RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ per sharpener.

1. CP of one sharpener $=\frac{150}{12}=12.5$ (₹)
2. SP $=21$ (₹), profit $=21-12.5=8.5$ (₹)
3. $\text{Profit}\%=\frac{8.5}{12.5}\times 100=68\%$

Trick: Compare per dozen: SP $=21\times 12=252$ vs CP 150 → $\frac{102}{150}\times 100=68\%$.

Answer: (D) 68

Concept: $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. Discount $=20\%$ of 2400 $=480$ (₹)
2. SP $=2400-480=1920$ (₹)

Trick: $20\%$ off means pay $80\%$: $2400\times 0.8=1920$.

Answer: (C) ₹1,920

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

1. Sum $=40+49+56+73+37=255\text{kg}$
2. Average $=\frac{255}{5}=51\text{kg}$

Trick: Assume 50: deviations $-10,-1,+6,+23,-13$ sum to $+5$; $50+\frac{5}{5}=51$.

Answer: (C) 51 kg

Concept: Express the actual SP in terms of CP using the loss condition, then use the second selling price.

1. $\frac{2}{7}S=76\%$ of CP $=0.76C$
2. $S=\frac{7\times 0.76C}{2}=2.66C$
3. New SP $=55\%$ of $S=0.55\times 2.66C=1.463C$
4. Profit $=46.3\%$

Trick: Let CP $=100$: $\frac{2}{7}S=76\Rightarrow S=266$; $55\%$ of 266 $=146.3$ → $46.3\%$ profit.

Answer: (D) $46.3\%$

Concept: SP $=$ MP after discount; CP $=\frac{\text{SP}\times 100}{100+\text{profit}\%}$.

1. SP $=280\times \frac{87.5}{100}=245$ (₹)
2. CP $=\frac{245\times 100}{140}=175$ (₹)

Trick: $12.5\%=\frac{1}{8}$: SP $=280\times \frac{7}{8}=245$; $40\%=\frac{2}{5}$: CP $=245\times \frac{5}{7}=175$.

Answer: (C) ₹175

Concept: Profit share $\propto$ capital $\times$ time.

1. Shares: M $+$ N $=3304$, M $-$ N $=472$ ⇒ M $=1888$, N $=1416$; ratio $=4:3$
2. Let N’s capital $=x$; M’s $=x+43000$
3. $\frac{6(x+43000)}{7x}=\frac{4}{3}\Rightarrow 18x+774000=28x$
4. $x=77400$; M $=77400+43000=1,20,400$ (₹)

Trick: Check (A): M $=120400$, N $=77400$; $120400\times 6:77400\times 7=722400:541800=4:3$ ✓

Answer: (A) ₹1,20,400

Concept: $A=P{(1+\frac{r}{100})}^n$.

1. $\frac{1367631}{1000000}=(1.11{)}^n$
2. ${1.11}^2=1.2321$, ${1.11}^3=1.367631$
3. So $n=3$ years

Trick: ${111}^3=1367631$ — the six digits after 1 are the giveaway for a cube.

Answer: (A) 3 years

Concept: Total work = LCM of days; find work done together and subtract.

1. Total $=\text{LCM}(9,18)=18$ units; A $=2$/day, B $=1$/day
2. Together $=3$ units/day; in 5 days $=15$ units
3. Remaining $=18-15=3$ units $=\frac{3}{18}=\frac{1}{6}$

Trick: $5(\frac{1}{9}+\frac{1}{18})=5\times \frac{1}{6}=\frac{5}{6}$ done, so $\frac{1}{6}$ left.

Answer: (A) $\frac{1}{6}$

Concept: Square: perimeter $=4a$, area $=a^2$.

1. $4a=160\Rightarrow a=40\text{m}$
2. Area $={40}^2=1600{\text{m}}^2$

Trick: Area of a square $={\left(\frac{\text{perimeter}}{4}\right)}^2$.

Answer: (A) 1600

Concept: Open the brackets carefully (sign of $-3$ changes every term), then combine like terms.

1. $2(\frac{3}{2}{x}^2-28x+14)=3x^2-56x+28$
2. $-3(x^2+6x-11)=-3x^2-18x+33$
3. Add: $3x^2-3x^2-56x-18x+28+33=-74x+61$

Yaad rakhein: A negative multiplier flips every sign inside: $-3\times (-11)=+33$.

Answer: (D) $-74x+61$

Concept: When everyone faces the centre, "right" means anticlockwise and "left" means clockwise.

1. C sits third to the left of B and K is immediately right of C, so B, K, C occupy a fixed block (going clockwise: B, _, K, C).
2. K is third to the right of J and A is second to the left of L; D is not a neighbour of C. The only arrangement that works, clockwise: A, B, D, K, C, L, J.
3. From D going right (anticlockwise): D → B → A → J → L, so B, A, J sit between them.

Trick: Draw the circle with arrows marking the right (anticlockwise) direction first; then place the fixed pairs.

Answer: (A) Three

Concept: Check the alphabet gaps between consecutive letters of each group.

1. SVR: S $+3$ = V, V $-4$ = R
2. KNJ: K $+3$ = N, N $-4$ = J; NQM: N $+3$ = Q, Q $-4$ = M
3. UXS: U $+3$ = X, X $-5$ = S ✗

Trick: Write the pattern as $(+3,-4)$; the group that breaks it is the odd one.

Answer: (D) UXS

Concept: Write the original and the rearranged numbers one under the other and compare each place.

1. Original: 3 6 7 1 9 5 4 2
2. Ascending: 1 2 3 4 5 6 7 9
3. No place has the same digit in both rows.

Trick: Compare position by position — do not count digits that merely appear in both.

Answer: (A) None

Concept: Look at the differences between consecutive terms.

1. Differences: $13-10=3$, $19-13=6$, $28-19=9$, $40-28=12$
2. They increase by 3 (multiples of 3), so next difference $=15$
3. $40+15=55$

Trick: Differences 3, 6, 9, 12, … → next is 15.

Answer: (C) 55

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

1. After swapping: $93+9-9÷3+8\times 23+6-3÷3+2$
2. Division/multiplication: $9÷3=3$, $8\times 23=184$, $3÷3=1$
3. $93+9-3+184+6-1+2=290$

Yaad rakhein: Rewrite the whole expression with the new signs before calculating — never swap in your head.

Answer: (C) 290

Concept: Look at the differences between consecutive terms.

1. $90-3=87$, $87-4=83$, $83-5=78$, $78-6=72$
2. Next subtract 7: $72-7=65$

Trick: Subtractions 3, 4, 5, 6, 7 — consecutive numbers.

Answer: (D) 65

Concept: Change each digit as told (odd $+1$, even $-2$) and then count repeats.

1. 3174658 → 3→4, 1→2, 7→8, 4→2, 6→4, 5→6, 8→6
2. New number: 4282466
3. 4 (twice), 2 (twice), 6 (twice), 8 (once) → three digits repeat

Yaad rakhein: Count the digits that repeat, not the number of repeated places.

Answer: (B) Three

Concept: Letters common to both words take the digits common to both codes.

1. NOPE = 6712, NOTE = 2178
2. Common letters N, O, E ↔ common digits 1, 2, 7
3. Left over: P = 6 and T = 8

Trick: Remove the common part; the unique letter gets the unique digit.

Answer: (B) 8

Concept: Find the shift of each letter position.

1. CFJP → FINT: $+3,+3,+4,+4$
2. ORZF → RUDJ: $+3,+3,+4,+4$ ✓
3. XALR: X$+3$ = A, A$+3$ = D, L$+4$ = P, R$+4$ = V → ADPV

Yaad rakhein: After Z the alphabet wraps round to A (X + 3 = A).

Answer: (B) ADPV

Concept: Track the position on x–y axes (East $+x$, North $+y$).

1. A(0, 0) → 9 km East → (9, 0)
2. Right → South 2 → (9, −2); right → West 4 → (5, −2)
3. Right → North 11 → (5, 9); left → West 5 → P(0, 9)
4. P is 9 km due North of A, so he must drive 9 km South.

Trick: Net east–west: $9-4-5=0$; net north–south: $11-2=9$ North → go back South.

Answer: (B) 9 km towards south