RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ per sharpener.

1. CP of one sharpener $=\frac{150}{12}=12.5$ (₹)
2. SP $=21$ (₹), profit $=21-12.5=8.5$ (₹)
3. $\text{Profit}\%=\frac{8.5}{12.5}\times 100=68\%$

Trick: Compare per dozen: SP $=21\times 12=252$ vs CP 150 → $\frac{102}{150}\times 100=68\%$.

Answer: (D) 68

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

1. Discount $=20\%$ of 2400 $=480$ (₹)
2. SP $=2400-480=1920$ (₹)

Trick: $20\%$ off means pay $80\%$: $2400\times 0.8=1920$.

Answer: (C) ₹1,920

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=40+49+56+73+37=255\text{kg}$
2. Average $=\frac{255}{5}=51\text{kg}$

Trick: Assume 50: deviations $-10,-1,+6,+23,-13$ sum to $+5$; $50+\frac{5}{5}=51$.

Answer: (C) 51 kg

Concept: Express the actual SP in terms of CP using the loss condition, then use the second selling price.

1. $\frac{2}{7}S=76\%$ of CP $=0.76C$
2. $S=\frac{7\times 0.76C}{2}=2.66C$
3. New SP $=55\%$ of $S=0.55\times 2.66C=1.463C$
4. Profit $=46.3\%$

Trick: Let CP $=100$: $\frac{2}{7}S=76\Rightarrow S=266$; $55\%$ of 266 $=146.3$ → $46.3\%$ profit.

Answer: (D) $46.3\%$

Concept: SP $=$ MP after discount; CP $=\frac{\text{SP}\times 100}{100+\text{profit}\%}$.

1. SP $=280\times \frac{87.5}{100}=245$ (₹)
2. CP $=\frac{245\times 100}{140}=175$ (₹)

Trick: $12.5\%=\frac{1}{8}$: SP $=280\times \frac{7}{8}=245$; $40\%=\frac{2}{5}$: CP $=245\times \frac{5}{7}=175$.

Answer: (C) ₹175

Concept: Profit share $\propto$ capital $\times$ time.

1. Shares: M $+$ N $=3304$, M $-$ N $=472$ ⇒ M $=1888$, N $=1416$; ratio $=4:3$
2. Let N’s capital $=x$; M’s $=x+43000$
3. $\frac{6(x+43000)}{7x}=\frac{4}{3}\Rightarrow 18x+774000=28x$
4. $x=77400$; M $=77400+43000=1,20,400$ (₹)

Trick: Check (A): M $=120400$, N $=77400$; $120400\times 6:77400\times 7=722400:541800=4:3$ ✓

Answer: (A) ₹1,20,400

Concept: $A=P{(1+\frac{r}{100})}^n$.

1. $\frac{1367631}{1000000}=(1.11{)}^n$
2. ${1.11}^2=1.2321$, ${1.11}^3=1.367631$
3. So $n=3$ years

Trick: ${111}^3=1367631$ — the six digits after 1 are the giveaway for a cube.

Answer: (A) 3 years

Concept: Total work = LCM of days; find work done together and subtract.

1. Total $=\text{LCM}(9,18)=18$ units; A $=2$/day, B $=1$/day
2. Together $=3$ units/day; in 5 days $=15$ units
3. Remaining $=18-15=3$ units $=\frac{3}{18}=\frac{1}{6}$

Trick: $5(\frac{1}{9}+\frac{1}{18})=5\times \frac{1}{6}=\frac{5}{6}$ done, so $\frac{1}{6}$ left.

Answer: (A) $\frac{1}{6}$

Concept: Square: perimeter $=4a$, area $=a^2$.

1. $4a=160\Rightarrow a=40\text{m}$
2. Area $={40}^2=1600{\text{m}}^2$

Trick: Area of a square $={\left(\frac{\text{perimeter}}{4}\right)}^2$.

Answer: (A) 1600

Concept: Open the brackets carefully (sign of $-3$ changes every term), then combine like terms.

1. $2(\frac{3}{2}{x}^2-28x+14)=3x^2-56x+28$
2. $-3(x^2+6x-11)=-3x^2-18x+33$
3. Add: $3x^2-3x^2-56x-18x+28+33=-74x+61$

Yaad rakhein: A negative multiplier flips every sign inside: $-3\times (-11)=+33$.

Answer: (D) $-74x+61$

Concept: For a fixed distance, speed is inversely proportional to time: $S_1{T}_1=S_2{T}_2$.

1. Distance $=14\times 20=280\text{km}$.
2. Required speed $=\frac{280}{5}=56\text{km/h}$.

Trick: Time becomes $\frac{1}{4}$, so speed becomes $4\times 14=56$.

Answer: (A) $56\text{km/h}$

Concept: Sum of first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Here $n=149$.
2. Average $=\frac{{149}^2}{149}=149$.

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

Answer: (C) $149$

Concept: $A=P{(1+\frac{R}{100})}^n$.

1. ${(1+\frac{R}{100})}^2=\frac{2809}{2500}$.
2. $1+\frac{R}{100}=\sqrt{\frac{2809}{2500}}=\frac{53}{50}$.
3. $\frac{R}{100}=\frac{3}{50}\Rightarrow R=6\%$.

Trick: $2500={50}^2$, $2809={53}^2$; $\frac{53-50}{50}\times 100=6\%$.

Answer: (B) $6\%$

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is $0$ or a multiple of $11$.

1. $877217$: odd places from right $7+2+7=16$; even places $1+7+8=16$.
2. Difference $=16-16=0\Rightarrow$ divisible by $11$.
3. Last digit $7$ is odd, so not divisible by $2$, $12$ or $16$.

Trick: Odd last digit rules out every even divisor at once.

Answer: (D) $11$

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

1. Equivalent of $35\%$ and $4\%$ $=35+4-\frac{35\times 4}{100}=37.6\%$.
2. Difference of discounts $=52\%-37.6\%=14.4\%$.
3. Amount $=\frac{14.4}{100}\times 1125=162$.

Trick: Directly: pay $0.48\times 1125=540$ vs $0.65\times 0.96\times 1125=702$; gap $=162$.

Answer: (A) ₹$162$

Concept: For divisibility by $11$, the difference of sums of alternate digits must be $0$ or a multiple of $11$.

1. $N51M54$: places 1, 3, 5 give $N+1+5=N+6$.
2. Places 2, 4, 6 give $5+M+4=M+9$.
3. $(M+9)-(N+6)=M-N+3=0\Rightarrow M-N=-3$ (e.g. $N=4,M=1$: $451154=11\times 41014$).

Trick: Write the alternate sums, equate the difference to $0$ first.

Answer: (B) $M-N=-3$

Concept: Net rate $=\frac{1}{A}+\frac{1}{B}-\frac{1}{C}=\frac{1}{\text{time together}}$.

1. $\frac{1}{93}+\frac{1}{31}=\frac{1+3}{93}=\frac{4}{93}$.
2. $\frac{1}{C}=\frac{4}{93}-\frac{1}{24}=\frac{96-93}{2232}=\frac{3}{2232}=\frac{1}{744}$.
3. Emptying pipe alone takes $744\text{min}$.

Trick: LCM method: tank $=744$ units; A $=8$, B $=24$, all $=31$ per min $\Rightarrow$ C $=1$ unit/min.

Answer: (B) $744\text{min}$

Concept: Volume of a cylinder $V=\pi r^{2}h$.

1. $V=\frac{22}{7}\times 9^2\times 35$.
2. $=22\times 81\times 5$.
3. $=8910{\text{cm}}^3$.

Trick: $h=35$ is a multiple of $7$ — cancel first: $\frac{35}{7}=5$.

Answer: (A) $8910{\text{cm}}^3$

Concept: Let Pranshu's speed be $p$ and Chandu's time be $t$ hours.

1. $\frac{56}{p}=t+3$ and $\frac{56}{2p}=t-4$.
2. Subtract: $\frac{56}{p}-\frac{28}{p}=7\Rightarrow \frac{28}{p}=7$.
3. $p=4\text{km/h}$ (check: $14\text{h}$ vs $7\text{h}$, Chandu $11\text{h}$).

Trick: Doubling speed halves time; the halved part equals the gap $3+4=7\text{h}$.

Answer: (C) $4\text{km/h}$

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. $\sqrt{25\times 196}$.
2. $=5\times 14=70$.

Trick: Root each number separately: $\sqrt{25}=5$, $\sqrt{196}=14$.

Answer: (D) $70$