RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Write everything in $\sin \theta$ and $\cos \theta$; use $a^3-b^3=(a-b)(a^2+ab+b^2)$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $1+\tan \theta +\cot \theta =\frac{\sin \theta \cos \theta +{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }$
2. First term $=\frac{(\sin \theta -\cos \theta )}{\sin \theta \cos \theta }=\frac{1}{\cos \theta }-\frac{1}{\sin \theta }=\sec \theta -\csc \theta$
3. $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +1+2\cos \theta +{\cos }^2\theta }{\sin \theta (1+\cos \theta )}=\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}=2\csc \theta$
4. Total $=\sec \theta -\csc \theta +2\csc \theta =\csc \theta +\sec \theta$

Trick: Put $\theta ={45}^{\circ }$: expression $=0+\frac{1}{\sqrt{2}+1}+(\sqrt{2}+1)=2\sqrt{2}$, and only $\csc {45}^{\circ }+\sec {45}^{\circ }=2\sqrt{2}$.

Answer: (A) $\csc \theta +\sec \theta$

Concept: Solve the inner bracket first (BODMAS), then add/subtract with a common denominator.

1. $1+\frac{2}{7}=\frac{9}{7}$, so $\frac{1}{9/7}=\frac{7}{9}$
2. $\frac{4}{3}-\frac{2}{3}=\frac{2}{3}=\frac{6}{9}$
3. $\frac{6}{9}+\frac{7}{9}=\frac{13}{9}$

Trick: Group the like fractions first: $\frac{4}{3}-\frac{2}{3}$ is instantly $\frac{2}{3}$.

Answer: (C) $\frac{13}{9}$

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=159$
2. Average $=\frac{159\times 160}{159}=159+1=160$

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

Answer: (D) $160$

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal powers.

1. $x^z=({112}^{0.57}{)}^z={112}^{0.57z}$
2. $y^6=({112}^{0.8}{)}^6={112}^{4.8}$
3. $0.57z=4.8\Rightarrow z=\frac{4.8}{0.57}=\frac{480}{57}\approx 8.42$

Trick: Just compare exponents: $z=\frac{6\times 0.8}{0.57}$.

Answer: (D) $8.42$

Concept: In "Buy $a$ get $b$ free", you pay for $a$ but receive $a+b$ items. $\text{Discount}\%=\frac{b}{a+b}\times 100$.

1. Items received $=12+8=20$; items paid for $=12$
2. Free items $=8$
3. Discount $=\frac{8}{20}\times 100=40\%$

Trick: Divide free items by TOTAL items (not by bought items): $\frac{8}{20}$, not $\frac{8}{12}$.

Answer: (C) $40\%$

Concept: Find each pipe’s full-tank time, then add rates (filling $+$, emptying $-$).

1. P: $\frac{3}{8}$ tank in $30$ h ⇒ full tank in $30\times \frac{8}{3}=80$ h; Q: $\frac{4}{5}$ in $16$ h ⇒ $20$ h
2. In $4$ h, P and Q fill $4(\frac{1}{80}+\frac{1}{20})=4\times \frac{5}{80}=\frac{1}{4}$ tank
3. R empties $\frac{1}{4}$ tank in $8$ h ⇒ full tank in $32$ h
4. Together: $\frac{1}{80}+\frac{1}{20}-\frac{1}{32}=\frac{2+8-5}{160}=\frac{1}{32}$ ⇒ $32$ h

Trick: Take LCM $160$ as tank capacity: P $=2$, Q $=8$, R $=-5$ units/h; net $5$ ⇒ $\frac{160}{5}=32$ h.

Answer: (D) $32$ hours

Concept: Let son $=s$, father $=f$ and form two equations.

1. $4s+f=46$ ⇒ $f=46-4s$
2. $s+2.8f=88$ ⇒ $s+2.8(46-4s)=88$
3. $s+128.8-11.2s=88$ ⇒ $10.2s=40.8$ ⇒ $s=4$
4. $f=46-16=30$

Trick: Check options: $f=30$ ⇒ $s=\frac{46-30}{4}=4$, and $4+2.8\times 30=88$ ✓.

Answer: (D) $30$

Concept: $V=\pi r^{2}h$, so new $V$ = old $V\times (\text{factor of }r{)}^2\times (\text{factor of }h)$.

1. $r$ factor $=1-0.46=0.54$; $h$ factor $=1+2.61=3.61$
2. New/old $={0.54}^2\times 3.61=0.2916\times 3.61=1.0527$
3. Increase $\approx 5.27\%\approx 5\%$

Trick: Note $3.61={1.9}^2$, so factor $=(0.54\times 1.9{)}^2={1.026}^2\approx 1.053$.

Answer: (A) $5\%$

Concept: Present value $=$ original $\times (1+\frac{a}{100})(1+\frac{b}{100})$; reverse it to get the original.

1. Combined factor $=1.04\times 1.25=1.3$
2. Original $=\frac{150800}{1.3}=116000$

Trick: Combined % $=4+25+\frac{4\times 25}{100}=30\%$ ⇒ divide by $1.3$.

Answer: (C) ₹1,16,000

Concept: Mode is the value that occurs most often.

1. Counts: $16\to 3$ times, $19\to 2$ times, all others once
2. Highest frequency is for $16$

Yaad rakhein: Arrange data: $10,12,14,16,16,16,18,19,19,20$ – the longest run is the mode.

Answer: (C) $16$

Concept: Two successive discounts $a\%$ and $b\%$ equal one discount of $(a+b-\frac{ab}{100})\%$.

1. $a=14$, $b=8$
2. Total $=14+8-\frac{14\times 8}{100}=22-1.12=20.88\%$
3. (Marked price ₹78,000 is not needed.)

Trick: Successive discount is always a little less than $a+b$ – here less than $22\%$.

Answer: (B) $20.88\%$

Concept: Sum of interior angles $=(n-2)\times {180}^{\circ }$; exterior angle $=\frac{{360}^{\circ }}{n}$.

1. $(n-2)\times 180=2520$ ⇒ $n-2=14$ ⇒ $n=16$
2. Exterior angle $=\frac{{360}^{\circ }}{16}={22.5}^{\circ }$

Trick: Exterior $={180}^{\circ }-\text{interior}$; interior $=\frac{2520}{16}={157.5}^{\circ }$ ⇒ ${22.5}^{\circ }$.

Answer: (B) ${22.5}^{\circ }$

Concept: Value of one ratio part $=\frac{\text{total}}{\text{sum of ratio terms}}$.

1. Sum of terms $=2+12+11=25$; one part $=\frac{570}{25}=22.8$
2. Largest − smallest $=12-2=10$ parts
3. Difference $=10\times 22.8=228$

Trick: Difference $=\frac{10}{25}\times 570=\frac{2}{5}\times 570=228$.

Answer: (B) $228$

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

1. $r=5\text{cm}$, $r^2=25$
2. $4\times 3.14\times 25=100\times 3.14=314{\text{cm}}^2$

Trick: $4\times 25=100$, so the answer is just $100\pi$.

Answer: (D) $314{\text{cm}}^2$

Concept: Under compound interest, if money becomes $k$ times in $T$ years, it becomes $k^n$ times in $nT$ years.

1. $3$ times in $15$ years
2. $27=3^3$ ⇒ time $=3\times 15=45$ years

Trick: CI multiplies: $3\to 9\to 27$ takes $15+15+15$ years. (In SI it would add, not multiply.)

Answer: (B) $45$ years

Concept: Let CP $=x$; write the new SP $=(1+\text{new profit}\%)\times$ new CP.

1. SP $=1.57x$
2. New CP $=x+20$, new SP $=1.57x-18$
3. $1.57x-18=1.475(x+20)=1.475x+29.5$
4. $0.095x=47.5$ ⇒ $x=500$

Trick: Check: CP $500$, SP $785$; new CP $520$, new SP $767$, profit $\frac{247}{520}=47.5\%$ ✓.

Answer: (B) ₹500

Concept: If a number increased by $r\%$ gives $N$, the number $=\frac{N}{1+\frac{r}{100}}$.

1. Number $\times 1.6=3610$
2. Number $=\frac{3610}{1.6}=2256.25$

Trick: Increase by $60\%$ = multiply by $\frac{8}{5}$, so reverse by $\times \frac{5}{8}$: $\frac{3610\times 5}{8}=2256.25$.

Answer: (A) $2256.25$

Concept: Profit $=$ total SP of items actually sold $-$ total CP of all items bought.

1. CP of $88$ lights $=$ ₹88
2. Lights left $=88-46=42$
3. SP $=42\times 3.10=$ ₹130.20
4. Profit $=130.20-88=$ ₹42.20

Trick: Broken items add to cost but earn nothing – count only the $42$ sold.

Answer: (C) ₹42.20

Concept: Open brackets carefully (sign of the outside number multiplies every term), then combine like terms.

1. $2(\frac{3}{2}{x}^2-20x+17)=3x^2-40x+34$
2. $-3(x^2+7x-15)=-3x^2-21x+45$
3. Sum $=0·x^2-61x+79=-61x+79$

Trick: The $x^2$ terms cancel ($3-3=0$); only the signs of $x$ and constant decide the option.

Answer: (C) $-61x+79$

Concept: Divisible by $11$ if (sum of digits at odd places) $-$ (sum at even places) is $0$ or a multiple of $11$.

1. Digits of $393064111$ from right: $1,1,1,4,6,0,3,9,3$
2. Odd places: $1+1+6+3+3=14$; even places: $1+4+0+9=14$
3. Difference $=0$ ⇒ divisible by $11$

Trick: Last digit $1$ is odd ⇒ not divisible by $10$, $14$ (even) or $15$ (needs $0/5$ at end) – only $11$ remains.

Answer: (A) $11$