RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: When all invest for the same time, profit is shared in the ratio of investments.

1. Total investment $=1200+1070+1310=3580$.
2. Profit $1790=\frac{3580}{2}$, i.e. profit is exactly half of the investment.
3. Tejas's share $=\frac{1310}{3580}\times 1790=\frac{1310}{2}=655$.

Trick: Spot that profit = half of total capital, so each share = half of own capital.

Answer: (D) ₹655

Concept: Use $(1+\cos \theta )(1-\cos \theta )={\sin }^2\theta$ and $(1+\sin \theta )(1-\sin \theta )={\cos }^2\theta$.

1. $X=\frac{2(1+\cos \theta )(1-\cos \theta )}{2(1+\sin \theta )(1-\sin \theta )}=\frac{{\sin }^2\theta }{{\cos }^2\theta }={\tan }^2\theta$.
2. $X+\frac{1}{X}+2={\tan }^2\theta +{\cot }^2\theta +2=(\tan \theta +\cot \theta {)}^2$.
3. $\tan \theta +\cot \theta =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\sec \theta \text{cosec}\theta$, so the value is ${\text{cosec}}^2\theta {\sec }^2\theta$.

Trick: Put $\theta ={45}^{\circ }$: $X=1$, value $=4$; only ${\text{cosec}}^2\theta {\sec }^2\theta =2\times 2=4$ matches.

Answer: (D) ${\text{cosec}}^2\theta {\sec }^2\theta$

Concept: Under compound interest, if a sum becomes $k$ times in $t$ years, it becomes $k^n$ times in $nt$ years.

1. 3 times in 17 years.
2. $81=3^4$, so the sum must triple 4 times.
3. Time $=4\times 17=68$ years.

Trick: Write the target as a power of the base multiple: $3\to 81=3^4\Rightarrow 4\times 17$.

Answer: (A) 68 years

Concept: Evaluate powers and roots first, then add/subtract (BODMAS).

1. $8^2=64$, $\sqrt{3^2}=3$, $2\sqrt{36}=2\times 6=12$.
2. $64+3-12-11=44$.

Trick: $\sqrt{a^2}=a$ for positive $a$ – no need to square and root again.

Answer: (D) 44

Concept: In 'Buy $x$ get $y$ free', the customer pays for $x$ but receives $x+y$ articles.

1. Articles received $=20+5=25$; articles free $=5$.
2. $\text{Discount}\%=\frac{5}{25}\times 100=20\%$.

Trick: $\text{Discount}\%=\frac{\text{free}}{\text{paid}+\text{free}}\times 100$ – not $\frac{5}{20}$ (that gives the tempting 25%).

Answer: (C) 20%

Concept: Expand the brackets; the $x^2$ terms cancel, leaving a linear equation.

1. $5(5x^2-5)=25x^2-25$ and $5(5x^2+7x-2)=25x^2+35x-10$.
2. Difference: $25x^2-25-25x^2-35x+10=-35x-15$.
3. $-35x-15=14\Rightarrow -35x=29\Rightarrow x=-\frac{29}{35}$.

Trick: Take 5 common: $5[(5x^2-5)-(5x^2+7x-2)]=5(-7x-3)=14$.

Answer: (A) $-\frac{29}{35}$

Concept: Make decimal places equal, find the LCM of the whole numbers, then put the decimal back.

1. $2.8=2.800=\frac{2800}{1000}$, $0.217=\frac{217}{1000}$.
2. $2800=2^4\times 5^2\times 7$, $217=7\times 31$.
3. $\text{LCM}(2800,217)=2^4\times 5^2\times 7\times 31=86800$.
4. LCM $=\frac{86800}{1000}=86.8$.

Trick: The LCM must be a multiple of 2.8: $2.8\times 31=86.8$, and $0.217\times 400=86.8$ ✓.

Answer: (D) 86.8

Concept: Form two linear equations with son's age $s$ and father's age $f$.

1. $4s+f=100$ … (i); $s+2.5f=97$ … (ii).
2. From (ii): $s=97-2.5f$. Put in (i): $388-10f+f=100$.
3. $9f=288\Rightarrow f=32$ (and $s=17$).

Trick: Check options in (i): $f=32\Rightarrow s=17$; then $17+2.5\times 32=97$ ✓.

Answer: (B) 32

Concept: Convert everyone into one type of worker using total work (man-days = woman-days).

1. Work $=4\times 12=48$ man-days $=8\times 15=120$ woman-days.
2. So $1\text{ man}=\frac{120}{48}=2.5$ women; $2\text{ men}=5$ women.
3. $2$ men $+3$ women $=8$ women; time $=\frac{120}{8}=15$ days.

Trick: Equate 48 M = 120 W → M : W efficiency = 5 : 2.

Answer: (B) 15 days

Concept: For a regular polygon, exterior angle $=\frac{{360}^{\circ }}{n}$ and interior + exterior $={180}^{\circ }$.

1. Exterior angle $={180}^{\circ }-{162}^{\circ }={18}^{\circ }$.
2. $n=\frac{{360}^{\circ }}{{18}^{\circ }}=20$.

Trick: Always go via the exterior angle – it gives $n$ in one division.

Answer: (C) 20

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