RRB NTPC Graduate CBT 1 2025 PYQs

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

1. Amount $=5P$ → SI $=4P$.
2. $R=\frac{4P\times 100}{P\times 6}=\frac{400}{6}=66\frac{2}{3}\%$.

Trick: "$n$ times" means interest is $(n-1)$ times the principal – never use $n$ directly.

Answer: (B) $66\frac{2}{3}\%$

Concept: Check the differences; if they form a simple pattern (multiples of 6 here), extend it.

1. $50-44=6$, $62-50=12$, $80-62=18$, $104-80=24$.
2. Differences $6,12,18,24$ → next difference $30$.
3. $?=104+30=134$.

Trick: Differences increasing by a constant = second-level AP; just add the next multiple.

Answer: (D) 134

Concept: Opposite directions → relative speed = sum of speeds; distance covered while crossing = sum of lengths.

1. Relative speed $=40+52=92\text{km/h}=92\times \frac{5}{18}=\frac{230}{9}\text{m/s}$.
2. Distance in 18 s $=\frac{230}{9}\times 18=460\text{m}$.
3. Other train $=460-250=210\text{m}$.

Trick: $92\times 18\times \frac{5}{18}=92\times 5=460$ – the 18 cancels, so no fraction work needed.

Answer: (D) 210 m

Concept: Evaluate the root, then do $\times$ and $÷$ from left to right, then solve for $y$.

1. $\sqrt{3025}=55$.
2. $26\times 5\times 715÷55=130\times 13=1690$ (since $715÷55=13$).
3. $y=1690-913=777$.

Trick: Divide first where it cancels cleanly: $715=55\times 13$ saves a big multiplication.

Answer: (A) 777

Concept: In an isosceles triangle the altitude from the apex bisects the base; use Pythagoras to get the height.

1. $BD=DC=\frac{12}{2}=6$ cm.
2. $AD=\sqrt{{10}^2-6^2}=\sqrt{64}=8$ cm.
3. Area $=\frac{1}{2}\times 12\times 8=48{\text{cm}}^2$.

Trick: 6-8-10 is a 3-4-5 triple – spot it and skip the square root.

Answer: (C) 48 cm²

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

1. Total $=85\times 407+23\times 365+42\times 280=34595+8395+11760=54750$.
2. Members $=85+23+42=150$.
3. Average $=\frac{54750}{150}=365$.

Trick: Take 365 as base: deviations $85\times 42=3570$ and $42\times (-85)=-3570$ cancel exactly, B is at the base → answer 365 without division.

Answer: (D) 365

Concept: For $n$ sorted values the median is the $\frac{n+1}{2}$-th value (odd $n$) or the mean of the two middle values (even $n$).

1. 13 values (already sorted): median = 7th value = 28.
2. Remove 55 → 12 values: median $=\frac{6\text{th}+7\text{th}}{2}=\frac{25+28}{2}=26.5$.
3. Difference $=28-26.5=1.5$ years.

Trick: Removing the largest value shifts the median down by half the gap between the two middle values – here $\frac{28-25}{2}=1.5$.

Answer: (B) 1.5 years

Concept: Step down from $x^4+\frac{1}{x^4}$ to $x-\frac{1}{x}$ using ${(x^2+\frac{1}{x^2})}^2=x^4+\frac{1}{x^4}+2$ and ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$.

1. $x^4+\frac{1}{x^4}=322$ → $x^2+\frac{1}{x^2}=\sqrt{324}=18$.
2. ${(x-\frac{1}{x})}^2=18-2=16$ → $x-\frac{1}{x}=4$ (positive as $x\ge 1$).
3. $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})=64+12=76$.

Trick: Remember the chain $322\to 18\to 4$: each step is "add or subtract 2, then square-root".

Answer: (A) 76

Concept: The ladder, wall and ground form a right triangle; the side opposite the angle at the top (between ladder and wall) is the foot-to-wall distance, so use $\sin$.

1. Angle with the wall $={30}^{\circ }$, hypotenuse (ladder) $=24$ m.
2. Distance from wall $=24\sin {30}^{\circ }=24\times \frac{1}{2}=12$ m.

Trick: Angle with the wall → $\sin$ gives the ground distance; angle with the ground → $\cos$ gives it. Here it is simply half the ladder.

Answer: (A) 12

Concept: CSA of a cone $=\pi rl$; when $r$ and $l$ change by $a\%$ and $b\%$, net change $=a+b+\frac{ab}{100}\%$.

1. $a=+8$, $b=-15$: net $=8-15+\frac{8\times (-15)}{100}=-7-1.2=-8.2\%$.
2. Check: $1.08\times 0.85=0.918$ → a $8.2\%$ decrease.

Trick: For a product of two quantities, use the successive-percentage formula; the sign of $\frac{ab}{100}$ is negative when one rises and the other falls.

Answer: (A) decreases by 8.2%

Concept: Let CP of each article be ₹1 and MP be ₹$m$. The customer pays for 120 articles at the discounted price but receives 123.

1. Cost to trader $=123\times 1=123$; revenue $=120\times 0.8m=96m$.
2. Profit 28% → $96m=1.28\times 123=157.44$ → $m=1.64$.
3. MP exceeds CP by $64\%$.

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{(1+\text{profit})}{(1-\text{discount})}\times \frac{\text{articles given}}{\text{articles paid}}=\frac{1.28}{0.8}\times \frac{123}{120}=1.6\times 1.025=1.64$.

Answer: (A) 64%

Concept: A tangent is perpendicular to the radius at the point of contact, so $△OBA$ is right-angled at B. Also $OB=OC$, so $△OBC$ is isosceles.

1. In $△OBA$: $\sin \angle OAB=\frac{OB}{OA}=\frac{3}{6}=\frac{1}{2}\Rightarrow \angle OAB={30}^{\circ }$, so $\angle AOB={60}^{\circ }$.
2. By symmetry $\angle AOC={60}^{\circ }$, hence $\angle BOC={120}^{\circ }$.
3. In isosceles $△OBC$: $\angle OBC=\angle OCB=\frac{{180}^{\circ }-{120}^{\circ }}{2}={30}^{\circ }$.

Trick: When the external point is at distance $2r$ from the centre, the two tangents make ${60}^{\circ }$ with each other, the central angle is ${120}^{\circ }$ and each base angle of $△OBC$ is ${30}^{\circ }$.

Answer: (A) ${30}^{\circ }$

Concept: $\frac{M_1{T}_1}{W_1}=\frac{M_2{T}_2}{W_2}$ (men × time ∝ work).

1. $\frac{9\times 3}{117}=\frac{M_2\times 1}{208}$
2. $M_2=\frac{27\times 208}{117}=\frac{27\times 208}{117}$; since $117=9\times 13$ and $208=16\times 13$: $M_2=\frac{27\times 16}{9}=48$.

Trick: One student makes $\frac{117}{27}=\frac{13}{3}$ charts per hour, so $208÷\frac{13}{3}=16\times 3=48$ students.

Answer: (B) 48 students

Concept: The mean proportional between $a$ and $b$ is $\sqrt{ab}$, i.e. $m^2=ab$.

1. ${18}^2=x\times 36$
2. $x=\frac{324}{36}=9$

Trick: $x:18=18:36$; the right ratio is $1:2$, so $x=\frac{18}{2}=9$.

Answer: (C) 9

Concept: Take the younger brother's present age as $x$ and express the others in terms of $x$; "3 years ago" reduces each age by 3.

1. Younger brother $=x$, Sharad $=2x$, other brother $=x-3$. Present sum $=4x-3$.
2. Three years ago: $(4x-3)-9=28\Rightarrow 4x=40\Rightarrow x=10$.
3. Sharad's age $=2x=20$ years.

Trick: Sum of ages 3 years ago $=28$ → present sum $=28+3\times 3=37$. Then $4x-3=37\Rightarrow x=10$.

Answer: (B) 20

Concept: Apply BODMAS: evaluate the square root first, then multiplication/division from left to right.

1. $\sqrt{2601}=51$ (since ${51}^2=2601$).
2. $22\times 3\times 816÷51=66\times \frac{816}{51}=66\times 16=1056$.
3. $y=1056-794=262$.

Trick: $816=51\times 16$ – spot such factors before multiplying, it keeps numbers small.

Answer: (B) 262

Concept: Profit is shared in the ratio of (capital × time).

1. Trisha: $8\times 12=96$; Riya: $5\times 12=60$; Poorva (equal to Riya, for 6 months): $5\times 6=30$.
2. Ratio $=96:60:30=16:10:5$.

Trick: Poorva's capital equals Riya's but for half the time, so her share is half of Riya's: $8:5:2.5=16:10:5$.

Answer: (C) 16 : 10 : 5

Concept: If the shopkeeper claims a loss of 45%, he charges only 55% of the cost of 1 kg. To still gain 50%, the goods he actually gives must cost $\frac{55}{150}$ of 1 kg.

1. Amount charged $=0.55\times (\text{CP of 1 kg})$.
2. Gain 50% ⇒ actual CP of goods given $=\frac{0.55}{1.5}$ kg $=\frac{55}{150}=\frac{11}{30}$ kg.
3. $\frac{11}{30}\times 1000=366.67\approx 366.7$ g.

Trick: $\text{True weight}=\frac{100-\text{claimed loss}\%}{100+\text{actual gain}\%}\times 1000=\frac{55}{150}\times 1000=366.7$ g.

Answer: (C) 366.7 gm

Concept: If $3A=4B=8C=k$, then $A=\frac{k}{3}$, $B=\frac{k}{4}$, $C=\frac{k}{8}$.

1. Take $k=\text{LCM}(3,4,8)=24$.
2. $A=8$, $B=6$, $C=3$ → $A:B:C=8:6:3$.

Trick: $A:B:C=\frac{1}{3}:\frac{1}{4}:\frac{1}{8}$; multiply by 24 → $8:6:3$. The biggest coefficient gives the smallest share.

Answer: (B) 8 : 6 : 3

Concept: A paper cup is an open cone, so paper needed = curved surface area $=\pi rl$, where slant height $l=\sqrt{h^2+r^2}$.

1. $l=\sqrt{{24}^2+7^2}=\sqrt{576+49}=\sqrt{625}=25$ cm.
2. CSA of one cup $=3.14\times 7\times 25=549.5{\text{cm}}^2$.
3. 1000 cups: $549500{\text{cm}}^2=\frac{549500}{10000}=54.95{\text{m}}^2$.

Trick: 7-24-25 is a Pythagorean triplet – no calculation needed for $l$. And $1{\text{m}}^2={10}^4{\text{cm}}^2$.

Answer: (D) $54.95{\text{m}}^2$