RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Divisibility by 7 (also 11, 13): make 3-digit groups from the right and take the alternating sum.

1. The number is odd and ends in 1 → not divisible by 8, 12 or 15
2. Groups from the right: 814 | 242 | 541 → $541-242+814=1113$
3. $1113=7\times 159$ → divisible by 7; indeed $814242541=7\times 116320363$

Trick: Last digit 1 kills every even option and 15 instantly; only 7 remains.

Answer: (A) 7

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; doubling speed halves the time.

1. Let Anmol's time be $T$; then $T-7$ = Nikhil's time
2. With double speed Anmol takes $\frac{T}{2}=(T-7)-8$
3. $\frac{T}{2}=15\Rightarrow T=30$ h
4. Speed $=\frac{30}{30}=1\text{km/hr}$

Trick: The gap $7+8=15$ h equals half of Anmol's time.

Answer: (B) 1 km/hr

Concept: Successive % changes multiply: $P\times \left(1\pm \frac{r}{100}\right)$ for each year.

1. $12500\times 1.52=19000$
2. $19000\times 1.30=24700$
3. $24700\times 0.85=20995$

Trick: Last digit check: only an answer ending in 5 can come from multiplying by 0.85 here.

Answer: (A) 20,995

Concept: Consecutive numbers lie close to $\sqrt{702}\approx 26.5$.

1. $26\times 27=702$
2. Greater number $=27$

Trick: Unit digit: $6\times 7=42$ ends in 2, matching 702.

Answer: (B) 27

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

1. $12.8=12.800$ and $0.004$ → 12800 and 4 (thousandths)
2. LCM(12800, 4) $=12800$
3. Answer $=12.800=12.8$

Trick: 12.8 is itself a multiple of 0.004 ($12.8÷0.004=3200$), so LCM = 12.8.

Answer: (D) 12.8

Concept: Discount% $=\frac{\text{free items}}{\text{total items received}}\times 100$.

1. Customer gets $30+10=40$ articles, pays for 30
2. Discount $=\frac{10}{40}\times 100=25\%$

Trick: Divide by items received (40), not items bought (30) – 33% is the trap.

Answer: (B) 25%

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

1. $10x^2-35-10x^2-16x+4=12$
2. $-16x-31=12$
3. $-16x=43\Rightarrow x=-\frac{43}{16}$

Trick: Watch the sign: $-2\times (-2)=+4$.

Answer: (A) $-\frac{43}{16}$

Concept: If $3A=6B=2C$, then $A:B:C=\frac{1}{3}:\frac{1}{6}:\frac{1}{2}$.

1. Multiply by 6: $A:B:C=2:1:3$
2. Total parts $=6$
3. $A=\frac{2}{6}\times 12300=4100$

Trick: Shares are inversely proportional to the multipliers 3, 6, 2.

Answer: (A) ₹4,100

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (left to right).

1. ${50}^{-6}÷{50}^{18}={50}^{-24}$
2. ${50}^{-24}\times {50}^{-5}={50}^{-29}$

Trick: Exponent $=-6-18-5=-29$.

Answer: (B) ${50}^{-29}$

Concept: Median = middle value after arranging in order; for $n$ odd it is the $\frac{n+1}{2}$th value.

1. Order: 8, 13, 14, 15, 23, 26, 35, 36, 50
2. $n=9$ → $\frac{9+1}{2}=5$th value
3. 5th value $=23$

Trick: Always sort first – 26 (the 5th value as printed) is the trap.

Answer: (B) 23

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. Odd places of N29M11: $N+9+1=N+10$.
2. Even places: $2+M+1=M+3$.
3. Difference $=(N+10)-(M+3)=N-M+7$.
4. Setting it $=0$ gives $M-N=7$ (e.g. $N=1,M=8$: $129811=11\times 11801$).

Trick: Write $N-M+7=0$ or $11$; only $M-N=7$ is in the options.

Answer: (A) $M-N=7$

Concept: $V=\pi r^{2}h$, so $V\propto r^{2}h$.

1. Take $r=10$, $h=10$: $r^{2}h=1000$.
2. New $r=8$ ($20\%$ less), new $h=14$ ($40\%$ more): $r^{2}h=64\times 14=896$.
3. Change $=\frac{896-1000}{1000}\times 100=-10.4\%$, i.e. a decrease.

Trick: Multiply factors: ${0.8}^2\times 1.4=0.896$ ⇒ $10.4\%$ decrease.

Answer: (A) $10.4\%$ decrease

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

1. Here $n=154$.
2. Sum $=154\times 155$.
3. Average $=\frac{154\times 155}{154}=155$.

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

Answer: (C) $155$

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

1. Net discount $=20+10-\frac{20\times 10}{100}=28\%$.
2. $28\%$ of MP = ₹23,800.
3. MP $=\frac{23800\times 100}{28}=85000$, i.e. ₹85,000.

Trick: Take MP $=100$: $100\to 80\to 72$, discount $28$ units.

Answer: (D) ₹85,000

Concept: If the bases are equal, the exponents can be equated: $(a^m{)}^n=a^{mn}$.

1. $x^z=({103}^{0.2}{)}^z={103}^{0.2z}$.
2. $y^6=({103}^{0.43}{)}^6={103}^{2.58}$.
3. So $0.2z=2.58$ ⇒ $z=\frac{2.58}{0.2}=12.9$.

Trick: $z=\frac{6\times 0.43}{0.2}=12.9$ directly.

Answer: (A) $12.9$

Concept: Write present ages as $F$ (father) and $S$ (son) and convert each statement into an equation.

1. 4 years ago: $F-4=2(S-4)+37$ ⇒ $F=2S+33$.
2. After $x$ years: $F+x=2(S+x)$.
3. $2S+33+x=2S+2x$ ⇒ $x=33$.

Trick: The extra '$+33$' over twice the son's age vanishes after $33$ years.

Answer: (B) $33$

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference: $EF=\frac{DC-AB}{2}$.

1. Let the midline through E and F be drawn parallel to AB and DC.
2. In $△ADC$ midpoint segment = $\frac{DC}{2}$; in $△ADB$ midpoint segment = $\frac{AB}{2}$; on the same line their difference is $EF$.
3. $EF=\frac{53-45}{2}=\frac{8}{2}=4\text{cm}$. (BC and AD are not needed.)

Trick: Midpoints of diagonals → half the difference; median → half the sum.

Answer: (A) $4$

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

1. Mohan + Nishant = 4280 and Mohan − Nishant = 1070 ⇒ Mohan = 2675, Nishant = 1605; ratio $2675:1605=5:3$.
2. Let Nishant's capital be $x$; Mohan's $=x+89000$.
3. $\frac{3(x+89000)}{2x}=\frac{5}{3}$ ⇒ $9x+801000=10x$ ⇒ $x=801000$.
4. Mohan's capital $=801000+89000=890000$, i.e. ₹8,90,000.

Trick: First turn the profit sum and difference into a ratio ($5:3$), then use capital × time.

Answer: (C) ₹8,90,000

Concept: Increasing a number by $100\%$ doubles it: new value $=x\times \frac{200}{100}=2x$.

1. $2x=2800$.
2. $x=\frac{2800}{2}=1400$.
3. Check: $1400+100\%\text{ of }1400=2800$.

Trick: $+100\%$ = double, so halve the result.

Answer: (A) $1400$

Concept: Expand the brackets and then combine like terms.

1. $x(7x-4)=7x^2-4x$.
2. $2(x^2-2)=2x^2-4$.
3. Sum $+11$: $7x^2+2x^2-4x-4+11=9x^2-4x+7$.

Trick: Put $x=1$: expression $=3+(-2)+11=12$, and $9-4+7=12$ ✓.

Answer: (A) $9x^2-4x+7$