RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Do division and multiplication first, then addition and subtraction; watch the signs.

1. $(-56)÷(-14)=4$ and $(-4)\times 6=-24$.
2. Expression becomes $(-9)-4+(-24)$.
3. $-9-4-24=-37$.

Trick: Negative ÷ negative = positive; the minus sign before it then makes it $-4$.

Answer: (B) -37

Concept: $\text{Overall average}=\frac{\sum (\text{members}\times \text{average})}{\text{total members}}$.

1. Totals: $42\times 275=11550$, $14\times 430=6020$, $21\times 498=10458$.
2. Grand total $=11550+6020+10458=28028$; members $=42+14+21=77$.
3. $\frac{28028}{77}=364$.

Trick: Members are in ratio $6:2:3$ (÷7): $\frac{6\times 275+2\times 430+3\times 498}{11}=\frac{1650+860+1494}{11}=\frac{4004}{11}=364$.

Answer: (C) 364

Concept: Income − Expenditure = Savings. Let incomes be $4x,5x$ and expenditures $3y,4y$, then solve two equations.

1. $4x-3y=10000$ … (i) and $5x-4y=12000$ … (ii).
2. (i)$\times 4$: $16x-12y=40000$; (ii)$\times 3$: $15x-12y=36000$. Subtract: $x=4000$.
3. Income of A $=4x=$ ₹$16,000$ (check: $y=2000$, B saves $20000-8000=12000$ ✓).

Trick: Eliminate $y$ by making its coefficients equal ($12y$); then $x$ comes out in one subtraction.

Answer: (B) 16,000

Concept: Use $a^2-b^2=(a+b)(a-b)$ instead of squaring decimals.

1. ${0.7}^2-{0.4}^2=(0.7+0.4)(0.7-0.4)=1.1\times 0.3=0.33$.
2. $\frac{0.33}{0.33}=1$.

Trick: When the denominator equals $(a+b)(a-b)$, the answer is simply $1$ – no decimal multiplication needed.

Answer: (B) 1

Concept: Work backwards: each sale multiplies the price by $(1\pm \text{rate})$, so divide to go back.

1. R paid ₹3900 after Q's 52% loss, so Q's cost $=\frac{3900}{0.48}=8125$.
2. Q's cost is P's selling price at 30% profit: P's cost $=\frac{8125}{1.30}=6250$.
3. Check: $6250\times 1.3=8125$, $8125\times 0.48=3900$ ✓.

Trick: Combined multiplier $=1.3\times 0.48=0.624$; $\frac{3900}{0.624}=6250$ in one step.

Answer: (A) 6250

Concept: $\text{Sum of observations}=\text{Mean}\times \text{Number of observations}$.

1. Total of 10 observations $=6\times 10=60$.
2. Sum of nine of them $=49$.
3. Tenth observation $=60-49=11$.

Trick: Mean questions almost always start with 'total = mean × n'.

Answer: (D) 11

Concept: Work per day is additive: $A=(A+B+C)-(B+C)$. Efficiency ratio gives individual rates.

1. A's one-day work $=\frac{1}{40}-\frac{1}{72}=\frac{9-5}{360}=\frac{1}{90}$.
2. B is 50% more efficient: B's one-day work $=\frac{1}{90}\times \frac{3}{2}=\frac{1}{60}$, so B alone takes 60 days.
3. In 16 days all three do $\frac{16}{40}=\frac{2}{5}$; remaining $=\frac{3}{5}$.
4. B needs $\frac{3}{5}\times 60=36$ days.

Trick: Take total work $=360$ units (LCM of 40, 72): all three 9/day, B+C 5/day, A 4/day, B 6/day; left $360-144=216$; $216÷6=36$.

Answer: (A) 36 days

Concept: $1+{\cot }^{2}A={\csc }^{2}A$ and $1+{\tan }^{2}A={\sec }^{2}A$.

1. Numerator $=4(1+{\cot }^{2}A)\tan A=4{\csc }^{2}A\tan A$; denominator $=3(1+{\tan }^{2}A)\csc A=3{\sec }^{2}A\csc A$.
2. Ratio $=\frac{4{\csc }^{2}A\tan A}{3{\sec }^{2}A\csc A}=\frac{4}{3}·\frac{\csc A\tan A}{{\sec }^{2}A}$.
3. $\csc A\tan A=\frac{1}{\sin A}·\frac{\sin A}{\cos A}=\sec A$, so the ratio $=\frac{4}{3}·\frac{\sec A}{{\sec }^{2}A}=\frac{4\cos A}{3}$.

Trick: Put $A={45}^{\circ }$: numerator $=4·2·1=8$, denominator $=3·2·\sqrt{2}$; ratio $=\frac{4}{3\sqrt{2}}=\frac{4\cos {45}^{\circ }}{3}$ ✓.

Answer: (B) $\frac{4\cos A}{3}$

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$ (since $a:x=x:b$).

1. $x=\sqrt{20\times 45}=\sqrt{900}$.
2. $x=30$ km/h.

Trick: $20\times 45=900={30}^2$ – spot perfect squares quickly.

Answer: (A) 30 km/h

Concept: First find the maximum marks from the given shortfall, then compute the friend's percentage.

1. 35% of maximum $=831+72=903$, so maximum $=\frac{903\times 100}{35}=2580$.
2. Friend's percentage $=\frac{774}{2580}\times 100=30\%$.

Trick: $2580\times 0.3=774$ – check options by multiplying rather than dividing.

Answer: (C) 30%

Concept: Arc length $=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $\frac{240}{360}=\frac{2}{3}$.
2. Circumference $=2\times \frac{22}{7}\times 7=44$ cm.
3. Arc $=\frac{2}{3}\times 44=29.33$ cm.

Trick: ${240}^{\circ }$ is two-thirds of a full turn, so take two-thirds of the circumference.

Answer: (C) 29.33 cm

Concept: Look at the differences between consecutive terms.

1. $81-79=2,79-76=3,76-71=5,71-64=7$ – differences are prime numbers.
2. Next prime is 11, so $?=64-11=53$.

Trick: Differences 2, 3, 5, 7, 11, 13… (primes) is a favourite pattern.

Answer: (B) 53

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

1. Amount $=5P$, so SI $=4P$ in 24 years.
2. $R=\frac{4P\times 100}{P\times 24}=\frac{400}{24}=16.67\%$.

Trick: '$n$ times in $T$ years' → $R=\frac{100(n-1)}{T}$; here $\frac{400}{24}$.

Answer: (A) 16.67

Concept: Volume of cylinder $=\pi r^{2}h$; free space = vessel volume − volume of tubes; 1 litre $=1000$ cm³.

1. Vessel: $\pi \times {15}^2\times 40=9000\pi$ cm³.
2. Three tubes: $3\times \pi \times 4^2\times 40=1920\pi$ cm³.
3. Free space $=7080\pi =7080\times 3.14=22231.2$ cm³.
4. Solution $=75\%$ of it $=16673.4$ cm³ $\approx 16.67$ litres.

Trick: Keep $\pi$ till the end: $0.75\times 7080\pi =5310\pi$, then multiply by 3.14 once.

Answer: (A) 16.67 liters

Concept: Take CP $=100$; $\text{MP}=\text{CP}(1+\text{markup})$, $\text{SP}=\text{CP}(1+\text{gain})$, discount is on MP.

1. MP $=100+180=280$; required SP $=100+152=252$.
2. Discount $=280-252=28$.
3. Discount% $=\frac{28}{280}\times 100=10\%$.

Trick: $(1+\text{markup})(1-d)=(1+\text{gain})$: $2.8(1-d)=2.52\Rightarrow 1-d=0.9$.

Answer: (D) 10%

Concept: For 2 years, $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $P\times {\left(\frac{4}{100}\right)}^2=47$.
2. $P\times \frac{16}{10000}=47\Rightarrow P=\frac{47\times 10000}{16}=29375$.

Trick: 2-year difference $=\frac{P{R}^2}{10000}$; 3-year difference $=\frac{P{R}^2(300+R)}{{10}^6}$.

Answer: (B) 29,375

Concept: For a fixed distance, time is inversely proportional to speed.

1. Speed becomes $\frac{4}{7}$ of usual ⇒ time becomes $\frac{7}{4}$ of usual time $t$.
2. Extra time $=\frac{7}{4}t-t=\frac{3}{4}t=15$ minutes.
3. $t=15\times \frac{4}{3}=20$ minutes.

Trick: Speed ratio $4:7$ ⇒ time ratio $7:4$; the difference of 3 parts $=15$ min ⇒ 1 part $=5$, usual $=4\times 5=20$.

Answer: (A) 20 minutes

Concept: Do division and multiplication before addition and subtraction (BODMAS).

1. $18÷6=3$ and $2\times 4=8$.
2. $16+3-8=11$.

Trick: Underline all ÷ and × first, solve them, then run left to right.

Answer: (B) 11

Concept: A sector of angle $\theta$ cut along the height takes $\frac{\theta }{{360}^{\circ }}$ of the cylinder's volume $\pi r^{2}h$.

1. Full volume $=\pi \times {12}^2\times 25=3600\pi$ cm³.
2. Removed piece $=\frac{45}{360}=\frac{1}{8}$ of it; larger piece $=\frac{7}{8}$.
3. Larger piece $=\frac{7}{8}\times 3600\pi =3150\pi$ cm³.

Trick: ${45}^{\circ }$ is one-eighth of a turn – keep $\frac{7}{8}$, remove $\frac{1}{8}$.

Answer: (A) $3150\pi$

Concept: Let the son's present age be $x$; express every age in $x$ and shift all three back by 3 years.

1. Son $=x$, daughter $=x-3$, Vanraj $=3x$; present sum $=5x-3$.
2. Three years ago the sum is 9 less: $5x-3-9=58\Rightarrow 5x=70\Rightarrow x=14$.
3. Vanraj $=3\times 14=42$ years.

Trick: For $n$ people '$k$ years ago', subtract $nk$ from the present total – here $3\times 3=9$.

Answer: (A) 42