RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: A car is treated as a point, so the distance covered while crossing the bridge = length of the bridge; $\text{Distance}=\text{Speed}\times \text{Time}$.

1. Speed $=99\times \frac{5}{18}=27.5\text{m/s}$.
2. Time $=1.2\times 60=72\text{s}$.
3. Length $=27.5\times 72=1980\text{m}$.

Trick: $99\text{km/h}$ for $1.2$ min: $\frac{99\times 1.2}{60}=1.98\text{km}=1980\text{m}$.

Answer: (B) $1980$

Concept: For the same time, equal interest means $P_1\times r_1=P_2\times r_2$, so principals are in the inverse ratio of rates.

1. $P_1\times 9\times 3=P_2\times 13\times 3\Rightarrow P_1:P_2=13:9$.
2. Total parts $=13+9=22$; one part $=\frac{20900}{22}=950$.
3. Sum at $9\%$ $=13\times 950=$ ₹12,350.

Trick: Money at the lower rate gets the larger share: $\frac{13}{22}\times 20900$.

Answer: (B) 12,350

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

1. $1310:1430:1040=131:143:104$.
2. Sum of ratio $=131+143+104=378$; one part $=\frac{1890}{378}=5$.
3. Yuvraj's share $=104\times 5=$ ₹520.

Trick: Profit is exactly $\frac{1890}{3780}=\frac{1}{2}$ of total capital, so Yuvraj gets half of ₹1,040 = ₹520.

Answer: (C) ₹520

Concept: For a square of side $a$: perimeter $=4a$, area $=a^2$.

1. $4a=336\Rightarrow a=84\text{m}$.
2. Area $={84}^2=7056{\text{m}}^2$.

Trick: ${84}^2=(80+4{)}^2=6400+640+16=7056$; also a square ends in 6 only if the side ends in 4 or 6 – options ending in 2 or 7 drop out.

Answer: (D) $7056$

Concept: Let the present ages of father and son be $F$ and $S$; form equations for 3 years ago and $x$ years later.

1. 3 years ago: $F-3=2(S-3)+21\Rightarrow F=2S+18$.
2. After $x$ years: $F+x=2(S+x)$.
3. $2S+18+x=2S+2x\Rightarrow x=18$.

Trick: If now $F=2S+k$, the father will be twice the son's age after exactly $k$ years. Here $k=18$.

Answer: (A) $18$

Concept: With a profit claim $g\%$ and a weight $e\%$ less, $\text{Profit}\%=(\frac{100+g}{100-e}-1)\times 100$.

1. Let CP of $100\text{g}$ be ₹100. He charges for 100 g at $57\%$ profit, i.e. SP $=157$.
2. He actually gives only $87\text{g}$, whose CP $=87$.
3. Profit $\%=\frac{157-87}{87}\times 100=\frac{7000}{87}\approx 80.46\%\approx 80\%$.

Trick: $\frac{157}{87}\approx 1.80$, so profit ≈ $80\%$.

Answer: (A) $80\%$

Concept: Check each option with divisibility rules, then confirm by division.

1. 50 needs the last digit 0 – the number ends in 3, so no.
2. 36 needs divisibility by 4 and 9: last two digits 43 are odd, so no.
3. $861616843÷43$: $86\to 2$, $1\to 0$, $16\to 0$, $161\to 3(r32)$, $326\to 7(r25)$, $258\to 6(r0)$, $4\to 0$, $43\to 1$. Quotient $=20037601$, remainder 0.
4. Hence 43 divides the number exactly.

Trick: Digit sum $=8+6+1+6+1+6+8+4+3=43$ – curiously, the number is $43\times 20037601$.

Answer: (A) $43$

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. In N18M05, odd places (from left): $N,8,0$; even places: $1,M,5$.
2. Difference $=(N+8+0)-(1+M+5)=N-M+2$.
3. Since $N,M$ are digits, take $N-M+2=0\Rightarrow M-N=2$ (e.g. $N=1,M=3$: $118305=11\times 10755$).

Trick: Write the digits with alternate signs: $+N-1+8-M+0-5=N-M+2$.

Answer: (B) $M-N=2$

Concept: Compare ratios as fractions (or decimals).

1. $\frac{37}{76}<\frac{1}{2}$ and $\frac{35}{72}<\frac{1}{2}$.
2. $\frac{43}{65}\approx 0.66$.
3. $\frac{50}{60}=\frac{5}{6}\approx 0.83$ – the greatest.

Trick: Use $\frac{1}{2}$ as a benchmark: two ratios are below it; of the other two, $\frac{5}{6}$ is larger than $\frac{43}{65}$ (which is close to $\frac{2}{3}$).

Answer: (D) $50:60$

Concept: A rectangle has all angles ${90}^{\circ }$; a rhombus has all sides equal.

1. A square has all angles ${90}^{\circ }$, so it is a rectangle.
2. A square has all four sides equal, so it is also a rhombus.
3. Thus a square is a special case of both – it lies in the overlap of rectangles and rhombuses.

Trick: Square = Rectangle ∩ Rhombus.

Answer: (C) A square is a special case of both a rectangle and a rhombus.

Concept: No overall profit/loss means the loss on one dress equals the gain on the other.

1. Let CP be $x$ and $1200-x$.
2. $0.09x=0.18(1200-x)\Rightarrow x=2(1200-x)$
3. $3x=2400\Rightarrow x=800$

Trick: Loss% : gain% = $9:18=1:2$, so CPs are in the inverse ratio $2:1$ → first = $\frac{2}{3}\times 1200=800$.

Answer: (D) ₹800

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

1. $5845=\frac{P\times 4.2\times \frac{10}{3}}{100}=0.14P\Rightarrow P=41750$
2. New SI $=\frac{41750\times 6.4\times 12.5}{100}=41750\times 0.8=33400$
3. Amount $=41750+33400=75150$

Trick: Rate × time: $4.2\times \frac{10}{3}=14\%$ and $6.4\times 12.5=80\%$. So new SI $=\frac{80}{14}\times 5845=33400$.

Answer: (D) ₹75,150

Concept: Find each value separately and subtract.

1. $65\%\text{ of }65=\frac{65\times 65}{100}=42.25$
2. $\frac{1}{5}\text{ of }45=9$
3. Difference $=42.25-9=33.25$

Trick: $a\%$ of $a$ $=\frac{a^2}{100}$: $\frac{4225}{100}=42.25$.

Answer: (A) 33.25

Concept: BODMAS: innermost bracket first, then multiplication, then the outer brackets.

1. $(5-9)=-4$
2. $2\times (-4)=-8$, so $\{11-8\}=3$
3. $459÷3=153$

Trick: Keep the sign of the bracket: $11+2\times (-4)=3$, not 52.

Answer: (A) 153

Concept: $\cot \theta =\frac{\text{base}}{\text{perpendicular}}$ and $\text{cosec}\theta =\frac{\text{hypotenuse}}{\text{perpendicular}}$.

1. $\cot \theta =\frac{x}{y}$ → base $=x$, perpendicular $=y$
2. Hypotenuse $=\sqrt{x^2+y^2}$ (Pythagoras)
3. $\text{cosec}\theta =\frac{\sqrt{x^2+y^2}}{y}$

Trick: Check: ${\text{cosec}}^2\theta =1+{\cot }^2\theta =\frac{y^2+x^2}{y^2}$.

Answer: (B) $\frac{\sqrt{x^2+y^2}}{y}$

Concept: Convert each pipe to a full-tank time; an emptying pipe has negative work.

1. P: $\frac{3}{4}$ in 27 h → full in 36 h; Q: $\frac{2}{3}$ in 12 h → full in 18 h
2. In 6 h: $6(\frac{1}{36}+\frac{1}{18})=\frac{1}{2}$ tank filled
3. R empties $\frac{1}{2}$ tank in 8 h → full tank in 16 h
4. Together: $\frac{1}{36}+\frac{1}{18}-\frac{1}{16}=\frac{4+8-9}{144}=\frac{1}{48}$

Trick: Take tank = LCM(36, 18, 16) = 144 units: P = 4, Q = 8, R = −9 per hour → net 3 → $\frac{144}{3}=48$ h.

Answer: (C) 48 hours

Concept: $\text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}$.

1. $\frac{1}{2}\times 12\times 8$
2. $=6\times 8=48{\text{cm}}^2$

Trick: 96 is the trap – it forgets the $\frac{1}{2}$ (that is the rectangle's area).

Answer: (C) $48{\text{cm}}^2$

Concept: After two discounts of $x\%$, SP $=\text{MP}{(1-\frac{x}{100})}^2$.

1. SP $=50000-18000=32000$
2. ${(1-\frac{x}{100})}^2=\frac{32000}{50000}=0.64$
3. $1-\frac{x}{100}=0.8\Rightarrow x=20$

Trick: Check: 20% then 20% → $20+20-\frac{20\times 20}{100}=36\%$, and 36% of 50000 = 18000.

Answer: (A) 20

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

1. Sum $=160\times 161$
2. Average $=\frac{160\times 161}{160}=161$

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

Answer: (C) 161

Concept: Add the one-hour work of all filling pipes.

1. $\frac{1}{18}+\frac{1}{21}+\frac{1}{16}$, LCM $=1008$
2. $=\frac{56+48+63}{1008}=\frac{167}{1008}$
3. Time $=\frac{1008}{167}=6\frac{6}{167}$ hours

Trick: Together they must beat the fastest pipe alone (16 h) yet take more than $16÷3\approx 5.3$ h, so only the answer near 6 h fits.

Answer: (A) $6\frac{6}{167}$ hours