RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Volume of material $=\pi h(R^2-r^2)$.

1. $R^2-r^2=64-49=15$
2. $V=3.14\times 20\times 15=942{\text{cm}}^3$

Trick: $R^2-r^2=(R+r)(R-r)=15\times 1$.

Answer: (D) $942$ cubic cm

Concept: Fourth proportional $x$ of $a,b,c$: $a:b=c:x\Rightarrow x=\frac{bc}{a}$.

1. $x=\frac{25\times 888}{37}$
2. $\frac{888}{37}=24$, so $x=25\times 24=600$

Yaad rakhein: $888=37\times 24$ — spot the factor first.

Answer: (D) $600$

Concept: Same distance: difference of times = delay.

1. $\frac{d}{30}-\frac{d}{40}=\frac{20}{60}=\frac{1}{3}$
2. $\frac{d}{120}=\frac{1}{3}\Rightarrow d=40\text{km}$

Trick: $d=\frac{S_1{S}_2}{S_1-S_2}\times t=\frac{40\times 30}{10}\times \frac{1}{3}=40$.

Answer: (A) $40\text{km}$

Concept: ${\cos }^2\theta +{\sin }^2\theta =1$ and $1+{\tan }^2\theta ={\sec }^2\theta$.

1. (D): $1+{\tan }^2\theta +1-{\sec }^2\theta ={\sec }^2\theta +1-{\sec }^2\theta =1$ ✓
2. (A) gives $2{\sec }^2\theta -1$, not 1
3. (B), (C) are always greater than 1

Yaad rakhein: ${\sec }^2\theta -{\tan }^2\theta =1$, ${\csc }^2\theta -{\cot }^2\theta =1$.

Answer: (D) $1+{\tan }^2\theta +(1-{\sec }^2\theta )$

Concept: $P=\frac{\text{SI}\times 100}{R\times T}$.

1. $R\times T=12\times 4=48\%$
2. $P=\frac{1320\times 100}{48}=2750$

Trick: $48\%$ of P $=1320$ → $1\%=27.5$ → $P=2750$.

Answer: (D) ₹$2,750$

Concept: A number is divisible by 9 if its digit sum is a multiple of 9.

1. Digit sum $=5+1+3+5+2+n=16+n$
2. Next multiple of 9 is 18 → $n=2$

Yaad rakhein: 9 test: digit sum ÷ 9; 3 test: digit sum ÷ 3.

Answer: (D) $2$

Concept: The age difference stays constant over time.

1. Let brother $=b$, Rahul $=b+10$
2. 5 years ago: $b+5=2(b-5)\Rightarrow b=15$
3. Rahul $=25$ years

Trick: 5 years ago Rahul = 2 × brother and difference 10 → brother 10, Rahul 20 then; now 25.

Answer: (B) $25$ years

Concept: $\text{Savings}=\text{Income}-\text{Expenditure}$; take income $=100$.

1. Before: $I=100$, $E=75$, $S=25$
2. After: $I=120$, $E=75\times 1.1=82.5$, $S=37.5$
3. Change $=\frac{12.5}{25}\times 100=50\%$ increase

Trick: Always assume income 100 in such questions.

Answer: (B) $50\%$ increase

Concept: Take the common factor $2^{x-1}$.

1. $2^x-2^{x-1}=2^{x-1}(2-1)=2^{x-1}$
2. $2^{x-1}=32=2^5\Rightarrow x=6$
3. $\frac{2x-1}{2x+3}=\frac{11}{15}$

Yaad rakhein: $a^x-a^{x-1}=a^{x-1}(a-1)$.

Answer: (B) $\frac{11}{15}$

Concept: $24=3\times 8$: last three digits divisible by 8 and digit sum divisible by 3.

1. Last three digits $44B$: $440$ and $448$ are multiples of 8 → $B=0$ or $8$
2. Digit sum $=26+A+B$ must be a multiple of 3
3. Take $B=8$: $34+A$ → $A=2,5,8$; max $A=8$
4. Max $A+B=8+8=16$

Trick: Choose the largest B first, then the largest A that fits the 3-test.

Answer: (D) $16$

Concept: Amount $A=P(1+\frac{RT}{100})$ under simple interest.

1. Total SI for 3 years $=3\times 5=15\%$ of $P$.
2. $A=1.15P=6900$
3. $P=\frac{6900}{1.15}=6000$

Trick: Amount = 115 parts, principal = 100 parts: $\frac{6900}{115}\times 100=6000$.

Answer: (B) ₹6,000

Concept: The difference between two ages never changes with time.

1. Let ages be $4k$ and $5k$; difference $=k$.
2. After 5 years difference is still $k=3$.
3. Ages $=12$ and $15$; sum $=27$.

Trick: Sum $=(4+5)\times \text{difference}=9\times 3=27$.

Answer: (A) 27

Concept: $\text{Reduction}\%=\frac{\text{Reduction}}{\text{Original price}}\times 100$

1. $R_1=\frac{3000}{10000}\times 100=30\%$
2. $R_2=\frac{2500}{10000}\times 100=25\%$
3. $R_1-R_2=5$

Trick: Same base: difference of reductions $=\frac{500}{10000}\times 100=5$.

Answer: (B) 5

Concept: Points on opposite sides: horizontal distances $=h\cot {30}^{\circ }$ and $h\cot {60}^{\circ }$ add up to the gap.

1. $AQ=h\sqrt{3}$, $QB=\frac{h}{\sqrt{3}}$
2. $h\sqrt{3}+\frac{h}{\sqrt{3}}=\frac{4h}{\sqrt{3}}=94$
3. $h=\frac{94\sqrt{3}}{4}=23.5\times 1.73=40.655\text{m}$

Trick: For angles ${30}^{\circ }$ and ${60}^{\circ }$ on opposite sides: $h=\frac{\sqrt{3}}{4}\times \text{gap}$.

Answer: (D) $40.655\text{m}$

Concept: Third proportional to $a$ and $b$ is $\frac{b^2}{a}$.

1. $x=\frac{(a+b{)}^2}{a^2-b^2}$
2. $a^2-b^2=(a+b)(a-b)$
3. $x=\frac{(a+b{)}^2}{(a+b)(a-b)}=\frac{a+b}{a-b}$

Trick: Always factorise $a^2-b^2$ first and cancel.

Answer: (A) $\frac{a+b}{a-b}$

Concept: BODMAS: brackets, then “of”, then division/multiplication, then addition.

1. $\frac{10/18}{10/9}=\frac{10}{18}\times \frac{9}{10}=\frac{1}{2}$
2. Bracket: $\frac{4}{9}\times \frac{18}{12}+\frac{5}{6}=\frac{2}{3}+\frac{5}{6}=\frac{3}{2}$
3. “of”: $\frac{27}{15}\times \frac{15}{9}=3$, so $\frac{9}{9}÷3=\frac{1}{3}$
4. $\frac{1}{2}÷\frac{3}{2}+\frac{1}{3}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

Trick: “of” is solved before $÷$ — doing $÷$ first would give a wrong value.

Answer: (D) $\frac{2}{3}$

Concept: Area of rhombus $=\frac{1}{2}{d}_1{d}_2$.

1. $d_1=5k$, $d_2=12k$
2. $\frac{1}{2}\times 5k\times 12k=30k^2=240$
3. $k^2=8\Rightarrow k=2\sqrt{2}$
4. $d_1=10\sqrt{2}\text{cm}$, $d_2=24\sqrt{2}\text{cm}$

Trick: Check: $\frac{1}{2}\times 10\sqrt{2}\times 24\sqrt{2}=240$ ✓; option “10 and 24” gives only 120.

Answer: (A) $10\sqrt{2}$ and $24\sqrt{2}$

Concept: Numbers in ratio $a:b:c$ (co-prime parts) $=ak,bk,ck$; LCM $=k\times \text{LCM}(a,b,c)$, HCF $=k$.

1. $10,13,11$ are pairwise co-prime, LCM $=1430$.
2. $1430k=2860\Rightarrow k=2$
3. HCF $=k=2$

Trick: HCF $=\frac{\text{given LCM}}{\text{LCM of ratio terms}}=\frac{2860}{1430}$.

Answer: (B) 2

Concept: $a:b=b:c\Rightarrow c=\frac{b^2}{a}$.

1. $a=4$, $b=6$
2. $c=\frac{6^2}{4}=\frac{36}{4}=9\text{cm}$

Trick: Middle term squared ÷ first term.

Answer: (D) 9 cm

Concept: Melting keeps volume same; hollow cylinder volume $=\pi (R^2-r^2)h$.

1. Solid: $\pi \times 6^2\times 14=504\pi$
2. Hollow: $h=7$, $R=2r$: $\pi (4r^2-r^2)\times 7=21\pi r^2$
3. $21r^2=504\Rightarrow r^2=24\Rightarrow r=2\sqrt{6}$
4. $R=2r=4\sqrt{6}\text{cm}$

Trick: Half height ⇒ cross-section area doubles: $3r^2=2\times 36=72$.

Answer: (B) $4\sqrt{6}$