RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Take total cost = 100; overall profit % = weighted sum of profits of each part.

1. Part 1: 35% at 22% profit → $35\times 0.22=7.7$.
2. Remaining 65%; 40% of it = 26% at 37% profit → $26\times 0.37=9.62$.
3. Rest = $65-26=39\%$ at 38% loss → $39\times 0.38=14.82$ loss.
4. Net = $7.7+9.62-14.82=2.5$, so $x=2.5$.

Trick: Split goods as 35 : 26 : 39 and multiply by +22, +37, −38 respectively.

Answer: (C) 2.5

Concept: Under compound interest, if money becomes $k$ times in $n$ years, it becomes $k^m$ times in $mn$ years.

1. Doubles in 8 years: $(1+r{)}^8=2$.
2. Eight times $=2^3$ → $(1+r{)}^{24}=2^3=8$.
3. Time = $3\times 8=24$ years.

Trick: 2× in 8 yrs → 4× in 16 yrs → 8× in 24 yrs (keep multiplying by 2 every 8 years).

Answer: (A) 24

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $840=\frac{8400\times 4\times T}{100}=336T$.
2. $T=\frac{840}{336}=2.5$ years.

Trick: SI is 10% of principal; at 4% per year that needs $\frac{10}{4}=2.5$ years.

Answer: (B) 2.5

Concept: ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$ and $a^m\times a^n=a^{m+n}$.

1. Write ${\left(\frac{4}{17}\right)}^8={\left(\frac{17}{4}\right)}^{-8}$. LHS $={\left(\frac{17}{4}\right)}^{9-8+15}={\left(\frac{17}{4}\right)}^{16}$.
2. ${\left(\frac{17}{4}\right)}^{16}={\left(\frac{4}{17}\right)}^{-16}$, so $5m+15=-16$.
3. $5m=-31\Rightarrow m=-\frac{31}{5}$.

Trick: Bring every term to the same base $\frac{4}{17}$: powers become $-9,+8,-15$ → sum $-16$.

Answer: (D) $-\frac{31}{5}$

Concept: Area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.

1. P(2,5), Q(8,1), R(4,9): $2(1-9)+8(9-5)+4(5-1)$.
2. $=-16+32+16=32$.
3. Area $=\frac{1}{2}\times 32=16$ square km.

Trick: Shoelace method – write the three points in a column, repeat the first, cross-multiply down and up, halve the difference.

Answer: (D) 16 square kilometers

Concept: In the right triangle formed by the pole, the ground and the line of sight, $\tan \theta =\frac{\text{height}}{\text{distance}}$.

1. $\tan {30}^{\circ }=\frac{h}{30}$.
2. $h=30\times \frac{1}{\sqrt{3}}=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m.

Trick: With ${30}^{\circ }$ the height is always the base divided by $\sqrt{3}$; with ${60}^{\circ }$ it is the base times $\sqrt{3}$.

Answer: (C) $10\sqrt{3}$

Concept: Average $=\frac{\text{sum of terms}}{\text{number of terms}}$.

1. $7+9+11=27$.
2. $\frac{27}{3}=9$.

Trick: For numbers in arithmetic progression the average is simply the middle term – here 9.

Answer: (A) 9

Concept: For grouped data, mean $=\frac{\sum f_i{x}_i}{\sum f_i}$ where $x_i$ is the class mark. Two unknowns need two equations: the total frequency and the mean.

1. Total: $6+19+x+5+y=59\Rightarrow x+y=29$.
2. Class marks 10, 30, 50, 70, 90: $\sum f_i{x}_i=60+570+50x+350+90y=980+50x+90y$.
3. Mean 50: $980+50x+90y=50\times 59=2950\Rightarrow 5x+9y=197$.
4. Put $x=29-y$: $145+4y=197\Rightarrow y=13,x=16$. So $7x+6y=112+78=190$.

Trick: Use the assumed mean 50: deviations $-40,-20,0,20,40$ give $-240-380+100+40y=0\Rightarrow y=13$ in one line.

Answer: (A) 190

Concept: Multiply each part of the ratio by its own percentage factor.

1. A: $4\times \frac{120}{100}=4.8$; B: $5\times \frac{90}{100}=4.5$.
2. New ratio $4.8:4.5=48:45=16:15$.

Trick: $4\times 12:5\times 9=48:45$ – drop the 100s, they cancel.

Answer: (A) 16 : 15

Concept: Downstream speed = speed in still water + speed of stream; time $=\frac{\text{distance}}{\text{speed}}$.

1. Downstream speed $=12+3=15\text{km/h}$.
2. Time $=\frac{60}{15}=4$ hours.

Trick: Down = add, Up = subtract. Downstream is always the faster, shorter trip.

Answer: (A) 4

Concept: Opposite angles of a cyclic quadrilateral add to ${180}^{\circ }$; co-interior angles between parallel sides also add to ${180}^{\circ }$.

1. PS ∥ QR, PQ is a transversal: $\angle P+\angle Q={180}^{\circ }\Rightarrow \angle Q={95}^{\circ }$.
2. Cyclic: $\angle P+\angle R={180}^{\circ }\Rightarrow \angle R={95}^{\circ }$.
3. Cyclic: $\angle Q+\angle S={180}^{\circ }\Rightarrow \angle S={85}^{\circ }$.

Trick: A cyclic trapezium is always isosceles – base angles equal in pairs: $\angle P=\angle S$, $\angle Q=\angle R$.

Answer: (D) $\angle Q={95}^{\circ },\angle R={95}^{\circ },\angle S={85}^{\circ }$

Concept: Round-trip time with boat speed $b$ and stream $s$: $\frac{d}{b+s}+\frac{d}{b-s}=\frac{2bd}{b^2-s^2}$.

1. With $b=7$, $d=40$: time $=\frac{560}{49-s^2}$ hours.
2. 560 minutes $=\frac{28}{3}$ h. Condition: $\frac{560}{49-(s+2{)}^2}-\frac{560}{49-s^2}=\frac{28}{3}$.
3. Try $s=3$: $\frac{560}{24}-\frac{560}{40}=\frac{70}{3}-14=\frac{28}{3}$. ✓

Trick: Options are small – plug them in. $49-s^2$ should divide 560 nicely: $s=3$ gives 40 and $s=5$ gives 24.

Answer: (D) 3

Concept: Multiply numerator and denominator by 10 to remove decimals, then divide.

1. $\frac{1.2}{0.3}=\frac{12}{3}=4$.
2. $\frac{2.4}{0.6}=\frac{24}{6}=4$.
3. Sum $=4+4=8$.

Trick: $0.3\times 4=1.2$ and $0.6\times 4=2.4$ – both quotients are 4 by inspection.

Answer: (D) 8

Concept: Profit% $=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$; compare on the same unit (one dozen).

1. CP of a dozen = ₹400. SP of a dozen $=58\times 12=$ ₹696.
2. Profit per dozen $=696-400=296$.
3. Profit% $=\frac{296}{400}\times 100=74\%$.

Trick: The '52 dozens' is a distraction – percentage profit is the same for one dozen.

Answer: (B) 74%

Concept: Profit is shared in the ratio of investments (same time). Let the total profit be $12k$ (sum of ratio parts).

1. A $=3k$, B $=5k$, C $=4k$.
2. After B gives ₹15,000 to A: $3k+15000=4k\Rightarrow k=15000$.
3. Total profit $=12k=12\times 15000=$ ₹1,80,000.

Trick: The transfer equals the gap between A and C, i.e. one ratio part – so one part = ₹15,000 and total = 12 parts.

Answer: (D) 1,80,000

Concept: Take CP = 100; marked price and selling price follow from the given percentages.

1. MP $=100+100\%=200$. Required SP for 2% gain $=102$.
2. Discount $=200-102=98$.
3. Discount% $=\frac{98}{200}\times 100=49\%$.

Trick: $\text{MP}\times (1-d)=\text{CP}\times (1+g)$: $2(1-d)=1.02\Rightarrow d=0.49$.

Answer: (A) 49%

Concept: Total salary ratio = (number ratio) × (salary ratio) for each group.

1. Trained : Untrained total $=5\times 4:3\times 1=20:3$.
2. Untrained share $=\frac{3}{23}\times 4,60,000$.
3. $4,60,000÷23=20,000$; $\times 3=$ ₹60,000.

Trick: 4,60,000 is exactly $23\times 20,000$ – the clean division confirms the 20:3 split.

Answer: (C) ₹60,000

Concept: Take successive differences and look for a simple pattern in them.

1. 23 → 26 → 32 → 41 → 53: differences $3,6,9,12$ (multiples of 3).
2. Next difference $=15$, so $53+15=68$.

Trick: Differences growing by a constant means the next term = last term + (last difference + that constant).

Answer: (B) 68

Concept: If the bisectors of $\angle A$ and $\angle B$ of quadrilateral ABCD meet at O, then $\angle AOB=\frac{1}{2}(\angle C+\angle D)$.

1. In $△AOB$: $\angle AOB={180}^{\circ }-\frac{\angle A+\angle B}{2}$.
2. Since $\angle A+\angle B={360}^{\circ }-(\angle C+\angle D)$, we get $\angle AOB=\frac{\angle C+\angle D}{2}$.
3. $\angle C+\angle D=2\times {41}^{\circ }={82}^{\circ }$.

Trick: Angle between two adjacent angle bisectors of a quadrilateral = half the sum of the other two angles.

Answer: (C) ${82}^{\circ }$

Concept: Express the cost price in terms of the actual selling price, then compare.

1. Let the actual selling price be $S$. Selling at $\frac{3}{10}S$ gives $25\%$ loss: $\frac{3}{10}S=\frac{3}{4}CP\Rightarrow CP=\frac{3}{10}S\times \frac{4}{3}=\frac{2}{5}S=0.4S$.
2. New selling price $=0.77S$. Profit $=0.77S-0.40S=0.37S$.
3. Profit$\%=\frac{0.37S}{0.40S}\times 100=92.5\%$.

Trick: Take $S=100$: CP $=30÷0.75=40$; selling at 77 → profit $37$ on $40$ $=92.5\%$.

Answer: (C) 92.5%