RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Take total work = LCM of the days; one-day work = total ÷ days.

1. Total work $=\text{LCM}(12,25)=300$ units; A $=25$, B $=12$ units/day.
2. Together $=37$ units/day; in 6 days $=6\times 37=222$ units.
3. Left $=300-222=78$ units $=\frac{78}{300}=\frac{13}{50}$.

Trick: Fraction done $=6(\frac{1}{12}+\frac{1}{25})=\frac{6\times 37}{300}=\frac{37}{50}$, so left $=\frac{13}{50}$.

Answer: (C) $\frac{13}{50}$

Concept: Follow BODMAS: division and multiplication before addition and subtraction; take care of signs.

1. $(-60)÷(-10)=6$ and $(-3)\times 9=-27$.
2. Expression $=(-9)-6+(-27)$.
3. $=-9-6-27=-42$.

Trick: Minus a positive or plus a negative both move left on the number line: $-9\to -15\to -42$.

Answer: (A) $-42$

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{First number}\times \text{Second number}$.

1. $18\times 216=72\times x$.
2. $x=\frac{18\times 216}{72}=\frac{3888}{72}=54$.
3. Check: $\text{HCF}(72,54)=18$, $\text{LCM}(72,54)=216$ ✓.

Trick: $\frac{216}{72}=3$, so $x=18\times 3=54$.

Answer: (C) $54$

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum of digits at even places) is $0$ or a multiple of 11.

1. Digits N, 3, 6, M, 7, 7: odd places $N+6+7=N+13$; even places $3+M+7=M+10$.
2. Difference $=(N+13)-(M+10)=N-M+3$.
3. For $N-M+3=0$: $M-N=3$ (e.g. $N=1,M=4$ gives 136477 $=11\times 12407$).

Trick: Alternate sum $=0$ or $\pm 11$; digits are 0–9, so $N-M+3=0$ is the workable case here.

Answer: (B) $M-N=3$

Concept: Let the second number be $x$; then the first is $5x$.

1. $5x\times x=245\Rightarrow x^2=49\Rightarrow x=7$.
2. First number $=35$, second $=7$.
3. Sum $=35+7=42$.

Trick: Sum $=6x$; $245=5\times 7^2$ gives $x=7$ directly, so sum $=42$.

Answer: (D) $42$

Concept: Take the distance MN as the LCM of the journey times; then the trains close the gap at the sum of their speeds.

1. A: 8:35 AM to 3:35 PM $=7\text{h}$; B: 10:35 AM to 3:35 PM $=5\text{h}$. Let MN $=35$ units: A $=5$, B $=7$ units/h.
2. By 10:35 A has run $2\text{h}$: $10$ units; gap $=25$ units.
3. Relative speed $=5+7=12$ units/h; time $=\frac{25}{12}\text{h}=2\text{h}5\text{min}$.
4. Meeting time $=$ 10:35 + 2 h 5 min $=$ 12:40 PM.

Trick: Always start both trains at the same clock time: move the early one ahead first.

Answer: (B) 12:40 PM

Concept: If $a:b=b:x$, then $x$ is the third proportional of $a$ and $b$: $x=\frac{b^2}{a}$.

1. $8:16=16:x$.
2. $x=\frac{{16}^2}{8}=\frac{256}{8}=32$.

Trick: Third proportional continues the ratio: $8\to 16\to 32$ (each ×2).

Answer: (C) $32$

Concept: $\text{Distance}=\text{Speed}\times \text{Time}$; for a fixed distance, speed is inversely proportional to time.

1. Distance $=10\times 16=160\text{km}$.
2. Required speed $=\frac{160}{8}=20\text{km/h}$.

Trick: Time halves (16 → 8 h) ⇒ speed doubles: $10\times 2=20$.

Answer: (C) $20\text{km/h}$

Concept: Profit is shared in the ratio of $\text{capital}\times \text{time}$.

1. Praveen's share $=\frac{5440+1360}{2}=3400$; Hemant's $=2040$. Ratio $=3400:2040=5:3$.
2. Let Hemant's capital $=H$: $\frac{5(H+78000)}{6H}=\frac{5}{3}$.
3. $15H+1170000=30H\Rightarrow H=78000$.
4. Praveen's capital $=78000+78000=$ ₹1,56,000.

Trick: Check option: ₹1,56,000 × 5 : ₹78,000 × 6 $=780000:468000=5:3$ ✓.

Answer: (A) ₹1,56,000

Concept: With the same time, equal SI means $P_1\times R_1=P_2\times R_2$, so principals are in the inverse ratio of rates.

1. $P_1\times 5\times 4=P_2\times 11\times 4\Rightarrow P_1:P_2=11:5$.
2. 16 parts $=22800\Rightarrow$ 1 part $=1425$.
3. Sum at 5% $=11\times 1425=15675$.

Trick: Principal at lower rate gets the larger share: rates $5:11$ ⇒ sums $11:5$.

Answer: (B) 15,675

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) $16$