RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Write every term with the same base; then $a^m\times a^n=a^{m+n}$ and ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$.

1. ${\left(\frac{7}{26}\right)}^{11}={\left(\frac{26}{7}\right)}^{-11}$
2. LHS $={\left(\frac{26}{7}\right)}^{5-11+19}={\left(\frac{26}{7}\right)}^{13}={\left(\frac{7}{26}\right)}^{-13}$
3. So $9m+17=-13\Rightarrow 9m=-30\Rightarrow m=-\frac{10}{3}$

Trick: Flip the base and the sign of the power flips: just add/subtract the powers of $\frac{26}{7}$ and change sign.

Answer: (A) $-\frac{10}{3}$

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $a=23,b=-11,c=-12$: $a+b+c=23-11-12=0$
2. Value $=3\times 23\times (-11)\times (-12)$
3. $=3\times 23\times 132=9108$

Trick: Always check the sum first — a zero sum turns a cube-sum into $3abc$.

Answer: (C) 9108

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

1. $R=\frac{16562\times 100}{9800\times 13}$
2. $=\frac{1656200}{127400}=13\%$

Trick: Divide SI by P first: $\frac{16562}{9800}=1.69$ total for 13 years, so $\frac{169}{13}=13\%$ per year.

Answer: (D) $13\%$

Concept: $\tan \theta +\cot \theta =\frac{1}{\sin \theta \cos \theta }$ and $\text{cosec}2\theta =\frac{1}{2\sin \theta \cos \theta }$.

1. $b=\tan \theta +\cot \theta =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }$
2. $\text{cosec}2\theta =\frac{1}{2\sin \theta \cos \theta }=\frac{b}{2}$
3. (Check: $a^2=b^2+2b$, so $\frac{a^2-b^2}{2}=b$, not $\frac{b}{2}$.)

Trick: Whenever $\tan \theta +\cot \theta$ appears, think $\frac{2}{\sin 2\theta }$, i.e. $2\text{cosec}2\theta$.

Answer: (D) $\frac{b}{2}$

Concept: If sides and diagonal satisfy Pythagoras, the angle between the sides is ${90}^{\circ }$ — the parallelogram is a rectangle.

1. $9^2+{12}^2=81+144=225={15}^2$
2. So the triangle on the diagonal is right-angled; the parallelogram is a rectangle
3. Area $=12\times 9=108{\text{cm}}^2$

Trick: 9-12-15 is the 3-4-5 triple $\times 3$ — spot it instantly.

Answer: (B) 108

Concept: Fourth proportional $y$ of $a,b,c$: $a:b=c:y\Rightarrow y=\frac{bc}{a}$.

1. $y=\frac{(x+1)(2x+3)}{x}=3x+3=3(x+1)$
2. Cancel $(x+1)$: $\frac{2x+3}{x}=3$
3. $2x+3=3x\Rightarrow x=3$

Trick: Option check: $x=3$ gives $3:4=9:12$ and $y=12=3\times 3+3$. ✔

Answer: (B) 3

Concept: Successive discounts multiply: $SP=MP\times (1-d_1)(1-d_2)$.

1. $SP=5000\times \frac{80}{100}\times \frac{70}{100}$
2. $=5000\times 0.56=2800$

Trick: Net discount of 20% and 30% $=20+30-\frac{20\times 30}{100}=44\%$; $56\%$ of ₹5,000 = ₹2,800.

Answer: (B) ₹2,800

Concept: $a^3-b^3=(a-b)(a^2+ab+b^2)$, so $\frac{a^3-b^3}{a^2+ab+b^2}=a-b$.

1. Here $a=1.4,b=1.2$
2. Value $=a-b=1.4-1.2=0.2$

Trick: Numerator cubes with minus, denominator $a^2+ab+b^2$ → answer is simply $a-b$.

Answer: (B) 0.2

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

1. $ab=0.15\times 0.000135=0.00002025$
2. $\sqrt{0.00002025}=0.0045$ (since ${45}^2=2025$)

Trick: Count decimals: 8 decimal places under the root → 4 decimal places in the answer.

Answer: (C) 0.0045

Concept: Form equations from present ages; each person gains the same years.

1. Let Priyal $=x$, brother $=x+5$
2. After 4 years: $(x+4)+(x+9)=39$
3. $2x+13=39\Rightarrow x=13$

Trick: Present sum $=39-8=31$; with difference 5, smaller age $=\frac{31-5}{2}=13$.

Answer: (B) 13 years

Concept: Solve brackets innermost first: ( ) → { } → [ ], then division.

1. $15-9=6$, $7\times 6=42$
2. $\{42-28\}=14$
3. $[12+14]=26$
4. $156÷26=6$

Trick: Inside-out order of brackets avoids sign mistakes.

Answer: (B) 6

Concept: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.

1. Time $=\frac{12}{6}=2$ hours

Trick: Every hour he covers 6 km, so 12 km needs 2 such hours.

Answer: (B) 2 hours

Concept: $SP=MP\times (1-d_1)(1-d_2)$.

1. After 25%: $2000\times 0.75=1500$
2. After 10%: $1500\times 0.90=1350$

Trick: Net discount $=25+10-\frac{25\times 10}{100}=32.5\%$; $67.5\%$ of 2000 = 1350.

Answer: (A) ₹1,350

Concept: Exterior angle = sum of the two interior opposite angles; in a triangle the larger angle faces the larger side.

1. $\angle PRT=P+Q$, $\angle RQS=P+R$
2. Sum: $2P+Q+R={180}^{\circ }+P={220}^{\circ }\Rightarrow P={40}^{\circ }$, $Q+R={140}^{\circ }$
3. $\angle PRT>\angle RQS\Rightarrow Q>R$, so $Q>{70}^{\circ }$: $Q$ is the largest angle → $PR$ is the largest side
4. Only option (D) has $PR$ largest; it holds with $P<R<Q$, giving $QR<PQ<PR$

Trick: Side opposite an angle carries the other two letters: angle $Q$ ↔ side $PR$.

Answer: (D) $QR<PQ<PR$

Concept: Median $=l+\frac{\frac{N}{2}-cf}{f}\times h$.

1. $N=100$, $\frac{N}{2}=50$; cumulative frequencies 20, 40, 60, … → median class 58–60
2. $l=58,cf=40,f=20,h=2$
3. Median $=58+\frac{50-40}{20}\times 2=58+1=59$

Trick: Equal frequencies in equal classes → median is simply the middle of the whole range: $\frac{54+64}{2}=59$.

Answer: (C) 59

Concept: Net work per hour = sum of filling rates − emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{15}-\frac{1}{15}=\frac{1}{4}$ per hour
2. Time for $\frac{1}{4}$ cistern $=\frac{1}{4}÷\frac{1}{4}=1$ hour

Trick: Second and third taps have equal rates and cancel — only the 4-hour tap matters.

Answer: (A) 1

Concept: Subtract equations having the same pen term to eliminate it.

1. $4n+6p=228$ and $7n+6p=333$
2. Subtract: $3n=105\Rightarrow n=35$

Trick: Same number of pens in both → extra 3 notebooks cost $333-228=105$.

Answer: (A) ₹ 35

Concept: Total string length = circumference; it is shared by the rectangle and square perimeters.

1. Length $=2\pi r=2\times \frac{22}{7}\times 21=132\text{cm}$
2. Rectangle: $l=2b\Rightarrow$ perimeter $=2(2b+b)=6b$; square perimeter $=s$
3. $6b+s=132$ and $6b-s=24\Rightarrow 12b=156\Rightarrow b=13\text{cm}$

Trick: Sum and difference given → bigger part $=\frac{132+24}{2}=78=6b$, so $b=13$.

Answer: (B) 13 cm

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

1. $\sqrt{0.16\times 0.49}=\sqrt{0.16}\times \sqrt{0.49}$
2. $=0.4\times 0.7=0.28$

Trick: Both are perfect squares — take roots separately and multiply.

Answer: (C) 0.28

Concept: Effective SP/CP ratio $=(1+\text{profit})\times \frac{\text{true weight}}{\text{false weight}}$.

1. He charges for 1000 g at $115\%$ but gives only 950 g
2. Ratio $=1.15\times \frac{1000}{950}=\frac{1150}{950}=1.2105$
3. Profit $\approx 21.05\%$

Trick: Use $\frac{1000\times (100+15)}{950}-100=21.05\%$ directly.

Answer: (C) $21.05\%$