RRB Technician Grade III 2024, 2025 PYQs

Concept: To test $n$ for primality, check divisibility by primes up to $\sqrt{n}$.

1. $437=19\times 23$, $161=7\times 23$, $221=13\times 17$ – all composite.
2. $\sqrt{373}<20$; 373 is not divisible by $2,3,5,7,11,13,17,19$.
3. Hence 373 is prime.

Yaad rakhein: Remember the "fake primes": $221=13\times 17$, $323=17\times 19$, $437=19\times 23$.

Answer: (D) 373

Concept: Solve the bracket first, then divide.

1. $\frac{4}{5}+\frac{1}{10}=\frac{8+1}{10}=\frac{9}{10}$.
2. $\frac{9}{10}÷\frac{9}{10}=1$.

Trick: Any non-zero number divided by itself is $1$ – spot equal values early.

Answer: (A) 1

Concept: Taking $p\%$ of X away leaves $(100-p)\%$ of X.

1. $X-90\%\text{ of }X=X-0.9X$.
2. $=0.1X$, i.e. X multiplied by $0.1$.

Trick: $100\%-90\%=10\%=0.1$.

Answer: (C) 0.1

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

1. Required SP $=153\times \frac{120}{100}=$ ₹183.60.
2. MP $\times \frac{85}{100}=183.6$.
3. MP $=\frac{183.6\times 100}{85}=$ ₹216.

Trick: One line: MP $=153\times \frac{120}{85}=\frac{153}{85}\times 120=1.8\times 120=216$.

Answer: (D) ₹216

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

1. SI $=\frac{85000\times 7.5\times 4}{100}=850\times 30=$ ₹25,500.
2. Amount $=85000+25500=$ ₹1,10,500.

Trick: Total rate $=7.5\times 4=30\%$ → Amount $=85000\times 1.3=110500$.

Answer: (D) ₹1,10,500

Concept: Crossing a pole, the train covers its own length: $L=\text{speed}\times \text{time}$.

1. $90\text{km/h}=90\times \frac{5}{18}=25$ m/s.
2. Length $=25\times 5=125$ m.

Yaad rakhein: km/h → m/s: multiply by $\frac{5}{18}$.

Answer: (D) 125 m

Concept: Average $=\frac{\text{sum}}{\text{count}}$.

1. First 10 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
2. Sum $=129$.
3. Average $=\frac{129}{10}=12.9$.

Yaad rakhein: Sum of first 10 primes $=129$ – worth memorising.

Answer: (D) 12.9

Concept: $a^2-b^2=(a-b)(a+b)$.

1. $(9982{)}^2-(18{)}^2=(9982-18)(9982+18)$.
2. $=9964\times 10000$.
3. So $10000x=9964\times 10000$ → $x=9964$.

Trick: When $a+b$ gives a round number like $10000$, use $a^2-b^2$ immediately.

Answer: (D) 9964

Concept: Numbers in an arithmetic progression have average $=\frac{\text{first}+\text{last}}{2}$.

1. Multiples of 5 between 12 and 39: 15, 20, 25, 30, 35.
2. Average $=\frac{15+35}{2}=25$.

Trick: For an odd number of AP terms, the average is the middle term: $25$.

Answer: (C) 25

Concept: Convert to one unit: $1\text{l}=1000$ ml.

1. 8 l 785 ml $=8785$ ml.
2. Each person gets $\frac{8785}{35}=251$ ml.

Trick: Check: $35\times 251=8785$ ✓ – unit digit $5\times 1=5$ matches.

Answer: (A) 251 ml

Concept: Shortfall marks correspond to the shortfall percentage of the maximum marks.

1. Shortfall $=780-728=52$ marks.
2. $5\%$ of max $=52$.
3. Max marks $=\frac{52\times 100}{5}=1040$.

Trick: $5\%=\frac{1}{20}$ → max $=52\times 20=1040$.

Answer: (D) 1040

Concept: If $a:b::b:k$, then $k=\frac{b^2}{a}$.

1. $k=\frac{{35}^2}{40}=\frac{1225}{40}=\frac{245}{8}$.
2. $k-1=\frac{245-8}{8}=\frac{237}{8}$.

Yaad rakhein: Product of means = product of extremes: $40k=35\times 35$.

Answer: (A) $\frac{237}{8}$

Concept: Write the statement as an equation and cancel the common term.

1. Amit + Amita $=$ Amita + Amrita $+15$.
2. Cancel Amita: Amit $-$ Amrita $=15$.
3. So Amrita is 15 years younger than Amit.

Trick: The common person (Amita) always cancels out – the difference is the given number.

Answer: (D) 15

Concept: Marks of a subject $=\text{its }\%\times \text{total marks}$.

1. From the chart: Physics $=25\%$, English $=10\%$.
2. Together $=35\%$.
3. Marks $=\frac{35}{100}\times 260=91$.

Trick: Add percentages first, then take one percentage of the total.

Answer: (B) 91

Concept: A regular hexagon splits into 6 equilateral triangles of the same side, so Area $=6\times \frac{\sqrt{3}}{4}{a}^2=\frac{3\sqrt{3}}{2}{a}^2$.

1. Here $a=6$ cm, $a^2=36$.
2. Area $=\frac{3\sqrt{3}}{2}\times 36=54\sqrt{3}{\text{cm}}^2$.

Yaad rakhein: Hexagon area $=\frac{3\sqrt{3}}{2}{a}^2$ – six equilateral triangles.

Answer: (B) $54\sqrt{3}$

Concept: Multiplication is done before addition and subtraction.

1. $6\times 6=36$.
2. $6+36-6=36$.

Trick: $+6$ and $-6$ cancel, leaving $6\times 6=36$.

Answer: (A) 36

Concept: ${\text{cosec}}^{2}A-1={\cot }^{2}A$ and $1+{\tan }^{2}A={\sec }^{2}A$.

1. Expression $={\sec }^{2}A+{\cot }^{2}A+{\sec }^{2}A-{\cot }^{2}A$.
2. ${\cot }^{2}A$ terms cancel.
3. Result $=2{\sec }^{2}A$.

Trick: Replace each bracket by its identity – opposite terms cancel.

Answer: (C) $2{\sec }^{2}A$

Concept: Write all numbers with the same base: $(a^m{)}^n=a^{mn}$.

1. $4^{50}=(2^2{)}^{50}=2^{100}$.
2. ${16}^{25}=(2^4{)}^{25}=2^{100}$.
3. All three equal $2^{100}$.

Trick: Convert to base 2 – compare only exponents.

Answer: (B) All are equal

Concept: A sum doubles when SI $=P$, i.e. $R\times T=100$.

1. $\frac{P\times 20\times T}{100}=P$.
2. $T=\frac{100}{20}=5$ years.

Yaad rakhein: Doubling time under SI $=\frac{100}{R}$ years.

Answer: (D) 5 years