RRB NTPC Graduate CBT 1 2024 PYQs

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) $74.85$

Concept: Find the part filled by A alone, then fill the rest at the net rate.

1. In 3 min A fills $\frac{3}{9}=\frac{1}{3}$; remaining $=\frac{2}{3}$
2. Net rate $=\frac{1}{9}-\frac{1}{10}=\frac{1}{90}$ per min
3. Time $=\frac{2}{3}÷\frac{1}{90}=\frac{2}{3}\times 90=60$ min

Trick: Tank $=90$ units: A $=10$, B $=-9$; after 3 min, $60$ units left at $1$ unit/min → $60$ min.

Answer: (B) $60$

Concept: Angle sum of a triangle $={180}^{\circ }$; an exterior angle equals the sum of the two interior opposite angles.

1. In right $△BDC$: $\angle C={90}^{\circ }-{22}^{\circ }={68}^{\circ }$
2. In $△AEC$: $\angle AEC={180}^{\circ }-{36}^{\circ }-{68}^{\circ }={76}^{\circ }$
3. $\angle AEB={180}^{\circ }-{76}^{\circ }={104}^{\circ }$ (linear pair on BC)

Trick: $\angle AEB$ is exterior to $△AEC$: ${36}^{\circ }+{68}^{\circ }={104}^{\circ }$.

Answer: (D) ${104}^{\circ }$

Concept: Water flowing per minute forms a cuboid: depth × width × distance moved in 1 minute.

1. Speed $=2.7\text{km/h}=\frac{2700}{60}=45\text{m/min}$.
2. Volume $=20\times 36\times 45$.
3. $=720\times 45=32400{\text{m}}^3$.

Trick: Convert km/h to m/min by multiplying by $\frac{50}{3}$: $2.7\times \frac{50}{3}=45$.

Answer: (B) 32,400

Concept: If $a:b::c:d$ then $ad=bc$; and if $a:y::y:b$ then $y^2=ab$ (mean proportional).

1. $\frac{20+x}{34+x}=\frac{11+x}{16+x}$ ⇒ $(20+x)(16+x)=(34+x)(11+x)$.
2. $320+36x=374+45x$ ⇒ $9x=-54$ ⇒ $x=-6$.
3. $3x=-18$ and $2x-6=-18$.
4. $y^2=(-18)(-18)=324$ ⇒ $y=18$ (since $y>0$).

Trick: After expanding, $x^2$ cancels, so it is just a linear equation.

Answer: (C) $18$

Concept: For $a{x}^3+b{x}^2+cx+d=0$: sum of roots $=-\frac{b}{a}$, product of roots $=-\frac{d}{a}$.

1. Here $a=4,b=-8,c=-13,d=19$.
2. Sum $=-\frac{-8}{4}=2$.
3. Product $=-\frac{19}{4}$.

Trick: Signs alternate for a cubic: sum $-\frac{b}{a}$, pair-sum $+\frac{c}{a}$, product $-\frac{d}{a}$.

Answer: (D) $2,\frac{-19}{4}$

Concept: If $m$ is the mean proportional between $a$ and $b$, then $m^2=ab$.

1. ${276}^2=368x$.
2. $x=\frac{276\times 276}{368}$.
3. $\frac{276}{368}=\frac{3}{4}$, so $x=\frac{3}{4}\times 276=207$.

Trick: Cancel first: $276=92\times 3$, $368=92\times 4$ – no need to square 276.

Answer: (A) $207$

Concept: In similar triangles, the ratio of perimeters equals the ratio of corresponding sides.

1. Perimeter of XYZ $=8+12+16=36\text{cm}$.
2. $\frac{AB}{XY}=\frac{6}{8}=\frac{3}{4}$.
3. Perimeter of ABC $=\frac{3}{4}\times 36=27\text{cm}$.

Trick: No need to find BC and CA separately – scale the whole perimeter.

Answer: (C) 27 cm

Concept: Open brackets from the innermost: ( ) → { } → [ ], then divide.

1. $(1+8)=9$; $4+7-9=2$.
2. $\{9+2-2\}=9$.
3. $[\frac{4}{2}\times 9]=18$.
4. $72÷18=4$.

Trick: Work inside-out and write each bracket's value before moving outward.

Answer: (A) 4

Concept: Half-yearly compounding: rate is halved and periods are doubled: $A=P{(1+\frac{r}{200})}^{2n}$.

1. Rate per half-year $=10\%$, periods $=2$.
2. $A=9900\times (1.1{)}^2=9900\times 1.21$.
3. $A=11979$, i.e. ₹11,979.

Trick: $10\%$ twice = $21\%$ total: $9900+21\%$ of $9900=9900+2079=11979$.

Answer: (D) ₹11,979

Concept: $\text{CP}=\frac{\text{SP}}{1+\frac{\text{gain}\%}{100}}$.

1. SP of 5 chocolates = ₹1, so SP of 1 $=\frac{1}{5}$.
2. CP of 1 $=\frac{1/5}{1.2}=\frac{1}{6}$.
3. So for ₹1 he bought $6$ chocolates.

Trick: With 20% gain, number bought per ₹1 $=5\times 1.2=6$.

Answer: (D) 6

Concept: If every group has the same average, the overall average is also that value.

1. Total $=12\times 60+18\times 60=30\times 60$.
2. Average $=\frac{30\times 60}{30}=60$.

Trick: Equal group averages → class average is the same, no calculation needed.

Answer: (A) 60