}{78}$, and $546=78\times 7$, so MP $=160\times 7=1120$.

Answer: (B) 1,120

Concept: Income = Expenditure + Savings. Savings 30% ⇒ expenditure is 70% of income.

1. $70\%\text{ of income}=840$
2. $\text{Income}=\frac{840\times 100}{70}=1200$

Trick: $70\%\to 840$ ⇒ $10\%\to 120$ ⇒ $100\%\to 1200$.

Answer: (D) 1,200

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

1. $D>0$: real and distinct; $D=0$: real and equal
2. $D<0$: $\sqrt{D}$ is imaginary, so roots are $\frac{-b\pm i\sqrt{|D|}}{2a}$ – a pair of complex conjugates (for real coefficients).

Trick: Negative D ⇒ no real roots ⇒ complex pair $p\pm iq$.

Answer: (B) complex conjugates

Concept: When two words share a letter, the digit common to both codes is the code of that letter.

1. CART → 4268 and RISE → 3527.
2. The only common letter is R.
3. The only common digit in the two codes is 2.

Trick: Common letter ↔ common digit; ignore order of digits.

Answer: (A) 2

Concept: Fix genders from marriages and 'brother' clues, then trace the path from B to A.

1. D is a brother (male), so his wife R is female – B's mother.
2. E is R's sister, so E is B's mother's sister (maasi).
3. A is married to E, so A is the husband of B's mother's sister.

Trick: Mother's sister's husband = maasa (मौसा).

Answer: (A) Mother's sister's husband

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

1. HUNT → 6438 and USER → 5723.
2. Common letter: U.
3. Common digit: 3.

Trick: Match common letter with common digit.

Answer: (A) 3

Concept: Check each number: the element just before it and just after it must both be numbers.

1. Run 4 3 5 6 7: inside numbers 3, 5, 6 → 3 qualify.
2. Run 9 2 7 5 9 (after %): inside numbers 2, 7, 5 → 3 qualify.
3. 1 3 (between * and %) and lone 7 have a symbol on one side.

Trick: In a run of k consecutive numbers, exactly $k-2$ qualify: $(5-2)+(5-2)=6$.

Answer: (B) 6

Concept: Look only at symbols; each must have a number immediately on both sides.

1. 3 * 2 → * qualifies.
2. 9 & 7 → & qualifies.
3. 9 @ 4 → last @ qualifies; other @, © and # touch another symbol or the end.

Trick: Scan for the pattern number–symbol–number.

Answer: (D) 3

Concept: Find the step pattern inside each set.

1. (22, 28, 31): $22+6=28,28+3=31$
2. (39, 45, 48): $39+6=45,45+3=48$
3. (27, 33, 36): $27+6=33,33+3=36$ ✓; others fail (e.g. 53, 59, 65 is +6, +6).

Trick: Pattern +6, +3 → also first and third differ by 9.

Answer: (D) (27, 33, 36)

Concept: Draw the minimum Venn diagram consistent with all statements.

1. Sparrows ⊂ Parrots ⊂ Macau, so every sparrow is a macau → I follows.
2. Some sparrows are doves; those sparrows are parrots, so some doves are parrots → II follows.

Trick: 'All A are B' + 'Some A are C' → 'Some C are B' is always definite.

Answer: (A) Both conclusions (I) and (II) follow

Concept: Find the shift from the right pair's rule and test the options.

1. Take # = UZV: U−3 = R, Z−3 = W, V−3 = S → RWS ✓
2. Apply the same −3 to INJ: I−3 = F, N−3 = K, J−3 = G → FKG
3. So # = UZV and % = FKG.

Trick: Positions: U(21)→R(18), I(9)→F(6): a uniform −3.

Answer: (C) # = UZV, % = FKG

Concept: Track coordinates (east +x, north +y) after every turn.

1. Y(0, 0) → E 22 → (22, 0); right = S 33 → (22, −33); left = E 25 → (47, −33)
2. Left = N 51 → (47, 18); left = W 47 → (0, 18); left = S 23 → Z(0, −5)
3. Z is 5 km directly south of Y, so he must go 5 km north.

Trick: Net E–W: $22+25-47=0$; net N–S: $-33+51-23=-5$ → 5 km north to return.

Answer: (A) 5 km towards the north

Concept: Number positions 1 (bottom) to 7 (top) and place fixed boxes first.

1. Two below T → T = 3; one above U → U = 6; one box between U and D → D = 4.
2. Free: 1, 2, 5, 7. G just above S → only S = 1, G = 2 works.
3. F below E → F = 5, E = 7.
4. Between E (7) and G (2): U, F, D, T → four boxes.

Trick: Boxes between positions p and q $=|p-q|-1=7-2-1=4$.

Answer: (D) Four