RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Work per hour adds when pipes run together.

1. LCM of 16, 23, 10 $=1840$ units (tank).
2. Per hour: A $=115$, B $=80$, C $=184$ → total $379$.
3. Time $=\frac{1840}{379}=4\frac{324}{379}$ hours.

Trick: Quick check: C alone takes 10 h, so all three together must take well under 10 h – only option D fits.

Answer: (D) $4\frac{324}{379}$ hours

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal exponents.

1. $x^z={134}^{0.78z}$ and $y^8={134}^{0.52\times 8}={134}^{4.16}$.
2. $0.78z=4.16\Rightarrow z=\frac{4.16}{0.78}=\frac{16}{3}\approx 5.33$.

Trick: $\frac{0.52}{0.78}=\frac{2}{3}$, so $z=8\times \frac{2}{3}=5.33$.

Answer: (B) 5.33

Concept: Divisibility by 11: difference of alternate-digit sums is 0 or a multiple of 11.

1. N51M91: odd places (from left) $N+1+9=N+10$; even places $5+M+1=M+6$.
2. Difference $=(N+10)-(M+6)=N-M+4$.
3. $N-M+4=0\Rightarrow M-N=4$ (e.g. 451891 $=11\times 41081$).

Trick: Option values like $M-N=1$ or $M=N$ give differences 3 and 4 – not multiples of 11.

Answer: (D) $M-N=4$

Concept: Same distance ⇒ time is inversely proportional to speed.

1. Distance $=60\times 8=480\text{km}$.
2. Malaika's time $=\frac{480}{96}=5$ hours.

Trick: $t_2=t_1\times \frac{s_1}{s_2}=8\times \frac{60}{96}=5$.

Answer: (A) 5 hours

Concept: Overall average = total runs ÷ total matches.

1. Runs $=23\times 40+10\times 13=920+130=1050$.
2. Average $=\frac{1050}{33}=31.818\dots \approx 31.82$.

Trick: Weighted average must lie between 13 and 40, nearer 40 (more matches).

Answer: (A) 31.82

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 95=7700$.
2. $r=\frac{7700\times 7}{44\times 95}=\frac{53900}{4180}$.
3. $r\approx 12.89\text{cm}$.

Trick: $\frac{7700}{2\pi h}=\frac{7700}{597.1}\approx 12.9$ – pick the closest.

Answer: (C) 12.89

Concept: Use one-day work rates and subtract.

1. Shifa $=\frac{1}{28}$; Afiya $=\frac{1}{20}-\frac{1}{28}=\frac{1}{70}$.
2. Mariyam $=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$.
3. Shifa + Mariyam $=\frac{1}{28}+\frac{1}{20}=\frac{12}{140}=\frac{3}{35}$ → $\frac{35}{3}$ days.

Trick: Mariyam = (S+A+M) − (S+A): the 10-day and 20-day groups directly.

Answer: (D) $\frac{35}{3}$

Concept: Net price factor = product of (1 − each discount).

1. $0.76\times 0.74=0.5624$.
2. $0.5624\times 0.82=0.461168$.
3. Single discount $=1-0.461168=0.538832\approx 53.88\%$.

Trick: Two at a time: $24+26-\frac{24\times 26}{100}=43.76$; then $43.76+18-\frac{43.76\times 18}{100}=53.88$.

Answer: (A) 53.88%

Concept: Find both quantities and subtract.

1. $65\%$ of $65=0.65\times 65=42.25$.
2. $\frac{2}{5}$ of $45=18$.
3. Difference $=42.25-18=24.25$.

Trick: ${65}^2=4225$, so 65% of 65 $=42.25$.

Answer: (B) 24.25

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

1. $\sqrt{28\times 252}=\sqrt{7056}$.
2. $=84$ (since $252=9\times 28$, $\sqrt{{28}^2\times 9}=28\times 3$).

Trick: $28\times 3=84$.

Answer: (C) 84

Concept: For two numbers, $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $4\times 208=52\times x$
2. $x=\frac{832}{52}=16$
3. Check: $\text{HCF}(52,16)=4$, $\text{LCM}=208$ ✓

Trick: $\frac{208}{52}=4$, so the other number $=4\times 4=16$.

Answer: (C) $16$

Concept: $\text{Area}=l\times b$ and $\text{Perimeter}=2(l+b)$.

1. $b=\frac{1904}{68}=28\text{m}$
2. $\text{Perimeter}=2(68+28)=2\times 96=192\text{m}$

Trick: $68\times 28$: unit digits $8\times 8$ end in 4, matching 1904 – quick check of $b$.

Answer: (A) $192\text{m}$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$

1. $\text{SI}=\frac{3000\times 8\times 6}{100}$
2. $=30\times 48=1440$

Trick: 8% for 6 years = 48% of 3000 = 1440.

Answer: (A) ₹1,440

Concept: If SP of $a$ items = CP of $b$ items and $b>a$, there is a gain of $\frac{b-a}{a}\times 100\%$.

1. Let CP of 1 book = ₹1, so SP of 25 books = ₹37.
2. CP of 25 books = ₹25 → gain = ₹12.
3. $\text{Gain}\%=\frac{12}{25}\times 100=48\%$

Trick: Gain% $=\frac{\text{difference}}{\text{SP count}}\times 100=\frac{37-25}{25}\times 100$.

Answer: (D) $48\%$ gain

Concept: Increasing a number by 80% makes it $180\%$ of itself, i.e. multiplied by $\frac{180}{100}$.

1. $x\times \frac{180}{100}=2880$
2. $x=\frac{2880\times 100}{180}=16\times 100=1600$

Trick: $2880÷1.8=1600$.

Answer: (A) $1600$

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

1. Sum of first two $=2\times 80=160$; third $=105$.
2. Remaining two $=655-160-105=390$.
3. Average $=\frac{390}{2}=195$

Trick: Subtract the known parts from the total, then divide by the count left.

Answer: (D) $195$

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ (compare per piece).

1. CP of 1 sharpener $=\frac{150}{12}=12.5$ (₹).
2. SP $=43$ → profit $=43-12.5=30.5$.
3. $\text{Profit}\%=\frac{30.5}{12.5}\times 100=244\%$

Trick: Per dozen: CP 150, SP $43\times 12=516$; profit $=\frac{366}{150}\times 100=244\%$.

Answer: (B) $244$

Concept: Expand the brackets, then combine like terms.

1. $x(3x-8)=3x^2-8x$
2. $3(x^2-5)=3x^2-15$
3. Sum: $3x^2-8x+3x^2-15+11=6x^2-8x-4$

Trick: Constant: $-15+11=-4$ and $x^2$ coefficient $+6$ → only option (D) fits.

Answer: (D) $6x^2-8x-4$

Concept: Form two equations from the two conditions and solve.

1. Now: $2P-Q=15$ … (i)
2. After 9 years: $2(Q+9)=2(P+9)-14\Rightarrow 2P-2Q=14$ … (ii)
3. (i) − (ii): $Q=1$; then $P=8$.

Trick: Subtracting the equations removes $2P$ directly and gives $Q$.

Answer: (C) $1$

Concept: Convert each ratio to a decimal (or compare how close the two terms are).

1. $\frac{48}{90}\approx 0.53$, $\frac{46}{51}\approx 0.90$
2. $\frac{22}{54}\approx 0.41$, $\frac{42}{68}\approx 0.62$
3. Greatest: $46:51$

Trick: When the antecedent is closest to the consequent, the ratio is nearest to 1 – here 46 is just 5 less than 51.

Answer: (B) $46:51$