RRB NTPC Undergraduate CBT 1 2024 PYQs

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

1. $4\times 208=52\times x$
2. $x=\frac{832}{52}=16$
3. Check: $\text{HCF}(52,16)=4$, $\text{LCM}=208$ ✓

Trick: $\frac{208}{52}=4$, so the other number $=4\times 4=16$.

Answer: (C) $16$

Concept: $\text{Area}=l\times b$ and $\text{Perimeter}=2(l+b)$.

1. $b=\frac{1904}{68}=28\text{m}$
2. $\text{Perimeter}=2(68+28)=2\times 96=192\text{m}$

Trick: $68\times 28$: unit digits $8\times 8$ end in 4, matching 1904 – quick check of $b$.

Answer: (A) $192\text{m}$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$

1. $\text{SI}=\frac{3000\times 8\times 6}{100}$
2. $=30\times 48=1440$

Trick: 8% for 6 years = 48% of 3000 = 1440.

Answer: (A) ₹1,440

Concept: If SP of $a$ items = CP of $b$ items and $b>a$, there is a gain of $\frac{b-a}{a}\times 100\%$.

1. Let CP of 1 book = ₹1, so SP of 25 books = ₹37.
2. CP of 25 books = ₹25 → gain = ₹12.
3. $\text{Gain}\%=\frac{12}{25}\times 100=48\%$

Trick: Gain% $=\frac{\text{difference}}{\text{SP count}}\times 100=\frac{37-25}{25}\times 100$.

Answer: (D) $48\%$ gain

Concept: Increasing a number by 80% makes it $180\%$ of itself, i.e. multiplied by $\frac{180}{100}$.

1. $x\times \frac{180}{100}=2880$
2. $x=\frac{2880\times 100}{180}=16\times 100=1600$

Trick: $2880÷1.8=1600$.

Answer: (A) $1600$

Concept: $\text{Sum}=\text{Average}\times \text{count}$.

1. Sum of first two $=2\times 80=160$; third $=105$.
2. Remaining two $=655-160-105=390$.
3. Average $=\frac{390}{2}=195$

Trick: Subtract the known parts from the total, then divide by the count left.

Answer: (D) $195$

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$ (compare per piece).

1. CP of 1 sharpener $=\frac{150}{12}=12.5$ (₹).
2. SP $=43$ → profit $=43-12.5=30.5$.
3. $\text{Profit}\%=\frac{30.5}{12.5}\times 100=244\%$

Trick: Per dozen: CP 150, SP $43\times 12=516$; profit $=\frac{366}{150}\times 100=244\%$.

Answer: (B) $244$

Concept: Expand the brackets, then combine like terms.

1. $x(3x-8)=3x^2-8x$
2. $3(x^2-5)=3x^2-15$
3. Sum: $3x^2-8x+3x^2-15+11=6x^2-8x-4$

Trick: Constant: $-15+11=-4$ and $x^2$ coefficient $+6$ → only option (D) fits.

Answer: (D) $6x^2-8x-4$

Concept: Form two equations from the two conditions and solve.

1. Now: $2P-Q=15$ … (i)
2. After 9 years: $2(Q+9)=2(P+9)-14\Rightarrow 2P-2Q=14$ … (ii)
3. (i) − (ii): $Q=1$; then $P=8$.

Trick: Subtracting the equations removes $2P$ directly and gives $Q$.

Answer: (C) $1$

Concept: Convert each ratio to a decimal (or compare how close the two terms are).

1. $\frac{48}{90}\approx 0.53$, $\frac{46}{51}\approx 0.90$
2. $\frac{22}{54}\approx 0.41$, $\frac{42}{68}\approx 0.62$
3. Greatest: $46:51$

Trick: When the antecedent is closest to the consequent, the ratio is nearest to 1 – here 46 is just 5 less than 51.

Answer: (B) $46:51$

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

1. Kunal's share of profit $=\frac{1800}{12000}=\frac{3}{20}$.
2. Kunal's investment $=\frac{3}{20}\times 70400=10560$ (₹).
3. Anshuman's investment $=70400-10560=59840$ (₹).
4. Difference $=59840-10560=49280$ (₹).

Trick: Difference $=70400\times \frac{12000-2\times 1800}{12000}=70400\times 0.7=49280$.

Answer: (C) ₹49,280

Concept: Write the decimals with the same number of decimal places, find the LCM of the whole numbers, then put the decimal back.

1. $2.9=2.900=\frac{2900}{1000}$ and $0.031=\frac{31}{1000}$.
2. $2900=2^2\times 5^2\times 29$ and $31$ is prime, so $\text{LCM}(2900,31)=2900\times 31=89900$.
3. LCM $=\frac{89900}{1000}=89.9$.

Trick: LCM of decimals = LCM of numbers without decimal, with decimal places equal to the larger count (here 3).

Answer: (D) $89.9$

Concept: Test each odd number for divisibility by primes up to its square root ($\sqrt{407}<21$).

1. Odd numbers: 397, 399, 401, 403, 405, 407.
2. $399=3\times 133$, $405=5\times 81$, $403=13\times 31$, $407=11\times 37$ – not prime.
3. 397 and 401 are not divisible by 2, 3, 5, 7, 11, 13, 17, 19 → prime.
4. So there are 2 primes.

Trick: Check only primes up to 19 here; use the 11-rule for 407 ($4-0+7=11$) and $13\times 31=403$.

Answer: (D) 2

Concept: Margin of victory = (winner% − loser%) of valid votes.

1. Let registered voters $=N$. Votes polled $=0.96N$.
2. Valid votes $=0.9\times 0.96N=0.864N$.
3. Winner 57%, loser 43% → margin $=14\%$ of valid $=0.14\times 0.864N=0.12096N$.
4. $0.12096N=378\Rightarrow N=3125$.

Trick: Check options: $3125\times 0.96\times 0.9\times 0.14=378$ ✓ – only a round answer fits.

Answer: (D) 3125

Concept: $\text{SP}=\text{CP}\times (1+\text{markup})\times (1-\text{discount})$.

1. Let $\text{CP}=100$. MP $=135$.
2. SP $=135\times 0.82=110.7$ → profit $=10.7\%$ of CP.
3. $10.7\%$ of CP $=1498\Rightarrow \text{CP}=\frac{1498\times 100}{10.7}=14000$.

Trick: Net effect of $+35\%$ and $-18\%$: $35-18-\frac{35\times 18}{100}=10.7\%$.

Answer: (C) ₹14,000

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

1. $2550=\frac{P\times 3.2\times 8.5}{100}=0.272P\Rightarrow P=9375$ (₹).
2. New SI $=\frac{9375\times 9.6\times 5.8}{100}=5220$ (₹).
3. Amount $=9375+5220=14595$ (₹).

Trick: Rate × time: old $=27.2$, new $=55.68$; new SI $=2550\times \frac{55.68}{27.2}=5220$.

Answer: (D) ₹14,595

Concept: Use $(a-b{)}^2=a^2-2ab+b^2$ with a round base.

1. ${24}^2=(25-1{)}^2=625-50+1$.
2. $=576$.

Trick: Squares ending in 4 end in 6; ${24}^2=576$ (since ${25}^2=625$, subtract $25+24=49$).

Answer: (A) 576

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

1. Distance $=147\times 8=1176\text{km}$.
2. Meghna's time $=\frac{1176}{42}=28$ hours.

Trick: $T_2=T_1\times \frac{S_1}{S_2}=8\times \frac{147}{42}=8\times 3.5=28$.

Answer: (D) 28 hours

Concept: Successive discounts multiply: SP $=\text{MP}\times (1-d_1)(1-d_2)(1-d_3)$.

1. $648200\times 0.94=609308$.
2. $609308\times 0.95=578842.6$.
3. $578842.6\times 0.96=555688.9\approx 555689$ (₹).

Trick: $0.94\times 0.95\times 0.96=0.85728$ → net discount $14.272\%$; SP $=648200\times 0.85728$.

Answer: (A) ₹5,55,689

Concept: Interior + exterior angle $={180}^{\circ }$; number of sides $n=\frac{{360}^{\circ }}{\text{exterior}}$.

1. Exterior $=\frac{1}{7}$ interior → ratio $1:7$.
2. Exterior $=\frac{{180}^{\circ }}{8}={22.5}^{\circ }$.
3. $n=\frac{{360}^{\circ }}{{22.5}^{\circ }}=16$.

Trick: If exterior $=\frac{1}{k}$ interior, then $n=2(k+1)=2\times 8=16$.

Answer: (D) 16