RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Two successive discounts $a\%$ and $b\%$ equal one discount of $(a+b-\frac{ab}{100})\%$.

1. $a=14$, $b=8$
2. Total $=14+8-\frac{14\times 8}{100}=22-1.12=20.88\%$
3. (Marked price ₹78,000 is not needed.)

Trick: Successive discount is always a little less than $a+b$ – here less than $22\%$.

Answer: (B) $20.88\%$

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

1. $(n-2)\times 180=2520$ ⇒ $n-2=14$ ⇒ $n=16$
2. Exterior angle $=\frac{{360}^{\circ }}{16}={22.5}^{\circ }$

Trick: Exterior $={180}^{\circ }-\text{interior}$; interior $=\frac{2520}{16}={157.5}^{\circ }$ ⇒ ${22.5}^{\circ }$.

Answer: (B) ${22.5}^{\circ }$

Concept: Value of one ratio part $=\frac{\text{total}}{\text{sum of ratio terms}}$.

1. Sum of terms $=2+12+11=25$; one part $=\frac{570}{25}=22.8$
2. Largest − smallest $=12-2=10$ parts
3. Difference $=10\times 22.8=228$

Trick: Difference $=\frac{10}{25}\times 570=\frac{2}{5}\times 570=228$.

Answer: (B) $228$

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=5\text{cm}$, $r^2=25$
2. $4\times 3.14\times 25=100\times 3.14=314{\text{cm}}^2$

Trick: $4\times 25=100$, so the answer is just $100\pi$.

Answer: (D) $314{\text{cm}}^2$

Concept: Under compound interest, if money becomes $k$ times in $T$ years, it becomes $k^n$ times in $nT$ years.

1. $3$ times in $15$ years
2. $27=3^3$ ⇒ time $=3\times 15=45$ years

Trick: CI multiplies: $3\to 9\to 27$ takes $15+15+15$ years. (In SI it would add, not multiply.)

Answer: (B) $45$ years

Concept: Let CP $=x$; write the new SP $=(1+\text{new profit}\%)\times$ new CP.

1. SP $=1.57x$
2. New CP $=x+20$, new SP $=1.57x-18$
3. $1.57x-18=1.475(x+20)=1.475x+29.5$
4. $0.095x=47.5$ ⇒ $x=500$

Trick: Check: CP $500$, SP $785$; new CP $520$, new SP $767$, profit $\frac{247}{520}=47.5\%$ ✓.

Answer: (B) ₹500

Concept: If a number increased by $r\%$ gives $N$, the number $=\frac{N}{1+\frac{r}{100}}$.

1. Number $\times 1.6=3610$
2. Number $=\frac{3610}{1.6}=2256.25$

Trick: Increase by $60\%$ = multiply by $\frac{8}{5}$, so reverse by $\times \frac{5}{8}$: $\frac{3610\times 5}{8}=2256.25$.

Answer: (A) $2256.25$

Concept: Profit $=$ total SP of items actually sold $-$ total CP of all items bought.

1. CP of $88$ lights $=$ ₹88
2. Lights left $=88-46=42$
3. SP $=42\times 3.10=$ ₹130.20
4. Profit $=130.20-88=$ ₹42.20

Trick: Broken items add to cost but earn nothing – count only the $42$ sold.

Answer: (C) ₹42.20

Concept: Open brackets carefully (sign of the outside number multiplies every term), then combine like terms.

1. $2(\frac{3}{2}{x}^2-20x+17)=3x^2-40x+34$
2. $-3(x^2+7x-15)=-3x^2-21x+45$
3. Sum $=0·x^2-61x+79=-61x+79$

Trick: The $x^2$ terms cancel ($3-3=0$); only the signs of $x$ and constant decide the option.

Answer: (C) $-61x+79$

Concept: Divisible by $11$ if (sum of digits at odd places) $-$ (sum at even places) is $0$ or a multiple of $11$.

1. Digits of $393064111$ from right: $1,1,1,4,6,0,3,9,3$
2. Odd places: $1+1+6+3+3=14$; even places: $1+4+0+9=14$
3. Difference $=0$ ⇒ divisible by $11$

Trick: Last digit $1$ is odd ⇒ not divisible by $10$, $14$ (even) or $15$ (needs $0/5$ at end) – only $11$ remains.

Answer: (A) $11$

Concept: Draw the least-overlap Venn diagram; a conclusion follows only if it is true in every possible diagram.

1. Bricks lie inside Spoons; Games lie fully outside Bricks.
2. I: Games may lie completely outside Spoons too ⇒ "Some games are spoons" is not certain.
3. II: "All spoons are games" is impossible to prove (bricks are spoons but not games) ⇒ false.

Trick: "All A are B" + "No A is C" gives only "Some B are not C" – nothing positive about B and C.

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

Concept: "Some A are B" converts to "Some B are A"; "No A is B" converts to "No B is A".

1. Some orchids are cacti ⇒ Some cacti are orchids: II follows.
2. No cactus is a marigold ⇒ No marigold is a cactus: I follows.

Yaad rakhein: Both conclusions are just the converses of the two statements – E and I type statements convert fully.

Answer: (D) Both conclusions (I) and (II) follow.

Concept: Replace letters by signs (A $÷$, B $\times$, C $+$, D $-$) and apply BODMAS.

1. (C): $41-13\times 3+35÷5$
2. $=41-39+7=9$ ✓
3. (A) gives $41+39÷35-5\approx 37.1$; (B) gives $533+3-7=529$

Trick: Look for the option where a big product nearly cancels $41$: $13\times 3=39$.

Answer: (C) 41 D 13 B 3 C 35 A 5

Concept: Number positions $1$ (top) to $7$ (bottom) and fix the definite clues first.

1. Two below B ⇒ B at $5$; one above F ⇒ F at $2$
2. One box between F and D ⇒ D at $4$
3. A just above G: only free pair is $6,7$ ⇒ A $6$, G $7$
4. C below E in $1,3$ ⇒ E $1$, C $3$

Trick: Fix "only $n$ above/below" clues first – they give exact positions.

Answer: (B) E

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

1. Q is immediate right of J; going left from Q, S is after $3$ people (opposite side)
2. H is $4$ places from J (other side of S)
3. Remaining seats: R and P adjacent with P to R’s right; I takes the last seat
4. From R going right: R → P → J ⇒ one person (P)

Trick: In a $7$-seat circle, "$3$ between" from one side means $2$ between from the other side.

Answer: (A) One

Concept: Rewrite the digits in ascending order and read the required positions.

1. Digits of $9823415$ ascending: $1234589$
2. Second from left $=2$; second from right $=8$
3. Sum $=2+8=10$

Trick: Second from left = second smallest digit; second from right = second largest digit.

Answer: (A) $10$

Concept: Letters common to both words share the common digits; the unique letter takes the unique digit.

1. GETS → $8,1,0,6$; EGIS → $0,1,6,7$
2. Common letters G, E, S ↔ common digits $0,1,6$
3. GETS: T ↔ $8$; EGIS: I ↔ $7$

Trick: Cancel the common part – what remains in EGIS is I and $7$.

Answer: (B) $7$

Concept: Decode each symbol from left to right and build a family tree.

1. D = O: D is wife of O
2. O # M: O is brother of M
3. M ¥ E: M is father of E; E % S: E is daughter of S ⇒ S is mother of E, i.e. wife of M
4. So D is wife of the brother of S’s husband

Trick: Two brothers’ wives are each other’s "husband’s brother’s wife" (devrani/jethani).

Answer: (C) Husband's brother's wife

Concept: Track each letter position separately using place values ($A=1\dots Z=26$, wrap around after Z).

1. 1st letters: S, D, O, Z, K (each $+11$)
2. 2nd letters: E, Q, C, O, A (each $+12$)
3. 3rd letters: T, E, P, A, L (each $+11$)
4. Next term: KAL

Trick: Add $11$ = subtract $15$ after wrap: Z($26$) $+11=37-26=11$ = K.

Answer: (B) KAL

Concept: Find how letter positions move from one cluster to the next.

1. LOPE → LEPO: 2nd and 4th letters swap
2. LEPO → EPOL: first letter moves to the end
3. BANG → BGNA (swap A, G) → GNAB (B to end) ✓

Trick: Check step 2 quickly: the last cluster must end with the first letter of the word.

Answer: (A) BANG - BGNA - GNAB