RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Overall no profit no loss ⇒ loss amount on the first frock = profit amount on the second frock.

1. Let CP of first $=x$, second $=1200-x$.
2. $\frac{7}{100}x=\frac{23}{100}(1200-x)$
3. $7x=27600-23x\Rightarrow 30x=27600\Rightarrow x=920$
4. Check: loss $=64.4$, profit on $280=64.4$ ✓

Trick: Alligation: $(-7)$ and $(+23)$ around $0$ give ratio $23:7$, so first $=1200\times \frac{23}{30}=920$.

Answer: (C) ₹920

Concept: Use $a^3+b^3=(a+b{)}^3-3ab(a+b)$ with $a={\sin }^2\theta$, $b={\cos }^2\theta$, $a+b=1$.

1. Numerator: ${\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta {\cos }^2\theta$, so numerator $=1-3{\sin }^2\theta {\cos }^2\theta +3{\sin }^2\theta {\cos }^2\theta =1$.
2. $(\sin \theta +\cos \theta {)}^2-1=2\sin \theta \cos \theta$.
3. Denominator $=\frac{1}{\sin \theta \cos \theta }\times 2\sin \theta \cos \theta =2$.
4. Value $=\frac{1}{2}$.

Trick: Put $\theta ={45}^{\circ }$: numerator $=\frac{1}{8}+\frac{1}{8}+\frac{3}{4}=1$, denominator $=\sqrt{2}·\sqrt{2}(2-1)=2$ → $\frac{1}{2}$.

Answer: (D) $\frac{1}{2}$

Concept: When a train overtakes a man moving in the same direction, it covers its own length $L$ at the relative speed (train − man).

1. Let train speed $=v\text{km/h}$.
2. $L=(v-9)\times 13.5=(v-12.6)\times 15$ (same units on both sides, so $\frac{5}{18}$ cancels).
3. $13.5v-121.5=15v-189\Rightarrow 1.5v=67.5\Rightarrow v=45$.
4. Length $=(45-9)\times \frac{5}{18}\times 13.5=135\text{m}$ ✓

Trick: Times ratio $13.5:15=9:10$ ⇒ relative speeds $10:9$; $(v-9)-(v-12.6)=3.6$ is 1 part, so $v-9=36$, $v=45$.

Answer: (C) $45\text{km/h}$

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

1. $r=16\text{cm}\Rightarrow r^2=256$.
2. $S=4\pi \times 256=1024\pi {\text{cm}}^2$.

Trick: Sphere surface area = 4 × (area of its great circle $\pi r^2$); $3\pi r^2=768\pi$ is the total surface area of a hemisphere – a common trap.

Answer: (D) $1024\pi {\text{cm}}^2$

Concept: Divisibility by 11: (sum of digits at odd places) − (sum at even places) must be $0$ or a multiple of $11$.

1. Digits: 7 4 7 5 7 9 7 2 3. Odd places: $7+7+7+7+3=31$; even places: $4+5+9+2=20$.
2. Difference $=31-20=11$ ⇒ divisible by 11 ($747579723=11\times 67961793$).
3. Digit sum $=51$ is not a multiple of 9; last three digits $723$ are odd, so not divisible by 8; $747579723÷13$ leaves remainder 7.

Trick: Check 11 by alternating sum quickly; check 9 by digit sum – one test each.

Answer: (B) 11

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

1. ${57}^{-8}÷{57}^{17}={57}^{-8-17}={57}^{-25}$.
2. ${57}^{-25}\times {57}^{-18}={57}^{-25-18}={57}^{-43}$.

Trick: Just add the exponents with signs: $-8-17-18=-43$.

Answer: (D) ${57}^{-43}$

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

1. $16\times 240=48\times x$
2. $x=\frac{3840}{48}=80$
3. Check: $\text{HCF}(48,80)=16$, $\text{LCM}=240$ ✓

Trick: Both numbers must be multiples of 16: 64, 80 qualify, but only $80$ gives LCM $240$.

Answer: (C) 80

Concept: If SP of $a$ articles = CP of $b$ articles with $a>b$, there is a loss: $\text{Loss}\%=\frac{a-b}{a}\times 100$.

1. Let CP of 1 book = ₹1 ⇒ SP of 40 books = CP of 28 books = ₹28.
2. CP of 40 books = ₹40 ⇒ loss = ₹12.
3. $\text{Loss}\%=\frac{12}{40}\times 100=30\%$ (not $\frac{100}{12}\%$, which wrongly divides by the loss).

Trick: More articles sold for the money ⇒ loss; divide the difference by the larger number.

Answer: (A) 30% loss

Concept: Take total work = LCM of the days and find each person's work per day.

1. Total work $=\text{LCM}(30,22,15)=330$ units.
2. Hema $=\frac{330}{30}=11$; H + R $=\frac{330}{22}=15$; H + R + S $=\frac{330}{15}=22$ units/day.
3. Smita $=22-15=7$ ⇒ Hema + Smita $=11+7=18$ units/day.
4. Days $=\frac{330}{18}=\frac{55}{3}=18\frac{1}{3}$.

Trick: Smita's rate = (all three) − (Hema + Raji); Raji's own rate is never needed.

Answer: (B) $\frac{55}{3}$ days

Concept: Equal simple interest for the same time ⇒ $P_1{R}_1=P_2{R}_2$, i.e. principal is inversely proportional to rate.

1. $P_1\times 14\times 2=P_2\times 10\times 2\Rightarrow P_1:P_2=10:14=5:7$.
2. Total parts $=12$ ⇒ 1 part $=\frac{21900}{12}=1825$.
3. $P_1=5\times 1825=9125$ (check: $9125\times 0.28=2555=12775\times 0.20$ ✓).

Trick: Swap the rates to get the ratio: at 14% : at 10% $=10:14$.

Answer: (D) ₹9,125

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