RRB NTPC Graduate CBT 2 2024 PYQs

Concept: Distance between consecutive buses is fixed; the traveller closes this gap at relative speed (sum of speeds, opposite directions).

1. Gap $=24\times \frac{8}{60}=3.2\text{km}$.
2. Traveller meets the next bus every 6 min: $(24+v)\times \frac{6}{60}=3.2$.
3. $24+v=32\Rightarrow v=8\text{km/h}$.

Trick: $v=24\times \frac{8-6}{6}=8\text{km/h}$.

Answer: (C) 8 km/hr

Concept: Take income as 100.

1. Food + rent + conveyance $=48+25+12=85\%$; remaining $=15\%$.
2. Education $=60\%$ of $15=9\%$.
3. Left $=15-9=6\%$.

Trick: Left $=15\times \frac{40}{100}=6\%$.

Answer: (B) 6%

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

1. ${(1+\frac{R}{100})}^2=\frac{12100}{10000}=\frac{121}{100}$.
2. $1+\frac{R}{100}=\frac{11}{10}\Rightarrow R=10\%$.

Trick: 121 = ${11}^2$, 100 = ${10}^2$ → rate 10%.

Answer: (D) 10

Concept: Write everything as powers of 2 and use $a^m\times a^n=a^{m+n}$.

1. Numerator: $4^3\times 4^4\times 2^4=2^6\times 2^8\times 2^4=2^{18}$.
2. Denominator: $4^2\times 8=2^4\times 2^3=2^7$.
3. Value $=2^{18-7}=2^{11}=2048$.

Trick: Convert to a single base (2) and just add/subtract exponents.

Answer: (D) 2048

Concept: Difference in SI comes only from the difference in rates.

1. Rate difference $=8-6=2\%$ for 3 years → $6\%$ of P.
2. $\frac{6}{100}\times P=360\Rightarrow P=6000$.

Trick: $P=\frac{\text{Diff}\times 100}{\Delta R\times T}=\frac{360\times 100}{2\times 3}$.

Answer: (B) ₹6,000

Concept: Average $=\frac{\text{Sum}}{\text{Count}}$

1. Sum $=6+8+10+12=36$.
2. Average $=\frac{36}{4}=9$.

Trick: For equally spaced numbers, average = middle value $=\frac{6+12}{2}=9$.

Answer: (A) 9

Concept: Simplify the bracket first, then take LCM.

1. $1+\frac{3}{7}=\frac{10}{7}$, so $\frac{1}{10/7}=\frac{7}{10}$.
2. $\frac{5}{4}+\frac{7}{10}-\frac{4}{3}$; LCM $=60$.
3. $=\frac{75+42-80}{60}=\frac{37}{60}$.

Trick: Denominator must be a factor of 60 (LCM of 4, 10, 3) – only $\frac{37}{60}$ fits.

Answer: (D) $\frac{37}{60}$

Concept: TSA of cone $=\pi r(l+r)$, where $l=\sqrt{r^2+h^2}$.

1. $l=\sqrt{8^2+{15}^2}=\sqrt{289}=17\text{cm}$.
2. TSA $=\pi \times 8\times (17+8)=200\pi {\text{cm}}^2$.

Trick: 8-15-17 is a Pythagorean triplet.

Answer: (C) $200\pi {\text{cm}}^2$

Concept: Stream speed $=\frac{\text{Downstream}-\text{Upstream}}{2}$

1. $=\frac{25-15}{2}=5\text{km/h}$.
2. (Boat speed $=\frac{25+15}{2}=20\text{km/h}$.)

Trick: Half the sum = boat; half the difference = stream.

Answer: (B) 5 km/hr

Concept: If $x\propto \frac{1}{y}$ then $xy=$ constant.

1. $xy=4\times 6=24$.
2. When $y=32$: $x=\frac{24}{32}=\frac{3}{4}$.

Trick: $x_1{y}_1=x_2{y}_2$ – no need to find the constant separately.

Answer: (A) $\frac{3}{4}$

Concept: Multiply the remaining fractions of price.

1. Final price $=100\times 0.90\times 0.85\times 0.80$.
2. $=100\times 0.612=61.2$.
3. Single discount $=100-61.2=38.8\%$.

Trick: Two at a time: $10,15\to 10+15-\frac{150}{100}=23.5$; then $23.5,20\to 43.5-4.7=38.8\%$.

Answer: (C) 38.8%

Concept: Adding the same number to every value raises the average by that number.

1. Old mean $=15$.
2. New mean $=15+2=17$.

Trick: Same change to each item → same change to the mean.

Answer: (A) 17

Concept: Brackets first, then ÷ and × from left to right, then + and −.

1. Bracket: $2.4÷1.2=2$.
2. $0.2\times 4.8÷2=0.96÷2=0.48$; $0.90÷2=0.45$.
3. $0.5+0.48-0.45=0.53$.

Trick: Clear all ÷/× terms first, then add/subtract in one line.

Answer: (A) 0.53

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$ first).

1. $\frac{43}{76}\approx 0.57$, $\frac{38}{51}\approx 0.75$.
2. $\frac{22}{61}\approx 0.36$, $\frac{30}{74}\approx 0.41$.
3. Greatest: $38:51$.

Trick: Only $38:51$ is well above $\frac{2}{3}$ ($38\times 3=114>102$).

Answer: (B) 38 : 51

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

1. Distances: $13\times 3=39\text{km}$ and $15\times 6=90\text{km}$; total $=129\text{km}$.
2. Total time $=13+15=28\text{h}$.
3. Average $=\frac{129}{28}=4\frac{17}{28}\text{km/h}$.

Trick: Average must lie between 3 and 6 – only $4\frac{17}{28}$ fits.

Answer: (A) $4\frac{17}{28}\text{km/hr}$

Concept: For successive discounts, multiply the marked price by $(1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. $\text{SP}=1200\times 0.95\times 0.85$
2. $1200\times 0.95=1140$, $1140\times 0.85=969$
3. Difference $=1200-969=231$

Trick: Equivalent discount $=5+15-\frac{5\times 15}{100}=19.25\%$; $19.25\%$ of $1200=231$.

Answer: (A) ₹231

Concept: Customer pays for 15 articles (at 10% off) but receives $15+4=19$ articles; trader's gain is on the cost of all 19.

1. Let $\text{CP}=1$ per article; cost of 19 articles $=19$.
2. Required SP $=19\times 1.35=25.65$
3. $15\times 0.9\times \text{MP}=25.65\Rightarrow \text{MP}=\frac{25.65}{13.5}=1.9$
4. MP is $\frac{1.9-1}{1}\times 100=90\%$ more than CP.

Trick: $\frac{\text{MP}}{\text{CP}}=\frac{19\times 135}{15\times 90}=\frac{19}{10}$ → $90\%$ more.

Answer: (A) $90\%$

Concept: A number is divisible by 35 if it is divisible by both 5 and 7 (co-prime factors).

1. Last digit 0 → divisible by 5.
2. $977340÷7=139620$ exactly → divisible by 7. Hence divisible by 35.
3. 24 fails: last three digits 340 not divisible by 8; 34 fails: $488670$ is not a multiple of 17; 37 also leaves a remainder.

Trick: Test 7 quickly: $977-340=637=7\times 91$ (alternate 3-digit groups) → divisible by 7.

Answer: (A) 35

Concept: Overall gain/loss % = net profit ÷ total CP × 100.

1. Let CP of X $=300$, Y $=700$ (ratio $3:7$), total $=1000$.
2. Profit on X $=14\%$ of $300=42$; loss on Y $=11\%$ of $700=77$.
3. Net $=42-77=-35$ → loss $=\frac{35}{1000}\times 100=3.5\%$

Trick: Weighted: $\frac{3\times 14-7\times 11}{10}=\frac{42-77}{10}=-3.5\%$.

Answer: (C) Loss, $3.5\%$

Concept: Simplify the inner bracket first, then combine the like fractions.

1. $1+\frac{5}{7}=\frac{12}{7}$, so $\frac{1}{12/7}=\frac{7}{12}$
2. $\frac{5}{2}-\frac{3}{2}=1$
3. Total $=1+\frac{7}{12}=\frac{19}{12}$

Trick: Pair $\frac{5}{2}$ and $-\frac{3}{2}$ first – they give 1 instantly.

Answer: (D) $\frac{19}{12}$