RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Test only odd numbers not ending in 5, using primes up to $\sqrt{491}\approx 22$ (2, 3, 5, 7, 11, 13, 17, 19).

1. Odd candidates: 477, 479, 481, 483, 487, 489, 491.
2. $477=9\times 53$, $481=13\times 37$, $483=3\times 161$, $489=3\times 163$ → composite.
3. 479, 487 and 491 have no factor up to 22 → prime.

Trick: Strike out multiples of 3 by digit sum first, then test 7, 11, 13 only for the rest.

Answer: (D) 3

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

1. Distance $=75\times 17=1275\text{km}$.
2. Required speed $=\frac{1275}{5}=255\text{km/h}$.

Trick: $S_2=75\times \frac{17}{5}=15\times 17=255$.

Answer: (A) $255\text{km/h}$

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

1. ${48}^{-17}÷{48}^{12}={48}^{-17-12}={48}^{-29}$.
2. ${48}^{-29}\times {48}^{-13}={48}^{-29-13}={48}^{-42}$.

Trick: Add powers being multiplied, subtract powers being divided: $-17-12-13=-42$.

Answer: (D) ${48}^{-42}$

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

1. Ratio $=1090:1170:1340=109:117:134$; total $=360$ parts.
2. 1 part $=\frac{1620}{360}=4.5$.
3. Rohan's share $=134\times 4.5=603$ rupees.

Trick: Profit is 45% of total capital (₹1,620 of ₹3,600), so Rohan gets $0.45\times 1340=603$.

Answer: (A) ₹603

Concept: Divide by each option (or use factor rules); only an exact quotient works.

1. 34 = 2 × 17 needs an even number – 587407 is odd → no.
2. $587407÷41=14327$ exactly ($41\times 14327=587407$).
3. $587407÷43$ leaves remainder 17; $587407÷47$ leaves remainder 33.

Trick: Eliminate even divisors instantly for an odd number; then test the rest.

Answer: (C) 41

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

1. 15% and 20% → $15+20-\frac{15\times 20}{100}=32\%$, so SP $=68\%$ of MP.
2. Single 40% → SP $=60\%$ of MP.
3. Difference $=8\%$ of MP $=400$ → MP $=\frac{400\times 100}{8}=5000$.

Trick: Take MP = 100: SPs are 60 and 68 → 8 units = ₹400.

Answer: (C) ₹5,000

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

1. $r=3\text{cm}$, $h=7\text{cm}$, $\pi =3.14$.
2. $V=3.14\times 3^2\times 7=3.14\times 63$.
3. $V=197.82{\text{cm}}^3$.

Trick: $\pi \times 63$: $63\times 3=189$, $63\times 0.14=8.82$ → $197.82$.

Answer: (D) $197.82{\text{cm}}^3$

Concept: For the same money (₹1), the number of articles sold is inversely proportional to the SP per article: $n_1(100-L)=n_2(100+G)$.

1. $6\times (100-30)=n\times (100+40)$.
2. $6\times 70=140n$ → $n=\frac{420}{140}=3$.

Trick: Loss 30% → 70; gain 40% → 140 = double, so articles become half: $\frac{6}{2}=3$.

Answer: (D) 3

Concept: Mode is the value that occurs most often.

1. Frequencies: 10→1, 11→3, 14→2, 17→1, 18→1, 19→1, 20→1.
2. 11 occurs 3 times – the highest frequency.

Trick: Sort first: 10, 11, 11, 11, 14, 14, … – the longest run is the mode.

Answer: (C) 11

Concept: Successive increases multiply: final $=x\times (1+\frac{4}{100})(1+\frac{25}{100})$.

1. $x\times 1.04\times 1.25=1.3x$.
2. $1.3x=140400$ → $x=\frac{140400}{1.3}=108000$.
3. Sales two years ago = ₹1,08,000.

Trick: $4\%$ then $25\%$ → net $4+25+\frac{4\times 25}{100}=30\%$ rise.

Answer: (D) ₹1,08,000

Concept: Write each ratio as a decimal (or fraction) and compare.

1. $46:75=\frac{46}{75}\approx 0.613$
2. $34:77\approx 0.442$, $39:89\approx 0.438$
3. $36:69\approx 0.522$
4. Largest value is $0.613$, i.e. $46:75$.

Trick: Only $46:75$ has the first term more than half of the second ($46>37.5$); compare others with $\frac{1}{2}$ to eliminate fast.

Answer: (A) $46:75$

Concept: Use ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $3{\sin }^2\theta +7{\cos }^2\theta =3({\sin }^2\theta +{\cos }^2\theta )+4{\cos }^2\theta =3+4{\cos }^2\theta$
2. $3+4{\cos }^2\theta =4\Rightarrow {\cos }^2\theta =\frac{1}{4}$
3. $\cos \theta =\frac{1}{2}$ (acute angle) $\Rightarrow \theta ={60}^{\circ }$

Trick: Split the bigger coefficient: $7=3+4$, so the identity removes ${\sin }^2\theta$ at once.

Answer: (B) ${60}^{\circ }$

Concept: If SP of $x$ articles = CP of $y$ articles, then $\text{Gain}\%=\frac{y-x}{x}\times 100$.

1. SP of 20 books = CP of 24 books, so SP is more than CP: it is a gain.
2. $\text{Gain}\%=\frac{24-20}{20}\times 100$
3. $=\frac{4}{20}\times 100=20\%$

Trick: Divide the difference by the number of articles sold (20), not by 24 – dividing by 4 or 24 gives the trap options.

Answer: (B) $20\%$ gain

Concept: Add one-day works: $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{\text{together}}$.

1. A + B in one day: $\frac{1}{22}+\frac{1}{11}=\frac{3}{22}$
2. A + B + C in one day: $\frac{1}{4}$
3. C in one day: $\frac{1}{4}-\frac{3}{22}=\frac{11-6}{44}=\frac{5}{44}$
4. C alone: $\frac{44}{5}=8\frac{4}{5}$ days

Trick: LCM method: total work $=44$ units; A $=2$, B $=4$, all $=11$ units/day, so C $=5$ units/day $\Rightarrow \frac{44}{5}$ days.

Answer: (C) $8\frac{4}{5}$ days

Concept: The largest circle inside a rectangle has diameter equal to the shorter side.

1. Diameter $d=168\text{m}$ (the shorter side)
2. Circumference $=\pi d=\frac{22}{7}\times 168$
3. $=22\times 24=528\text{m}$

Trick: The circle can never be wider than the smaller side – the 249 m side is extra data.

Answer: (B) $528$

Concept: Angle sum of a triangle is ${180}^{\circ }$; an exterior angle equals the sum of the two opposite interior angles.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{76}^{\circ }={14}^{\circ }$
2. In $△AEC$: $\angle AEC={180}^{\circ }-{74}^{\circ }-{14}^{\circ }={92}^{\circ }$
3. B, E, C are collinear: $\angle AEB={180}^{\circ }-{92}^{\circ }={88}^{\circ }$

Trick: Directly: $\angle AEB$ is exterior to $△AEC$, so $\angle AEB={74}^{\circ }+{14}^{\circ }={88}^{\circ }$.

Answer: (A) ${88}^{\circ }$

Concept: Form two linear equations from the two statements and solve.

1. $4P-Q=41\Rightarrow Q=4P-41$
2. After 10 years: $3(Q+10)=7(P+10)-33\Rightarrow 3Q=7P+7$
3. $3(4P-41)=7P+7\Rightarrow 5P=130\Rightarrow P=26$
4. $Q=4\times 26-41=63$

Trick: Check options: $Q=63\Rightarrow P=26$; then $3\times 73=219$ and $7\times 36-33=219$ ✔.

Answer: (B) $63$

Concept: In 'buy 4 get 2 free', the customer pays only for 4 out of every 6 items.

1. Six shirts = 4 paid + 2 free
2. Amount $=4\times 1250=5000$, i.e. ₹5,000

Trick: Effective discount $=\frac{2}{6}=33\frac{1}{3}\%$; $\frac{2}{3}$ of ₹7,500 = ₹5,000.

Answer: (B) ₹5,000

Concept: Find each pipe's full-tank time, then use filled part = rate × time.

1. P fills $\frac{7}{9}$ in 14 h $\Rightarrow$ full tank in $14\times \frac{9}{7}=18$ h; Q fills $\frac{5}{9}$ in 20 h $\Rightarrow$ 36 h
2. In 2 h, P + Q fill $2(\frac{1}{18}+\frac{1}{36})=\frac{1}{6}$ tank
3. R empties $\frac{1}{6}$ tank in 6 h $\Rightarrow$ full tank in 36 h
4. Together: $\frac{1}{18}+\frac{1}{36}-\frac{1}{36}=\frac{1}{18}\Rightarrow 18$ h

Trick: Q's filling rate and R's emptying rate are equal (36 h each), so they cancel – only P works: 18 h.

Answer: (B) 18 hours

Concept: After a $70\%$ increase, the number becomes $170\%$ of itself.

1. $1.7x=3230$
2. $x=\frac{3230}{1.7}=1900$

Trick: $3230÷17=190$, so $x=190\times 10=1900$.

Answer: (A) $1900$