RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Assume CP $=100$; find MP and required SP.

1. MP $=100+150=250$; required SP for 130% gain $=230$.
2. Discount $=250-230=20$.
3. Discount% $=\frac{20}{250}\times 100=8\%$.

Trick: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$ with CP $=100$ – no algebra needed.

Answer: (D) 8%

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

1. Total $=30\times 434+72\times 324+75\times 280=13020+23328+21000=57348$.
2. Members $=30+72+75=177$.
3. Average $=\frac{57348}{177}=324$.

Trick: Take 324 as base: deviations $30\times 110=+3300$ and $75\times (-44)=-3300$ cancel exactly, so the average is 324.

Answer: (C) 324

Concept: Rearrange and eliminate; or verify the options directly.

1. Equations: $3x-4y-6z=-16$, $-4x+y+z=-5$, $x-3y-2z=-2$.
2. From (2): $y+z=4x-5$. Try $x=2$: $y+z=3$; with $y=-2$, $z=5$.
3. Check (1): $6+8-30=-16$ ✓; (3): $2+6-10=-2$ ✓.

Trick: Put each option in the simplest equation $-4x+y+z=-5$: options A and B give $-9$, option D gives $-6$; only option C gives $-5$.

Answer: (C) $x=2,y=-2,z=5$

Concept: $\text{M.D.}=\frac{\sum |x_i-\overline{x}|}{n}$.

1. Sum $=58+65+42+66+49+59+44+45+62+60=550$; mean $\overline{x}=\frac{550}{10}=55$.
2. $|x_i-55|$: $3,10,13,11,6,4,11,10,7,5$; total $=80$.
3. M.D. $=\frac{80}{10}=8$.

Trick: Deviations above 55 sum to $3+10+11+4+7+5=40$ and below sum to $13+6+11+10=40$; they always balance, so just double one side.

Answer: (B) 8

Concept: $M_1{H}_1{D}_1=M_2{H}_2{D}_2$ (same work).

1. $15\times 6\times 20=18\times 5\times D$.
2. $1800=90D\Rightarrow D=20$ days.

Trick: $15\times 6=90$ and $18\times 5=90$ – man-hours per day are equal, so days stay 20.

Answer: (D) 20

Concept: Tangents from an external point are equal; OA bisects BC perpendicularly at M; $\angle OBA={90}^{\circ }$.

1. $AB=AC=30$, $BM=MC=12$ $\Rightarrow AM=\sqrt{{30}^2-{12}^2}=\sqrt{756}\approx 27.5$.
2. In right $△OBA$ with altitude BM: $B{M}^2=OM\times AM\Rightarrow OM=\frac{144}{27.5}\approx 5.24$.
3. $r=OB=\sqrt{B{M}^2+O{M}^2}=\sqrt{144+27.4}=\sqrt{171.4}\approx 13.1\approx 13$ cm.

Trick: Direct formula: $r={\displaystyle \frac{AB·BM}{AM}}={\displaystyle \frac{30\times 12}{\sqrt{756}}}={\displaystyle \frac{360}{27.5}}\approx 13$.

Answer: (C) 13

Concept: Angle in the minor arc subtends the major arc; central angle $=2\times$ inscribed angle; $\angle APB+\angle AMB={180}^{\circ }$ (tangent–radius angles are ${90}^{\circ }$).

1. C lies on the minor arc, so $\angle ACB={124}^{\circ }$ stands on the major arc AB: reflex $\angle APB={248}^{\circ }$.
2. Minor $\angle APB={360}^{\circ }-{248}^{\circ }={112}^{\circ }$.
3. In quadrilateral PAMB, $\angle PAM=\angle PBM={90}^{\circ }$ $\Rightarrow \angle AMB={180}^{\circ }-{112}^{\circ }={68}^{\circ }$.

Trick: For a point C on the minor arc: $\angle AMB=2\angle ACB-{180}^{\circ }={248}^{\circ }-{180}^{\circ }={68}^{\circ }$.

Answer: (D) ${68}^{\circ }$

Concept: Keep the ratio sugar : flour $=3:4$.

1. $\frac{3}{4}=\frac{12}{x}\Rightarrow 3x=48$.
2. $x=16$ cups of flour.

Trick: 12 is $4\times 3$, so flour $=4\times 4=16$.

Answer: (A) 16

Concept: For consecutive even numbers the average is the middle value.

1. Let numbers be $x,x+2,\dots ,x+10$; average $=x+5=175\Rightarrow x=170$.
2. Numbers: 170, 172, 174, 176, 178, 180 – smallest 170.

Trick: Six even numbers centred on 175: three below (174, 172, 170) and three above.

Answer: (A) 170

Concept: Volume of hemisphere $=\frac{2}{3}\pi r^3$.

1. Diameter 14 cm $\Rightarrow r=7$ cm.
2. $V=\frac{2}{3}\times \frac{22}{7}\times 7^3=\frac{2}{3}\times 22\times 49=\frac{2156}{3}$.
3. $V=718.666\dots \approx 718.67{\text{cm}}^3$.

Trick: With $r=7$, $\pi r^3=22\times 49=1078$; hemisphere $=\frac{2}{3}\times 1078$.

Answer: (C) 718.67

Concept: Add filling rates, subtract emptying rates (per hour).

1. Net rate $=\frac{1}{4}+\frac{1}{37}-\frac{1}{37}=\frac{1}{4}$ per hour.
2. Full tank in 4 hours $\Rightarrow$ one-fourth in $\frac{4}{4}=1$ hour.

Trick: The 37-hour inlet and 37-hour outlet cancel exactly – only the 4-hour tap matters.

Answer: (A) 1

Concept: Twins have equal ages. Form equations from the given ratio of past ages.

1. Let Tejasvi = Trishana = $t$ years. Then Meena = $t+t=2t$.
2. Four years ago: $\frac{t-4}{2t-4}=\frac{1}{3}\Rightarrow 3t-12=2t-4\Rightarrow t=8$.
3. Trishana after 4 years $=8+4=12$ years.

Trick: Check: 4 years ago ages were $4$ and $12$, ratio $1:3$ ✓.

Answer: (C) 12 years

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

1. Amount $=5P\Rightarrow$ SI $=4P$.
2. $R=\frac{4P\times 100}{P\times 15}=\frac{400}{15}=26.666\dots \approx 26.67\%$.

Trick: "5 times" means interest is 4 times the principal, not 5.

Answer: (A) 26.67

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Distance $=20\times 90+72\times 66=1800+4752=6552$ m.
2. Time $=90+66=156$ min.
3. Average $=\frac{6552}{156}=42$ m/min.

Trick: Never average the two speeds directly when times differ; use total distance ÷ total time.

Answer: (D) 42

Concept: $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}$ with class marks $x_i$.

1. Class marks: 15, 25, 35, 45, 55; frequencies: 42, 26, 45, 46, $X$.
2. $\sum f_i{x}_i=630+650+1575+2070+55X=4925+55X$; $\sum f_i=159+X$.
3. $35(159+X)=4925+55X\Rightarrow 5565+35X=4925+55X\Rightarrow 20X=640\Rightarrow X=32$.

Trick: Use assumed mean 35: deviations $-20,-10,0,10,20$ give $-840-260+0+460+20X=0\Rightarrow X=32$.

Answer: (D) 32

Concept: Follow BODMAS: division and multiplication before subtraction and addition.

1. $(-60)÷(-6)=10$ and $(-2)\times 8=-16$.
2. Expression $=-9-10+(-16)=-35$.

Trick: Negative ÷ negative is positive; the minus before it then makes it $-10$.

Answer: (C) $-35$

Concept: With parallel lines, the angles made by a transversal are related (sum ${180}^{\circ }$); the exterior angle of a triangle equals the sum of the two opposite interior angles.

1. AO ∥ DE and DC is the transversal, so the angle at O, $\angle BOC={180}^{\circ }-{80}^{\circ }={100}^{\circ }$.
2. In $△OBC$, $\angle ABC={135}^{\circ }$ is the exterior angle at B: ${135}^{\circ }=\angle BOC+x$.
3. $x={135}^{\circ }-{100}^{\circ }={35}^{\circ }$.

Trick: Exterior angle = sum of the two far interior angles – no need to find $\angle OBC$ separately.

Answer: (D) ${35}^{\circ }$

Concept: Successive changes multiply: $\text{Final}=\text{CP}\times \frac{170}{100}\times \frac{33}{100}$.

1. Q's cost $=1.7\text{CP}$; R's cost $=1.7\times 0.33\text{CP}=0.561\text{CP}$.
2. $0.561\text{CP}=3927\Rightarrow \text{CP}=\frac{3927}{0.561}=7000$.

Trick: $3927=7\times 561$, so $\text{CP}=7\times 1000=7000$ – spot the factor 561.

Answer: (D) 7000

Concept: Equal amounts after $n$ and $n+1$ years at CI mean the shares are in the ratio $(1+r):1$.

1. $V(1.1{)}^{17}=A(1.1{)}^{18}\Rightarrow V=1.1A$.
2. $V+A=1407\Rightarrow 2.1A=1407\Rightarrow A=670$.
3. Ajay received ₹670 (Vijay ₹737).

Trick: One extra year at 10% → ratio $11:10$; $1407÷21=67$, so shares $737:670$.

Answer: (B) 670

Concept: Melting keeps the volume same: $\frac{4}{3}\pi R^3=\pi r^{2}L$.

1. Sphere: $R=12$ cm; wire: $r=2$ mm $=0.2$ cm.
2. $\frac{4}{3}\times {12}^3=0.04\times L\Rightarrow 2304=0.04L$.
3. $L=57600$ cm $=576$ m.

Trick: $L=\frac{4R^3}{3r^2}$; convert mm to cm before squaring.

Answer: (C) 576