RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $\text{Saving}=\text{Income}-\text{Expenditure}$; find old and new savings separately.

1. Old saving $=7.5\%\times 80200=6015$; expenditure $=80200-6015=74185$.
2. New income $=80200\times 1.34=107468$.
3. New expenditure $=74185\times 1.40=103859$.
4. New saving $=107468-103859=3609$; change $=3609-6015=-2406$.

Trick: Expenditure rose faster (40% vs 34%) on a much bigger base, so savings must fall — only the "decrease" options survive.

Answer: (B) decrease by ₹2,406

Concept: On melting and recasting, total volume stays the same. Volume of cube $=a^3$.

1. Total volume $=789+865+1721=3375{\text{cm}}^3$.
2. $a^3=3375\Rightarrow a=\sqrt[3]{3375}=15\text{cm}$.

Trick: Cubes to remember: ${15}^3=3375$, ${14}^3=2744$, ${12}^3=1728$.

Answer: (A) 15 cm

Concept: Write both $x$ and $y$ as powers of the same base 81 and compare exponents.

1. $x^z=({81}^{0.65}{)}^z={81}^{0.65z}$ and $y^7=({81}^{0.26}{)}^7={81}^{1.82}$.
2. Equal bases ⇒ $0.65z=1.82$.
3. $z=\frac{1.82}{0.65}=2.8$.

Trick: $(a^m{)}^n=a^{mn}$ — once bases match, just equate exponents.

Answer: (A) 2.8

Concept: $\text{MP}\times (1-\text{discount})=\text{SP}=\text{CP}\times (1+\text{profit})$.

1. SP $=1540\times 1.10=1694$.
2. MP $=\frac{1694}{0.70}=2420$.

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{100+P}{100-D}=\frac{110}{70}=\frac{11}{7}$, so MP $=1540\times \frac{11}{7}=2420$.

Answer: (B) ₹2,420

Concept: $\text{Distance}=\text{Speed}\times \text{Time}$ (car length negligible, so distance = bridge length).

1. Time $=3.8\text{min}=\frac{3.8}{60}\text{h}$.
2. Length $=48\times \frac{3.8}{60}=0.8\times 3.8=3.04\text{km}$.

Trick: $48\text{km/h}=0.8\text{km}$ per minute — multiply by the minutes directly.

Answer: (B) 3.04 km

Concept: No overall profit/loss ⇒ loss on first shirt = gain on second shirt (in ₹).

1. Let first shirt cost $x$, second $1200-x$.
2. $0.08x=0.24(1200-x)\Rightarrow 0.32x=288$.
3. $x=900$.

Trick: Loss% : Gain% $=8:24=1:3$, so costs are in ratio $3:1$ ⇒ first $=\frac{3}{4}\times 1200=900$.

Answer: (C) 900

Concept: Volume of sphere $=\frac{4}{3}\pi r^3$, with $r=\frac{d}{2}$.

1. $r=\frac{12}{2}=6\text{cm}$.
2. $V=\frac{4}{3}\pi \times 6^3=\frac{4}{3}\pi \times 216=288\pi {\text{cm}}^3$.

Trick: Do not use $d$ in place of $r$ — that gives $8$ times the volume ($2304\pi$).

Answer: (B) $288\pi {\text{cm}}^3$

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

1. Let CP of 1 book = ₹1 ⇒ SP of 20 books = ₹36, CP of 20 books = ₹20.
2. Gain $=36-20=16$; gain% $=\frac{16}{20}\times 100=80\%$.

Trick: Divide by the number on the SP side (20), not the CP side (36).

Answer: (C) 80% gain

Concept: Check options: sum must be 44 and LCM must be 459.

1. $459=27\times 17$ and 27, 17 are co-prime, so LCM$(27,17)=459$.
2. $27+17=44$ ✓.
3. Others fail: e.g. LCM$(20,24)=120$, LCM$(32,12)=96$.

Trick: Factorise the LCM: $459=3^3\times 17$ — the split $27,17$ adding to 44 jumps out.

Answer: (C) 27, 17

Concept: Sum of first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Here $n=147$.
2. Average $=\frac{{147}^2}{147}=147$.

Yaad rakhein: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (B) 147

Concept: Required speed $=\frac{\text{remaining distance}}{\text{remaining time}}$.

1. First part: $34\times 4=136\text{km}$ in 4 h.
2. Second part: 180 km at 30 km/h takes $\frac{180}{30}=6$ h.
3. Covered $316\text{km}$ in 10 h ⇒ left $843-316=527\text{km}$ in $11-10=1$ h.
4. Speed $=\frac{527}{1}=527\text{km/h}$.

Trick: With exactly 1 hour left, the speed equals the remaining distance.

Answer: (A) 527 km/hr

Concept: Let second number $=x$, first $=5x$; then $5x·x=3645$.

1. $5x^2=3645\Rightarrow x^2=729\Rightarrow x=27$.
2. First $=135$; sum $=135+27=162$.

Trick: Sum $=6x$, so the answer must be a multiple of 6 — only 162 is.

Answer: (C) 162

Concept: Take total work = LCM of days; find units done together and the remainder.

1. Total work $=\text{LCM}(9,19)=171$ units; A $=19$, B $=9$ units/day.
2. Together $=28$ units/day; in 6 days $=168$ units.
3. Left $=171-168=3$ units $=\frac{3}{171}=\frac{1}{57}$.

Trick: Fraction done $=6(\frac{1}{9}+\frac{1}{19})=\frac{168}{171}=\frac{56}{57}$, so left $=\frac{1}{57}$.

Answer: (D) $\frac{1}{57}$

Concept: If $8A=3B=6C=k$, then $A:B:C=\frac{1}{8}:\frac{1}{3}:\frac{1}{6}$.

1. Multiply by LCM 24: $A:B:C=3:8:4$; total parts $=15$.
2. Share of A $=\frac{3}{15}\times 35600=7120$.

Trick: Check the ratio: $8\times 3=3\times 8=6\times 4=24$ — all equal, so $3:8:4$ is right.

Answer: (C) ₹7,120

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

1. $\sqrt{324\times 16}=\sqrt{324}\times \sqrt{16}=18\times 4=72$.

Trick: Take square roots separately when both numbers are perfect squares.

Answer: (B) 72

Concept: Simplify the complex fraction first, then combine.

1. $1+\frac{2}{7}=\frac{9}{7}$, so $\frac{1}{9/7}=\frac{7}{9}$.
2. $\frac{5}{3}-\frac{2}{3}=1$.
3. Value $=1+\frac{7}{9}=\frac{16}{9}$.

Trick: Group like denominators ($\frac{5}{3}-\frac{2}{3}$) before anything else.

Answer: (A) $\frac{16}{9}$

Concept: Divisible by 5 ⇔ last digit 0 or 5; by 3 ⇔ digit sum divisible by 3.

1. Last digit is 5 ⇒ divisible by 5.
2. Digit sum $=4+8+1+3+5+5=26$, not divisible by 3 ⇒ not by 3, 6 or 15.
3. It is odd, so not by 6 either.

Yaad rakhein: 15 needs both 3 and 5; 6 needs both 2 and 3.

Answer: (D) 5

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

1. $\text{SI}=\frac{400\times 16\times 4.5}{100}=4\times 16\times 4.5$.
2. $=64\times 4.5=288$.

Trick: $P=400$ ⇒ SI $=4\times R\times T$ directly.

Answer: (D) 288

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. Sum $=40+49+56+74+38=257$.
2. Average $=\frac{257}{5}=51.4\text{kg}$.

Trick: Deviation from 50: $-10-1+6+24-12=7$; $50+\frac{7}{5}=51.4$.

Answer: (C) 51.4

Concept: Compute each quantity and subtract.

1. $65\%$ of $65=\frac{65\times 65}{100}=42.25$.
2. $\frac{1}{5}$ of $55=11$.
3. Difference $=42.25-11=31.25$.

Trick: ${65}^2=4225$ (numbers ending in 5: $6\times 7=42$, then 25).

Answer: (C) 31.25