RRB NTPC Undergraduate CBT 1 2025 PYQs

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

1. Walking time $=\frac{40}{5}=8$ h.
2. Running time $=\frac{30}{10}=3$ h.
3. Ratio $=8:3$.

Trick: Find each time separately, then compare – never compare speeds directly.

Answer: (D) $8:3$

Concept: $\text{Saving}=\text{Income}-\text{Expenditure}$.

1. Expenditure $=10000$ ⇒ 5 parts $=10000$, so income $=8\times 2000=16000$.
2. New income $=16000\times 1.4=22400$; new expenditure $=10000\times 0.9=9000$.
3. New saving $=22400-9000=13400$.

Trick: 1 part $=\frac{10000}{5}=2000$ – convert the ratio to rupees first.

Answer: (A) ₹13,400

Concept: A regular $n$-gon splits into $n$ equal triangles: $\text{Area}=n\times \frac{1}{2}\times a\times r$ ($r$ = distance from centre to side).

1. One triangle $=\frac{1}{2}\times 11\times 9=49.5{\text{cm}}^2$.
2. $n\times 49.5=247.5$.
3. $n=\frac{247.5}{49.5}=5$.

Trick: $\text{Area}=\frac{1}{2}\times \text{Perimeter}\times r$ ⇒ perimeter $=55$, so $n=\frac{55}{11}=5$.

Answer: (B) 5

Concept: Number $=k\times \text{LCM}+3$.

1. $\text{LCM}(6,10,14)=210$.
2. $9999÷210$ gives quotient 47, so largest multiple $=47\times 210=9870$.
3. Add remainder: $9870+3=9873$.

Trick: Check: $9873-3=9870$ is divisible by 6, 10 and 14.

Answer: (A) 9873

Concept: Take the lower power common: $6^x-6^{x-1}=6^{x-1}(6-1)$.

1. $5\times 6^{x-1}=180$ ⇒ $6^{x-1}=36=6^2$.
2. So $x-1=2$, $x=3$.
3. $\frac{2x-1}{2x+3}=\frac{5}{9}$.

Trick: $a^x-a^{x-1}=a^{x-1}(a-1)$ – always factor out the smaller power.

Answer: (B) $\frac{5}{9}$

Concept: Salary $=\text{Expenditure}(E)+\text{Savings}(S)$.

1. Given $E=2(S-E)$ ⇒ $3E=2S$ ⇒ $E:S=2:3$.
2. Salary $=2+3=5$ parts.
3. $E\%=\frac{2}{5}\times 100=40\%$.

Trick: Convert the condition to a ratio, then $\frac{\text{part}}{\text{total}}\times 100$.

Answer: (A) $40\%$

Concept: Successive discounts $a\%,b\%$ = single $(a+b-\frac{ab}{100})\%$.

1. Option 1: $30\%$ of 8000 $=2400$.
2. Option 2: $20+15-\frac{300}{100}=32\%$ ⇒ $32\%$ of 8000 $=2560$.
3. Difference $=2560-2400=160$, Option 2 is better.

Trick: Difference $=(32-30)\%$ of $8000=2\%\times 8000=160$.

Answer: (A) Option 2 and by ₹160

Concept: Use LCM as tank capacity; emptying pipe has negative efficiency.

1. Tank $=\text{LCM}(10,15,30)=30$ units; A $=+3$, B $=+2$, C $=-1$ per hour.
2. First 6 h (all three): $4\times 6=24$ units.
3. Remaining $6$ units by B + C at $1$ unit/h ⇒ $6$ h.
4. Total time $=6+6=12$ h.

Trick: Efficiency method avoids fractions: units ÷ net rate = time.

Answer: (C) 12 hours

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference.

1. Here AB ∥ DC, so $EF=\frac{DC-AB}{2}$.
2. $EF=\frac{72-14}{2}=\frac{58}{2}=29$ cm.
3. BC and AD are extra data (not needed).

Yaad rakhein: Midpoints of sides → half the sum; midpoints of diagonals → half the difference.

Answer: (C) 29

Concept: Median of an odd number of terms = middle term after arranging in order.

1. Since $x>0$: $\frac{x}{5}<\frac{x}{4}<\frac{x}{3}<\frac{2x}{5}<\frac{2x}{3}<\frac{3x}{4}<x$.
2. 7 terms ⇒ median = 4th term $=\frac{2x}{5}$.
3. $\frac{2x}{5}=8.4$ ⇒ $x=21$.

Trick: Compare as decimals: $0.2,0.25,0.33,0.4,0.67,0.75,1$.

Answer: (D) 21

Concept: Write both situations in terms of the cost price $x$.

1. Original SP $=1.32x$.
2. New CP $=x+20$, new SP $=1.32x-42$, profit $27.5\%$.
3. $1.32x-42=1.275(x+20)=1.275x+25.5$ ⇒ $0.045x=67.5$.
4. $x=1500$.

Trick: Option check: CP 1500 ⇒ SP 1980; $1938=1.275\times 1520$ ✓.

Answer: (A) ₹1,500

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

1. Net rate $=\frac{1}{4}+\frac{1}{20}-\frac{1}{20}=\frac{1}{4}$ per hour.
2. The 20-h filler and 20-h emptier cancel out.
3. Time for $\frac{1}{4}$ tank $=\frac{1/4}{1/4}=1$ hour.

Trick: Equal fill and empty times cancel – only the 4-h tap matters.

Answer: (D) 1

Concept: Solve innermost bracket first (BODMAS).

1. $26-11+3=18$.
2. $-9\times 18=-162$.
3. $30+(-162)=-132$.

Trick: Sign rule: $30-162$ ⇒ answer negative, magnitude $132$.

Answer: (A) $-132$

Concept: $24=8\times 3$ (coprime) ⇒ number must be divisible by both 8 and 3.

1. By 8: last three digits $16B$ ⇒ $160$ or $168$, so $B=0$ or $8$.
2. By 3: digit sum $=19+A+B$ must be a multiple of 3.
3. For max, take $B=8$: $27+A$ ⇒ $A=9$ is allowed.
4. Max $A+B=9+8=17$.

Trick: For a maximum, try the largest B first, then the largest valid A.

Answer: (B) 17

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a+b{)}^2={88}^2=7744$.
2. $4ab=4\times 1927=7708$.
3. $(a-b{)}^2=36$ ⇒ $a-b=6$.

Trick: Numbers are $47$ and $41$: $47\times 41=1927$ ✓.

Answer: (C) 6

Concept: For a fixed sum, the product is maximum when the two numbers are as close as possible.

1. $41$ is odd, so the closest pair is $20$ and $21$.
2. $\text{HCF}(20,21)=1$ ⇒ they are co-prime.
3. Product $=20\times 21=420$.

Yaad rakhein: Consecutive numbers are always co-prime.

Answer: (C) 420

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=10630-9248.10=1381.90$.
2. $\frac{1381.90}{10630}\times 100=13\%$.

Trick: $10\%$ of 10630 $=1063$; $3\%=318.9$; total $1381.9$ ✓.

Answer: (A) $13\%$

Concept: $V=\pi r^{2}h$; halving the radius makes volume $\frac{1}{4}$.

1. Original volume $V=\pi R^{2}H$.
2. Each small cylinder $=\pi {\left(\frac{R}{2}\right)}^{2}H=\frac{V}{4}$.
3. Three of them use $\frac{3V}{4}$.
4. Unused $=V-\frac{3V}{4}=\frac{1}{4}$ of $V$.

Trick: Volume ∝ $r^2$ (same height): $\frac{1}{2}$ radius ⇒ $\frac{1}{4}$ volume.

Answer: (B) $\frac{1}{4}$

Concept: Total value = sum of (denomination × number of notes).

1. Let ₹100 notes $=x$, ₹500 notes $=x+4$.
2. $500(x+4)+100x=24800$ ⇒ $600x=22800$.
3. $x=38$.

Trick: Check: $42\times 500+38\times 100=21000+3800=24800$ ✓.

Answer: (B) 38

Concept: Proportion of red $=\frac{\text{red}}{\text{red}+\text{blue}}$.

1. Bag A: $\frac{2}{5}=40\%$.
2. Bag B: $\frac{4}{5}=80\%$.
3. Bag C: $\frac{3}{5}=60\%$ ⇒ Bag B is highest.

Trick: All totals are 5 parts, so just compare the red parts: 2, 4, 3.

Answer: (C) Bag B