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