RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Classes are inclusive (35–44, 45–54 …), so first make them continuous by adjusting $0.5$; then $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$.

1. Highest frequency $f_1=30$ (class 55–64) → modal class $54.5$–$64.5$, $l=54.5$, $h=10$.
2. $f_0=20$ (previous class), $f_2=25$ (next class).
3. $\text{Mode}=54.5+\frac{30-20}{60-20-25}\times 10=54.5+\frac{10}{15}\times 10=54.5+6.67=61.17$.

Trick: Inclusive classes → subtract $0.5$ from the lower limit; forgetting this gives $61.67$ (option B), the trap.

Answer: (A) 61.17

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ – never the average of the three speeds.

1. Total distance $=336+925+432=1693\text{km}$.
2. Times: $\frac{336}{24}=14$, $\frac{925}{5}=185$, $\frac{432}{12}=36$ → total $235\text{h}$.
3. Average speed $=\frac{1693}{235}=7\frac{48}{235}\text{km/h}$ (since $7\times 235=1645$, remainder $48$).

Trick: All options have denominator $235$ = total time; only the remainder $1693-1645=48$ decides the answer.

Answer: (C) $7\frac{48}{235}$

Concept: Length of an arc $=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $r=8$, $\theta ={45}^{\circ }$ → fraction of circle $=\frac{45}{360}=\frac{1}{8}$.
2. Circumference $=2\times \frac{22}{7}\times 8=\frac{352}{7}$.
3. Arc $=\frac{1}{8}\times \frac{352}{7}=\frac{44}{7}$ units.

Trick: ${45}^{\circ }$ is one-eighth of a turn; $2\pi r÷8=\frac{\pi r}{4}=\frac{22}{7}\times 2=\frac{44}{7}$.

Answer: (B) $\frac{44}{7}$

Concept: Translate the sentence directly into an equation: $20\%$ of $x$ $=20\%$ of $620+220$.

1. $20\%$ of $620=124$.
2. $0.2x=124+220=344$.
3. $x=\frac{344}{0.2}=1720$.

Trick: $20\%=\frac{1}{5}$, so $x=5\times 344=1720$; or note the number exceeds $620$ by $\frac{220}{0.2}=1100$.

Answer: (C) 1720

Concept: Take a convenient equal number of each type (LCM of 9 and 1 = 9), find total CP and SP, then $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$.

1. Buy 9 of each: first type costs ₹8 for all 9; second type costs $9\times 8=$ ₹72. Total CP $=8+72=$ ₹80.
2. Sold 18 chocolates at ₹8 each: SP $=18\times 8=$ ₹144.
3. Profit $=144-80=64$; Profit$\%=\frac{64}{80}\times 100=80\%$.

Trick: Second type is sold at cost (no gain); the whole profit comes from the cheap type: ₹8 → ₹72. Overall $\frac{64}{80}=80\%$.

Answer: (D) 80%

Concept: Add the filling rates and subtract the emptying rate (work per hour).

1. Rates: $\frac{1}{4}+\frac{1}{31}-\frac{1}{31}=\frac{1}{4}$ of the cistern per hour.
2. The second tap and the emptying tap cancel each other exactly.
3. One-fourth of the cistern at $\frac{1}{4}$ per hour → $1$ hour.

Trick: Same time to fill and to empty ($31$ h) → the pair does nothing; only the $4$-hour tap counts.

Answer: (D) 1

Concept: Each number must be a multiple of every gcd it is part of; choose the smallest such numbers and verify the gcds.

1. $b$ is a multiple of $12$ and $18$ → multiple of $\text{LCM}=36$; $a$ is a multiple of $12$ and $6$ → of $12$; $c$ is a multiple of $18$ and $6$ → of $18$.
2. Smallest choice: $a=12,b=36,c=18$. Check: $gcd(12,36)=12$, $gcd(36,18)=18$, $gcd(18,12)=6$ ✓.
3. $a+b+c=66=6\times 11$ → minimum $k=11$.

Trick: Minimum sum $=12+36+18=66$; any smaller $a,b,c$ would violate a gcd, so $k=\frac{66}{6}=11$.

Answer: (A) 11

Concept: Sum $=$ average $\times$ count; the excluded number is the difference of the two sums.

1. Sum of 5 numbers $=5\times 20=100$.
2. Sum of remaining 4 numbers $=4\times 18=72$.
3. Excluded number $=100-72=28$.

Trick: Excluded $=$ new average $+$ (old count) $\times$ (drop in average) $=18+5\times 2=28$.

Answer: (C) 28

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$; if $m$ is the mean proportional between $a$ and $y$, then $m^2=ay$.

1. $\sqrt{18\times 32}=\sqrt{576}=24$.
2. Increase by $25\%$: $24\times \frac{5}{4}=30$.
3. ${30}^2=18y\Rightarrow y=\frac{900}{18}=50$.

Trick: $18\times 32=576={24}^2$ – spot it as $(25-1{)}^2$; then $y=\frac{{30}^2}{18}=50$.

Answer: (B) 50

Concept: An equation is quadratic only if, after simplification, the highest power of $x$ is exactly $2$ (coefficient of $x^2\ne 0$).

1. (A) $(x-1)(x+2)=x^2+3\Rightarrow x^2+x-2=x^2+3\Rightarrow x-5=0$ – the $x^2$ terms cancel → linear, not quadratic.
2. (B) $x^2+2x+1=0$ and (D) $3x^2-5x+2=0$ are already in the form $a{x}^2+bx+c=0$.
3. (C) $x(x-3)=2\Rightarrow x^2-3x-2=0$ – quadratic.

Trick: Whenever $x^2$ appears on both sides, expand and check whether it cancels.

Answer: (A) (x – 1)(x + 2) = x² + 3

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%