RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: The largest cube has side = smallest dimension $=6\text{cm}$. Cutting a cube from a corner removes 3 faces but exposes 3 equal new faces.

1. Original TSA $=2(lb+bh+hl)=2(80+48+60)=2\times 188=376$.
2. Removed surface $=3\times 6^2=108$; newly exposed surface $=3\times 6^2=108$.
3. Net change $=0$, so remaining TSA $=376{\text{cm}}^2$.

Trick: A cube cut from a corner never changes the total surface area.

Answer: (B) $376{\text{cm}}^2$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Times: $\frac{585}{65}=9$, $\frac{368}{92}=4$, $\frac{168}{56}=3$ → total $16$ h.
2. Total distance $=585+368+168=1121$ km.
3. Average $=\frac{1121}{16}=70.0625\approx 70.06\text{km/h}$.

Trick: Never average the speeds; each part divides exactly, so find hours first.

Answer: (D) $70.06$

Concept: Convert averages to totals.

1. $T+S+M=3\times 45=135$.
2. $T+S=2\times 44=88$, $S+M=2\times 47=94$.
3. $S=(T+S)+(S+M)-(T+S+M)=88+94-135=47$.

Trick: Common person = sum of the two pair-totals − total of all three.

Answer: (D) $47$

Concept: Overheads (decoration) are added to cost price. $\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100$.

1. CP $=255+5=260$.
2. SP $=154$ → loss $=260-154=106$.
3. $\text{Loss}\%=\frac{106}{260}\times 100=\frac{530}{13}\%\approx 40.77\%$.

Trick: SP < CP means loss – this removes both 'profit' options at once.

Answer: (C) $\frac{530}{13}\%$ loss

Concept: Write $\cot$ as $\frac{1}{\tan }$ and simplify.

1. $\tan A-\cot B=\tan A-\frac{1}{\tan B}=\frac{\tan A\tan B-1}{\tan B}$.
2. $\tan B-\cot A=\frac{\tan A\tan B-1}{\tan A}$.
3. Ratio $=\frac{\tan A}{\tan B}=\tan A\cot B$.

Trick: Put $A={60}^{\circ },B={45}^{\circ }$: $\frac{\sqrt{3}-1}{1-\frac{1}{\sqrt{3}}}=\sqrt{3}=\tan {60}^{\circ }\cot {45}^{\circ }$.

Answer: (B) $\tan A\cot B$

Concept: Expand the brackets and combine like terms.

1. $x(4x-4)=4x^2-4x$.
2. $5(x^2-5)=5x^2-25$.
3. Sum $+12$: $9x^2-4x-25+12=9x^2-4x-13$.

Trick: Constant check: $-25+12=-13$ removes two options instantly.

Answer: (C) $9x^2-4x-13$

Concept: Take square roots of perfect squares, including decimals.

1. $\sqrt{196}=14$.
2. $\sqrt{0.0225}=0.15$ (since ${15}^2=225$).
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$).
4. $14+0.15-2.4=11.75$.

Trick: Decimal places of the root = half the decimal places of the number.

Answer: (C) $11.75$

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

1. Customer pays for 6 and gets $6+4=10$ trousers.
2. Free $=4$ out of $10$.
3. Discount $=\frac{4}{10}\times 100=40\%$.

Trick: Buy $x$ get $y$ free → discount $=\frac{y}{x+y}\times 100$.

Answer: (B) $40\%$

Concept: Let the numbers be $n$ and $n+2$ and test squares.

1. $n^2+(n+2{)}^2=164\Rightarrow 2n^2+4n-160=0\Rightarrow n^2+2n-80=0$.
2. $(n+10)(n-8)=0\Rightarrow n=8$.
3. Numbers $8,10$ ($64+100=164$); sum $=18$.

Trick: Half of 164 is 82 – the squares lie around $9^2=81$, so try 8 and 10.

Answer: (D) $18$

Concept: In 'Buy 3 get 1 free', 4 items cost the price of 3; then apply the extra discount.

1. Price of 3 items $=3\times 3000=9000$.
2. 20% discount $=0.2\times 9000=1800$.
3. Amount paid $=9000-1800=7200$.

Trick: Directly: $9000\times 0.8=7200$.

Answer: (A) ₹7,200

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