RRB NTPC Undergraduate CBT 1 2025 PYQs

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) $4\sqrt{6}$

Concept: Heron: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$; altitude $=\frac{2\times \text{Area}}{\text{base}}$.

1. $s=\frac{51+35+26}{2}=56$
2. $\text{Area}=\sqrt{56\times 5\times 21\times 30}=\sqrt{176400}=420$
3. $h=\frac{2\times 420}{35}=24\text{cm}$

Trick: The triangle is obtuse, so this altitude falls outside, on the extended base (see figure).

Answer: (D) 24

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

1. Digit sum $=4+5+8+3=20$
2. Next multiple of 9 is 27; $27-20=7$
3. $4583+7=4590$ ✓

Trick: Remainder of $4583÷9$ = remainder of $20÷9=2$; add $9-2=7$.

Answer: (B) 7

Concept: Take total work = LCM of days and count each person’s units for the days he worked.

1. Work $=\text{LCM}(8,22,33)=264$; rates: Mike 33, Connor 12, Floyd 8 units/day.
2. Let total time $=T$: Mike 4 days, Connor $(T-2)$ days, Floyd $T$ days.
3. $132+12(T-2)+8T=264$
4. $20T=156\Rightarrow T=7.8$ days

Trick: Draw a bar for each worker’s working period — it prevents mixing up who left when.

Answer: (A) 7.8

Concept: Extra selling price corresponds to the extra profit percentage of CP.

1. Extra profit $=48\%-16\%=32\%$ of CP $=₹80$
2. CP of one article $=\frac{80}{0.32}=₹250$
3. Number of articles $=\frac{8000}{250}=32$

Trick: $32\%\to 80$ ⇒ $100\%\to 250$.

Answer: (B) 32

Concept: At $10\%$ for 2 years, CI $=21\%$ of $P$ and SI $=20\%$ of $P$.

1. $0.21P=220.5\Rightarrow P=1050$
2. SI $=\frac{1050\times 10\times 2}{100}=210$

Trick: $\text{SI}=\text{CI}\times \frac{20}{21}=220.5\times \frac{20}{21}=210$.

Answer: (B) 210

Concept: Remainder on division by 8 depends only on the last three digits.

1. Last three digits: 488
2. $488=8\times 61$, remainder 0

Trick: Divisibility by 8 → check last 3 digits; by 4 → last 2 digits.

Answer: (A) 0

Concept: If $a+b+c=0$ then $a^3+b^3+c^3=3abc$.

1. $20+(-16)+(-4)=0$
2. Sum $=3\times 20\times (-16)\times (-4)=3\times 1280=3840$

Trick: Check: $8000-4096-64=3840$ ✓.

Answer: (C) 3840

Concept: Compare SP actually received with CP of the quantity actually given.

1. SP per “kg” $=32\times 0.8=₹25.6$
2. He gives 800 g, whose CP $=25\times 0.8=₹20$
3. Profit $\%=\frac{25.6-20}{20}\times 100=28\%$

Trick: Factor method: $\frac{0.8\times 32}{0.8\times 25}=\frac{25.6}{20}=1.28$ ⇒ 28% profit.

Answer: (C) 28% profit

Concept: Savings = Income − Expenditure; take income $=100$.

1. Old: income 100, expenditure 70, savings 30.
2. New: income 120, expenditure $70\times 1.25=87.5$, savings $32.5$.
3. Change $=\frac{2.5}{30}\times 100=8\frac{1}{3}\%$ increase

Trick: Always take income as 100 — the whole sum becomes mental maths.

Answer: (D) $8\frac{1}{3}\%$ increase

Concept: Mode = the value occurring most often.

1. Frequencies: 2 → 4 times, 3 → 3, 4 → 2, 5 → 1.
2. Highest frequency is for 2 hours.

Trick: Mode = “most” — just count tallies.

Answer: (A) 2 hours