RRB NTPC Undergraduate CBT 2 2025 PYQs

Concept: $\text{SP}=\text{CP}\times (1+\frac{\text{Profit}\%}{100})$; find the SP of one item, then see how many fit into ₹400.

1. CP of 1 item $=\frac{400}{20}=$ ₹20.
2. SP of 1 item at $25\%$ profit $=20\times \frac{125}{100}=$ ₹25.
3. Number of items sold for ₹400 $=\frac{400}{25}=16$.

Trick: Same money both sides → $\text{Items sold}=\text{Items bought}\times \frac{100}{100+P}=20\times \frac{100}{125}=16$.

Answer: (B) 16

Concept: When a transversal cuts two parallel lines, corresponding angles are equal, and angles on a straight line add up to ${180}^{\circ }$.

1. At M, $\angle XMA$ and $\angle AMN$ form a linear pair on line XY: $\angle AMN={180}^{\circ }-{125}^{\circ }={55}^{\circ }$.
2. $\angle AMN$ (at M, between MA and MN) and $\angle CNY$ (at N, between NC and NY) are corresponding angles for $AB\parallel CD$.
3. So $\angle CNY=\angle AMN={55}^{\circ }$.

Trick: With parallel lines every angle at the two crossings is either $x$ or ${180}^{\circ }-x$; here $x={125}^{\circ }$, so the answer must be ${55}^{\circ }$.

Answer: (C) ${55}^{\circ }$

Concept: LCM = product of the highest powers of all prime factors present in the numbers.

1. $3=3$, $77=7\times 11$, $33=3\times 11$.
2. Primes involved: 3, 7, 11 (each to power 1).
3. LCM $=3\times 7\times 11=231$.

Trick: 3 already divides 33, so LCM(3, 77, 33) = LCM(77, 33) $=\frac{77\times 33}{11}=231$.

Answer: (B) 231

Concept: Amount = Principal + Simple Interest, so $\text{SI}=\text{Amount}-\text{Principal}$ – the rate is not needed.

1. Principal = ₹620, Amount after 2 years = ₹910.
2. SI $=910-620=$ ₹290.
3. (If asked, rate $=\frac{290\times 100}{620\times 2}\approx 23.4\%$ p.a.)

Trick: Do not get distracted by "certain rate" – interest for the whole period is just $A-P$.

Answer: (A) 290

Concept: Express all shares in terms of the smallest share, add the parts and divide the total.

1. Let C $=x$. Then B $=3x$ and A $=2\times 3x=6x$.
2. Total $=6x+3x+x=10x=720\Rightarrow x=72$.
3. So C gets ₹72 (B = ₹216, A = ₹432).

Trick: A : B = 2 : 1 and B : C = 3 : 1 → A : B : C = 6 : 3 : 1; C's share $=\frac{1}{10}\times 720=72$.

Answer: (C) ₹72

Concept: Dividend $=\text{Divisor}\times \text{Quotient}+\text{Remainder}$.

1. Remainder $=12$; divisor $=8\times 12=96$.
2. Divisor $=3\times$ quotient $\Rightarrow$ quotient $=\frac{96}{3}=32$.
3. Dividend $=96\times 32+12=3072+12=3084$.

Trick: $96\times 32=96\times 30+96\times 2=2880+192=3072$; add the remainder 12 → 3084 (ends in 4, so options A/C/D are out).

Answer: (B) 3084

Concept: Variance ${\sigma }^2=\frac{\sum x_i^2}{n}-{\overline{x}}^2$ (population variance, $n=6$).

1. $\sum x=4+8+6+5+3+9=35$, so $\overline{x}=\frac{35}{6}$.
2. $\sum x^2=16+64+36+25+9+81=231$.
3. ${\sigma }^2=\frac{231}{6}-{\left(\frac{35}{6}\right)}^2=\frac{231\times 6-1225}{36}=\frac{1386-1225}{36}=\frac{161}{36}$.

Trick: Keep the mean as a fraction ($\frac{35}{6}$) instead of a decimal; the denominator 36 appears directly, which eliminates options B and D.

Answer: (C) $\frac{161}{36}$

Concept: If gain is $40\%$ on CP, then $\text{SP}=\frac{140}{100}\times \text{CP}$, so $\text{Profit}=\frac{40}{140}\times \text{SP}=\frac{2}{7}\times \text{SP}$.

1. Sale proceeds (SP) = ₹3,955.
2. CP $=3955\times \frac{100}{140}=\frac{3955}{1.4}=$ ₹2,825.
3. Profit $=3955-2825=$ ₹1,130.

Trick: Profit $=\frac{2}{7}\times 3955=2\times 565=1130$ (since $3955÷7=565$).

Answer: (A) 1,130

Concept: Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$; the square's area is twice this.

1. Triangle area $=\frac{1}{2}\times 23\times 12=138{\text{cm}}^2$.
2. Square area $=2\times 138=276{\text{cm}}^2$.
3. (Its side would be $\sqrt{276}\approx 16.6$ cm, but only the area is asked.)

Trick: Twice the triangle's area is simply $\text{base}\times \text{height}=23\times 12=276$ – no halving needed.

Answer: (D) $276{\text{cm}}^2$

Concept: Divide the whole equation by $\sin y$ to get $\cot y$, then invert and rationalise.

1. $5\sin y+\cos y=\sqrt{13}\sin y\Rightarrow \cos y=(\sqrt{13}-5)\sin y$.
2. $\tan y=\frac{\sin y}{\cos y}=\frac{1}{\sqrt{13}-5}$.
3. Rationalise: $\frac{1}{\sqrt{13}-5}\times \frac{\sqrt{13}+5}{\sqrt{13}+5}=\frac{\sqrt{13}+5}{13-25}=\frac{\sqrt{13}+5}{-12}=\frac{-5-\sqrt{13}}{12}$.

Trick: Since $\sqrt{13}<5$, $\cos y$ and $\sin y$ have opposite signs, so $\tan y$ must be negative with denominator $25-13=12$ – only option C matches.

Answer: (C) $\frac{-5-\sqrt{13}}{12}$

Concept: Fourth proportional to $a,b,c$ is $x$ with $a:b=c:x$, so $x=\frac{b\times c}{a}$. With powers of the same base, add/subtract exponents.

1. $x=\frac{2^{1.5}\times 2^{2.5}}{2^{0.5}}$
2. $=2^{1.5+2.5-0.5}=2^{3.5}$

Trick: Fourth proportional = (2nd × 3rd) ÷ 1st. For same base: multiply → add powers, divide → subtract powers.

Answer: (D) $2^{3.5}$

Concept: Apply BODMAS: division and multiplication (left to right) before addition.

1. $(-56)÷(-3)=\frac{56}{3}$, then $\frac{56}{3}\times 3=56$
2. $(-37)+56=19$

Trick: Negative ÷ negative = positive. The $÷3$ and $\times 3$ cancel, leaving $-37+56$.

Answer: (C) 19

Concept: Required percentage $=\frac{\text{part}}{\text{whole}}\times 100$; the number itself cancels out.

1. $\frac{7\%\text{ of }N}{9\%\text{ of }N}\times 100=\frac{7}{9}\times 100$
2. $=77.777\dots \%\approx 77.78\%$

Trick: Such questions never depend on the number — just divide the two percentages: $\frac{7}{9}=0.7778$.

Answer: (D) 77.78%

Concept: Write every base as a product of primes; the total number of prime factors is the sum of all exponents (counted with repetition).

1. $6^7=(2\times 3{)}^7=2^7\times 3^7$
2. $N=5^8\times 2^7\times 3^7\times {11}^4$
3. Total prime factors $=8+7+7+4=26$

Trick: Only 6 is composite here — split it first. (Distinct primes would be 4; total prime factors counts repetition.)

Answer: (A) 26

Concept: Convert each percentage relation into a ratio, then combine by making the common term (B) equal.

1. $25\%A=40\%B\Rightarrow \frac{A}{B}=\frac{40}{25}=\frac{8}{5}$
2. $30\%B=45\%C\Rightarrow \frac{B}{C}=\frac{45}{30}=\frac{3}{2}$
3. Make B equal: $A:B=24:15$, $B:C=15:10$ → $A:B:C=24:15:10$

Trick: $pA=qB$ gives $A:B=q:p$ (the numbers swap sides).

Answer: (D) 24:15:10

Concept: Sum $=$ average $\times$ number of terms.

1. Sum of three numbers $=15\times 3=45$
2. Third number $=45-(10+20)=15$

Trick: If two numbers average to the given mean (here $\frac{10+20}{2}=15$), the third must equal the mean itself.

Answer: (B) 15

Concept: Floor + ceiling $=2lb$; four walls $=2h(l+b)$. Volume $=lbh$, and $1{\text{m}}^3=1000$ litres.

1. $2\times 15\times 12=2h(15+12)\Rightarrow 360=54h\Rightarrow h=\frac{20}{3}\text{m}$
2. Volume $=15\times 12\times \frac{20}{3}=1200{\text{m}}^3$
3. $=1200\times 1000=1200\times {10}^3$ litres

Trick: Floor + ceiling = walls ⇒ $h=\frac{lb}{l+b}$. Then always convert m³ → litres by $\times 1000$.

Answer: (D) $1200\times {10}^3$ liters

Concept: Work per day $=\frac{1}{\text{days}}$. Efficiency of B $=1.5\times$ efficiency of A.

1. A's one-day work $=\frac{1}{37}-\frac{1}{74}=\frac{1}{74}$
2. B's one-day work $=\frac{3}{2}\times \frac{1}{74}=\frac{3}{148}$
3. Work done in 28 days $=\frac{28}{37}$, remaining $=\frac{9}{37}$
4. B alone: $\frac{9}{37}÷\frac{3}{148}=\frac{9}{37}\times \frac{148}{3}=12$ days

Trick: Take total work $=148$ units: group does 4/day, A does 2/day, B does 3/day. Remaining $148-28\times 4=36$ units $\Rightarrow 36÷3=12$ days.

Answer: (C) 12 days

Concept: Division and multiplication first (left to right), then subtraction.

1. $12÷12=1$, then $1\times 8=8$
2. $11-8=3$

Trick: Never subtract first: $11-12$ is the classic trap. Resolve the $÷$ and $\times$ chain before touching $-$.

Answer: (A) 3

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (never the average of speeds).

1. Times: $\frac{30}{20}=1.5$ h, $\frac{50}{25}=2$ h, $\frac{20}{10}=2$ h → total $5.5$ h
2. Average speed $=\frac{100}{5.5}=18.18\approx 18\text{km/h}$

Trick: Different distances at different speeds → compute each time separately; the slow 10 km/h stretch pulls the average well below 20.

Answer: (A) 18 km/h