RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Expenditure = Income − Savings.

1. Savings 40% ⇒ expenditure $=60\%$ of income.
2. $0.6\times I=960\Rightarrow I=\frac{960}{0.6}=1600$.

Trick: $60\%=960\Rightarrow 10\%=160\Rightarrow 100\%=1600$.

Answer: (C) ₹1,600

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

1. Onward time $=\frac{d}{11}+\frac{2d}{44}=\frac{3d}{22}$.
2. Return 3d at v: time $=\frac{3d}{v}$. Total distance $=6d$.
3. $\frac{6d}{\frac{3d}{22}+\frac{3d}{v}}=34\Rightarrow \frac{1}{22}+\frac{1}{v}=\frac{1}{17}$.
4. $\frac{1}{v}=\frac{22-17}{374}=\frac{5}{374}\Rightarrow v=74.8\text{km/h}$.

Trick: Onward average $=\frac{3d}{3d/22}=22$; equal legs ⇒ $\frac{2\times 22v}{22+v}=34$.

Answer: (C) $74.8$

Concept: Volume of cone $=\frac{1}{3}\pi r^{2}h$.

1. $\frac{1}{3}\pi r^2\times 30=160\pi$.
2. $10r^2=160\Rightarrow r^2=16\Rightarrow r=4\text{cm}$.
3. Diameter $=2r=8\text{cm}$.

Trick: Cancel π first: $r^2=\frac{3\times 160}{30}=16$.

Answer: (C) 8 cm

Concept: Average per article $=\frac{\text{total cost}}{\text{total articles}}$.

1. Shirts $8\times 550=4400$; pants $10\times 1000=10000$; shoes $12\times 850=10200$.
2. Total $=24600$ for $8+10+12=30$ articles.
3. Average $=\frac{24600}{30}=820$.

Trick: Count articles carefully: pairs of shoes are counted as 12 articles.

Answer: (A) ₹820

Concept: Convert averages to sums: the overlapping member B is counted twice.

1. $A+B+C=3\times 45=135$.
2. $A+B=80$, $B+C=84$.
3. $(A+B)+(B+C)-(A+B+C)=B=80+84-135=29$.

Trick: Middle person = sum of two pair-totals − grand total.

Answer: (C) 29

Concept: $SI=\frac{P\times R\times T}{100}$; the interest difference comes from the principal difference.

1. Principal difference $=9000-5000=4000$.
2. $\frac{4000\times R\times 3}{100}=360$.
3. $120R=360\Rightarrow R=3\%$.

Trick: Difference in SI depends only on difference in principal.

Answer: (C) 3%

Concept: Successive changes multiply: new price = original × (1 − decrease) × (1 + increase).

1. $8000\times 0.88=7040$.
2. $7040\times 1.25=8800$.

Trick: Net change $=-12+25-\frac{12\times 25}{100}=+10\%$ → $8000\times 1.1=8800$.

Answer: (C) ₹8,800

Concept: Mean of a frequency distribution $=\frac{\sum fx}{\sum f}$.

1. $\sum fx=1274+2028+4180+2196+8232=17910$.
2. $\sum f=91+78+95+36+98=398$.
3. Mean $=\frac{17910}{398}=45$.

Trick: Check: $398\times 45=17910$ exactly.

Answer: (A) 45

Concept: Take total work = LCM of days and compare daily work.

1. LCM$(8,24,3)=24$ units.
2. A $=\frac{24}{8}=3$, B $=\frac{24}{24}=1$, A+B+C $=\frac{24}{3}=8$ units/day.
3. C $=8-3-1=4$ units/day.
4. C alone $=\frac{24}{4}=6$ days.

Trick: $\frac{1}{3}-\frac{1}{8}-\frac{1}{24}=\frac{8-3-1}{24}=\frac{1}{6}$.

Answer: (C) 6 days

Concept: Evaluate powers and brackets first (BODMAS).

1. ${17}^3=4913$, ${10}^2=100$, ${\left(\frac{26}{13}\right)}^2=2^2=4$, $20\times 12=240$.
2. $4913-100+4-12+240$.
3. $=5045$.

Trick: Unit digit check: $3-0+4-2+0=5$ → only 5055 or 5045; size check gives 5045.

Answer: (D) 5045

Concept: Arithmetic mean $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=79+92+39+57+91+28+74+90+44=594$.
2. Number of observations $=9$.
3. Mean $=\frac{594}{9}=66$.

Trick: Pair to round numbers: $(79+91)+(92+28)+(57+44)+(39+74)+90$.

Answer: (C) 66

Concept: If speeds are in the lowest ratio $a:b$, the number of distinct meeting points is $a+b$ (opposite directions) and $|a-b|$ (same direction).

1. Siddharth's speed $=\frac{2295}{85}=27\text{m/s}$; Yash's speed $=\frac{2295}{45}=51\text{m/s}$
2. Ratio $27:51=9:17$
3. Opposite directions: $9+17=26$ distinct points

Trick: Ratio of speeds = inverse ratio of lap times: $85:45=17:9$ directly.

Answer: (C) $26$

Concept: Form two equations from the two conditions and solve.

1. $\frac{S+15}{F}=\frac{2}{3}\Rightarrow 2F=3S+45$
2. $F-5=4(S-5)\Rightarrow F=4S-15$
3. $2(4S-15)=3S+45\Rightarrow 5S=75\Rightarrow S=15$
4. $F=4\times 15-15=45$

Trick: Check options: (45, 15): $\frac{30}{45}=\frac{2}{3}$ and $40=4\times 10$ ✓

Answer: (A) 45; 15

Concept: In a proportion $a:b::c:d$, product of extremes = product of means, i.e. $ad=bc$.

1. $9.8\times x=14.7\times 15$
2. $x=\frac{14.7\times 15}{9.8}=1.5\times 15$
3. $x=22.5$

Trick: $\frac{14.7}{9.8}=1.5$, so $x=1.5\times 15=22.5$.

Answer: (C) $22.5$

Concept: $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$. First split 35 km in the ratio covered : remaining $=4:3$.

1. Covered $=\frac{4}{7}\times 35=20\text{km}$, remaining $=15\text{km}$
2. Time $=4\text{h}15\text{min}=4.25\text{h}$
3. Speed $=\frac{20}{4.25}=4.705\dots \approx 4.7\text{km/h}$

Trick: $\frac{20}{4.25}=\frac{80}{17}\approx 4.7$.

Answer: (D) $4.7$

Concept: $\text{CSA}=2\pi rh$ and $\text{TSA}=2\pi r(h+r)$.

1. $2\pi rh=\frac{1}{2}\times 2\pi r(h+r)$
2. $2h=h+r\Rightarrow r=h$
3. $r=128\text{cm}$

Trick: CSA = half of TSA means the two circular ends together equal the CSA, i.e. $2\pi r^2=2\pi rh\Rightarrow r=h$.

Answer: (A) $128$

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

1. $P=\frac{P\times R\times 4}{100}$
2. $R=\frac{100}{4}=25\%$

Trick: Money doubles in $T$ years at SI → $R=\frac{100}{T}\%$.

Answer: (A) $25\%$

Concept: Divide each option by 71 (or multiply 71 by nearby numbers).

1. $71\times 79=5609$ ✓
2. $71\times 86=6106$, so $6121$ leaves remainder 15
3. $71\times 84=5964$, $6010-5964=46$; $5603=5609-6$ leaves remainder 65

Trick: $71\times 80=5680$; subtract 71 → $5609$.

Answer: (A) $5609$

Concept: Net rate = filling rate − emptying rate (in tank per minute).

1. Net rate $=\frac{1}{9}-\frac{1}{90}=\frac{10-1}{90}=\frac{9}{90}=\frac{1}{10}$
2. Time $=10$ minutes

Trick: LCM method: tank $=90$ units; fill $+10$, empty $-1$ → $9$ units/min → $\frac{90}{9}=10$.

Answer: (D) $10$

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $36.3=3\times 62-2\text{Mean}$
2. $2\text{Mean}=186-36.3=149.7$
3. $\text{Mean}=74.85$

Trick: $\text{Mean}=\frac{3\text{Median}-\text{Mode}}{2}$.

Answer: (A) $74.85$