RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Multiply the numerator first, then divide.

1. $1.5\times 0.2=0.30$.
2. $\frac{0.3}{0.3}=1$.

Trick: $1.5\times 0.2$: $15\times 2=30$ with two decimal places → $0.30$.

Answer: (B) 1

Concept: If CP of $a$ articles $=$ SP of $b$ articles, profit$\%=\frac{a-b}{b}\times 100$.

1. Let CP of each article be ₹1: $25=x\times \text{SP}$.
2. Profit $25\%\Rightarrow \text{SP}=1.25$.
3. $x=\frac{25}{1.25}=20$.

Trick: $\frac{25-x}{x}=\frac{1}{4}\Rightarrow 100-4x=x\Rightarrow x=20$.

Answer: (C) 20

Concept: Mode is the value that occurs most often.

1. Frequencies: 40 → 2, 41 → 1, 42 → 2, 43 → 2, 45 → 2, 46 → 2, 51 → 3, others → 1.
2. 51 appears 3 times, the maximum.

Trick: Tally only the values that repeat; one value with 3 occurrences stands out quickly.

Answer: (B) 51

Concept: Solve the innermost bracket first, then multiply, then add.

1. $(26-11+6)=21$.
2. $-9\times 21=-189$.
3. $30+(-189)=-159$.

Trick: Left-to-right inside the bracket: $26-11=15$, then $+6=21$ – do not subtract 17 from 26.

Answer: (C) -159

Concept: Leak's rate $=$ filling rate of A $-$ net filling rate.

1. A fills $\frac{1}{10}$ per hour; with leak net is $\frac{1}{12}$ per hour.
2. Leak empties $\frac{1}{10}-\frac{1}{12}=\frac{6-5}{60}=\frac{1}{60}$ per hour.
3. Leak alone empties the tank in 60 hours.

Trick: Leak time $=\frac{10\times 12}{12-10}=60$.

Answer: (A) 60 hours

Concept: Mean proportional $m$ between $a$ and $b$: $m^2=ab$.

1. $ab=900$ and $a-b=45$.
2. $(a+b{)}^2=(a-b{)}^2+4ab=2025+3600=5625\Rightarrow a+b=75$.
3. $b=\frac{75-45}{2}=15$ (and $a=60$; $60\times 15=900$ ✓).

Trick: Factor pairs of 900 with difference 45: $60\times 15$ – spot it directly.

Answer: (B) 15

Concept: $A=P{(1+\frac{r}{100})}^n$.

1. $\frac{74263}{70000}=1.0609$.
2. ${1.03}^2=1.0609$, so $n=2$.

Trick: ${1.03}^2=1.0609$, ${1.03}^3\approx 1.0927$ – the amount barely exceeds $6\%$, so 2 years.

Answer: (B) 2

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

1. $SI=\frac{25200\times 6\times 4}{100}$.
2. $=252\times 24=6048$.

Trick: $6\%\times 4$ years $=24\%$ of 25200 $=\frac{24}{100}\times 25200=6048$.

Answer: (C) ₹6,048

Concept: Take total work $=$ LCM of days; work per day $=$ total $÷$ days.

1. Total work $=36$ units; A $=3$ units/day, B $=1$ unit/day.
2. In 4 days together: $4\times (3+1)=16$ units done; $20$ remain.
3. B alone: $\frac{20}{1}=20$ days.

Trick: Fraction left $=1-4(\frac{1}{12}+\frac{1}{36})=1-\frac{16}{36}=\frac{20}{36}$ → B takes $\frac{20}{36}\times 36=20$ days.

Answer: (A) 20

Concept: For numbers $3x,2x,13x$ with co-prime ratio terms, HCF $=x$ and LCM $=3\times 2\times 13\times x=78x$.

1. $78x=6318\Rightarrow x=\frac{6318}{78}=81$.
2. HCF $=x=81$.

Trick: $78\times 80=6240$, $6318-6240=78$ → one more, so $x=81$.

Answer: (D) 81

Concept: Weighted average $=\frac{\sum (\text{score}\times \text{weight})}{\sum \text{weights}}$.

1. $75\times 2+80\times 3+90\times 5=150+240+450=840$
2. Total weight $=2+3+5=10$
3. Weighted average $=\frac{840}{10}=84$

Trick: Take 80 as base: deviations $-5\times 2+0\times 3+10\times 5=40$; $40÷10=4$, so $80+4=84$.

Answer: (D) 84

Concept: A number leaving the same remainder $r$ on division by several numbers is of the form $k\times \text{LCM}+r$.

1. LCM$(7,6,10)=210$
2. Largest 4-digit multiple of 210: $9999÷210=47$ rem $129$, so $9999-129=9870$
3. Required number $=9870+5=9875$

Trick: Check options: $9875-5=9870=210\times 47$ – only this one works.

Answer: (B) 9875

Concept: Amount $A=P{(1+\frac{R}{100})}^T$.

1. $A=4000\times {\left(\frac{11}{10}\right)}^2=4000\times \frac{121}{100}$
2. $A=40\times 121=4840$

Trick: 10% for 2 years = effective $21\%$; $4000+840=4840$.

Answer: (D) ₹4840

Concept: Work per hour of each tap is $\frac{1}{\text{time}}$; an emptying tap is negative.

1. Net part filled per hour $=\frac{1}{4}+\frac{1}{11}-\frac{1}{11}=\frac{1}{4}$
2. The second tap and the emptying tap cancel each other exactly.
3. Full cistern in 4 h, so one-fourth in $4\times \frac{1}{4}=1$ hour.

Trick: Equal fill and empty times (11 h each) cancel – only the 4-hour tap matters.

Answer: (C) 1

Concept: Three successive discounts of $d\%$ leave ${(1-\frac{d}{100})}^3$ of the price.

1. Remaining fraction $=0.9\times 0.9\times 0.9=0.729$
2. Price paid $=72.9\%$
3. Single equivalent discount $=100-72.9=27.1\%$

Trick: Two 10% discounts $=19\%$; then $19+10-\frac{19\times 10}{100}=27.1\%$.

Answer: (C) $27.1\%$

Concept: $9^{x-1}=\frac{9^x}{9}$; take $9^x$ common.

1. $9^x-\frac{9^x}{9}=72\Rightarrow 9^x·\frac{8}{9}=72$
2. $9^x=81=9^2\Rightarrow x=2$
3. $\frac{2x-1}{2x+3}=\frac{4-1}{4+3}=\frac{3}{7}$

Trick: $a^n-a^{n-1}=a^{n-1}(a-1)$; here $9^{x-1}\times 8=72\Rightarrow 9^{x-1}=9$, so $x=2$ at once.

Answer: (D) $\frac{3}{7}$

Concept: Opposite directions: relative speed $=$ sum; same direction: difference. Distance to cross $=$ sum of lengths.

1. Total length $=200+300=500$ m; crossing in 15 s $\Rightarrow$ relative speed $=\frac{500}{15}=\frac{100}{3}$ m/s $=120$ km/h
2. Speed of Y $=120-72=48$ km/h
3. Same direction: relative speed $=72-48=24$ km/h $=\frac{20}{3}$ m/s
4. Time $=500÷\frac{20}{3}=75$ s

Trick: Relative speeds are $120$ and $24$ (ratio $5:1$), so time becomes $15\times 5=75$ s.

Answer: (A) 75 seconds

Concept: Form a linear equation from the given relation.

1. Let the younger be $x$; the other $=2x-4$
2. $x+2x-4=44\Rightarrow 3x=48\Rightarrow x=16$
3. Ages 16 and 28; difference $=12$ years

Trick: Difference $=(2x-4)-x=x-4$; with $x=16$ it is 12.

Answer: (C) 12 years

Concept: Apply each change on the latest price: $-25\%$ then $+20\%$.

1. After 25% discount: $4000\times \frac{3}{4}=3000$
2. After 20% increase: $3000\times \frac{6}{5}=3600$

Trick: Net change $=-25+20-\frac{25\times 20}{100}=-10\%$; $4000\times 0.9=3600$.

Answer: (B) ₹3,600

Concept: Order: bar (vinculum) → ( ) → { } → [ ].

1. Bar: $2.2-1.8=0.4$; so $(3.2-0.4)=2.8$
2. $\{3.3-2.8\}=0.5$
3. $[6.5-0.5]=6$
4. $10.5-6=4.5$

Trick: Solve the bar first – treating $3.2-2.2-1.8$ left to right gives a wrong value.

Answer: (B) 4.5