RRB NTPC Graduate CBT 1 2025 PYQs

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) $54.95{\text{m}}^2$

Concept: Total surface area of a solid hemisphere $=2\pi r^2+\pi r^2=3\pi r^2$.

1. $3\times \frac{22}{7}\times r^2=200\Rightarrow r^2=\frac{200\times 7}{66}=\frac{1400}{66}=\frac{700}{33}$.
2. $r=\sqrt{\frac{700}{33}}$ cm.

Trick: Hemisphere: CSA $=2\pi r^2$, TSA $=3\pi r^2$, volume $=\frac{2}{3}\pi r^3$. Do not forget the flat circular base in TSA.

Answer: (D) $\sqrt{\frac{700}{33}}$

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

1. Total $=69\times 468+65\times 120+88\times 276=32292+7800+24288=64380$.
2. Members $=69+65+88=222$.
3. Average $=\frac{64380}{222}=290$.

Trick: Take 288 as base: deviations $69\times 180-65\times 168-88\times 12=12420-10920-1056=444$; $\frac{444}{222}=2$ → $288+2=290$.

Answer: (C) 290

Concept: Trains moving towards each other: relative speed = sum of speeds; distance to cross = sum of lengths.

1. Relative speed $=49+59=108\text{km/h}=108\times \frac{5}{18}=30\text{m/s}$.
2. Distance $=349+371=720$ m.
3. Time $=\frac{720}{30}=24$ s.

Trick: $108\text{km/h}$ is exactly $30\text{m/s}$ ($\times \frac{5}{18}$); memorise 18 → 5, 36 → 10, 54 → 15, 72 → 20, 90 → 25, 108 → 30.

Answer: (C) 24

Concept: Ladder, wall and ground form a right triangle; height on the wall $=\text{ladder}\times \sin \theta$.

1. $h=28\sin {60}^{\circ }=28\times \frac{\sqrt{3}}{2}=14\sqrt{3}$.
2. $14\times 1.74=24.36$ m.

Trick: In a 30°-60°-90° triangle the sides are $1:\sqrt{3}:2$; opposite ${60}^{\circ }$ is $\frac{\sqrt{3}}{2}$ of the hypotenuse.

Answer: (C) 24.36 m

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

1. Five times ⇒ SI $=4P$.
2. $R=\frac{4P\times 100}{P\times 4}=100\%$ per annum.

Trick: $R\times T=(n-1)\times 100$ → $R\times 4=400$ → $R=100\%$.

Answer: (B) 100%

Concept: Fixed provisions: men × days = constant (inverse variation).

1. $100\times 27=50\times D$
2. $D=\frac{2700}{50}=54$ days.

Trick: Half the men ⇒ double the days: $27\times 2=54$.

Answer: (D) 54

Concept: Take CP = 100; MP = CP + 160% = 260; required SP = 130 (30% gain). Discount is measured on MP.

1. Discount $=260-130=130$.
2. Discount% $=\frac{130}{260}\times 100=50\%$.

Trick: $\frac{\text{SP}}{\text{MP}}=\frac{130}{260}=\frac{1}{2}$, so half the marked price is knocked off → 50% discount.

Answer: (B) 50%

Concept: If $a+\frac{1}{a}=k$, then $a^2+\frac{1}{a^2}=k^2-2$ and $a^3+\frac{1}{a^3}=k^3-3k$; also $a^6+\frac{1}{a^6}={(a^3+\frac{1}{a^3})}^2-2$.

1. $a^3+\frac{1}{a^3}=3^3-3\times 3=27-9=18$.
2. $a^6+\frac{1}{a^6}={18}^2-2=324-2=322$.

Trick: For $a+\frac{1}{a}=3$: squares chain 3 → 7 → 47 (for $a^4$), cubes 18, sixth power ${18}^2-2=322$. Cross-check: $7^3-3\times 7=343-21=322$ ✓.

Answer: (D) 322

Concept: Same base → add the powers: $a^m\times a^n=a^{m+n}$, and $\left(\frac{9}{28}\right)={\left(\frac{28}{9}\right)}^{-1}$.

1. Write everything with base $\frac{28}{9}$: ${\left(\frac{9}{28}\right)}^{16}={\left(\frac{28}{9}\right)}^{-16}$.
2. LHS $={\left(\frac{28}{9}\right)}^{11-16+17}={\left(\frac{28}{9}\right)}^{12}$.
3. RHS $={\left(\frac{9}{28}\right)}^{2m+8}={\left(\frac{28}{9}\right)}^{-(2m+8)}$.
4. Equate powers: $-(2m+8)=12\Rightarrow 2m=-20\Rightarrow m=-10$.

Trick: Flip the fraction = change the sign of the power. Bring every term to one base first, then just compare exponents.

Answer: (A) $-10$

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

1. Let $a=555,b=444$.
2. Numerator $=(a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$.
3. Denominator $=a\times a+b\times b=a^2+b^2$.
4. Ratio $=\frac{2(a^2+b^2)}{a^2+b^2}=2$.

Trick: Big numbers in a "simplify" question are a hint that an identity will cancel everything – never multiply them out.

Answer: (B) 2

Concept: $a{x}^2+bx+c=0$ is quadratic only when the coefficient of $x^2$ is non-zero: $a\ne 0$.

1. Here $a=2m-3$.
2. Condition: $2m-3\ne 0\Rightarrow m\ne \frac{3}{2}$.
3. The coefficient of $x$ or the constant may be anything, so options with $m=-1$ or $m=-\frac{3}{2}$ are not required.

Trick: "Must be true" for a quadratic = only one check: leading coefficient $\ne 0$.

Answer: (D) $m\ne \frac{3}{2}$