RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=14\text{cm}$, so $r^2=196$.
2. $4\pi r^2=4\times \frac{22}{7}\times 196$.
3. $=4\times 22\times 28=2464{\text{cm}}^2$.

Trick: When $r$ is a multiple of 7, cancel 7 first: $\frac{196}{7}=28$, then $88\times 28=2464$.

Answer: (D) $2464$

Concept: Area of a quadrilateral $=\frac{1}{2}\times \text{diagonal}\times (\text{sum of perpendiculars on it})$.

1. $324=\frac{1}{2}\times AC\times (12+15)$.
2. $324=\frac{27}{2}\times AC$.
3. $AC=\frac{648}{27}=24\text{cm}$.

Trick: $AC=\frac{2\times \text{Area}}{h_1+h_2}=\frac{648}{27}=24$.

Answer: (A) $24$

Concept: $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$.

1. $10.1=10+0.1$, so $a=10,b=0.1$.
2. $a^3=1000,3a^{2}b=3\times 100\times 0.1=30$.
3. $3a{b}^2=3\times 10\times 0.01=0.3,b^3=0.001$.
4. Sum $=1000+30+0.3+0.001=1030.301$.

Trick: ${10.1}^3$ must end in $0.001$ ($1^3$), and be about $1030$ — only $1030.301$ fits.

Answer: (B) $1030.301$

Concept: Income after $n$ years $=P{(1+\frac{r}{100})}^n$.

1. $70000\times (1.15{)}^2$.
2. $(1.15{)}^2=1.3225$.
3. $70000\times 1.3225=92575$.

Trick: Two successive 15% rises $=15+15+\frac{15\times 15}{100}=32.25\%$; $32.25\%$ of $70000=22575$.

Answer: (C) ₹92,575

Concept: If $d$ is the fourth proportional to $a,b,c$, then $a:b=c:d$, i.e. $ad=bc$.

1. $P:4=3:6$.
2. $6P=4\times 3=12$.
3. $P=2$.

Trick: Product of extremes = product of means: $P\times 6=4\times 3$.

Answer: (D) $2$

Concept: $x$ is what % of $y$ $=\frac{x}{y}\times 100$.

1. $\frac{1/20}{1/40}=\frac{1}{20}\times \frac{40}{1}=2$.
2. $2\times 100=200\%$.

Trick: Same numerator: the smaller denominator is bigger; $40÷20=2$ times $=200\%$.

Answer: (C) $200\%$

Concept: $\text{Sum}=\text{Average}\times \text{count}$.

1. Sum of 8 numbers $=8\times 16=128$.
2. Sum of 6 numbers $=6\times 16=96$.
3. Remaining two $=128-96=32$; average $=\frac{32}{2}=16$.

Trick: If a part has the same average as the whole, the rest must also have that average.

Answer: (C) $16$

Concept: Lateral surface area of a cuboid $=2h(l+b)$.

1. Edge of each cube $=\sqrt[3]{592704}=84\text{cm}$.
2. Cuboid: $l=5\times 84=420,b=84,h=84$.
3. $\text{LSA}=2\times 84\times (420+84)=168\times 504$.
4. $=84672{\text{cm}}^2$.

Trick: $\text{LSA}=2\times 6a^2=12a^2=12\times 7056=84672$, since $2h(l+b)=2a(5a+a)$.

Answer: (C) $84672$

Concept: Simplify the denominator first, then change $÷$ into $\times$ by the reciprocal.

1. $\frac{7}{8}+\frac{4}{6}=\frac{21}{24}+\frac{16}{24}=\frac{37}{24}$.
2. $\frac{1}{37/24}=\frac{24}{37}$.
3. $\frac{24}{37}÷\frac{6}{29}=\frac{24}{37}\times \frac{29}{6}=\frac{116}{37}$.
4. $\frac{116}{37}=3\frac{5}{37}$.

Trick: $37\times 3=111$, remainder $5$ → $3\frac{5}{37}$.

Answer: (A) $3\frac{5}{37}$

Concept: Form two equations in chair price $c$ and table price $t$ and solve.

1. $4c+4t=7000\Rightarrow c+t=1750$.
2. $6c+5t=8800\Rightarrow c+5(c+t)=8800\Rightarrow c+8750=8800$.
3. $c=50,t=1700$.
4. $16c+2t=800+3400=4200$.

Trick: Write $6c+5t=c+5(c+t)$ to get $c$ in one step.

Answer: (B) ₹4,200

Concept: Take $\text{CP}=x$ and write both selling prices in terms of $x$.

1. Actual SP at 17% loss $=0.83x$.
2. New CP $=0.95x$; new SP at 20% gain $=1.2\times 0.95x=1.14x$.
3. New SP is ₹310 more: $1.14x-0.83x=310\Rightarrow 0.31x=310$.
4. $x=1000$.

Trick: Difference in SP as % of CP: $114\%-83\%=31\%$ = ₹310 → $100\%$ = ₹1,000.

Answer: (C) ₹1,000

Concept: For a fixed distance, time is inversely proportional to speed.

1. Speed becomes $\frac{1}{3}$, so time becomes $3T$.
2. Extra time $=3T-T=2T=50\text{min}$.
3. $T=25\text{min}$.

Trick: Speed $\frac{1}{n}$ → late by $(n-1)T$; here $2T=50$.

Answer: (A) $25$

Concept: For 3 years: $\text{CI}-\text{SI}=P{\left(\frac{r}{100}\right)}^2(3+\frac{r}{100})$.

1. $r=4\%$: ${\left(\frac{4}{100}\right)}^2=\frac{16}{10000}$, and $3+0.04=3.04$.
2. $152=P\times \frac{16}{10000}\times 3.04=P\times 0.004864$.
3. $P=\frac{152}{0.004864}=31250$.

Trick: Check the option: $31250\times {0.04}^2=50$; $50\times 3.04=152$ ✓.

Answer: (D) ₹31,250

Concept: Net filling rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{12}-\frac{1}{24}=\frac{1}{24}$ tank per hour.
2. Time for $\frac{2}{3}$ tank $=\frac{2}{3}\times 24=16$ hours.

Trick: LCM method: tank $=24$ units; filler $+2$, emptier $-1$, net $1$ unit/h; $\frac{2}{3}$ of 24 = 16 units → 16 h.

Answer: (A) $16$

Concept: Complete the squares: a sum of squares equals 0 only when each square is 0.

1. $x^2-16x+64+y^2+38y+361=64+361-425$.
2. $(x-8{)}^2+(y+19{)}^2=0$.
3. $x=8,y=-19$.
4. $x^2+y^2=64+361=425$.

Trick: Half of 16 is 8 and half of 38 is 19; $8^2+{19}^2=425$ matches the constant, so the answer is the constant itself.

Answer: (D) $425$

Concept: Split the middle term: find two numbers with sum $2a+3$ and product $a^2+3a=a(a+3)$.

1. The numbers are $a$ and $a+3$ (sum $2a+3$, product $a(a+3)$).
2. $x^2+(2a+3)x+a(a+3)=(x+a)(x+a+3)$.
3. Roots: $x=-a$ and $x=-a-3$.

Trick: Put $a=0$: $x^2+3x=0$ gives $0,-3$; only option (A) gives $0,-3$.

Answer: (A) $x=-a,-a-3$

Concept: Work done in $t$ hours by both $=t(\frac{1}{a}+\frac{1}{b})$.

1. One hour together $=\frac{1}{6}+\frac{1}{8}=\frac{4+3}{24}=\frac{7}{24}$.
2. In 2 hours $=2\times \frac{7}{24}=\frac{7}{12}$.

Trick: LCM 24 units: pumps fill 4 + 3 = 7 units/h; 2 h → 14 units $=\frac{14}{24}=\frac{7}{12}$.

Answer: (C) $\frac{7}{12}$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $f_1$ is the highest frequency.

1. Highest frequency 35 → modal class 20–30: $l=20,h=10$.
2. $f_0=29,f_1=35,f_2=25$.
3. $\text{Mode}=20+\frac{35-29}{70-29-25}\times 10=20+\frac{60}{16}$.
4. $=20+3.75=23.75$.

Trick: The mode must lie inside the modal class (20 to 30) — only $23.75$ and $22.58$ qualify; compute to choose.

Answer: (C) $23.75$

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

1. Distance 1 $=77\times 3.5=269.5\text{km}$.
2. Distance 2 $=(77+9)\times 1=86\text{km}$.
3. Total $=355.5\text{km}$ in $4.5\text{h}$.
4. Average $=\frac{355.5}{4.5}=79\text{km/h}$.

Trick: Weighted average: $77+\frac{9\times 1}{4.5}=77+2=79$.

Answer: (D) $79$

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $a=12+3\sqrt{4}=12+6=18$
2. $b=16-4\sqrt{4}=16-8=8$
3. $\sqrt{18\times 8}=\sqrt{144}=12$

Trick: Simplify $\sqrt{4}=2$ first; the mean proportional always lies between the two numbers ($8<12<18$).

Answer: (D) $12$