RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: For a fixed distance, speed is inversely proportional to time: $S_1{T}_1=S_2{T}_2$.

1. Distance $=14\times 20=280\text{km}$.
2. Required speed $=\frac{280}{5}=56\text{km/h}$.

Trick: Time becomes $\frac{1}{4}$, so speed becomes $4\times 14=56$.

Answer: (A) $56\text{km/h}$

Concept: Sum of first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Here $n=149$.
2. Average $=\frac{{149}^2}{149}=149$.

Trick: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (C) $149$

Concept: $A=P{(1+\frac{R}{100})}^n$.

1. ${(1+\frac{R}{100})}^2=\frac{2809}{2500}$.
2. $1+\frac{R}{100}=\sqrt{\frac{2809}{2500}}=\frac{53}{50}$.
3. $\frac{R}{100}=\frac{3}{50}\Rightarrow R=6\%$.

Trick: $2500={50}^2$, $2809={53}^2$; $\frac{53-50}{50}\times 100=6\%$.

Answer: (B) $6\%$

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is $0$ or a multiple of $11$.

1. $877217$: odd places from right $7+2+7=16$; even places $1+7+8=16$.
2. Difference $=16-16=0\Rightarrow$ divisible by $11$.
3. Last digit $7$ is odd, so not divisible by $2$, $12$ or $16$.

Trick: Odd last digit rules out every even divisor at once.

Answer: (D) $11$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. Equivalent of $35\%$ and $4\%$ $=35+4-\frac{35\times 4}{100}=37.6\%$.
2. Difference of discounts $=52\%-37.6\%=14.4\%$.
3. Amount $=\frac{14.4}{100}\times 1125=162$.

Trick: Directly: pay $0.48\times 1125=540$ vs $0.65\times 0.96\times 1125=702$; gap $=162$.

Answer: (A) ₹$162$

Concept: For divisibility by $11$, the difference of sums of alternate digits must be $0$ or a multiple of $11$.

1. $N51M54$: places 1, 3, 5 give $N+1+5=N+6$.
2. Places 2, 4, 6 give $5+M+4=M+9$.
3. $(M+9)-(N+6)=M-N+3=0\Rightarrow M-N=-3$ (e.g. $N=4,M=1$: $451154=11\times 41014$).

Trick: Write the alternate sums, equate the difference to $0$ first.

Answer: (B) $M-N=-3$

Concept: Net rate $=\frac{1}{A}+\frac{1}{B}-\frac{1}{C}=\frac{1}{\text{time together}}$.

1. $\frac{1}{93}+\frac{1}{31}=\frac{1+3}{93}=\frac{4}{93}$.
2. $\frac{1}{C}=\frac{4}{93}-\frac{1}{24}=\frac{96-93}{2232}=\frac{3}{2232}=\frac{1}{744}$.
3. Emptying pipe alone takes $744\text{min}$.

Trick: LCM method: tank $=744$ units; A $=8$, B $=24$, all $=31$ per min $\Rightarrow$ C $=1$ unit/min.

Answer: (B) $744\text{min}$

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

1. $V=\frac{22}{7}\times 9^2\times 35$.
2. $=22\times 81\times 5$.
3. $=8910{\text{cm}}^3$.

Trick: $h=35$ is a multiple of $7$ — cancel first: $\frac{35}{7}=5$.

Answer: (A) $8910{\text{cm}}^3$

Concept: Let Pranshu's speed be $p$ and Chandu's time be $t$ hours.

1. $\frac{56}{p}=t+3$ and $\frac{56}{2p}=t-4$.
2. Subtract: $\frac{56}{p}-\frac{28}{p}=7\Rightarrow \frac{28}{p}=7$.
3. $p=4\text{km/h}$ (check: $14\text{h}$ vs $7\text{h}$, Chandu $11\text{h}$).

Trick: Doubling speed halves time; the halved part equals the gap $3+4=7\text{h}$.

Answer: (C) $4\text{km/h}$

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

1. $\sqrt{25\times 196}$.
2. $=5\times 14=70$.

Trick: Root each number separately: $\sqrt{25}=5$, $\sqrt{196}=14$.

Answer: (D) $70$

Concept: Solve innermost bracket first, then multiplication, then addition, and finally the outer division (BODMAS).

1. $(3-7)=-4$
2. $4\times (-4)=-16$
3. $17+(-16)=1$
4. $167÷1=167$

Trick: If the curly bracket becomes $1$, the answer is the numerator itself.

Answer: (A) $167$

Concept: Successive % changes multiply: $P_3=P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. $20000\times 1.3=26000$
2. $26000\times 1.5=39000$
3. Decrease $68\%$ leaves $32\%$: $39000\times 0.32=12480$

Trick: $20000\times 1.3\times 1.5\times 0.32=20000\times 0.624=12480$ – the answer must be a multiple of $10$ here.

Answer: (A) $12,480$

Concept: Expand each bracket and combine like terms.

1. $2(\frac{5}{2}{x}^2-22x+11)=5x^2-44x+22$
2. $-5(x^2+9x-11)=-5x^2-45x+55$
3. Add: $(5x^2-5x^2)+(-44x-45x)+(22+55)=-89x+77$

Trick: The $x^2$ terms cancel, so only a linear expression remains – check sign of $x$: $-44-45=-89$.

Answer: (C) $-89x+77$

Concept: If $\sin \theta =x$, take perpendicular $=x$, hypotenuse $=1$; then base $=\sqrt{1-x^2}$ and $\tan \theta =\frac{x}{\sqrt{1-x^2}}$.

1. $\frac{\sqrt{1+x}\tan \theta }{x}=\frac{\sqrt{1+x}}{x}\times \frac{x}{\sqrt{1-x^2}}$
2. $=\frac{\sqrt{1+x}}{\sqrt{(1-x)(1+x)}}$
3. $=\frac{1}{\sqrt{1-x}}$

Trick: Put $x=\frac{1}{2}$ ($\theta ={30}^{\circ }$): LHS $=\sqrt{1.5}\times \frac{1}{\sqrt{3}}\times 2=\sqrt{2}$ and $\frac{1}{\sqrt{1-0.5}}=\sqrt{2}$.

Answer: (A) $\frac{1}{\sqrt{1-x}}$

Concept: For the same time period, profit is shared in the ratio of investments.

1. Ratio $=1730:1420:1350=173:142:135$
2. Sum $=173+142+135=450$
3. Rupesh's share $=\frac{135}{450}\times 1490=0.3\times 1490=447$

Trick: $\frac{135}{450}=\frac{3}{10}$, so just take $30\%$ of ₹$1490$.

Answer: (B) ₹$447$

Concept: $V=\pi r^{2}h$. Take $r=100$, $h=100$; new $r=77$, new $h=251$.

1. Old $V\propto {100}^2\times 100={10}^6$
2. New $V\propto {77}^2\times 251=5929\times 251=1488179$
3. Increase $=\frac{1488179-1000000}{1000000}\times 100\approx 48.8\%\approx 49\%$

Trick: Multiplying factor $={0.77}^2\times 2.51=0.5929\times 2.51\approx 1.488$ → about $49\%$ increase.

Answer: (B) $49\%$

Concept: Form equations from the statements; the first one alone gives the difference.

1. $2R=2Y+24\Rightarrow R-Y=12$
2. Check with second: $4Y=R-3$; with $R=Y+12$: $3Y=9\Rightarrow Y=3$, $R=15$
3. Difference $=15-3=12$ years

Trick: Divide the first equation by $2$ – the difference is directly $\frac{24}{2}=12$.

Answer: (B) $12$

Concept: Check each option: sum must be $78$ and LCM must be $180$.

1. $58+20=78$ but LCM $(58,20)=580$ ✗
2. $53+25=78$, LCM $=1325$ ✗
3. $60+18=78$, $60=2^2·3·5$, $18=2·3^2$, LCM $=2^2·3^2·5=180$ ✓
4. $65+13=78$, LCM $=65$ ✗

Trick: Both numbers must divide $180$; only $60$ and $18$ are divisors of $180$.

Answer: (C) $60,18$

Concept: Profit ratio $=$ (capital $\times$ time) ratio.

1. Kunal $+$ Abhishek $=5308$, difference $=2654$ → Kunal $=3981$, Abhishek $=1327$; ratio $3:1$
2. Let Abhishek $=A$, Kunal $=A+33000$: $\frac{7(A+33000)}{3A}=3$
3. $7A+231000=9A\Rightarrow A=115500$
4. Kunal $=115500+33000=148500$

Trick: Profit ratio $3:1$ with times $7:3$ → capital ratio $\frac{3}{7}:\frac{1}{3}=9:7$; $2$ parts $=33000$ → $9$ parts $=148500$.

Answer: (D) ₹$1,48,500$

Concept: Bells ring together again after the LCM of their intervals.

1. $36=2^2·3^2$, $45=3^2·5$
2. LCM $=2^2·3^2·5=180$ minutes $=3$ hours
3. 9:00 a.m. $+3$ h $=$ 12:00 noon

Trick: LCM $=\frac{36\times 45}{\text{HCF}(9)}=180$ min.

Answer: (C) 12 noon