RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Selling price $=\text{MP}\times \frac{100-\text{Discount}\%}{100}$.

1. MP = ₹900, discount = $13\%$.
2. SP $=900\times \frac{87}{100}=783$.

Trick: Discount $=13\%$ of $900=117$; $900-117=783$.

Answer: (C) ₹783

Concept: Sum of numbers = average × count.

1. Sum of three numbers $=20\times 3=60$.
2. Third number $=60-(18+28)=60-46=14$.

Trick: Deviation method: $18\to -2$, $28\to +8$; third must be $20-6=14$.

Answer: (C) 14

Concept: Divide the total by the sum of ratio parts to get the value of one part.

1. Total parts $=2+5+4=11$; one part $=\frac{638}{11}=58$.
2. Largest − smallest $=(5-2)$ parts $=3\times 58=174$.

Trick: Directly: $\frac{3}{11}\times 638=174$.

Answer: (D) 174

Concept: Take total work = LCM of man-days and woman-days, then find one person's daily work.

1. 2 men × 18 days = 36 man-days; 4 women × 6 days = 24 woman-days. Total work = LCM$(36,24)=72$ units.
2. 1 man $=\frac{72}{36}=2$ units/day; 1 woman $=\frac{72}{24}=3$ units/day.
3. 9 men + 6 women $=9\times 2+6\times 3=36$ units/day → time $=\frac{72}{36}=2$ days.

Trick: 9 men do $\frac{9}{36}=\frac{1}{4}$ and 6 women do $\frac{6}{24}=\frac{1}{4}$ per day → $\frac{1}{2}$ per day → 2 days.

Answer: (B) 2 days

Concept: Write each statement as a linear equation and solve.

1. Let Mary = $M$, Diya = $D$. Then $2M=6D+20\Rightarrow M=3D+10$.
2. Also $5D=M-6\Rightarrow M=5D+6$.
3. $3D+10=5D+6\Rightarrow D=2$, $M=16$.
4. Difference $=16-2=14$ years.

Trick: Check: $2\times 16=32=6\times 2+20$ ✓ and $5\times 2=10=16-6$ ✓.

Answer: (D) 14

Concept: Required speed $=\frac{\text{remaining distance}}{\text{remaining time}}$.

1. First part: $11\times 36=396\text{km}$.
2. Next 136 km at 34 km/h takes $\frac{136}{34}=4$ h. Covered $=396+136=532\text{km}$ in $15$ h.
3. Remaining: $574-532=42\text{km}$ in $18-15=3$ h → speed $=\frac{42}{3}=14\text{km/h}$.

Trick: Always track both distance and time used; the leftover of each gives the speed.

Answer: (B) $14\text{km/h}$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$ and Amount $=P+\text{SI}$.

1. $7\frac{1}{3}=\frac{22}{3}$ years: $6270=\frac{P\times 4.5\times 22}{3\times 100}=0.33P\Rightarrow P=19000$.
2. New SI $=\frac{19000\times 8.6\times 11.5}{100}=190\times 98.9=18791$.
3. Amount $=19000+18791=37791$.

Trick: Rate × time: $4.5\times \frac{22}{3}=33\%$ gives ₹6,270, so $P=\frac{6270}{0.33}$; then $8.6\times 11.5=98.9\%$.

Answer: (B) ₹37,791

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{18\times 32}=\sqrt{576}=24$.
2. Check: $18:24=24:32$ (both $=3:4$).

Trick: $18=2\times 9$, $32=2\times 16$ → $\sqrt{4\times 144}=2\times 12=24$.

Answer: (A) 24

Concept: Use ${\sin }^{2}B+{\cos }^{2}B=1$ and ${\csc }^{2}A-{\cot }^{2}A=1$.

1. $\frac{x^2}{a^2}={\csc }^{2}A{\sin }^{2}B$ and $\frac{y^2}{b^2}={\csc }^{2}A{\cos }^{2}B$.
2. Sum $={\csc }^{2}A({\sin }^{2}B+{\cos }^{2}B)={\csc }^{2}A$.
3. $\frac{z^2}{c^2}={\cot }^{2}A$, so the expression $={\csc }^{2}A-{\cot }^{2}A=1$.

Trick: Put $A={90}^{\circ }$, $B=0^{\circ }$: $x=0$, $y=b$, $z=0$ → value $=1$.

Answer: (D) $1$

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

1. $P=1200$, $R=4\%$ per month, $T=2$ months.
2. $\text{SI}=\frac{1200\times 4\times 2}{100}=96$.

Trick: $4\%$ per month for 2 months $=8\%$; $8\%$ of $1200=96$.

Answer: (A) ₹96

Concept: Total = average × number of matches; overall average = combined total ÷ total matches.

1. Runs in 23 matches $=23\times 40=920$.
2. Runs in next 10 matches $=10\times 14=140$.
3. Total $=920+140=1060$ in $33$ matches.
4. Average $=\frac{1060}{33}=32.1212\dots \approx 32.12$.

Trick: Deviation method: $40-\frac{10\times (40-14)}{33}=40-\frac{260}{33}=40-7.88=32.12$.

Answer: (A) $32.12$

Concept: When pipes work together, their one-hour works add: $\frac{1}{T}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

1. One-hour work $=\frac{1}{16}+\frac{1}{26}+\frac{1}{16}=\frac{1}{8}+\frac{1}{26}$.
2. $=\frac{13+4}{104}=\frac{17}{104}$.
3. Time $=\frac{104}{17}=6\frac{2}{17}$ hours.

Trick: LCM method: tank $=208$ units; A, B, C fill $13,8,13$ units/h $=34$ units/h; $\frac{208}{34}=6\frac{2}{17}$ h.

Answer: (C) $6\frac{2}{17}$ hours

Concept: Evaluate powers and roots first (BODMAS), then add/subtract left to right.

1. $6^2=36$, $\sqrt{3^2}=3$, $7\sqrt{81}=7\times 9=63$.
2. Value $=36+3-63-16$.
3. $=39-79=-40$.

Trick: $\sqrt{a^2}=a$ for positive $a$ – no need to square and root again.

Answer: (B) $-40$

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

1. $24\times 288=72\times x$.
2. $x=\frac{24\times 288}{72}=24\times 4=96$.
3. Check: $\text{HCF}(72,96)=24$, $\text{LCM}=288$ ✓.

Trick: $\frac{288}{72}=4$, so other number $=24\times 4$.

Answer: (D) $96$

Concept: Write every number in prime factors; LCM takes the highest power of each prime.

1. $2^3\times 7^2\times 12=2^5\times 3\times 7^2$.
2. $2^2\times {12}^2\times 18=2^2\times 2^4{3}^2\times 2·3^2=2^7\times 3^4$.
3. $7^3\times {12}^2\times {18}^2=7^3\times 2^4{3}^2\times 2^2{3}^4=2^6\times 3^6\times 7^3$.
4. LCM $=2^7\times 3^6\times 7^3$.

Trick: $12=2^2·3$, $18=2·3^2$ – convert first, then pick max powers.

Answer: (C) $2^7\times 3^6\times 7^3$

Concept: Expand each bracket and combine like terms.

1. $3(\frac{7}{3}{x}^2-26x+13)=7x^2-78x+39$.
2. $-7(x^2+9x-11)=-7x^2-63x+77$.
3. Add: $(7-7)x^2+(-78-63)x+(39+77)=-141x+116$.

Trick: The $x^2$ terms cancel ($7x^2-7x^2$), so only check the sign of the $x$-term: $-78-63=-141$.

Answer: (D) $-141x+116$

Concept: With false weight, the trader receives the price of 1 unit but gives only $(1-0.08)$ unit.

1. Let CP of 1 kg $=₹100$; he charges $₹152$.
2. He actually gives $0.92$ kg, costing $₹92$.
3. Profit $\%=\frac{152-92}{92}\times 100=\frac{6000}{92}=65.2\%$.
4. Nearest whole number $=65\%$.

Trick: $\text{Profit}\%=(\frac{1.52}{0.92}-1)\times 100\approx 65\%$.

Answer: (D) $65\%$

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 29=1100$.
2. $r=\frac{1100\times 7}{2\times 22\times 29}=\frac{7700}{1276}$.
3. $r\approx 6.034\approx 6.03\text{cm}$.

Trick: $\frac{1100}{44}=25$, so $r=\frac{25\times 7}{29}=\frac{175}{29}\approx 6.03$.

Answer: (C) $6.03$

Concept: BODMAS: brackets and powers first, then × and ÷ from left to right, then + and −.

1. $3\times 4^2=48$.
2. $13\times 48=624$, then $624÷9=\frac{208}{3}$.
3. $\frac{208}{3}+59-75=\frac{208}{3}-16=\frac{208-48}{3}=\frac{160}{3}$.

Trick: Only the options with denominator 3 can be right since $\frac{624}{9}=\frac{208}{3}$.

Answer: (C) $\frac{160}{3}$

Concept: Area of four walls of a room $=2(l+b)\times h$.

1. $l+b=10.5+8.5=19\text{m}$.
2. Area $=2\times 19\times 11$.
3. $=418{\text{m}}^2$.

Trick: Four walls = perimeter of floor × height: $38\times 11=418$.

Answer: (B) $418$