RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Average of the first $n$ even numbers $=n+1$.

1. Sum of first $n$ even numbers $=n(n+1)$.
2. Average $=\frac{n(n+1)}{n}=n+1$.
3. Here $n=167$, so average $=167+1=168$.

Trick: First $n$ even numbers → average $n+1$; first $n$ odd numbers → average $n$.

Answer: (D) $168$

Concept: Angle sum of a triangle is ${180}^{\circ }$; an exterior angle equals the sum of the two opposite interior angles.

1. In right $△BDC$: $\angle BCD={180}^{\circ }-{90}^{\circ }-{40}^{\circ }={50}^{\circ }$.
2. In $△AEC$: $\angle EAC={37}^{\circ }$, $\angle ECA={50}^{\circ }$.
3. $\angle AEB$ is the exterior angle of $△AEC$ at E: $\angle AEB={37}^{\circ }+{50}^{\circ }={87}^{\circ }$.

Trick: $\angle C={90}^{\circ }-\angle DBC$ directly, then add $\angle CAE$.

Answer: (B) ${87}^{\circ }$

Concept: Convert each part to a number and subtract.

1. $65\%$ of $65=\frac{65\times 65}{100}=42.25$
2. $\frac{3}{5}$ of $55=33$
3. Difference $=42.25-33=9.25$

Trick: ${65}^2=4225$, so $65\%$ of $65$ is simply $42.25$.

Answer: (A) $9.25$

Concept: Margin $=$ (winner% $-$ loser%) of valid votes.

1. Let registered voters $=N$. Votes cast $=0.92N$; valid $=0.8\times 0.92N=0.736N$.
2. Winner $75\%$, loser $25\%$ → margin $=50\%$ of valid $=0.368N$.
3. $0.368N=322\Rightarrow N=\frac{322}{0.368}=875$.

Trick: Check options: $875\times 0.92\times 0.8\times 0.5=322$ ✔.

Answer: (A) $875$

Concept: Total surface area of a cube $=6a^2$.

1. $a=40\text{cm}$, so $a^2=1600$.
2. TSA $=6\times 1600=9600{\text{cm}}^2$.

Yaad rakhein: Cube has 6 equal square faces; lateral surface area $=4a^2$.

Answer: (C) $9600{\text{cm}}^2$

Concept: For equal time, profit is shared in the ratio of investments.

1. Ratio $P:Q:R=2030:1700:1010=203:170:101$.
2. Q's 170 parts $=$ ₹975 → 1 part $=\frac{975}{170}$.
3. $P-R=203-101=102$ parts $=\frac{975\times 102}{170}=975\times 0.6=585$.

Trick: $\frac{102}{170}=\frac{3}{5}$, so the difference is $\frac{3}{5}$ of Q's share.

Answer: (A) ₹585

Concept: Third proportional to $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{24}^2}{32}=\frac{576}{32}$
2. $x=18$

Yaad rakhein: Third proportional $=\frac{b^2}{a}$; mean proportional $=\sqrt{ab}$.

Answer: (D) $18$

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

1. $7p+8c=230$ …(1)
2. $14(p-1)+4(c+7)=150\Rightarrow 14p+4c=136\Rightarrow 7p+2c=68$ …(2)
3. (1) $-$ (2): $6c=162\Rightarrow c=27$; then $7p=230-216=14\Rightarrow p=2$.
4. $8p+8c=8(2+27)=232$

Trick: Halve the second equation to match the $7p$ term and subtract directly.

Answer: (C) ₹232

Concept: Test odd numbers for divisibility by primes up to $\sqrt{403}\approx 20$ (2, 3, 5, 7, 11, 13, 17, 19).

1. Odd numbers: 389, 391, 393, 395, 397, 399, 401, 403.
2. $391=17\times 23$, $393=3\times 131$, $395=5\times 79$, $399=3\times 133$, $403=13\times 31$ → composite.
3. 389, 397, 401 are prime → 3 primes.

Trick: Digit sum divisible by 3 (393, 399) and ending in 5 (395) are removed at once.

Answer: (B) $3$

Concept: If $2A=9B=6C=k$, then $A:B:C=\frac{1}{2}:\frac{1}{9}:\frac{1}{6}$.

1. Multiply by LCM 18: $A:B:C=9:2:3$; total $=14$ parts.
2. A's share $=\frac{9}{14}\times 28000=18000$.

Trick: Shares are inversely proportional to the multipliers 2, 9, 6.

Answer: (D) ₹18,000

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{5}{3}{x}^2-28x+15)=5x^2-84x+45$
2. $-5(x^2+6x-15)=-5x^2-30x+75$
3. Add: $(5x^2-5x^2)+(-84x-30x)+(45+75)=-114x+120$

Trick: The $x^2$ terms cancel; only track the $x$ terms ($-84-30$) and constants ($45+75$).

Answer: (D) $-114x+120$

Concept: Total $=$ average $\times$ number; then share in ratio.

1. Total price $=3\times 5310=15930$.
2. Ratio sum $=3+5+7=15$.
3. Cheapest $=\frac{3}{15}\times 15930=3186$.

Trick: Average of the parts $=5$ parts $=5310$ → 1 part $=1062$ → 3 parts $=3186$.

Answer: (B) $3,186$

Concept: One-day work of C $=$ (A+B+C) work $-$ A's work $-$ B's work.

1. $\frac{1}{C}=\frac{1}{2}-\frac{1}{22}-\frac{1}{10}$
2. LCM 110: $\frac{55-5-11}{110}=\frac{39}{110}$
3. C alone $=\frac{110}{39}=2\frac{32}{39}$ days.

Trick: All three finish in 2 days, so C alone must take more than 2 days but fewer than A or B — only $2\frac{32}{39}$ fits.

Answer: (D) $2\frac{32}{39}$ days

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

1. Equivalent discount $=10+15-\frac{150}{100}=23.5\%$.
2. Difference in discount $=25\%-23.5\%=1.5\%$.
3. Difference in SP $=1.5\%$ of $38800=582$.

Trick: No need to find both SPs — just take the gap between the two equivalent discounts on the MP.

Answer: (D) ₹582

Concept: Let Ram $=R$, Ramya $=r$ and form two equations.

1. $2R=9r+25$ and $5r=R-4\Rightarrow R=5r+4$.
2. $2(5r+4)=9r+25\Rightarrow 10r+8=9r+25\Rightarrow r=17$.
3. $R=5(17)+4=89$; difference $=89-17=72$.

Trick: Substitute $R=5r+4$ straight away — one variable remains.

Answer: (A) $72$

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$.

1. Discount $=10\%$ of $400=40$.
2. SP $=400-40=360$.

Trick: 10% off → multiply MP by $0.9$: $400\times 0.9=360$.

Answer: (A) ₹360

Concept: Count days excluding the deposit day and including the withdrawal day; $\text{SI}=\frac{P\times R\times T}{100}$.

1. 2024 is a leap year: February $29-21=8$ days, March 31, April 22 → $61$ days.
2. $T=\frac{61}{366}$ year.
3. $\text{SI}=\frac{4000\times 6.75\times 61}{100\times 366}=\frac{16470}{366}=45$

Yaad rakhein: Count only one of the two end dates; with 365 days the SI is $45.12\approx 45$, the same option.

Answer: (A) ₹45

Concept: If the bases are equal, the exponents are equal.

1. Both $x$ and $y$ equal $7^{0.07}$, so $x=y$.
2. $x^z=y^3\Rightarrow (7^{0.07}{)}^z=(7^{0.07}{)}^3$.
3. $0.07z=0.21\Rightarrow z=3$.

Trick: $x=y$ (and $x\ne 1$), so $x^z=x^3$ gives $z=3$ at once.

Answer: (B) $3$

Concept: Distance $=$ speed $\times$ time (same units). The car's own length is negligible.

1. $3.6$ min $=\frac{3.6}{60}=0.06$ h.
2. Length $=79\times 0.06=4.74\text{km}$.

Trick: Minutes → hours: divide by 60; $3.6÷60=0.06$.

Answer: (C) $4.74\text{km}$

Concept: Product of two numbers $=$ HCF $\times$ LCM.

1. $55\times x=5\times 495=2475$
2. $x=\frac{2475}{55}=45$
3. Check: HCF(55, 45) $=5$ ✔

Trick: The other number must be a multiple of the HCF 5 — rules out 64, 8, 9 instantly.

Answer: (A) $45$