RRB NTPC Undergraduate CBT 1 2024 PYQs

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) $418$

Concept: Exterior angle $={180}^{\circ }-\text{interior}$; $n=\frac{{360}^{\circ }}{\text{exterior}}$; diagonals $=\frac{n(n-3)}{2}$.

1. Exterior angle $={180}^{\circ }-{140}^{\circ }={40}^{\circ }$.
2. $n=\frac{360}{40}=9$ sides.
3. Diagonals $=\frac{9\times 6}{2}=27$.

Trick: Each vertex sends $n-3$ diagonals; every diagonal is counted twice, hence divide by 2.

Answer: (C) $27$

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

1. Ritik's profit $=19000-7600=₹11400$.
2. Ratio R : S $=11400:7600=3:2$.
3. $5$ parts $=₹77600$ ⇒ $1$ part $=₹15520$.
4. Difference $=3-2=1$ part $=₹15520$.

Trick: Difference of investments $=77600\times \frac{3-2}{5}$.

Answer: (B) ₹15,520

Concept: Build the chain: registered → polled → valid; margin = winner% − loser% of valid votes.

1. Let registered voters $=x$. Polled $=0.96x$.
2. Valid $=0.85\times 0.96x=0.816x$.
3. Margin $=(80\%-20\%)$ of valid $=0.6\times 0.816x=0.4896x$.
4. $0.4896x=1224$ ⇒ $x=2500$.

Trick: Check from options: $2500\times 0.96\times 0.85\times 0.6=1224$ ✓.

Answer: (C) $2500$

Concept: $24=3\times 8$ (co-prime), so the number must be divisible by both 3 and 8.

1. 7340390: last three digits $390$ are not divisible by 8 ✗; 7991987 is odd ✗.
2. 6830908: $908÷8=113.5$ ✗.
3. 7132968: $968÷8=121$ ✓ and digit sum $7+1+3+2+9+6+8=36$ (divisible by 3) ✓.

Trick: Test 8 first (last three digits), then 3 (digit sum).

Answer: (C) 71,32,968

Concept: Overhead (decoration) is added to CP; $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP $=322+6=₹328$; SP $=₹398$ ⇒ profit.
2. Profit $=398-328=₹70$.
3. Profit $\%=\frac{70}{328}\times 100=\frac{7000}{328}=\frac{875}{41}\%$ $(\approx 21.34\%)$.

Trick: SP > CP ⇒ profit, so the loss options drop out at once.

Answer: (D) $\frac{875}{41}\%$ profit

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. Places of N73M10: odd places N, 3, 1 → $N+4$; even places 7, M, 0 → $M+7$.
2. $(M+7)-(N+4)=M-N+3$ must be $0$ or $\pm 11$.
3. $M-N+3=0$ ⇒ $M-N=-3$ (e.g. $N=5,M=2$: 573210 $=11\times 52110$).

Trick: Alternate sum from the right: $0-1+M-3+7-N=M-N+3$.

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

Concept: Increasing by $100\%$ means the number becomes double.

1. $x+100\%\text{ of }x=2x$.
2. $2x=2730$.
3. $x=1365$.

Trick: $100\%$ increase ⇒ halve the result.

Answer: (C) $1365$

Concept: For the same distance, time is inversely proportional to speed.

1. Distance $=150\times 8=1200\text{km}$.
2. Gauri's time $=\frac{1200}{60}=20$ hours.

Trick: $t_2=t_1\times \frac{v_1}{v_2}=8\times \frac{150}{60}=20$.

Answer: (A) $20$ hours

Concept: Successive discounts $a\%$ and $b\%$ equal $(a+b-\frac{ab}{100})\%$.

1. After $20\%$: $1000\to 800$.
2. After $5\%$ more: $800\times 0.95=760$.
3. Total discount $=1000-760=₹240=24\%$.

Trick: $20+5-\frac{20\times 5}{100}=24\%$.

Answer: (C) $24\%$

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

1. $19p+13c=254$ …(1)
2. $14(p-6)+2(c+9)=106$ ⇒ $14p+2c=172$ ⇒ $c=86-7p$ …(2)
3. Put in (1): $19p+1118-91p=254$ ⇒ $72p=864$ ⇒ $p=12$, $c=2$.
4. $12p+15c=144+30=₹174$.

Trick: Check (1): $19\times 12+13\times 2=228+26=254$ ✓.

Answer: (C) ₹174