RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: In $a:b=c:d$, product of extremes $=$ product of means.

1. $(6+2):(8-2)=(12+4):(d-4)\Rightarrow 8:6=16:(d-4)$
2. $8(d-4)=96\Rightarrow d-4=12$
3. $d=16$

Trick: $8:6$ doubled is $16:12$, so $d-4=12$ directly.

Answer: (C) 16

Concept: If $\sin \alpha =\cos \beta$ (acute), then $\alpha +\beta ={90}^{\circ }$.

1. $6A+12A={90}^{\circ }\Rightarrow A=5^{\circ }$
2. $9A={45}^{\circ }$
3. $\tan {45}^{\circ }+\cot {45}^{\circ }=1+1=2$

Trick: $9A$ is exactly half of $18A={90}^{\circ }$, so $9A={45}^{\circ }$ without finding $A$.

Answer: (B) 2

Concept: Total discount $\%$ = weighted sum of the discounts on each part.

1. On 70% of price: $15\%\times 0.7=10.5\%$
2. On remaining 30%: $5\%\times 0.3=1.5\%$
3. Total $=10.5+1.5=12\%$

Trick: The ₹50,000 is a distraction – percentages alone give the answer.

Answer: (D) $12\%$

Concept: CSA $=2\pi rh$, TSA $=2\pi r(h+r)$.

1. CSA of first $=2\pi \times 4\times 7=56\pi$
2. Second cylinder: $r=2$; TSA $=2\pi \times 2(h+2)=56\pi \Rightarrow h+2=14\Rightarrow h=12$
3. CSA of second $=2\pi \times 2\times 12=48\pi {\text{cm}}^2$

Trick: CSA of second $=$ TSA $-$ two circles $=56\pi -2\pi (2{)}^2=48\pi$.

Answer: (B) $48\pi {\text{cm}}^2$

Concept: A number is divisible by 4 if its last two digits form a multiple of 4.

1. Last two digits of $2X2$ are $X2$.
2. Options: 12 ✓, 42 ✗, 02 ✗, 22 ✗
3. So $X=1$ (212 $=4\times 53$).

Trick: Two-digit multiples of 4 ending in 2: 12, 32, 52, 72, 92 – tens digit is always odd.

Answer: (A) 1

Concept: When SP of $m$ articles $=$ CP of $n$ articles, profit$\%=\frac{n-m}{m}\times 100$.

1. CP of 60 pens is recovered as CP of 72 pens: gain $=72-60=12$ pens
2. Profit$\%=\frac{12}{60}\times 100=20\%$

Trick: Profit is always on cost: $\frac{12}{60}$, not $\frac{12}{72}$ (which gives the wrong $16.67\%$).

Answer: (D) $20\%$

Concept: Appending digit $d$ gives $92419\times 10+d=924190+d$; find the remainder of 924190 on division by 13.

1. $924190÷13$: $13\times 71091=924183$, remainder $=7$
2. Need $7+d$ divisible by 13 $\Rightarrow d=6$
3. $924196=13\times 71092$ ✓

Trick: Remainder $r$ after appending $\Rightarrow$ digit $=13-r$ (if $r\ne 0$).

Answer: (A) 6

Concept: In a trapezium with parallel sides $a$ and $b$, the segment joining the midpoints of the diagonals is parallel to them and equals $\frac{a-b}{2}$.

1. $AB=74$, $DC=26$ (parallel sides)
2. $EF=\frac{74-26}{2}=\frac{48}{2}=24$ cm
3. The sides $BC=91$ and $AD=39$ are not needed.

Trick: Midpoints of non-parallel sides $\to \frac{a+b}{2}$; midpoints of diagonals $\to \frac{a-b}{2}$.

Answer: (D) 24

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$.

1. Distance $=72\times 3+60\times 1.5=216+90=306$ km
2. Time $=3+1.5=4.5$ h
3. Average speed $=\frac{306}{4.5}=68$ km/h

Trick: Weighted by time $3:1.5=2:1$: $\frac{2\times 72+60}{3}=68$.

Answer: (C) 68 km/h

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

1. $SI=\frac{1000\times 5\times 2}{100}=100$

Trick: 5% of 1000 is 50 per year; two years give ₹100.

Answer: (A) ₹100

Concept: Third proportional to $a$ and $b$ is $c$ with $a:b=b:c$, i.e. $c=\frac{b^2}{a}$.

1. $c=\frac{{10}^2}{5}=\frac{100}{5}=20$

Trick: $5\to 10$ is doubling, so $10\to 20$.

Answer: (D) 20

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

1. $24+(-7)+(-17)=0$
2. $3abc=3\times 24\times (-7)\times (-17)=3\times 24\times 119$
3. $=72\times 119=8568$

Trick: Check the sum first – if zero, never cube the numbers.

Answer: (D) 8568

Concept: They depart together again after the LCM of the intervals.

1. $12=2^2\times 3$, $18=2\times 3^2$, $27=3^3$
2. LCM $=2^2\times 3^3=108$ min $=1$ h $48$ min
3. $6:00+1:48=7:48$ am

Trick: LCM must be a multiple of 27 and 4: $108$.

Answer: (C) 7:48 am

Concept: Let the numbers be $L$ and $S$ with $L+S=72$; set up the ratio equation.

1. $\frac{L+8}{S-6}=\frac{3}{2}\Rightarrow 2L+16=3S-18$
2. Put $L=72-S$: $144-2S+16=3S-18\Rightarrow 5S=178$
3. $S=35.6$

Trick: Total after change $=72+8-6=74$; smaller changed $=\frac{2}{5}\times 74=29.6$, so $S=29.6+6=35.6$.

Answer: (A) 35.6

Concept: Difference of percentages of the same number equals a percentage of it.

1. $40\%-25\%=15\%$ of the number $=36$
2. Number $=\frac{36\times 100}{15}=240$

Trick: $15\%\to 36$, so $5\%\to 12$ and $100\%\to 240$.

Answer: (A) 240

Concept: Compute the fraction of marked price actually paid for 10 toys; the scheme with the highest payment is least beneficial.

1. (A) $0.56\times 0.88=0.4928\to$ pays $49.3\%$
2. (B) $0.82\times 0.57\times 0.77\approx 0.36\to$ pays $36\%$
3. (C) $\frac{4\times 0.92+6\times 0.62}{10}=\frac{3.68+3.72}{10}=0.74\to$ pays $74\%$
4. (D) pays for 7 of 10 $\to 70\%$

Trick: Only (C) and (D) are close; $74\%>70\%$, so (C) costs the most.

Answer: (C) C

Concept: Decrease in average $=\frac{\text{total decrease}}{k}$.

1. Total decrease $=48\times 2=96$
2. New average $=x-12\Rightarrow \frac{96}{k}=12$
3. $k=8$

Trick: Average falls by 12 while sum falls by 96 $\Rightarrow 96÷12=8$ numbers.

Answer: (A) 8

Concept: In equal time, distance $\propto$ speed.

1. Faster car: $140$ km corresponds to ratio part 7, so one part $=20$ km
2. Slower car: $5\times 20=100$ km

Trick: Distances in the same time are in the ratio $5:7$ directly.

Answer: (C) 100 km

Concept: Take tank capacity $=$ LCM of times as units; add inlet rates, subtract the leak.

1. Capacity $=\text{LCM}(18,9,27)=54$ units
2. M $=3$/h, N $=6$/h, leak $=-2$/h; net $=3+6-2=7$/h
3. Time $=\frac{54}{7}=7\frac{5}{7}$ hours

Trick: Without the leak it takes 6 h; the leak slows it, so the answer must exceed 6 h – rules out nothing but confirms direction.

Answer: (B) $7\frac{5}{7}$ hours

Concept: Diagonals of a rhombus bisect each other at right angles; side $=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$.

1. Half-diagonals $=112$ m and $210$ m
2. Side $=\sqrt{{112}^2+{210}^2}=\sqrt{12544+44100}=\sqrt{56644}=238$ m
3. Perimeter $=4\times 238=952$ m

Trick: $112:210=8:15$, a Pythagorean triplet $(8,15,17)$ scaled by 14 $\Rightarrow$ side $=17\times 14=238$.

Answer: (A) 952