RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs = half-diagonals.

1. Side $={\displaystyle \frac{100}{4}}=25$ m; half of given diagonal $=7$ m.
2. Half of other diagonal $=\sqrt{{25}^2-7^2}=\sqrt{576}=24$ m.
3. Other diagonal $=2\times 24=48$ m.

Yaad rakhein: $(7,24,25)$ is a Pythagorean triplet.

Answer: (A) 48 meters

Concept: Convert each ratio to a decimal and compare.

1. ${\displaystyle \frac{5}{8}}=0.625$, ${\displaystyle \frac{3}{5}}=0.6$, ${\displaystyle \frac{7}{11}}\approx 0.636$, ${\displaystyle \frac{4}{7}}\approx 0.571$.
2. Greatest $={\displaystyle \frac{7}{11}}$.

Trick: Compare ${\displaystyle \frac{5}{8}}$ and ${\displaystyle \frac{7}{11}}$ by cross-multiplying: $5\times 11=55<7\times 8=56$, so ${\displaystyle \frac{7}{11}}$ is bigger.

Answer: (C) 7 : 11

Concept: Multiply by $\sin \theta$ to get a perfect square in $\sin \theta$.

1. $4\sin \theta +{\displaystyle \frac{1}{\sin \theta }}=4$ ⟹ $4{\sin }^2\theta -4\sin \theta +1=0$ ⟹ $(2\sin \theta -1{)}^2=0$.
2. $\sin \theta ={\displaystyle \frac{1}{2}}$ ⟹ $\theta ={30}^{\circ }$.
3. $\sin {90}^{\circ }+\cos {90}^{\circ }=1+0=1$.

Trick: $x+{\displaystyle \frac{1}{x}}=2$ type (here $2\sin \theta +{\displaystyle \frac{1}{2\sin \theta }}=2$) forces $x=1$, i.e. $2\sin \theta =1$.

Answer: (C) 1

Concept: Combined average $={\displaystyle \frac{\text{total of all scores}}{\text{total students}}}$.

1. Boys total $=8\times 56=448$; girls total $=4\times 80=320$.
2. Average $={\displaystyle \frac{768}{12}}=64$.

Trick: Ratio $8:4=2:1$, so the average lies ${\displaystyle \frac{1}{3}}$ of the way from 56 to 80: $56+8=64$.

Answer: (D) 64

Concept: Divisible by $24$ ⟺ divisible by both $8$ and $3$.

1. By 8: last three digits $84B$ must be divisible by 8 ⟹ $B=0$ (840) or $B=8$ (848).
2. By 3: digit sum $=4+6+A+4+8+4+B=26+A+B$ must be a multiple of 3.
3. With $B=0$: $A=1$ works ($27$) ⟹ $A+B=1$, the minimum.

Trick: Try the smallest $B$ first; the smallest $A$ that fixes the digit sum gives the minimum.

Answer: (C) 1

Concept: Solve brackets first; "of" means multiplication and is done before $÷$.

1. ${\displaystyle \frac{12}{33}}÷{\displaystyle \frac{12}{2}}={\displaystyle \frac{12}{33}}\times {\displaystyle \frac{2}{12}}={\displaystyle \frac{2}{33}}$.
2. ${\displaystyle \frac{3}{7}}\times {\displaystyle \frac{14}{9}}+{\displaystyle \frac{3}{11}}={\displaystyle \frac{2}{3}}+{\displaystyle \frac{3}{11}}={\displaystyle \frac{31}{33}}$; so ${\displaystyle \frac{2}{33}}÷{\displaystyle \frac{31}{33}}={\displaystyle \frac{2}{31}}$.
3. ${\displaystyle \frac{31}{25}}\text{ of }{\displaystyle \frac{25}{8}}={\displaystyle \frac{31}{8}}$; ${\displaystyle \frac{2}{8}}÷{\displaystyle \frac{31}{8}}={\displaystyle \frac{2}{31}}$.
4. Total $={\displaystyle \frac{2}{31}}+{\displaystyle \frac{2}{31}}={\displaystyle \frac{4}{31}}$.

Trick: Both halves reduce to ${\displaystyle \frac{2}{31}}$ – the 33s and 8s cancel neatly.

Answer: (A) $\frac{4}{31}$

Concept: Total cost includes overheads (transport, repair).

1. Total CP $=7000+400+600=8000$.
2. Profit $=9800-8000=1800$.
3. Profit% $={\displaystyle \frac{1800}{8000}}\times 100=22.5\%$.

Trick: ${\displaystyle \frac{1800}{8000}}={\displaystyle \frac{9}{40}}=22.5\%$.

Answer: (A) 22.5%

Concept: Profit is shared in the ratio of (investment $\times$ time).

1. A : B : C $=5\times 12:7\times 12:5\times 8=60:84:40=15:21:10$.
2. Total parts $=46$; B + C $=31$ parts.
3. Share $={\displaystyle \frac{31}{46}}\times 92000=31\times 2000=62000$.

Trick: $92000÷46=2000$ exactly – a clean sign the ratio is right.

Answer: (C) ₹62,000

Concept: Divisible by 6 ⟺ even (by 2) and digit sum divisible by 3.

1. Only 7,53,216 and 8,91,452 are even.
2. $7+5+3+2+1+6=24$ ✓ (multiple of 3); $8+9+1+4+5+2=29$ ✗.

Trick: Eliminate odd numbers first, then check digit sums only for the rest.

Answer: (B) 7,53,216

Concept: Same distance both ways; total time $={\displaystyle \frac{d}{5}}+{\displaystyle \frac{d}{10}}$.

1. ${\displaystyle \frac{d}{5}}+{\displaystyle \frac{d}{10}}={\displaystyle \frac{3d}{10}}=3$.
2. $d=10$ km.

Trick: Average speed $={\displaystyle \frac{2\times 5\times 10}{15}}={\displaystyle \frac{20}{3}}$ km/h; round trip $={\displaystyle \frac{20}{3}}\times 3=20$ km ⟹ one way 10 km.

Answer: (A) 10 km

Concept: Check every even digit: the digit just before it must be a perfect square (1, 4, 9) and the digit just after it must be odd.

1. Series: 9 2 3 1 4 5 6 8 5 4 3 2 1 9 3 4 5 6 7 8 1 2 9 8 3 4 5 2 6 8 2.
2. 9–2–3 ✓, 1–4–5 ✓, 1–2–9 ✓, 9–8–3 ✓.
3. Other even digits are preceded by 5, 6, 3, 7, 2 or 8 (not squares), so they fail.
4. Total = $4$, i.e. more than three.

Trick: Scan only the evens and look one step left for 1, 4 or 9 – this cuts the checking to a few digits.

Answer: (B) More than three

Concept: Plot the path on coordinates (East = $+x$, North = $+y$) and read the final position relative to the start.

1. Y$(0,0)$ → South 14 → $(0,-14)$ → left (East) 35 → $(35,-14)$ → left (North) 68 → $(35,54)$.
2. Left (West) 41 → $(-6,54)$ → left (South) 54 → $(-6,0)$ → right (West) 53 → Z$(-59,0)$.
3. Z is $59$ km due west of Y, so Amit must drive $59$ km towards East to reach Y.

Trick: Net N–S: $-14+68-54=0$; net E–W: $+35-41-53=-59$. Zero in one axis means the answer is a straight line along the other.

Answer: (D) 59 km towards east

Concept: Compare the same position of consecutive clusters to find the shift.

1. 1st letters: S → T → U → V → W ($+1$ each).
2. 2nd letters: Q → R → S → T → U ($+1$ each).
3. 3rd letters: U → V → W → X → Y ($+1$ each).
4. Next cluster = WUY.

Trick: Each cluster is simply the previous one with every letter moved one step ahead.

Answer: (B) WUY

Concept: Apply the rule to each digit separately, then pick the even digits of the new number.

1. 5973642: odd digits $-2$ → 5→3, 9→7, 7→5, 3→1; even digits $+2$ → 6→8, 4→6, 2→4.
2. New number = 3751864.
3. Even digits: 8, 6, 4 → sum $=8+6+4=18$.

Trick: Odd $-2$ stays odd and even $+2$ stays even, so only the original even digits (6, 4, 2) matter: $(6+4+2)+2\times 3=18$.

Answer: (D) 18

Concept: Total $=$ rank from top $+$ rank from bottom $-1$ (the person is counted twice).

1. Rank from top $=29$, from bottom $=16$.
2. Total students $=29+16-1=44$.

Trick: Add the two ranks and subtract 1 – never forget the $-1$.

Answer: (C) 44

Concept: Interchange the signs as instructed, then apply BODMAS.

1. 603 + 9 × 18 − 6 ÷ 56 becomes $603÷9-18\times 6+56$.
2. $603÷9=67$ and $18\times 6=108$.
3. $67-108+56=15$.

Trick: Rewrite the whole expression with new signs first; only then calculate – mixing the two steps causes errors.

Answer: (D) 15

Concept: Letters common to both words get the common digits; the leftover letter gets the leftover digit.

1. NORM → 7341 and ROME → 3748 share the letters R, O, M and the digits 7, 3, 4.
2. In NORM the extra letter N ↔ extra digit 1; in ROME the extra letter E ↔ extra digit 8.

Trick: Cancel the common letters and common digits – whatever remains pairs up directly.

Answer: (D) 8

Concept: "Employees who also speak German" = the regions common to the Employees circle and the German-speakers circle (whether male or not).

1. Employees ∩ German only (not male) $=28$.
2. Employees ∩ German ∩ Male $=17$.
3. Required $=28+17=45$.

Trick: The overlap of two circles is always made of two pieces – the part inside the third circle and the part outside it. Add both.

Answer: (B) 45

Concept: Arrange the letters alphabetically, locate the two positions, then count letters strictly between them in the alphabet.

1. LANGUID → A, D, G, I, L, N, U.
2. Second from left = D; fourth from right (of 7) = 4th from left = I.
3. Between D(4) and I(9): E, F, G, H $=9-4-1=4$ letters.

Trick: Letters between two positions $=$ difference of their place values $-1$.

Answer: (C) 4

Concept: Number the seats 1–7 from left to right (all face north, so right = higher number) and place people from the most definite clue.

1. Only two to the right of Rakesh → Rakesh at 5. Vivek third to the left of Rakesh → Vivek at 2; Nidhi immediately left → Nidhi at 1.
2. Palak is third to the right of Tina and not at an end → Tina 3, Palak 6 (Tina 4 would put Palak at 7, an end).
3. Mayank is not next to Rakesh → Mayank 7, Surya 4.
4. Order: Nidhi, Vivek, Tina, Surya, Rakesh, Palak, Mayank. Tina (3) and Vivek (2) are adjacent → none between.

Trick: Start with "only two to the right of" – it fixes a seat instantly and the rest falls in place.

Answer: (C) None