RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: For the same time, equal simple interest means $P_1{R}_1=P_2{R}_2$, i.e. principal is inversely proportional to rate.

1. Let ₹$x$ be at $11\%$ and ₹$(20300-x)$ at $3\%$, both for 5 years.
2. $\frac{x\times 11\times 5}{100}=\frac{(20300-x)\times 3\times 5}{100}\Rightarrow 11x=60900-3x$
3. $14x=60900\Rightarrow x=4350$

Trick: Ratio of sums $=3:11$ (inverse of rates). Sum at $11\%=\frac{3}{14}\times 20300=4350$.

Answer: (A) 4,350

Concept: To test a number $n$ for primality, check divisibility by primes up to $\sqrt{n}$; here $\sqrt{443}<21$, so test 2, 3, 5, 7, 11, 13, 17, 19.

1. Even numbers and those ending in 5 are out; left: 431, 433, 437, 439, 441, 443.
2. $441={21}^2$ (divisible by 3) and $437=19\times 23$ — not prime.
3. 431, 433, 439, 443 pass all tests — prime.

Trick: After removing evens and multiples of 5, check only 3, 7, 11, 13, 17, 19 on the remaining odd numbers.

Answer: (A) 4

Concept: $x\%$ of $y=\frac{x}{100}\times y$; 'greater by' means take the difference.

1. $65\%\text{ of }65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{1}{5}\text{ of }25=5$
3. Difference $=42.25-5=37.25$

Trick: ${65}^2=4225$ (number ending in 5: $6\times 7=42$, then 25) → $42.25$.

Answer: (B) 37.25

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

1. Present: $2P-Q=41$ … (i)
2. After 7 years: $9(Q+7)=6(P+7)-24\Rightarrow 9Q+63=6P+18\Rightarrow 2P-3Q=15$ … (ii)
3. (i) − (ii): $2Q=26\Rightarrow Q=13$, and $P=27$.

Trick: Check options: $Q=13\Rightarrow P=27$; after 7 years $6\times 34-9\times 20=204-180=24$ ✓.

Answer: (D) 13

Concept: By BODMAS, do division and multiplication first, then addition and subtraction, keeping track of signs.

1. $(-60)÷(-10)=6$
2. $(-3)\times 7=-21$
3. $(-9)-6+(-21)=-9-6-21=-36$

Trick: Negative ÷ negative = positive; then '$-$ positive' subtracts.

Answer: (C) $-36$

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

1. $14.2=14.20=\frac{1420}{100}$ and $0.05=\frac{5}{100}$
2. $\text{LCM}(1420,5)=1420$ (since 1420 is a multiple of 5)
3. LCM $=\frac{1420}{100}=14.2$

Trick: If the bigger number is a multiple of the smaller ($14.2=284\times 0.05$), the bigger number itself is the LCM.

Answer: (A) 14.2

Concept: Mean proportional of $a$ and $b$ is $x$ where $a:x=x:b$, so $x=\sqrt{ab}$.

1. $x=\sqrt{400\times 4}$
2. $=\sqrt{1600}=40$

Trick: $\sqrt{400}\times \sqrt{4}=20\times 2=40$.

Answer: (C) 40

Concept: Take CP $=100$ units and apply mark-up and discount one after another.

1. MP $=100\times 1.44=144$ units
2. SP $=144\times 0.75=108$ units, so profit $=8$ units
3. $8\text{ units}=126\Rightarrow 1\text{ unit}=15.75$
4. MP $=144\times 15.75=2268$

Trick: Net profit $\%=44-25-\frac{44\times 25}{100}=8\%$; MP $=\frac{144}{8}\times 126=18\times 126=2268$.

Answer: (D) 2,268

Concept: Use the angle-sum of a triangle, alternate angles between parallel lines, and equal angles opposite equal sides.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{36}^{\circ }-{94}^{\circ }={50}^{\circ }$
2. $BC\parallel AD$ with transversal $AC$: $\angle CAD=\angle BCA={50}^{\circ }$ (alternate angles)
3. $AC=CD\Rightarrow \angle CDA=\angle CAD={50}^{\circ }$
4. In $△ACD$: $\angle ACD={180}^{\circ }-{50}^{\circ }-{50}^{\circ }={80}^{\circ }$

Trick: $\angle ACD={180}^{\circ }-2\times \angle BCA$ when $AC=CD$ and $BC\parallel AD$.

Answer: (C) ${80}^{\circ }$

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

1. Distance $=84\times 20=1680\text{km}$
2. Required speed $=\frac{1680}{10}=168\text{km/hr}$

Trick: Time halved ($20\to 10$) ⇒ speed doubled: $84\times 2=168$.

Answer: (D) $168\text{km/hr}$

Concept: Profit is shared in the ratio of investment.

1. Sum of ratio terms $=19+88+18=125$.
2. One part $=\frac{2200}{125}=17.6$.
3. Difference of Y and Z $=(88-18)\times 17.6=70\times 17.6=1232$.

Trick: Work with the ratio difference only: $70$ parts $\times 17.6$.

Answer: (D) ₹1,232

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; for dates count the days (exclude the deposit day, include the withdrawal day). 2024 is a leap year (366 days).

1. Days: February $29-9=20$, March $31$, April $10$ → total $61$ days.
2. $T=\frac{61}{366}=\frac{1}{6}$ year.
3. $\text{SI}=\frac{4000\times 7.5\times 1}{100\times 6}=\frac{300}{6}=50$.

Trick: $61$ days of a leap year is exactly $\frac{1}{6}$ year, so SI $=\frac{1}{6}$ of yearly interest ₹300.

Answer: (D) ₹50

Concept: Mode is the value that occurs most often in the data.

1. Count each value: 18 occurs 4 times.
2. 10, 15, 13, 20, 17, 19 each occur only once.
3. So the most frequent value is 18.

Trick: Sort the data: 10, 13, 15, 17, 18, 18, 18, 18, 19, 20 – the repeated block is the mode.

Answer: (A) 18

Concept: If $a^2+b^2=c^2$, the triangle is right-angled and area $=\frac{1}{2}\times a\times b$.

1. ${14}^2+{48}^2=196+2304=2500={50}^2$, so it is a right triangle.
2. Legs are $14$ and $48$ (hypotenuse $50$).
3. Area $=\frac{1}{2}\times 14\times 48=336{\text{cm}}^2$.

Trick: $14,48,50=2\times (7,24,25)$ – a known Pythagorean triplet.

Answer: (C) $336$

Concept: Use divisibility rules to eliminate, then divide directly.

1. $444348269$ is odd, so it is not divisible by $12$ or $10$.
2. Digit sum $=4+4+4+3+4+8+2+6+9=44$, not a multiple of $9$.
3. $444348269÷19=23386751$ exactly, since $19\times 23386751=444348269$.

Trick: Eliminate by last digit and digit sum first – only one option remains to test.

Answer: (D) $19$

Concept: Profit ratio = capital × time.

1. Let N's share $=x$; M's $=x+552$; $2x+552=1656\Rightarrow x=552$, M's $=1104$. Ratio $=2:1$.
2. Let N's capital $=n$, M's $=n+64000$.
3. $\frac{2(n+64000)}{3n}=\frac{2}{1}\Rightarrow 2n+128000=6n\Rightarrow n=32000$.
4. M's capital $=32000+64000=96000$.

Trick: Check option: M ₹96,000 for 2 months $=192000$; N ₹32,000 for 3 months $=96000$ → $2:1$ ✓.

Answer: (B) ₹96,000

Concept: Eliminate even divisors, then test the remaining options.

1. $889897$ is odd, so $52$ and $64$ are out.
2. Digit sum $=49$, not divisible by $3$, so $69=3\times 23$ is out.
3. $889897÷59=15083$ exactly ($59\times 15083=889897$).

Trick: Odd number → no even divisor; digit sum not a multiple of 3 → no multiple of 3.

Answer: (D) $59$

Concept: Work done in one day = $\frac{1}{\text{days}}$; add rates when working together.

1. One-day work $=\frac{1}{16}+\frac{1}{18}=\frac{9+8}{144}=\frac{17}{144}$.
2. In 4 days $=\frac{4\times 17}{144}=\frac{17}{36}$.
3. Work left $=1-\frac{17}{36}=\frac{19}{36}$.

Trick: LCM method: total work $=144$ units, A $=9$, B $=8$ per day; 4 days $=68$; left $=76$ → $\frac{76}{144}=\frac{19}{36}$.

Answer: (B) $\frac{19}{36}$

Concept: Literate females = total literate − literate males.

1. Total literate $=36\%$ of $350000=126000$.
2. Literate males $=22\%$ of $175000=38500$.
3. Females $=350000-175000=175000$; literate females $=126000-38500=87500$.
4. Percentage $=\frac{87500}{175000}\times 100=50\%$.

Trick: Males and females are equal, so overall $36\%$ is their average: $\frac{22+x}{2}=36\Rightarrow x=50$.

Answer: (A) $50\%$

Concept: Subtract one-day work rates to isolate each person.

1. Smitha's rate $=\frac{1}{10}-\frac{1}{25}=\frac{5-2}{50}=\frac{3}{50}$.
2. Hema + Smitha $=\frac{1}{30}+\frac{3}{50}=\frac{5+9}{150}=\frac{14}{150}=\frac{7}{75}$.
3. Days $=\frac{75}{7}$.

Trick: LCM of 30, 25, 10 $=150$ units: H $=5$, H+R $=6$, H+R+S $=15$ → S $=9$; H+S $=14$ → $\frac{150}{14}=\frac{75}{7}$.

Answer: (A) $\frac{75}{7}$ days