RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Translate the sentence into an equation: $20\%$ of the number $=20\%$ of $620+170$.

1. $20\%$ of $620=\frac{20}{100}\times 620=124$.
2. So $20\%$ of $N=124+170=294$.
3. $N=294\times \frac{100}{20}=294\times 5=1470$.

Trick: $20\%=\frac{1}{5}$, so the number is simply $5\times 294$.

Answer: (A) 1470

Concept: Correct sum $=$ wrong sum $-$ (wrong values) $+$ (correct values); then divide by $n$.

1. Wrong sum $=80\times 49=3920$.
2. Correction $=(92+83)-(69+74)=175-143=32$.
3. Correct sum $=3920+32=3952$; correct mean $=\frac{3952}{80}=49.4$.

Trick: New mean $=$ old mean $+\frac{\text{net correction}}{n}=49+\frac{32}{80}=49.4$.

Answer: (C) 49.4

Concept: Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$ with class marks $x_i=10,30,50,70,90$.

1. Total frequency: $4+19+x+5+y=64\Rightarrow x+y=36$.
2. $\sum f_i{x}_i=40+570+50x+350+90y=960+50x+90y$.
3. $\frac{960+50x+90y}{64}=55\Rightarrow 50x+90y=2560\Rightarrow 5x+9y=256$.
4. Subtract $5(x+y)=180$: $4y=76\Rightarrow y=19,x=17$.
5. $7x+5y=119+95=214$.

Trick: Two unknowns need two equations — one always comes from the total frequency.

Answer: (A) 214

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$.

1. Group A: $48\times 477=22896$; B: $46\times 236=10856$; C: $36\times 348=12528$.
2. Total $=22896+10856+12528=46280$; members $=48+46+36=130$.
3. Average $=\frac{46280}{130}=356$.

Trick: Never average the averages; always weight by group size.

Answer: (C) 356

Concept: Amount $=P{(1+\frac{r}{100})}^n$; equate the two amounts.

1. $V(1.1{)}^{14}=N(1.1{)}^{15}\Rightarrow V=1.1N=\frac{11}{10}N$.
2. $V+N=1407\Rightarrow \frac{11N}{10}+N=\frac{21N}{10}=1407$.
3. $N=\frac{1407\times 10}{21}=670$.

Trick: One extra year at $10\%$ means shares are in ratio $11:10$; $1407÷21=67$, so Nitin $=10\times 67=670$.

Answer: (D) 670

Concept: Co-interior (same-side interior) angles between parallel lines are supplementary.

1. Let the other interior angle on the same side be $x$.
2. ${70}^{\circ }+x={180}^{\circ }$.
3. $x={110}^{\circ }$.

Yaad rakhein: Alternate / corresponding angles are equal; co-interior angles add to ${180}^{\circ }$.

Answer: (A) ${110}^{\circ }$

Concept: Take CP $=100$; then MP $=100+110=210$ and required SP $=126$.

1. Discount $=\text{MP}-\text{SP}=210-126=84$.
2. Discount$\%=\frac{84}{210}\times 100=40\%$.

Trick: $\frac{\text{SP}}{\text{MP}}=\frac{126}{210}=\frac{3}{5}=60\%$ → discount $=40\%$.

Answer: (D) $40\%$

Concept: Look at the differences — they are perfect cubes.

1. $325-200=125=5^3$, $200-136=64=4^3$, $136-109=27=3^3$, $109-101=8=2^3$.
2. Next difference $=1^3=1$.
3. $101-1=100$.

Trick: Decreasing differences 125, 64, 27, 8 scream "cubes"; the last one is $1$.

Answer: (B) 100

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

1. $a+b+c=\frac{1}{14}+\frac{1}{14}-\frac{1}{7}=\frac{2}{14}-\frac{2}{14}=0$.
2. So $a^3+b^3+c^3=3abc=3\times \frac{1}{14}\times \frac{1}{14}\times (-\frac{1}{7})$.
3. $=-\frac{3}{14\times 14\times 7}=-\frac{3}{1372}$.

Trick: Whenever three fractions are given for a cubes question, first check if their sum is zero.

Answer: (D) $-\frac{3}{1372}$

Concept: Equal roots $\iff$ discriminant $b^2-4ac=0$.

1. Here $a=1,b=k,c=361$.
2. $k^2-4\times 1\times 361=0\Rightarrow k^2=1444$.
3. $k=\pm \sqrt{1444}=\pm 38$.

Trick: $x^2+kx+361=(x\pm 19{)}^2$ needs $k=\pm 2\times 19=\pm 38$.

Answer: (C) $\pm 38$

Concept: $1{\text{m}}^3=1000$ litres; time $=\frac{\text{volume}}{\text{rate}}$.

1. Volume $=168{\text{m}}^3=168\times 1000=168000$ litres.
2. Time $=\frac{168000}{70}=2400$ minutes.
3. $2400÷60=40$ hours.

Trick: Rate in litres/hour $=70\times 60=4200$; $168000÷4200=40$.

Answer: (D) 40

Concept: Margin $=$ (winner's $\%$ $-$ loser's $\%$) of valid votes.

1. Let registered voters $=R$. Polled $=0.86R$; valid $=80\%$ of polled $=0.688R$.
2. Winner $55\%$, loser $45\%$ → margin $=10\%$ of valid $=0.0688R$.
3. $0.0688R=3010\Rightarrow R=\frac{3010}{0.0688}=43750$.

Trick: Chain the percentages: $R\times 0.86\times 0.8\times 0.10=3010$.

Answer: (A) 43750

Concept: In $a:b::c:d$, $\frac{a}{b}=\frac{c}{d}$.

1. $9:81::2.5:x\Rightarrow \frac{9}{81}=\frac{2.5}{x}\Rightarrow x=2.5\times 9=22.5$.
2. $3:6::5:y\Rightarrow \frac{3}{6}=\frac{5}{y}\Rightarrow y=10$.
3. $x:y=22.5:10=45:20=9:4$.

Trick: $9\to 81$ is $\times 9$, $3\to 6$ is $\times 2$; apply the same multipliers to $2.5$ and $5$.

Answer: (C) $9:4$

Concept: Express all present ages in one variable, then go back 3 years for each person.

1. Son $=s$, Vanraj $=3s$, daughter $=s-3$.
2. 3 years ago: $(3s-3)+(s-3)+(s-6)=5s-12=73$.
3. $5s=85\Rightarrow s=17$; Vanraj $=3\times 17=51$.

Trick: Present sum $=73+3\times 3=82=5s-3\Rightarrow s=17$.

Answer: (A) 51

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

1. $15\times 72.6=1089$.
2. $\sqrt{1089}=33$ (since ${33}^2=1089$).

Trick: $72.6=\frac{726}{10}$ and $15\times 726=10890$ → $1089={33}^2$; remember ${33}^2=1089$.

Answer: (A) 33

Concept: Take total work $=\text{LCM}(21,24)=168$ units.

1. (A+B) per day $=\frac{168}{21}=8$ units; A per day $=\frac{168}{24}=7$ units.
2. B per day $=8-7=1$ unit → B alone finishes in $168$ days.
3. Three-fourth of the work: $\frac{3}{4}\times 168=126$ days.

Trick: B's time $=\frac{21\times 24}{24-21}=168$; then take $\frac{3}{4}$.

Answer: (B) 126

Concept: If HCF $=h$, write the numbers as multiples of $h$ with coprime cofactors.

1. First number $=5h$, second $=3h$ ($5$ and $3$ are coprime, so HCF is indeed $h$).
2. $3\times$ first $=15h$ and $5\times$ second $=15h$ → their HCF $=15h$.
3. $15h=3\times 5\times h$, where $h$ is a two-digit prime → product of three prime numbers.

Trick: Test with $h=11$: numbers 55 and 33 → $165$ and $165$ → HCF $165=3\times 5\times 11$.

Answer: (D) three prime numbers

Concept: Take the marked price (MP) as a convenient number; find the cost price from the loss case, then apply it to the new selling price.

1. Let MP $=100$. First SP $=\frac{2}{5}\times 100=40$, which gives a $21\%$ loss, so $\text{CP}=\frac{40}{0.79}=\frac{4000}{79}$.
2. New SP $=56\%$ of $100=56$.
3. $\text{Profit}\%=\frac{56-\frac{4000}{79}}{\frac{4000}{79}}\times 100=(\frac{56\times 79}{4000}-1)\times 100=(\frac{4424}{4000}-1)\times 100=10.6\%$.

Trick: $\frac{{\text{SP}}_2}{{\text{SP}}_1}=\frac{56}{40}=1.4$; so the new multiplier on CP is $0.79\times 1.4=1.106$, i.e. $10.6\%$ profit – no need to find CP at all.

Answer: (C) 10.6%

Concept: Runs scored $=$ run rate $\times$ overs. Find the runs already scored, subtract from the target and divide by the remaining overs.

1. Runs so far: $15\times 5+15\times 8+10\times 4=75+120+40=235$.
2. Overs used $=15+15+10=40$, so $10$ overs remain.
3. Runs still needed $=300-235=65$; required rate $=\frac{65}{10}=6.5$ runs per over.

Trick: Always total the overs first – here $40$ of $50$ are gone, so the whole question reduces to "65 in 10".

Answer: (C) 6.5 runs per over

Concept: For parallel lines cut by a transversal, alternate interior angles are equal and angles on a straight line add to ${180}^{\circ }$.

1. At line $l$ the given angle is ${130}^{\circ }$, so the interior angle made with the transversal is ${180}^{\circ }-{130}^{\circ }={50}^{\circ }$.
2. $l\parallel n$, so $x$ (alternate interior angle at $n$) $={50}^{\circ }$.
3. $n\parallel m$ and the angle at $m$ is ${50}^{\circ }$, so $y={50}^{\circ }$ (alternate interior).
4. $x+y={50}^{\circ }+{50}^{\circ }={100}^{\circ }$.

Trick: With a "bent" transversal through a middle parallel line, the angle at the bend always splits into the two alternate angles from the outer lines.

Answer: (B) ${100}^{\circ }$