RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Take total work = LCM of the days; work per day = total ÷ days.

1. Total work = LCM(21, 28) = 84 units.
2. (A+B) per day = $\frac{84}{21}=4$, A per day = $\frac{84}{28}=3$, so B per day = $4-3=1$ unit.
3. B alone finishes the whole work in 84 days; three-fourth of it in $\frac{3}{4}\times 84=63$ days.

Trick: B alone = $\frac{21\times 28}{28-21}=84$ days; then take $\frac{3}{4}$ of it.

Answer: (C) 63

Concept: First find the maximum marks from the given percentage condition.

1. 30% of maximum = $441+75=516$.
2. Maximum marks = $\frac{516\times 100}{30}=1720$.
3. Friend's percentage = $\frac{688}{1720}\times 100=40\%$.

Trick: $1720\times \frac{2}{5}=688$, so 40% directly.

Answer: (D) 40%

Concept: Factorise both equations and pick the common root.

1. $x^2-21x+110=0\Rightarrow (x-10)(x-11)=0\Rightarrow x=10,11$.
2. $110+x-x^2=0\Rightarrow x^2-x-110=0\Rightarrow (x-11)(x+10)=0\Rightarrow x=11,-10$.
3. Common root = 11.

Trick: Subtracting the two equations: $-22x+242=0\Rightarrow x=11$ at once.

Answer: (A) 11

Concept: Scale 1 : 829483 means 1 cm on the map = 829483 cm on the ground.

1. Actual distance = $4\times 829483=3,317,932$ cm.
2. Convert: 1 km = 100000 cm, so $\frac{3317932}{100000}=33.17932$ km.
3. Rounded to two decimals: 33.18 km.

Trick: $829483\text{ cm}\approx 8.29\text{ km}$; $4\times 8.295\approx 33.18$.

Answer: (D) 33.18

Concept: Add equations to eliminate one variable, or simply verify the options.

1. Adding (1) and (2): $-q-2r=-3\Rightarrow q+2r=3$.
2. (2)×2 − (3): $2p-6q+2r-2p+5q-5r=12-17\Rightarrow -q-3r=-5\Rightarrow q+3r=5$.
3. Subtracting: $r=2$, $q=3-4=-1$; from (2): $p=6+3q-r=6-3-2=1$.
4. Check (1): $-1-2-6=-9$ ✓.

Trick: Put each option in the easiest equation $p-3q+r=6$: only (B) gives $1+3+2=6$.

Answer: (B) $p=1,q=-1,r=2$

Concept: Division and multiplication first, then addition/subtraction from left to right.

1. $(-56)÷(-14)=4$ and $(-4)\times 8=-32$.
2. Expression = $-9-4+(-32)=-13-32=-45$.

Trick: Negative ÷ negative = positive; watch the minus sign before the division result.

Answer: (D) −45

Concept: For a moderately skewed distribution, $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $48=3\text{Median}-2\times 72=3\text{Median}-144$.
2. $3\text{Median}=192\Rightarrow \text{Median}=64$.

Trick: Median always lies between mode and mean, one-third of the way from the mean: $72-\frac{72-48}{3}=64$.

Answer: (B) 64

Concept: Take total work = LCM; efficiency ratio converts to units per day.

1. Total work = LCM(36, 48) = 144 units. (A+B+C) = 4/day, (B+C) = 3/day → A = 1/day.
2. B is 75% more efficient than A → B = $1.75$/day; C = $3-1.75=1.25$/day. To avoid decimals multiply by 4: work = 576, A = 4, B = 7, C = 5, total 16/day.
3. Work done in 15 days = $15\times 16=240$; remaining = $576-240=336$.
4. B alone: $\frac{336}{7}=48$ days.

Trick: 15 of 36 days done → $\frac{21}{36}=\frac{7}{12}$ of work left; B does $\frac{7}{576}$ per day → $\frac{7}{12}\times \frac{576}{7}=48$.

Answer: (A) 48 days

Concept: If HCF(a, b) = 28, write $a=28x$, $b=28y$ with HCF(x, y) = 1.

1. $28x+28y=868\Rightarrow x+y=\frac{868}{28}=31$.
2. 31 is prime, so any pair with $x+y=31$ is automatically co-prime (a common factor of x and y would divide 31).
3. x can be 1, 2, …, 30 → 30 ordered pairs (a, b).

Trick: Number of co-prime ordered pairs with sum n = $\varphi (n)$; for prime 31 it is 30.

Answer: (D) 30

Concept: Volume of a cylinder $V=\pi r^{2}h$.

1. $\frac{22}{7}\times r^2\times 14=1584\Rightarrow 44r^2=1584$.
2. $r^2=36\Rightarrow r=6$ m.
3. Diameter $=2r=12$ m.

Trick: $\pi h=\frac{22}{7}\times 14=44$; so $r^2=\frac{1584}{44}=36$.

Answer: (B) 12 m

Concept: Overall average = total amount ÷ total members.

1. Total = $46\times 496+15\times 106+11\times 406=22816+1590+4466=28872$.
2. Members = $46+15+11=72$.
3. Average = $\frac{28872}{72}=401$.

Trick: Deviations from 400: $46\times 96-15\times 294+11\times 6=4416-4410+66=72$; $\frac{72}{72}=1$ → 401.

Answer: (D) 401

Concept: $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

1. $5xyz=150\Rightarrow xyz=30$.
2. $x^3+y^3+z^3=10\times (38-31)+3\times 30=70+90=160$.
3. $2(x^3+y^3+z^3)=320$.

Trick: Check consistency: $(x+y+z{)}^2=38+2\times 31=100$ ✓, so the data fits.

Answer: (B) 320

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