RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Successive discounts are applied one after another on the reduced price.

1. After 10%: $5000\times \frac{90}{100}=4500$
2. After 5%: $4500\times \frac{95}{100}=4275$

Trick: Net discount $=10+5-\frac{10\times 5}{100}=14.5\%$, so SP $=5000\times 0.855=4275$.

Answer: (D) ₹4,275

Concept: Total literates = male literates + female literates.

1. Females $=374000-165000=209000$
2. Total literates $=21\%\text{ of }374000=78540$
3. Male literates $=40\%\text{ of }165000=66000$
4. Female literates $=78540-66000=12540$; $\frac{12540}{209000}\times 100=6\%$

Trick: Always find the female base first; percentage is of females, not of total population.

Answer: (B) $6\%$

Concept: Work per minute: filling pipes add, the leak (outlet) subtracts.

1. $\frac{1}{16}+\frac{1}{32}-\frac{1}{x}=\frac{1}{12}$
2. $\frac{3}{32}-\frac{1}{12}=\frac{9-8}{96}=\frac{1}{96}$
3. So $\frac{1}{x}=\frac{1}{96}\Rightarrow x=96$ minutes

Trick: LCM method: capacity $=96$ units; A $=6$, B $=3$, together $=8$ units/min ⇒ outlet $=9-8=1$ unit/min ⇒ $96$ min.

Answer: (D) 96 minutes

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

1. $\sqrt{490\times 10}=\sqrt{4900}$
2. $=70$

Trick: Check: $490:70=70:10$ (both equal 7).

Answer: (A) $70$

Concept: Solve innermost bracket first, then × , then +, then ÷ outside.

1. $(5-7)=-2$
2. $9\times (-2)=-18$
3. $26-18=8$
4. $616÷8=77$

Trick: Watch the negative sign: $26+9\times (-2)$ is $8$, not $70$.

Answer: (C) $77$

Concept: Average = sum of observations ÷ number of observations.

1. Sum $=40+49+56+74+39=258$
2. Average $=\frac{258}{5}=51.6\text{kg}$

Trick: Assume mean $50$: deviations $-10,-1,+6,+24,-11$ sum to $+8$; $50+\frac{8}{5}=51.6$.

Answer: (D) $51.6$

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

1. Ratio $=1340:1100:2440=67:55:122$
2. B's 55 parts $=814\Rightarrow 1$ part $=14.8$
3. C − A $=(122-67)=55$ parts $=814$

Trick: C − A investment $=2440-1340=1100=$ B's investment, so the profit difference equals B's share directly.

Answer: (D) ₹814

Concept: Check options: sum must be 84 and LCM must be 120.

1. All options sum to 84, so test LCM.
2. $60=2^2\times 3\times 5,24=2^3\times 3$
3. LCM $=2^3\times 3\times 5=120$ ✓

Trick: Both numbers must divide 120 – only 60 and 24 are factors of 120.

Answer: (B) 60, 24

Concept: Make decimal places equal, find LCM of the numbers without decimals, then put the decimal back.

1. $5.1=5.100$ and $0.102$ → $5100$ and $102$ (thousandths)
2. $5100=102\times 50$, so LCM $=5100$
3. Restore 3 decimal places: $5.100=5.1$

Trick: If one number is a multiple of the other, the larger is the LCM: $5.1=0.102\times 50$.

Answer: (B) $5.1$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; Amount = P + SI.

1. $8400=\frac{P\times 2.5\times 4.2}{100}\Rightarrow P=\frac{840000}{10.5}=80000$
2. New SI $=\frac{80000\times 8.1\times 10.25}{100}=66420$
3. Amount $=80000+66420=146420$

Trick: $4\frac{1}{5}=4.2$ and $10\frac{1}{4}=10.25$ – convert mixed fractions to decimals first.

Answer: (C) ₹1,46,420

Concept: Total = average × count; split total in the given ratio.

1. Total $=16860\times 3=50580$
2. Ratio sum $=3+5+7=15$
3. Costliest $=\frac{7}{15}\times 50580=23604$

Trick: Average of 15 parts over 3 items ⇒ 1 part $=\frac{16860}{5}=3372$; $7\times 3372=23604$.

Answer: (A) 23604

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ per eraser.

1. CP of one eraser $=\frac{150}{12}=12.5$
2. SP $=42$, profit $=29.5$
3. $\frac{29.5}{12.5}\times 100=236\%$

Trick: Per dozen: CP $=150$, SP $=42\times 12=504$; profit $=354$; $\frac{354}{150}\times 100=236\%$.

Answer: (A) 236

Concept: Area $=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2}$.

1. $s=\frac{11+15+16}{2}=21$
2. $s-a=10,s-b=6,s-c=5$
3. Area $=\sqrt{21\times 10\times 6\times 5}=\sqrt{6300}\approx 79.4{\text{cm}}^2$

Trick: ${79}^2=6241$ and ${80}^2=6400$; $6300$ is nearer $6241$, so the area rounds to 79.

Answer: (A) $79{\text{cm}}^2$

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$. Let Pranay's speed $=p$, Aman's $=a$.

1. $\frac{60}{p}-\frac{60}{a}=6$ … (1)
2. $\frac{60}{a}-\frac{60}{2p}=9$ … (2)
3. Add: $\frac{60}{p}-\frac{30}{p}=15\Rightarrow \frac{30}{p}=15\Rightarrow p=2\text{km/h}$

Trick: Option check: $p=2$ ⇒ Pranay takes 30 h, Aman 24 h, doubled speed 15 h $=24-9$ ✓.

Answer: (B) $2\text{km/hr}$

Concept: Find both values separately, then subtract.

1. $65\%\text{ of }55=\frac{65\times 55}{100}=35.75$
2. $\frac{2}{5}\text{ of }15=6$
3. Difference $=35.75-6=29.75$

Trick: Quick: $0.65\times 55=33+2.75=35.75$ (60% plus 5%).

Answer: (C) $29.75$

Concept: Let present ages be father $F$, son $S$.

1. $F-5=2(S-5)+11\Rightarrow F=2S+6$
2. After $x$ years: $F+x=2(S+x)$
3. $2S+6+x=2S+2x\Rightarrow x=6$

Trick: Father is always '2S + 6'; the extra 6 is exactly the years needed to make it double.

Answer: (D) 6

Concept: $\text{SI}=\frac{P\times R\times T}{100}$ with R and T in the same unit (here months).

1. $P=1200,R=4\%$ per month, $T=4$ months
2. $\text{SI}=\frac{1200\times 4\times 4}{100}=192$

Trick: Rate is per month, so do not convert months into years.

Answer: (D) ₹192

Concept: Product of consecutive numbers $n(n+1)$ is just above $n^2$.

1. $\sqrt{600}\approx 24.5$
2. Try $24\times 25=600$ ✓
3. Greater number $=25$

Trick: Take the square root and check the two integers around it.

Answer: (B) 25

Concept: Distance = speed × time; the bicycle's length is negligible, so distance = bridge length.

1. $3.1\text{min}=\frac{3.1}{60}\text{h}$
2. Length $=24\times \frac{3.1}{60}=0.4\times 3.1=1.24\text{km}$

Trick: $24\text{km/h}=0.4\text{km}$ per minute — multiply by minutes.

Answer: (B) $1.24\text{km}$

Concept: Rates add when pipes work together.

1. $\frac{1}{20}+\frac{1}{25}+\frac{1}{10}=\frac{5+4+10}{100}=\frac{19}{100}$
2. Time $=\frac{100}{19}=5\frac{5}{19}$ hours

Trick: Capacity $=100$ units (LCM); rates $5+4+10=19$ units/h ⇒ $\frac{100}{19}$ h.

Answer: (B) $5\frac{5}{19}$ hours