RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Work per day $=\frac{1}{\text{days}}$; subtract the joint work, then divide the rest by the remaining worker's rate.

1. Together in 2 days: $2(\frac{1}{13}+\frac{1}{4})=2\times \frac{17}{52}=\frac{17}{26}$.
2. Remaining work $=1-\frac{17}{26}=\frac{9}{26}$.
3. Naman alone: $\frac{9}{26}\times 13=4.5$ days.

Trick: Total work $=\text{LCM}(13,4)=52$ units; rates 4 and 13; 2 days $=34$ units; left $18$; $\frac{18}{4}=4.5$.

Answer: (A) $4.5$

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$; work left to right.

1. ${83}^{-6}÷{83}^{16}={83}^{-6-16}={83}^{-22}$.
2. ${83}^{-22}\times {83}^{-8}={83}^{-30}$.

Trick: Just combine the powers: $-6-16+(-8)=-30$.

Answer: (C) ${83}^{-30}$

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Sum $=40+49+56+73+38=256$.
2. Average $=\frac{256}{5}=51.2$ kg.

Trick: Assume 50: deviations $-10,-1,+6,+23,-12$ total $+6$; $50+\frac{6}{5}=51.2$.

Answer: (A) $51.2$

Concept: Exterior + interior angle $={180}^{\circ }$; number of sides $n=\frac{{360}^{\circ }}{\text{exterior angle}}$.

1. Ratio $1:8$ ⇒ exterior $=\frac{{180}^{\circ }}{9}={20}^{\circ }$.
2. $n=\frac{{360}^{\circ }}{{20}^{\circ }}=18$.

Trick: If E : I $=1:k$, then $n=2(k+1)$ ⇒ $2\times 9=18$.

Answer: (C) $18$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$ with rate and time in the same unit.

1. Rate $3\%$ per month, time 12 months ⇒ total rate $=36\%$.
2. $\text{SI}=\frac{1200\times 36}{100}=432$.

Trick: $3\%$ per month $=36\%$ per year – do not take it as $3\%$ per annum (that gives ₹36).

Answer: (C) ₹432

Concept: Check each odd number for divisibility by primes up to $\sqrt{311}\approx 17$.

1. $303=3\times 101$, $305=5\times 61$, $309=3\times 103$ – composite.
2. $307$ and $311$ are not divisible by 2, 3, 5, 7, 11, 13, 17 – prime.
3. Even numbers are composite. Count $=2$.

Trick: Test only up to the square root; 3 divides 303 and 309 (digit sums 6 and 12).

Answer: (B) $2$

Concept: $V=\pi r^{2}h$; with $h=r$, $V=\pi r^3$.

1. $\frac{22}{7}{r}^3=8624$.
2. $r^3=\frac{8624\times 7}{22}=392\times 7=2744$.
3. $r=\sqrt[3]{2744}=14$, so $h=14$ cm.

Trick: $2744={14}^3$ – remember cubes up to 20.

Answer: (C) 14 cm

Concept: Count days excluding the deposit day and including the withdrawal day; 2024 is a leap year (366 days).

1. Feb: $29-7=22$, March: 31, April: 8 ⇒ 61 days.
2. $\text{SI}=4000\times \frac{7.5}{100}\times \frac{61}{366}=300\times \frac{1}{6}=50$.

Trick: $\frac{61}{366}=\frac{1}{6}$ exactly in a leap year – a common exam pairing.

Answer: (B) $50$

Concept: Form two linear equations in mother's age $M$ and daughter's age $D$.

1. $3M=5D+26$ and $M=2D+2$.
2. $3(2D+2)=5D+26$ ⇒ $D=20$.
3. $M=42$; difference $=42-20=22$.

Trick: Check options: $M-D=22$ with $M=2D+2$ gives $D=20$, $M=42$; $3\times 42=126=5\times 20+26$ ✓.

Answer: (D) $22$

Concept: $V=\frac{4}{3}\pi r^3$, with $r=\frac{d}{2}$.

1. $r=11$ cm, $r^3=1331$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 1331=\frac{117128}{21}$.
3. $V\approx 5577.52{\text{cm}}^3$.

Trick: $\frac{88}{21}\approx 4.1905$; $4.1905\times 1331\approx 5577.5$.

Answer: (D) $5577.52$

Concept: Increasing a number by $100\%$ makes it $\frac{200}{100}=2$ times.

1. Let the number be $x$. Then $x\times \frac{200}{100}=2820$.
2. $2x=2820\Rightarrow x=1410$.

Trick: Up by $100\%$ = double, so just halve: $2820÷2=1410$.

Answer: (C) 1410

Concept: Form two linear equations in the cost of a pen ($p$) and a pencil ($q$) and solve.

1. $8p+10q=132$ … (i)
2. $17(p-2)+6(q+7)=136\Rightarrow 17p+6q+8=136\Rightarrow 17p+6q=128$ … (ii)
3. (i)$\times 3$: $24p+30q=396$; (ii)$\times 5$: $85p+30q=640$. Subtract: $61p=244\Rightarrow p=4$.
4. From (i): $10q=132-32=100\Rightarrow q=10$.
5. $12p+9q=48+90=138$.

Trick: Check options quickly: $p=4,q=10$ satisfies both equations ($32+100=132$; $34+102=136$).

Answer: (A) ₹138

Concept: Use the angle-sum property of triangles and the linear pair at E.

1. In right $△BDC$ ($\angle BDC={90}^{\circ }$): $\angle C={180}^{\circ }-{90}^{\circ }-{16}^{\circ }={74}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-\angle CAE-\angle C={180}^{\circ }-{51}^{\circ }-{74}^{\circ }={55}^{\circ }$.
3. $\angle AEB$ and $\angle AEC$ form a linear pair: $\angle AEB={180}^{\circ }-{55}^{\circ }={125}^{\circ }$.

Trick: Exterior angle of $△AEC$ at E: $\angle AEB=\angle CAE+\angle C={51}^{\circ }+{74}^{\circ }={125}^{\circ }$.

Answer: (D) ${125}^{\circ }$

Concept: If $3A=9B=4C=k$, then $A:B:C=\frac{1}{3}:\frac{1}{9}:\frac{1}{4}$.

1. Multiply by LCM $36$: $A:B:C=12:4:9$; total parts $=25$.
2. Share of A $=\frac{12}{25}\times 75400=12\times 3016=36192$.

Trick: Answer must be a multiple of $12$ (and $75400÷25=3016$): only $36192=12\times 3016$ fits.

Answer: (B) ₹36,192

Concept: Divisibility by 7 (also 11, 13): split the number into groups of three from the right and take the alternating sum.

1. $774|359|327$: $327-359+774=742$.
2. $742=7\times 106$, so the number is divisible by 7 ($774359327=7\times 110622761$).
3. It is odd, so not divisible by 2, 12 or 16.

Trick: Odd number → 2, 12, 16 are out at once; only 7 is left to check.

Answer: (A) 7

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum of $545714151$: $5+4+5+7+1+4+1+5+1=33$, and $33$ is divisible by 3. ✔
2. Last digit is 1 (odd, not 0), so it is not divisible by 10 or by 6 (needs 2 as well).
3. By 7: $151-714+545=-18$, not a multiple of 7.

Trick: Odd number → 6 and 10 are out; digit sum test settles 3.

Answer: (D) 3

Concept: Perimeter of a sector $=2r+l$, where arc length $l=\frac{\theta }{{360}^{\circ }}\times 2\pi r$.

1. $l=\frac{18}{360}\times 2\times 3.14\times 9=\frac{1}{20}\times 56.52=2.826\text{cm}$.
2. Perimeter $=2\times 9+2.826=20.826\approx 20.83\text{cm}$.

Trick: ${18}^{\circ }$ is $\frac{1}{20}$ of a full turn; arc $=\frac{2\pi r}{20}=\frac{\pi r}{10}$.

Answer: (C) $20.83\text{cm}$

Concept: Work step by step from registered voters $R$ to the winning margin.

1. Votes polled $=85\%$ of $R=0.85R$; valid votes $=96\%$ of polled $=0.96\times 0.85R=0.816R$.
2. Winner $65\%$, loser $35\%$ of valid votes, so margin $=30\%$ of valid $=0.3\times 0.816R=0.2448R$.
3. $0.2448R=2142\Rightarrow R=\frac{2142}{0.2448}=8750$.

Trick: $R\times \frac{85}{100}\times \frac{96}{100}\times \frac{30}{100}=2142$; only 8750 makes this exact.

Answer: (B) 8750

Concept: The third proportional $x$ of $a$ and $b$ satisfies $a:b=b:x$, so $x=\frac{b^2}{a}$.

1. $x=\frac{{56}^2}{16}=\frac{3136}{16}=196$.

Trick: $\frac{56}{16}=3.5$, so $x=56\times 3.5=196$.

Answer: (A) 196

Concept: Take total work $=$ LCM of $26,13,10=130$ units and find each person's daily units.

1. Priya $=\frac{130}{26}=5$ units/day; Priya + Varsha $=\frac{130}{13}=10$, so Varsha $=5$.
2. All three $=\frac{130}{10}=13$ units/day, so Nisha $=13-10=3$ units/day.
3. Priya + Nisha $=5+3=8$ units/day; time $=\frac{130}{8}=\frac{65}{4}=16\frac{1}{4}$ days.

Trick: LCM method avoids fractions: total ÷ combined daily units.

Answer: (B) $\frac{65}{4}$