RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Apply BODMAS: brackets first, then of, then $÷$/$\times$ left to right, then $+$.

1. $\frac{16/21}{16/4}=\frac{16}{21}\times \frac{4}{16}=\frac{4}{21}$
2. Bracket: $\frac{4}{5}\times \frac{10}{12}+\frac{4}{7}=\frac{2}{3}+\frac{4}{7}=\frac{26}{21}$, so $\frac{4}{21}÷\frac{26}{21}=\frac{2}{13}$
3. of first: $\frac{26}{14}\times \frac{14}{8}=\frac{26}{8}$, so $\frac{4}{8}÷\frac{26}{8}=\frac{4}{26}=\frac{2}{13}$
4. Total $=\frac{2}{13}+\frac{2}{13}=\frac{4}{13}$

Trick: of is solved before $÷$ — that is why the second part gives $\frac{2}{13}$, not a big number.

Answer: (A) $\frac{4}{13}$

Concept: Sum of the interior angles of a quadrilateral $={360}^{\circ }$. Express all angles in terms of $\angle Q$.

1. Let $\angle Q=x$. Then $\angle P=2x$, $\angle R=4x$, $\angle S=\frac{\angle R}{2}=2x$
2. $2x+x+4x+2x=9x={360}^{\circ }\Rightarrow x={40}^{\circ }$
3. $\angle S=2x={80}^{\circ }$

Trick: Write every angle as a multiple of the smallest one; sum of multiples $=9$, so one part $=\frac{{360}^{\circ }}{9}={40}^{\circ }$.

Answer: (C) ${80}^{\circ }$

Concept: BODMAS – brackets first, then division, then subtraction.

1. ${(2+\frac{3}{2})}^2={\left(\frac{7}{2}\right)}^2=\frac{49}{4}$
2. $[4-{\left(\frac{6}{3}\right)}^2]=4-2^2=0$
3. Division before subtraction: $0÷\frac{1}{2}=0$
4. $\frac{49}{4}-0=\frac{49}{4}$

Trick: The $÷\frac{1}{2}$ acts only on the square bracket next to it, and that bracket is $0$ – so the answer is just the first term.

Answer: (C) $\frac{49}{4}$

Concept: Total distance and total time are fixed; let the time at $40\text{km/h}$ be $t$.

1. $40t+160(9.5-t)=560$
2. $1520-120t=560\Rightarrow t=8\text{h}$
3. Time at $160\text{km/h}=1.5\text{h}$, distance $=160\times 1.5=240\text{km}$

Trick: If all 9.5 h were at $40\text{km/h}$: $380\text{km}$. Extra $180\text{km}$ comes from the extra $120\text{km/h}$ → $\frac{180}{120}=1.5\text{h}$ at the fast speed.

Answer: (B) $240\text{km}$

Concept: Same principal and time, so the SI difference depends only on the rate difference: $\text{SI}=\frac{P\times R\times T}{100}$.

1. Rate difference $=6.5\%-3.5\%=3\%$
2. Difference $=\frac{340000\times 3\times 2}{100}=20400$

Trick: Never compute both interests separately — use the rate difference directly.

Answer: (C) ₹20,400

Concept: If $a:b=b:c$ (continued proportion), then $b^2=ac$, so $c=\frac{b^2}{a}$.

1. $16:24=24:c$
2. $c=\frac{{24}^2}{16}=\frac{576}{16}=36$

Trick: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$.

Answer: (C) 36

Concept: Take original price $=100$ and follow each step carefully — the last increase is on the original price.

1. After $20\%$ increase: $120$
2. Decrease of $x\%$ of $120$: $120-1.2x$
3. Add $10\%$ of original $=10$: $130-1.2x=100$
4. $1.2x=30\Rightarrow x=25$

Trick: Base 100 makes every % a plain number.

Answer: (C) 25

Concept: Divisible by 2 → last digit even; divisible by 3 → digit sum divisible by 3.

1. Digit sum $=3+P+2+Q=5+P+Q$ must be a multiple of 3
2. Smallest $P+Q$ with $5+P+Q$ a multiple of 3 is $1$ (sum $=6$)
3. $P=1$, $Q=0$ (even) works: $3120=2\times 1560=3\times 1040$

Trick: Check smallest option first: $P+Q=1$ makes digit sum $6$ ✓.

Answer: (D) 1

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{36x}=12$
2. $36x=144\Rightarrow x=4$

Trick: $x=\frac{(\text{mean}{)}^2}{36}=\frac{144}{36}$.

Answer: (C) 4

Concept: Net change for $+a\%$ then $-b\%$: $a-b-\frac{ab}{100}$. The original price does not matter.

1. $20-20-\frac{20\times 20}{100}=-4\%$
2. Check: $500\to 600\to 480$, fall $=20$ = $4\%$ of 500

Trick: Equal % up then down always gives a loss of $\frac{x^2}{100}\%$.

Answer: (C) 4% decrease

Concept: Write both selling prices in terms of CP $x$ and equate the ₹294 gap.

1. Original SP $=1.15x$
2. New CP $=0.75x$, new SP $=1.30\times 0.75x=0.975x$
3. $1.15x-0.975x=294\Rightarrow 0.175x=294$
4. $x=\frac{294}{0.175}=1680$

Trick: Gap in SP $=17.5\%$ of CP $=294$ → CP $=\frac{294\times 100}{17.5}$.

Answer: (D) ₹1,680

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (not the mean of speeds).

1. Times: $\frac{90}{45}=2$, $\frac{120}{60}=2$, $\frac{150}{75}=2$ hours
2. Total distance $=360\text{km}$, total time $=6\text{h}$
3. Average $=\frac{360}{6}=60\text{km/h}$

Trick: Equal times on each part → average speed = simple mean $\frac{45+60+75}{3}=60$.

Answer: (D) $60\text{km/h}$

Concept: Take tank $=\text{LCM}(12,15,20)=60$ units; filling pipes add, emptying pipe subtracts.

1. Rates: A $=5$, B $=4$, C $=-3$ units/h
2. First 2 h (A+B): $9\times 2=18$ units
3. Remaining $42$ units at $5+4-3=6$ units/h → $7$ h
4. Total $=2+7=9$ hours

Trick: LCM-unit method avoids fractions completely.

Answer: (D) 9 hours

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \%}{100}$. Find CP from the loss, then apply the profit.

1. $\text{CP}=\frac{1786.40}{0.88}=2030$
2. Required SP $=2030\times 1.14=2314.20$

Trick: Directly: new SP $=1786.40\times \frac{114}{88}$.

Answer: (D) ₹2,314.20

Concept: Net work per hour $=$ sum of filling rates $-$ emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{8}-\frac{1}{8}=\frac{1}{4}$ per hour
2. Time for $\frac{1}{4}$ tank $=\frac{1/4}{1/4}=1$ hour

Trick: The 8-hour filler and 8-hour emptier cancel each other — only the 4-hour tap works.

Answer: (A) 1

Concept: When coefficients are swapped, add the two equations.

1. $(5c+3t)+(3c+5t)=12500+15500$
2. $8c+8t=28000\Rightarrow c+t=3500$
3. $2c+2t=7000$

Trick: Add equations → $c+t$ directly; no need to find each price.

Answer: (A) ₹7,000

Concept: When solids are melted and recast, the volume stays the same.

1. Volume of one cube $=2^3=8{\text{cm}}^3$
2. Cuboid volume $=16\times 8=128{\text{cm}}^3$
3. Check: $2b\times b\times b=2b^3=128\Rightarrow b=4$, so $8\times 4\times 4$

Trick: The shape details are not needed for the volume — just add the volumes.

Answer: (B) $128{\text{cm}}^3$

Concept: Form two equations: one for the present sum, one for 2 years ago.

1. $P+Q=46$
2. $P-2=2(Q-2)\Rightarrow P=2Q-2$
3. $3Q-2=46\Rightarrow Q=16$, $P=30$

Trick: Check options: $P=30$, $Q=16$ → 2 years ago $28$ and $14$ ✓.

Answer: (B) 30 years

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. Side $=\sqrt{16\times 25}=\sqrt{400}=20\text{m}$
2. Check: $16:20=20:25$ ✓

Trick: $\sqrt{16}\times \sqrt{25}=4\times 5$.

Answer: (D) 20 m

Concept: Convert time to minutes and find pages per minute.

1. 1 h 20 min $=80$ min; rate $=\frac{2400}{80}=30$ pages/min
2. In 45 min: $30\times 45=1350$

Trick: $2400\times \frac{45}{80}=2400\times \frac{9}{16}=1350$.

Answer: (B) 1350