RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: For 2 years, $\frac{CI}{SI}=\frac{200+R}{200}$.

1. $C{I}_2=P[(1.1{)}^2-1]=0.21P=136.50\Rightarrow P=650$.
2. $SI=\frac{650\times 10\times 2}{100}=₹130$.

Trick: $SI=CI\times \frac{200}{200+R}=136.5\times \frac{200}{210}=130$.

Answer: (C) 130

Concept: $a^{-n}=\frac{1}{a^n}$; dividing by $8^{-2}$ means multiplying by $64$.

1. $0.5\times 0.125=0.0625=\frac{1}{16}$.
2. $\frac{1}{16}÷\frac{1}{64}=\frac{1}{16}\times 64=4$.

Trick: In powers of 2: $2^{-1}\times 2^{-3}÷2^{-6}=2^{-1-3+6}=2^2=4$.

Answer: (C) 4

Concept: Factor out the smaller power: $4^x-4^{x-1}=4^{x-1}(4-1)$.

1. $3\times 4^{x-1}=48\Rightarrow 4^{x-1}=16=4^2$.
2. $x-1=2\Rightarrow x=3$.
3. $\frac{2x-1}{2x+3}=\frac{5}{9}$.

Trick: $a^n-a^{n-1}=a^{n-1}(a-1)$ – spot this pattern instantly in such questions.

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

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{28}{2}=14$ cm.
2. $4\times \frac{22}{7}\times 14\times 14=4\times 22\times 28=2464{\text{cm}}^2$.

Yaad rakhein: With $r$ a multiple of 7, cancel 7 first – no decimals at all.

Answer: (C) 2464 cm²

Concept: Take the route length $=D$; speeds are $\frac{D}{9}$ and $\frac{D}{11}$ (9 h and 11 h journeys).

1. By 2:00 PM Patna Express has covered $4\times \frac{D}{9}=\frac{4D}{9}$; gap left $=\frac{5D}{9}$.
2. Relative speed $=\frac{D}{9}+\frac{D}{11}=\frac{20D}{99}$.
3. Time $=\frac{5D}{9}÷\frac{20D}{99}=\frac{5}{9}\times \frac{99}{20}=2.75$ h $=2$ h $45$ min.
4. $2:00+2:45=4:45$ PM.

Trick: Take $D=99$ km: speeds 11 and 9 km/h, gap at 2 PM $=55$ km, time $=\frac{55}{20}=2.75$ h.

Answer: (A) 04:45 PM

Concept: $MP\times (1-d)=CP\times (1+p)$; find MP in terms of CP, then apply the new discount.

1. $0.75MP=1.1CP\Rightarrow MP=\frac{1.1}{0.75}CP=\frac{22}{15}CP$.
2. With $10\%$ discount: SP $=0.9\times \frac{22}{15}CP=1.32CP$.
3. Profit $=32\%$.

Trick: Take CP $=75$: MP $=110$; SP at 10% off $=99$; profit $=24$ on 75 $=32\%$.

Answer: (D) 32%

Concept: Translate the two statements into equations.

1. Keeran $=k$, Ramu $=k+8$.
2. Seven years ago: $k+1=3(k-7)\Rightarrow k+1=3k-21\Rightarrow k=11$.
3. Ramu $=11+8=19$ years.

Trick: Seven years ago the difference (8) was twice Keeran's age → Keeran was 4, Ramu 12; add 7 → 19.

Answer: (C) 19 years

Concept: Discount of ₹2 on ₹30 is a $\frac{2}{30}=\frac{1}{15}$ discount on the marked price.

1. Let CP $=100$ → MP $=120$.
2. SP $=120\times \frac{14}{15}=112$.
3. Gain $=12\%$.

Trick: Take CP $=25$: MP $=30$, pay $28$ → gain $3$ on $25$ $=12\%$.

Answer: (D) 12% gain

Concept: Convert the mixed fraction to a decimal and add.

1. $1\frac{1}{4}=1.25$.
2. $0.75+1.25=2.00$ cups.

Yaad rakhein: $\frac{1}{4}=0.25$, $\frac{3}{4}=0.75$ – they pair up to make 1.

Answer: (C) 2.0

Concept: $169={13}^2$ and $2.56={1.6}^2$, so numerator and denominator cancel neatly.

1. $(169{)}^2={13}^4$, $(2.56{)}^2={1.6}^4$ → these cancel with the denominator.
2. Left: $\sqrt{\frac{(0.25{)}^2}{(0.5{)}^2}}=\frac{0.25}{0.5}=0.5$.

Trick: Spot perfect squares ($169$, $2.56$) before multiplying anything.

Answer: (D) 0.5

Concept: When capitals are invested for the same time, profit is divided in the ratio of capitals.

1. Ratio of investments $=2300:4600=1:2$.
2. Total parts $=1+2=3$; one part $=\frac{37272}{3}=12424$.
3. P has 1 part, so P's share = ₹12,424.

Trick: Q invested exactly double, so P gets one-third of the profit – just divide by 3.

Answer: (A) 12,424

Concept: Area of rectangle $=l\times b$; area of circle $=\pi r^2$; circumference $=2\pi r$.

1. Area of field $=77\times 98=7546{\text{m}}^2$.
2. $\frac{22}{7}{r}^2=7546\Rightarrow r^2=7546\times \frac{7}{22}=343\times 7=2401\Rightarrow r=49$ m.
3. Perimeter $=2\times \frac{22}{7}\times 49=308$ m.

Trick: $77=7\times 11$ and $98=2\times 49$ – the 11 cancels the 22 and 7s pair up nicely; the answer must be a multiple of 44 ($2\times 22$), and only 308 is.

Answer: (B) 308

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; add the times of the two legs.

1. Let one-way distance be $d$ km. $\frac{d}{60}+\frac{d}{80}=3.5$.
2. LCM 240: $\frac{4d+3d}{240}=3.5\Rightarrow 7d=840\Rightarrow d=120$ km.
3. Check: $\frac{120}{60}+\frac{120}{80}=2+1.5=3.5$ h.

Trick: Same distance both ways – average speed $=\frac{2\times 60\times 80}{140}=\frac{480}{7}$ km/h; total distance $=\frac{480}{7}\times 3.5=240$, so one way $=120$ km.

Answer: (B) 120 km

Concept: In SI the amount grows by the same interest every year.

1. Interest for 2 years (from year 2 to year 4) $=6400-5200=1200$, so yearly SI $=600$.
2. Principal $=5200-2\times 600=4000$.
3. Rate $=\frac{600}{4000}\times 100=15\%$.

Trick: Difference of amounts ÷ difference of years = SI per year; subtract it back to reach P.

Answer: (D) 15 percent

Concept: If sum $=S$ and difference $=D$, larger $=\frac{S+D}{2}$, smaller $=\frac{S-D}{2}$.

1. Elder $=\frac{28+8}{2}=18$ years.
2. Younger $=\frac{28-8}{2}=10$ years; check $18+10=28$, $18-10=8$.

Trick: Larger number is always half of (sum + difference).

Answer: (C) 18 years

Concept: Take total work = LCM of days and work in units per day.

1. Total work $=24$ units; A does $2$ units/day, B does $1$ unit/day.
2. In 2 days together: $(2+1)\times 2=6$ units. Remaining $=18$ units.
3. B alone: $\frac{18}{1}=18$ days.

Trick: 2 days of both = 6 days of B alone; so B needs $24-6=18$ days more.

Answer: (D) 18

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

1. $(a-b{)}^2={40}^2-4\times 175=1600-700=900$.
2. $a-b=\sqrt{900}=30$.
3. Check: numbers are 35 and 5 ($35\times 5=175$).

Trick: Product 175 $=35\times 5$ and $35+5=40$ – spot the pair directly.

Answer: (D) 30

Concept: Semi-annual compounding: rate per half-year $=\frac{R}{2}$, periods $=2\times$ years.

1. 1 year 6 months $=3$ half-years.
2. $\frac{1728}{1000}={\left(\frac{12}{10}\right)}^3$, so growth per half-year $=1.2$, i.e. $20\%$.
3. Annual rate $=2\times 20=40\%$.

Yaad rakhein: $1728={12}^3$ and $1000={10}^3$ – a cube means three periods of 20%.

Answer: (B) 40%

Concept: Numbers leaving the same remainder $r$ on division by several divisors are of the form (multiple of LCM) $+r$.

1. LCM of 12, 8, 11 $=264$.
2. Largest 4-digit multiple of 264: $9999÷264=37$ remainder $231$, so $264\times 37=9768$.
3. Required number $=9768+3=9771$.

Trick: Subtract 3 from each option and test divisibility by 8 and 11: $9768$ ends in 768 (divisible by 8) and $9-7+6-8=0$ (divisible by 11).

Answer: (B) 9771

Concept: Placing $n$ identical cubes in a row gives a cuboid of length $n\times$ side.

1. Side of each cube $=\frac{256}{4}=64$ cm.
2. Volume $={64}^3=64\times 64\times 64=262144{\text{cm}}^3$.

Yaad rakhein: ${64}^3=2^{18}=262144$; the options differ only in digit order, so compute the last digits: $4\times 4\times 4=64$ ends in 4 and ${64}^2=4096$, $4096\times 64=262144$.

Answer: (D) $262144{\text{cm}}^3$