RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Form two linear equations from the statements and solve.

1. $3M=9D+24\Rightarrow M=3D+8$.
2. $4D=M-2\Rightarrow M=4D+2$.
3. $3D+8=4D+2\Rightarrow D=6$, $M=26$; difference $=20$.

Trick: Difference $=M-D$: from $M=4D+2$, it is $3D+2=20$.

Answer: (A) 20

Concept: LCM of decimals: make equal decimal places, find LCM of the numbers, then put back the decimal.

1. $5.3=\frac{53}{10}$, $0.4=\frac{4}{10}$.
2. $\text{LCM}(53,4)=212$.
3. LCM $=\frac{212}{10}=21.2$.

Trick: LCM must be a multiple of both: $21.2=5.3\times 4=0.4\times 53$.

Answer: (C) 21.2

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Digit sum $=5+6+3+6+0+7=27$, divisible by 3 ✓.
2. Not by 8 (last three digits 607 odd), not by 10 (ends in 7).

Trick: Digit sum 27 is also divisible by 9, so the number is divisible by 9 as well.

Answer: (B) 3

Concept: Winning margin = (winner's % − loser's %) of valid votes.

1. Votes polled $=0.8R$; valid $=81\%$ of polled $=0.648R$.
2. Margin $=(70\%-30\%)=40\%$ of valid $=0.2592R$.
3. $0.2592R=1458\Rightarrow R=5625$.

Trick: $R\times \frac{80}{100}\times \frac{81}{100}\times \frac{40}{100}=1458$; $5625={75}^2$ makes it clean.

Answer: (A) 5625

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

1. ${95}^{-17}÷{95}^9={95}^{-26}$.
2. ${95}^{-26}\times {95}^{-15}={95}^{-41}$.

Trick: Just add/subtract exponents: $-17-9-15=-41$.

Answer: (B) ${95}^{-41}$

Concept: Form two equations in pen price $p$ and pencil price $c$.

1. $2p+15c=286$.
2. $17(p-4)+3(c+2)=128\Rightarrow 17p+3c=190$.
3. Multiply second by 5: $85p+15c=950$; subtract: $83p=664\Rightarrow p=8$, $c=18$.
4. $11p+8c=88+144=232$.

Trick: Make the $c$ coefficients equal ($15c$) and eliminate.

Answer: (C) ₹232

Concept: For the same time, profit ratio = investment ratio.

1. Vivek : Salman profit $=3700:6300=37:63$.
2. Vivek $=0.37\times 33900=12543$; Salman $=21357$.
3. Difference $=0.26\times 33900=8814$.

Trick: Difference of shares $=\frac{63-37}{100}\times 33900=8814$ directly.

Answer: (A) ₹8,814

Concept: Equal SI for equal time ⇒ principals are in inverse ratio of rates.

1. $x\times 8\times 6=(20300-x)\times 6\times 6$.
2. $8x=6(20300-x)\Rightarrow 14x=121800$.
3. $x=8700$.

Trick: Sums at $8\%:6\%$ are $6:8=3:4$; $\frac{3}{7}\times 20300=8700$.

Answer: (D) 8,700

Concept: If SP of $a$ items = CP of $b$ items, gain $\%=\frac{b-a}{a}\times 100$.

1. Let CP of 1 book $=$ ₹1. SP of 10 books $=$ ₹16, CP of 10 books $=$ ₹10.
2. Gain $=6$ on $10$ ⇒ $\frac{6}{10}\times 100=60\%$ gain.

Trick: SP count less than CP count ⇒ gain; $\frac{16-10}{10}=60\%$.

Answer: (C) 60% gain

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

1. $r=8\text{cm}$, $r^2=64$.
2. $4\times 3.14\times 64=12.56\times 64=803.84{\text{cm}}^2$.

Yaad rakhein: $4\pi r^2$ — four times the area of its great circle.

Answer: (D) $803.84{\text{cm}}^2$

Concept: In a partnership, profit is shared in the ratio of (capital $\times$ time).

1. Total profit = ₹1,494 and M gets ₹498 more, so M $=\frac{1494+498}{2}=996$, N $=498$; ratio M : N $=2:1$.
2. Let N's capital $=x$; M's $=x+17000$. Then $\frac{6(x+17000)}{5x}=\frac{2}{1}$.
3. $6x+102000=10x\Rightarrow x=25500$.
4. M's capital $=25500+17000=42500$.

Trick: Check options: M = 42,500 → N = 25,500; $\frac{42500\times 6}{25500\times 5}=\frac{255000}{127500}=2$ ✓.

Answer: (A) ₹42,500

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with R and T in the same time unit.

1. Here $P=1200$, $R=4\%$ per month, $T=10$ months.
2. $\text{SI}=\frac{1200\times 4\times 10}{100}=480$.
3. (If the rate were 4% per annum, $T=\frac{10}{12}$ year would give $\frac{1200\times 4\times 10}{1200}=40$ – but the question says per month.)

Trick: 4% per month for 10 months = 40% of 1200 = 480.

Answer: (C) ₹480

Concept: Expand each bracket, then combine like terms.

1. $x(7x-5)=7x^2-5x$
2. $3(x^2-2)=3x^2-6$
3. Sum: $7x^2-5x+3x^2-6+14=10x^2-5x+8$

Trick: Put $x=1$: $1(2)+3(-1)+14=13$; only $10-5+8=13$ ✓.

Answer: (C) $10x^2-5x+8$

Concept: Find each pipe's full-tank time, take LCM as tank capacity, and add efficiencies (emptying pipe negative).

1. P fills $\frac{4}{5}$ in 12 h → full in $12\times \frac{5}{4}=15$ h; Q fills $\frac{2}{7}$ in 30 h → full in $30\times \frac{7}{2}=105$ h.
2. Tank $=105$ units: P $=7$, Q $=1$ unit/h. In 5 h: $5\times 8=40$ units.
3. R empties 40 units in 8 h → R $=-5$ unit/h.
4. P + Q + R $=7+1-5=3$ unit/h → time $=\frac{105}{3}=35$ h.

Yaad rakhein: Filling pipe +, emptying pipe −; time $=\frac{\text{capacity}}{\text{net efficiency}}$.

Answer: (B) 35 hours

Concept: First $n$ even numbers: $2,4,\dots ,2n$; their sum $=n(n+1)$, so average $=n+1$.

1. Here $n=163$; last term $=2\times 163=326$.
2. Average $=\frac{2+326}{2}=164$.
3. Or directly: $n+1=163+1=164$.

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

Answer: (D) 164

Concept: Simplify the inner (continued) fraction first, then add/subtract using the LCM.

1. $1+\frac{4}{9}=\frac{13}{9}$, so $\frac{1}{1+\frac{4}{9}}=\frac{9}{13}$.
2. Expression $=\frac{5}{3}+\frac{9}{13}-\frac{2}{5}$; LCM of 3, 13, 5 $=195$.
3. $=\frac{325+135-78}{195}=\frac{382}{195}$.

Trick: Denominator must be the LCM 195 – only option B has it.

Answer: (B) $\frac{382}{195}$

Concept: Consecutive natural numbers are $n$ and $n+1$.

1. $n(n+1)=110\Rightarrow n^2+n-110=0$.
2. $(n+11)(n-10)=0\Rightarrow n=10$ (natural).
3. Numbers: 10 and 11; greater $=11$.

Trick: $\sqrt{110}\approx 10.5$ → the numbers lie on either side: $10\times 11=110$.

Answer: (C) 11

Concept: For a decimal, take the square root of the digits and halve the number of decimal places.

1. $\sqrt{121}=11$
2. $\sqrt{0.0225}=0.15$ (since ${15}^2=225$)
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$)
4. $11+0.15-2.4=8.75$

Trick: Decimal places in the square = 2 × decimal places in the root: 0.0225 (4) → 0.15 (2).

Answer: (C) 8.75

Concept: Use $1+{\tan }^2\beta ={\sec }^2\beta$ and $\tan \beta \text{cosec}\beta =\frac{\sin \beta }{\cos \beta }·\frac{1}{\sin \beta }=\sec \beta$.

1. Numerator: ${\tan }^2\beta {\text{cosec}}^2\beta ={\sec }^2\beta$.
2. Denominator: $1+{\tan }^2\beta ={\sec }^2\beta$.
3. $\frac{{\sec }^2\beta }{{\sec }^2\beta }=1$.

Trick: Put $\beta ={45}^{\circ }$: $\frac{1\times 2}{1+1}=1$.

Answer: (C) 1

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $72\times 432=144\times x$
2. $x=\frac{72\times 432}{144}=\frac{432}{2}=216$.
3. Check: HCF(144, 216) $=72$, LCM $=432$ ✓.

Trick: $\frac{144}{72}=2$, so $x=\frac{432}{2}=216$.

Answer: (A) 216