RRB NTPC Graduate CBT 1 2024 PYQs

Concept: When the cistern is full and both pipes are open, the net rate = emptying rate − filling rate.

1. Filling pipe: $\frac{1}{10}$ per hour; emptying pipe: $\frac{1}{8}$ per hour.
2. Net emptying = $\frac{1}{8}-\frac{1}{10}=\frac{5-4}{40}=\frac{1}{40}$ per hour.
3. Time to empty = $40$ hours.

Trick: Time $=\frac{a\times b}{a-b}=\frac{10\times 8}{10-8}=40$ hours.

Answer: (D) 40 hours

Concept: Distance = speed × time; split the journey into its parts.

1. Times: $13$ min at $x$, $25$ min at $2x$, rest $50-13-25=12$ min at $x$.
2. So speed $x$ for $13+12=25$ min and $2x$ for $25$ min.
3. $x·\frac{25}{60}+2x·\frac{25}{60}=30\Rightarrow \frac{75x}{60}=30$
4. $x=\frac{30\times 60}{75}=24$

Trick: Same 25 min at $x$ and $2x$ = 25 min at $3x$: $3x\times \frac{25}{60}=30\Rightarrow x=24$.

Answer: (D) $24$

Concept: Required total = desired average × number of subjects.

1. Needed total = $80\times 5=400$.
2. Present total = $78+85+69+90=322$.
3. Fifth subject = $400-322=78$.

Trick: Deviations from 80: $-2+5-11+10=+2$; fifth score $=80-2=78$.

Answer: (C) 78

Concept: Median $=l+\frac{\frac{N}{2}-cf}{f}\times h$.

1. $N=12+16+30+20+17+38+15=148$, $\frac{N}{2}=74$.
2. Cumulative frequencies: $12,28,58,78,\dots$ → median class $50$–$60$.
3. $l=50,cf=58,f=20,h=10$.
4. Median $=50+\frac{74-58}{20}\times 10=50+8=58$.

Trick: Median class = first class whose cumulative frequency crosses N/2.

Answer: (C) $58$ years

Concept: Area = length × breadth; perimeter = 2(length + breadth).

1. Let longer side $=L$, shorter $=L-21$.
2. $L(L-21)=7\times 2(2L-21)$ ⇒ $L^2-21L=28L-294$.
3. $L^2-49L+294=0$ ⇒ $(L-42)(L-7)=0$.
4. $L=7$ gives a negative side, so $L=42$ cm.

Trick: Option check: $42\times 21=882$ and $7\times 2(63)=882$ ✓.

Answer: (D) $42$

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

1. $\frac{22}{7}\times r^2\times 2.2=8470$
2. $r^2=\frac{8470\times 7}{22\times 2.2}=\frac{59290}{48.4}=1225$
3. $r=\sqrt{1225}=35$ cm

Trick: Unit digit of $r^2$ is 5, so $r$ ends in 5 → only 35 fits.

Answer: (A) $35$

Concept: Find total investment from the profit percentage, then share it in the given ratio.

1. $25\%$ of total investment $=$ ₹28,800 ⇒ total $=\frac{28800\times 100}{25}=$ ₹1,15,200.
2. Ratio sum $=7+3+20=30$.
3. Anjali $=\frac{3}{30}\times 115200=$ ₹11,520.

Trick: Total $=28800\times 4$; Anjali gets $\frac{1}{10}$ of it.

Answer: (C) ₹11,520

Concept: If $X\propto \frac{1}{Y}$ and $Y\propto \frac{1}{Z}$, then $X\propto Z$.

1. $X=kZ$
2. $\frac{1}{16}=k\times \frac{1}{32}$ ⇒ $k=2$
3. When $Z=3$: $X=2\times 3=6$

Trick: Inverse of inverse = direct: $Z$ grows $\times 96$ ($\frac{1}{32}\to 3$), so $X$ also grows $\times 96$: $\frac{96}{16}=6$.

Answer: (A) $6$

Concept: For $n$ (odd) observations, median = ${\left(\frac{n+1}{2}\right)}^{th}$ value in ascending order.

1. Ascending order: 13, 18, 25, 27, 31, 39, 56, 80, 87, 92, 98
2. $n=11$ ⇒ median = $\frac{11+1}{2}=6^{th}$ value $=39$.

Trick: Always sort first – picking the middle of the given list gives 39 here only by luck.

Answer: (C) 39

Concept: B's one-day work = (A + B)'s one-day work − A's one-day work.

1. B's rate $=\frac{1}{2}-\frac{1}{38}=\frac{19-1}{38}=\frac{18}{38}=\frac{9}{19}$ work/day.
2. B alone does 1 work in $\frac{19}{9}$ days.
3. 9 times the work: $9\times \frac{19}{9}=19$ days.

Trick: B does 9 units in 19 days directly from rate $\frac{9}{19}$.

Answer: (D) $19$

Concept: If A is $r\%$ more than B, then B is $\frac{r}{100+r}\times 100\%$ less than A.

1. Let B $=100$, then A $=116$.
2. B is less by $\frac{16}{116}\times 100=13.793...\%$
3. $\approx 13.79\%$

Trick: Use $\frac{r}{100+r}$: $\frac{16}{116}=\frac{4}{29}\approx 13.79\%$.

Answer: (A) $13.79\%$

Concept: If $a:b::b:c$, then $b^2=ac$ ($b$ is the mean proportional).

1. $b^2=a\times 9a=9a^2$ ⇒ $b=3a$.
2. $6=3a$ ⇒ $a=2$.
3. $c=9a=18$.

Trick: $c=9a$ means ratio $a:b:c=1:3:9$; $b=6$ ⇒ $c=18$.

Answer: (B) $18$

Concept: Convert mixed fractions to improper fractions and cancel before multiplying.

1. $1\frac{4}{9}=\frac{13}{9}$, $5\frac{2}{9}=\frac{47}{9}$
2. $\frac{13}{9}\times \frac{47}{9}\times 324=13\times 47\times 4=2444$
3. $2444+9\frac{3}{4}=2444+9.75=2453.75$

Trick: $324=81\times 4$, so both 9s cancel at once.

Answer: (D) 2453.75

Concept: Sum of numbers = average × count.

1. Sum $=80\times 8=640$.
2. Least number $=\frac{1}{16}\times 640=40$.

Trick: $\frac{1}{16}$ of sum $=\frac{8\times 80}{16}=\frac{80}{2}=40$.

Answer: (A) $40$

Concept: Evaluate both parts and subtract.

1. $60\%$ of $40=\frac{60}{100}\times 40=24$
2. $\frac{2}{5}$ of $55=22$
3. Difference $=24-22=2$

Trick: $60\%=\frac{3}{5}$: $\frac{3}{5}\times 40=24$ quickly.

Answer: (B) $2$

Concept: Use ${\cot }^2\theta ={\csc }^2\theta -1$.

1. Sum $=\frac{{\cot }^2\theta +(\csc \theta +1{)}^2}{\cot \theta (\csc \theta +1)}$
2. Numerator $={\csc }^2\theta -1+{\csc }^2\theta +2\csc \theta +1=2\csc \theta (\csc \theta +1)$
3. Sum $=\frac{2\csc \theta }{\cot \theta }=\frac{2/\sin \theta }{\cos \theta /\sin \theta }=\frac{2}{\cos \theta }=2\sec \theta$

Trick: Put $\theta ={45}^{\circ }$: $\frac{1}{\sqrt{2}+1}+(\sqrt{2}+1)=2\sqrt{2}=2\sec {45}^{\circ }$.

Answer: (C) $2\sec \theta$

Concept: Follow BODMAS: brackets from inside out, then division/multiplication left to right, then subtraction.

1. $-60-72-92=-224$
2. $\{2+(-6)\times (-9)\}=2+54=56$, so $[8\times 56]=448$
3. $(-18)\times (-224)=4032$ and $4032÷448=9$
4. $48-9=39$

Trick: Two negatives multiply to a positive – check signs first.

Answer: (B) $39$

Concept: Subtract the two equations to remove the common term (19 tables).

1. $(19c+19t)-(15c+19t)=5700-5300\Rightarrow 4c=400\Rightarrow c=100$
2. $19t=5700-19\times 100=3800\Rightarrow t=200$
3. $12c+13t=1200+2600=3800$

Trick: Same number of tables in both – the difference is purely due to 4 chairs.

Answer: (B) ₹$3800$

Concept: For similar triangles, the ratio of areas = square of the ratio of corresponding sides.

1. Side ratio $=3:10$
2. Area ratio $=3^2:{10}^2=9:100$

Trick: Sides → square for area, cube for volume (similar solids).

Answer: (A) $9:100$

Concept: Upstream speed $=b-s$, downstream speed $=b+s$.

1. Let $b=8k,s=5k$: upstream $=3k$, downstream $=13k$.
2. Upstream speed $=\frac{37.2}{27/60}=\frac{248}{3}\text{km/h}$, so $k=\frac{248}{9}$ and downstream $=\frac{3224}{9}\text{km/h}$.
3. Time up $=79.4\times \frac{3}{248}=0.9605\text{h}$; time down $=85.8\times \frac{9}{3224}=0.2395\text{h}$.
4. Total $=0.9605+0.2395=1.2\text{h}$

Trick: Express everything in units of $k$; only $3k$ is needed from the data.

Answer: (B) $1.2$ hours