RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Successive increases multiply: final $=P\times (1+\frac{a}{100})(1+\frac{b}{100})$.

1. After year 1: $2,90,000\times \frac{105}{100}=3,04,500$.
2. After year 2: $3,04,500\times \frac{130}{100}=3,95,850$.

Trick: Net single change $=a+b+\frac{ab}{100}=5+30+1.5=36.5\%$; $2,90,000\times 1.365=3,95,850$.

Answer: (A) 3,95,850

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $l$ = lower limit of the modal class (highest frequency), $f_1$ its frequency, $f_0$, $f_2$ the frequencies before and after, $h$ = class width.

1. Highest frequency 36 → modal class 20–30: $l=20$, $f_1=36$, $f_0=16$, $f_2=34$, $h=10$.
2. Mode $=20+\frac{36-16}{72-16-34}\times 10=20+\frac{20}{22}\times 10$.
3. $=20+9.09=29.09$.

Trick: The mode always lies inside the modal class (20–30) – options 30.15 and 31.25 are out at once.

Answer: (D) 29.09

Concept: Speed of boat in still water $=\frac{\text{downstream speed}+\text{upstream speed}}{2}$.

1. Downstream speed $=\frac{120}{6}=20\text{km/h}$; upstream speed $=\frac{120}{10}=12\text{km/h}$.
2. Boat speed $=\frac{20+12}{2}=16\text{km/h}$ (current $=\frac{20-12}{2}=4\text{km/h}$, matching the data).

Trick: Boat = average of the two speeds; current = half their difference.

Answer: (C) 16 km/hr

Concept: When a transversal cuts two parallel lines, each pair of alternate interior angles is equal.

1. The two alternate interior angles lie on opposite sides of the transversal, between the parallel lines, forming a "Z".
2. Given one of them $={65}^{\circ }$, the other alternate interior angle is also ${65}^{\circ }$.
3. ${110}^{\circ }$ would be the co-interior (same-side) angle (${180}^{\circ }-{65}^{\circ }$), not the alternate one.

Trick: Z-angles (alternate) equal, F-angles (corresponding) equal, C-angles (co-interior) add to ${180}^{\circ }$.

Answer: (B) 65°

Concept: Alternate interior angles made by a transversal with two parallel lines are equal.

1. Transversal PQ meets AB at M and CD at N. $\angle AMN$ (at M, towards A) and $\angle MND$ (at N, towards D) lie between the parallels on opposite sides of MN – alternate interior angles.
2. Hence $\angle AMN=\angle MND={47}^{\circ }$.
3. ${133}^{\circ }$ is the adjacent angle ($\angle BMN$ or $\angle MNC$), not the one asked.

Trick: A and D are on opposite sides of the transversal → the "Z" pair → equal.

Answer: (C) 47°

Concept: Equal time ⇒ $\frac{{\text{distance}}_1}{{\text{speed}}_1}=\frac{{\text{distance}}_2}{{\text{speed}}_2}$. Use it to find each track length.

1. Speeds: Ankur $=\frac{5}{3}$ m/s, Mahima $=\frac{25}{12}$ m/s. Let track X $=x$ m, track Y $=y$ m, $x+y=325$.
2. Same time: $\frac{9x}{5/3}=\frac{5y}{25/12}\Rightarrow \frac{27x}{5}=\frac{12y}{5}\Rightarrow 27x=12y\Rightarrow y=\frac{9x}{4}$.
3. $x+\frac{9x}{4}=325\Rightarrow \frac{13x}{4}=325\Rightarrow x=100$ m (so $y=225$ m).
4. Mahima on track X: time $=\frac{100}{25/12}=\frac{100\times 12}{25}=48$ s.

Trick: Convert 0.325 km to 325 m at once and keep speeds as fractions – the numbers cancel neatly.

Answer: (B) 48

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$ with $l$ = lower limit of the modal class, $f_1$ its frequency, $f_0$ and $f_2$ the neighbouring frequencies, $h$ = class width.

1. Highest frequency 40 → modal class 65–75: $l=65$, $f_1=40$, $f_0=30$, $f_2=34$, $h=10$.
2. Mode $=65+\frac{40-30}{80-30-34}\times 10=65+\frac{10}{16}\times 10$.
3. $=65+6.25=71.25$.

Trick: $\frac{10}{16}=0.625$; the decimal part ".25" of the answer comes straight from $6.25$.

Answer: (C) 71.25

Concept: Form equations from the count of coins and their total value.

1. Let ₹5 coins $=x$, ₹2 coins $=x+50$, ₹1 coins $=z$. Count: $x+(x+50)+z=112\Rightarrow 2x+z=62$.
2. Value: $5x+2(x+50)+z=212\Rightarrow 7x+z=112$.
3. Subtracting: $5x=50\Rightarrow x=10$, so $z=62-20=42$.
4. Check: $10+60+42=112$ coins; $50+120+42=212$ ₹ ✓.

Trick: Subtract the count equation from the value equation – the ₹1 coins vanish, leaving only $x$.

Answer: (A) 42

Concept: Take the tank capacity as the LCM of the times; each pipe's work per hour is then a whole number of units.

1. LCM(4, 24, 6) $=24$ units. Pipe A fills $\frac{24}{4}=6$/h, pipe B fills $\frac{24}{24}=1$/h, pipe C empties $\frac{24}{6}=4$/h.
2. Net per hour $=6+1-4=3$ units.
3. Half tank $=12$ units → time $=\frac{12}{3}=4$ hours.

Trick: Fraction method: $\frac{1}{4}+\frac{1}{24}-\frac{1}{6}=\frac{3}{24}=\frac{1}{8}$ → full in 8 h, half in 4 h.

Answer: (D) 4 hours

Concept: For a cone with base radius $r$, height $h$ and slant height $l$, the radius of the largest inscribed sphere is $R=\frac{rh}{l+r}$.

1. $r=42\sqrt{3}$, $l=2r=84\sqrt{3}$ ⇒ $h=\sqrt{l^2-r^2}=\sqrt{3r^2}=r\sqrt{3}=126$ cm.
2. $R=\frac{42\sqrt{3}\times 126}{84\sqrt{3}+42\sqrt{3}}=\frac{42\sqrt{3}\times 126}{126\sqrt{3}}=42$ cm.
3. Volume $=\frac{4}{3}\times \frac{22}{7}\times {42}^3=\frac{88}{21}\times 74088=310464{\text{cm}}^3$.

Trick: Slant height = diameter means the cone is equilateral (semi-vertical angle ${30}^{\circ }$); then the inscribed sphere has $R=\frac{h}{3}$ directly.

Answer: (D) $310464{\text{cm}}^3$

Concept: Numbers $1$ to $n$ form an AP, so their average is $\frac{\text{first}+\text{last}}{2}=\frac{n+1}{2}$.

1. Average of $1$ to $60$ $=\frac{1+60}{2}=30.5$.
2. Check: sum $=\frac{60\times 61}{2}=1830$, $\frac{1830}{60}=30.5$.

Yaad rakhein: Average of first $n$ natural numbers $=\frac{n+1}{2}$ — no need to add anything.

Answer: (D) 30.5

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$ (weighted mean).

1. Totals: $39\times 377=14703$, $33\times 401=13233$, $52\times 233=12116$.
2. Sum $=14703+13233+12116=40052$; members $=39+33+52=124$.
3. Average $=\frac{40052}{124}=323$.

Trick: Take 323 as base: deviations $39\times 54=2106$, $33\times 78=2574$, $52\times (-90)=-4680$; they sum to $0$, so the average is exactly 323.

Answer: (C) 323

Concept: Write every base as a power of the same prime, then add/subtract exponents: $a^m\times a^n=a^{m+n}$, $a^m÷a^n=a^{m-n}$.

1. ${125}^{1.3}=5^{3\times 1.3}=5^{3.9}$, ${25}^{1.8}=5^{2\times 1.8}=5^{3.6}$.
2. $5^{3.9+3.6-2.5}=5^5$ ⇒ $x=5$.
3. $18x-5=90-5=85$.

Trick: $125=5^3$, $25=5^2$ — just multiply the decimal exponents by 3 and 2 mentally.

Answer: (A) 85

Concept: To join ratios, make the common term equal by cross-multiplying step by step.

1. $a:b=1:3$, $b:c=5:7$ ⇒ $a:b:c=5:15:21$.
2. $c:d=9:5$ with $c=21$: multiply first by 9 and second by 21 ⇒ $a:b:c:d=45:135:189:105$.
3. Divide by 3: $15:45:63:35$.

Trick: $a:b:c:d=(1·5·9):(3·5·9):(3·7·9):(3·7·5)=45:135:189:105$ — one line.

Answer: (C) 15 : 45 : 63 : 35

Concept: $V=\frac{4}{3}\pi r^3$ with $r=\frac{\text{diameter}}{2}$.

1. $r=\frac{37.8}{2}=18.9$ cm; ${18.9}^3=6751.269$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 6751.269=\frac{88\times 6751.269}{21}=\frac{594111.672}{21}$.
3. $V\approx 28291.03{\text{cm}}^3$.

Trick: $18.9=\frac{189}{10}=\frac{7\times 27}{10}$, so the 7 cancels with $\pi =\frac{22}{7}$: $V=\frac{88\times 49\times {27}^3}{3\times 1000}=\frac{88\times 49\times 6561}{1000}\approx 28291.03$.

Answer: (B) 28291.03

Concept: Discount is on MP, profit is on CP: $\text{SP}=\text{MP}(1-d)=\text{CP}(1+p)$.

1. Let CP $=100$. MP $=100+150=250$.
2. Required SP for 60% gain $=160$.
3. Discount $=250-160=90$ ⇒ $\frac{90}{250}\times 100=36\%$.

Trick: $\frac{\text{SP}}{\text{MP}}=\frac{160}{250}=0.64$ ⇒ discount $=36\%$.

Answer: (A) $36\%$

Concept: Complete the square: $p^2-2p+4=(p-1{)}^2+3\ge 3$ and $-p^2+2p-4=-[(p-1{)}^2+3]\le -3$.

1. The first expression is always $\ge 3$ and the second always $\le -3$, so $max\{·\}$ is always the first one.
2. Minimum of $(p-1{)}^2+3$ on $0\le p\le 1$ occurs at $p=1$: value $=3$.

Trick: The two expressions are negatives of each other, so the max is simply the positive one, $(p-1{)}^2+3$.

Answer: (D) 3

Concept: If a sum becomes $n$ times in $T$ years at SI, interest $=(n-1)P$ and $R=\frac{(n-1)\times 100}{T}$.

1. 5 times ⇒ SI $=4P$ in 12 years.
2. $R=\frac{4\times 100}{12}=33.33\%$ per annum.

Trick: "$n$ times in $T$ years" ⇒ $R=\frac{100(n-1)}{T}$ — plug and go.

Answer: (C) 33.33

Concept: Monthly saving = salary − expenses; yearly saving = saving × number of months actually saved.

1. Monthly saving $=18000-5000-2000=11000$.
2. In the birthday month the whole saving is spent, so only 11 months count.
3. Yearly saving $=11000\times 11=121000$.

Trick: Read the twist — "12 months" gives 132000 (option A), the trap answer.

Answer: (B) 121000

Concept: LCM of monomials = each variable raised to its highest power among the terms.

1. Rewrite: $x^{3.5}{y}^{1.5}$; $\frac{y^{2.5}}{x^{-4.5}}=x^{4.5}{y}^{2.5}$; $\frac{x}{y^{-3}}=x{y}^3$.
2. Highest powers: $x^{4.5}$ and $y^3$ ⇒ LCM $=x^{4.5}{y}^3$.
3. $x^{4.5}{y}^3=x^3{y}^3·x^{1.5}=(xy{)}^3\sqrt{x^3}$.

Trick: A negative power in the denominator jumps to the numerator with a positive sign: $\frac{1}{x^{-a}}=x^a$.

Answer: (A) $(xy{)}^3\sqrt{x^3}$