RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Simple interest is directly proportional to time: $SI=\frac{P\times R\times T}{100}$.

1. SI for 5 years $-$ SI for 3 years $=$ SI for 2 years $=$ ₹784.
2. SI for 1 year $=\frac{784}{2}=$ ₹392.
3. SI for 8 years $=392\times 8=$ ₹3136.

Trick: In SI, the interest for each year is the same, so the difference for two periods is just the interest for the difference of years. No need to find the principal!

Answer: (A) 3136

Concept: In a trapezium, the segment joining the mid-points of the diagonals is parallel to the parallel sides and equals half their difference: $EF=\frac{1}{2}(AB-DC)$.

1. Here $AB\parallel DC$, $AB=60$ cm, $DC=18$ cm.
2. $EF=\frac{1}{2}(60-18)=\frac{42}{2}=21$ cm.
3. The lengths $BC=65$ and $AD=10$ are extra information and are not needed.

Yaad rakhein: Mid-points of the non-parallel sides → half the sum (median); mid-points of the diagonals → half the difference.

Answer: (B) 21

Concept: Evaluate powers and square roots first, then add/subtract left to right (BODMAS).

1. $9^2=81$, $\sqrt{4^2}=4$, $\sqrt{16}=4$ so $7\sqrt{16}=28$.
2. $81+4-28-7=85-35=50$.

Trick: $\sqrt{a^2}=a$ for positive $a$ – do not expand $4^2=16$ and then take the root; just cancel.

Answer: (C) 50

Concept: Take total work = LCM of the days and convert to units/day.

1. Total work $=\text{LCM}(10,12,15)=60$ units. A $=6$, B $=5$, C $=4$ units/day.
2. A and B for 3 days: $(6+5)\times 3=33$ units. Remaining $=60-33=27$.
3. B and C do $5+4=9$ units/day $\Rightarrow \frac{27}{9}=3$ days.

Trick: LCM method turns every work problem into simple whole-number counting.

Answer: (B) 3 days

Concept: Volume of a spherical shell $=\frac{4}{3}\pi (R^3-r^3)$; mass $=$ density $\times$ volume.

1. $R^3-r^3={13}^3-{12}^3=2197-1728=469$.
2. Volume $=\frac{4}{3}\times \frac{22}{7}\times 469=\frac{88\times 67}{3}=\frac{5896}{3}{\text{cm}}^3$ (since $469=7\times 67$).
3. Mass $=39\times \frac{5896}{3}=13\times 5896=76648$ g $=76.648$ kg.

Trick: Cancel before multiplying: $\frac{39}{3}=13$ and $\frac{469}{7}=67$ – the whole thing reduces to $88\times 67\times 13$.

Answer: (A) 76.648

Concept: To cross a walking person the train covers its own length at relative speed (same direction $\Rightarrow$ subtract).

1. Let train speed $=v$ km/h. Length $L=(v-2)\times \frac{5}{18}\times 9=(v-4)\times \frac{5}{18}\times 10$.
2. $9(v-2)=10(v-4)\Rightarrow 9v-18=10v-40\Rightarrow v=22$ km/h.
3. $L=(22-2)\times \frac{5}{18}\times 9=20\times 2.5=50$ m.

Trick: The length is the same both times, so just equate $9(v-2)=10(v-4)$ – the $\frac{5}{18}$ cancels.

Answer: (B) 50 meters

Concept: If $a:b=20:36$ (fourth proportional), then $a:b=5:9$. Geometric mean $=\sqrt{ab}$.

1. $a:b::20:36\Rightarrow \frac{a}{b}=\frac{5}{9}$.
2. $a+b=14\Rightarrow a=5,b=9$.
3. GM $=\sqrt{5\times 9}=\sqrt{45}=3\sqrt{5}$.

Trick: Fourth proportional of $a,b,c$ is $\frac{bc}{a}$ – so $\frac{a}{b}=\frac{c}{d}=\frac{20}{36}$ directly gives the ratio.

Answer: (B) $3\sqrt{5}$

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={67}^2-4\times 1012=4489-4048=441$.
2. $a-b=\sqrt{441}=21$.
3. Check: numbers are 44 and 23 ($44\times 23=1012$).

Trick: Never solve the quadratic – the identity gives the difference in one line.

Answer: (A) 21

Concept: Form one equation in the number of girls.

1. Girls $=g$, boys $=g+10$.
2. Pencils: $3g+2(g+10)=170\Rightarrow 5g=150\Rightarrow g=30$, boys $=40$.
3. Girls : boys $=30:40=3:4$.

Trick: Check options: only $3:4$ gives a difference of 10 with totals $(30,40)$ that produce $90+80=170$ pencils.

Answer: (C) 3:4

Concept: LCM method: total work $=\text{LCM}(12,15)=60$ units.

1. A $=5$ units/day, B $=4$ units/day.
2. 4 days together: $9\times 4=36$ units; remaining $=24$.
3. B alone: $\frac{24}{4}=6$ days.

Trick: Alternatively, work done in 4 days $=\frac{4}{12}+\frac{4}{15}=\frac{3}{5}$; B finishes $\frac{2}{5}\times 15=6$ days.

Answer: (B) 6

Concept: A number is divisible by 11 if (sum of digits at odd places) $-$ (sum at even places) is 0 or a multiple of 11.

1. 2145: $(2+4)-(1+5)=6-6=0$ ✓ divisible.
2. 5281: $(5+8)-(2+1)=10$ ✗; 4735: $(4+3)-(7+5)=-5$ ✗; 3522: $(3+2)-(5+2)=-2$ ✗.
3. $2145=11\times 195$.

Trick: Alternate-sum test is fastest; do it mentally from the right.

Answer: (C) 2145

Concept: List primes $a,b$ with $a^2+b$ prime and less than 20.

1. $a=2$: $4+b<20\Rightarrow b<16$; $4+b$ prime for $b=3(7),7(11),13(17)$.
2. $a=3$: $9+b<20\Rightarrow b<11$; $9+b$ prime only for $b=2(11)$ ($b=3,5,7$ give 12, 14, 16).
3. $a\ge 5$: $a^2\ge 25>20$, impossible. So $b\in \{2,3,7,13\}$ – 4 values.

Trick: Parity check: $a^2+b$ must be odd, so exactly one of $a,b$ is 2 – this cuts the search to two short lists.

Answer: (D) 4

Concept: Mode is the value that occurs most often.

1. Frequencies: 40 → 2, 47 → 2, 50 → 2, 52 → 3, 53 → 2, others once.
2. 52 appears 3 times – the highest frequency.

Trick: Put a tick above each value as you scan the list once; the value with most ticks is the mode.

Answer: (B) 52

Concept: Greatest length that measures all pieces exactly $=$ HCF of the lengths (in cm).

1. Lengths: 735 cm, 945 cm, 1050 cm.
2. $735=3\times 5\times 7^2$, $945=3^3\times 5\times 7$, $1050=2\times 3\times 5^2\times 7$.
3. HCF $=3\times 5\times 7=105$ cm $=1$ m 5 cm.

Trick: Differences $945-735=210$, $1050-945=105$; HCF must divide 105, and 105 divides all three – done.

Answer: (C) 1 m 5 cm

Concept: Melting keeps volume constant: $\frac{4}{3}\pi r^3=\frac{1}{3}\pi r^{2}h$.

1. Same radius cancels: $4r=h=84\Rightarrow r=21$ cm.
2. Volume of sphere $=\frac{4}{3}\times \frac{22}{7}\times {21}^3=\frac{4}{3}\times \frac{22}{7}\times 9261$.
3. $=4\times 22\times 441=38808{\text{cm}}^3$.

Yaad rakhein: Sphere recast into a cone of the same radius always has $h=4r$.

Answer: (B) $38808{\text{cm}}^3$

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

1. $p=\frac{{18}^2}{6}=\frac{324}{6}=54$.
2. $q=\frac{{54}^2}{p}=\frac{{54}^2}{54}=54$.

Trick: Since $p$ came out equal to 54, $q=\frac{54\times 54}{54}=54$ – no multiplication needed.

Answer: (D) 54

Concept: $SP=MP\times (1-d\%)$ and $CP=\frac{SP}{1+\text{gain}\%}$.

1. $SP=4884\times 0.9=4395.6$; $CP=\frac{4395.6}{1.08}=4070$.
2. Without discount $SP=MP=4884$; gain $=4884-4070=814$.
3. Gain$\%=\frac{814}{4070}\times 100=20\%$.

Trick: $\frac{MP}{CP}=\frac{1+g}{1-d}=\frac{1.08}{0.9}=1.2\Rightarrow 20\%$ – the price ₹4,884 is not even needed.

Answer: (A) 20%

Concept: Overall average $=\frac{\text{total marks}}{\text{total subjects}}$.

1. Total $=72\times 5+84\times 3=360+252=612$.
2. Average $=\frac{612}{8}=76.5$.

Trick: Weighted shift: $72+\frac{3}{8}(84-72)=72+4.5=76.5$.

Answer: (C) 76.5

Concept: $\frac{a}{e}=\frac{a}{b}\times \frac{b}{c}\times \frac{c}{d}\times \frac{d}{e}$.

1. $\frac{a}{e}=\frac{2}{3}\times \frac{4}{5}\times \frac{3}{2}\times \frac{5}{4}$.
2. Cancel: $\frac{2}{3}\times \frac{3}{2}=1$, $\frac{4}{5}\times \frac{5}{4}=1$ $\Rightarrow \frac{a}{e}=1$.
3. $a:e=1:1$.

Trick: Multiply all antecedents and all consequents: $\frac{2·4·3·5}{3·5·2·4}=1$.

Answer: (C) 1 : 1

Concept: $SP=CP\times (1+\text{gain}\%)$.

1. CP per notebook $=\frac{1000}{50}=$ ₹20.
2. SP $=20\times 1.2=$ ₹24.

Trick: 20% gain means multiply by $\frac{6}{5}$: $20\times \frac{6}{5}=24$.

Answer: (B) ₹24