RRB Technician Grade III 2024, 2025 PYQs

Concept: If $x\%$ of total $=$ value, then total $=\frac{\text{value}\times 100}{x}$.

1. $94\%$ of $T=22560$
2. $T=\frac{22560\times 100}{94}$
3. $=240\times 100=24000$

Trick: $1\%=\frac{22560}{94}=240$, so $100\%=24000$.

Answer: (A) 24,000

Concept: $V=\frac{4}{3}\pi r^3$, so volume varies as $r^3$.

1. New radius $=\frac{r}{2}$ (reduced by $50\%$).
2. $V_{\text{new}}=\frac{4}{3}\pi {\left(\frac{r}{2}\right)}^3=\frac{1}{8}\times \frac{4}{3}\pi r^3$
3. New volume $=\frac{1}{8}$ of original.

Trick: Scale factor $k$ on length → $k^3$ on volume: ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$.

Answer: (C) $\frac{1}{8}$ times

Concept: Form equations from the age conditions; each year the combined age rises by 2.

1. Son $=x$, father $=7x$.
2. $7x+3=5(x+3)\Rightarrow 2x=12\Rightarrow x=6$; father $=42$.
3. Present sum $=48$; needed increase $=62-48=14$ → $\frac{14}{2}=7$ years.

Trick: Test options: ages 6 and 42 → after 7 years: $13+49=62$ ✓.

Answer: (C) 7 years

Concept: Total profit $=$ profit per kg $\times$ quantity sold.

1. CP per kg $=\frac{21}{3}=$ ₹7
2. SP per kg $=\frac{50}{5}=$ ₹10 → profit ₹3 per kg
3. Quantity $=\frac{102}{3}=34$ kg

Trick: Always bring CP and SP to the same quantity (per kg) before comparing.

Answer: (B) 34 kg

Concept: Wages are shared in the ratio of work done by each person.

1. In 3 days A does $\frac{3}{6}=\frac{1}{2}$ of the work.
2. B does $\frac{3}{8}$; C does $1-\frac{1}{2}-\frac{3}{8}=\frac{1}{8}$.
3. B's share $=\frac{3}{8}\times 3200=1200$

Trick: Work ratio A : B : C $=4:3:1$ → B gets $\frac{3}{8}$ of ₹3,200.

Answer: (B) ₹1,200

Concept: If $a:x::x:b$, then $x^2=ab$ ($x$ is the mean proportional).

1. $x^2=64\times 169$
2. $x=\sqrt{64}\times \sqrt{169}=8\times 13$
3. $x=104$

Trick: Mean proportional $=\sqrt{ab}$; both 64 and 169 are perfect squares.

Answer: (A) 104

Concept: Two numbers are co-prime if their HCF is 1.

1. $455=5\times 7\times 13,49=7^2$ → common 7.
2. $363=3\times {11}^2,77=7\times 11$ → common 11; $52,24$ → common 4.
3. $81=3^4,16=2^4$ → no common factor, HCF $=1$.

Yaad rakhein: Co-prime numbers need not be prime themselves (81 and 16 are both composite).

Answer: (B) 81, 16

Concept: $\text{SP}=\text{MP}\times (1-\frac{\text{discount}\%}{100})$

1. Discount $=25\%$ of 100 $=25$
2. SP $=100-25$
3. $=75$

Trick: When MP is ₹100, SP equals $(100-\text{discount}\%)$ directly.

Answer: (C) ₹75

Concept: Under CI the amount multiplies by the same factor in equal time intervals.

1. $2$ times in 12 years.
2. $8=2^3$ → three doublings → $3\times 12=36$ years.
3. Extra time $=36-12=24$ years.

Trick: $\times 2$ (12 y) → $\times 4$ (24 y) → $\times 8$ (36 y). Question asks 'how many more' — subtract the first 12.

Answer: (A) 24

Concept: Total of ratio parts $=5+7+15=27$.

1. A's 5 parts $=1500$ → 1 part $=300$
2. Total $=27\times 300$
3. $=8100$

Trick: Total $=\frac{1500}{5}\times 27=8100$.

Answer: (C) ₹8,100

Concept: Diagonals of a rhombus bisect each other at ${90}^{\circ }$, and opposite sides are parallel.

1. Let diagonals meet at O; $\angle QOR={90}^{\circ }$.
2. In $△QOR$: $\angle OQR={180}^{\circ }-{90}^{\circ }-{40}^{\circ }={50}^{\circ }$.
3. $PS\parallel QR$, so $\angle PSQ=\angle SQR={50}^{\circ }$ (alternate angles).

Trick: Angle between a side and one diagonal $+$ angle between the adjacent side and the other diagonal $={90}^{\circ }$: ${90}^{\circ }-{40}^{\circ }={50}^{\circ }$.

Answer: (B) ${50}^{\circ }$

Concept: Mode is the value with the highest frequency.

1. Frequencies: 44, 48, 58, 46, 39, 37.
2. Highest frequency $=58$.
3. It belongs to shoe size 7.

Yaad rakhein: Mode = most demanded item — just pick the largest count.

Answer: (D) 7

Concept: Read the values from the graph and simplify the ratio.

1. Friday $=12$ plants.
2. Saturday $=10$ plants.
3. $12:10=6:5$

Trick: Divide both terms by their HCF (2).

Answer: (D) $6:5$

Concept: A perfect square is the square of an integer.

1. ${14}^2=196$ ✓
2. ${14}^2=196<222<225={15}^2$; ${11}^2=121<124,141<144={12}^2$
3. So only 196 is a perfect square.

Trick: Squares never end in 2, 3, 7 or 8 — 222 is out at once.

Answer: (B) 196

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

1. $\frac{50(1+{\cot }^{2}A)}{25(1+{\tan }^{2}A)}=\frac{2{\csc }^{2}A}{{\sec }^{2}A}$
2. $=2\times \frac{{\cos }^{2}A}{{\sin }^{2}A}$
3. $=2{\cot }^{2}A$

Trick: Put $A={45}^{\circ }$: expression $=\frac{100}{50}=2$, and only $2{\cot }^{2}A$ gives 2.

Answer: (A) $2{\cot }^{2}A$

Concept: Let chips $=c$, biscuits $=b$; form two equations.

1. $c+b=50$ and $6c+5b=273$
2. Subtract $5\times$(first): $c=273-250=23$
3. $b=27$; difference $=27-23=4$

Trick: If all 50 were ₹5 packets: ₹250; extra ₹23 comes from ₹1 extra per chips packet → 23 chips.

Answer: (B) 4

Concept: If $\sqrt{a}=3$, then $a=9$.

1. One number $=3^2=9$
2. $9^2+y^2=145\Rightarrow y^2=145-81=64$
3. $y=8$

Trick: Check: $81+64=145$ ✓.

Answer: (C) 8

Concept: Amount under compound interest: $A=P{(1+\frac{R}{100})}^n$.

1. Time = from age 16 to 18 = $2$ years.
2. $A=200000\times {\left(1.08\right)}^2=200000\times 1.1664$.
3. $A=233280$, i.e. ₹2,33,280.

Trick: For 2 years at 8%, the effective rate is $8+8+\frac{8\times 8}{100}=16.64\%$; $16.64\%$ of 200000 = 33280.

Answer: (D) ₹2,33,280

Concept: Depreciated value: $V=P{(1-\frac{R}{100})}^n$.

1. $P=162000,R=10\%,n=2$.
2. $V=162000\times {\left(0.9\right)}^2=162000\times 0.81$.
3. $V=131220$, i.e. ₹1,31,220.

Trick: Take 10% off twice: $162000\to 145800\to 131220$.

Answer: (C) ₹1,31,220

Concept: Form equations from the two given conditions.

1. Let scanner = $s$, printer = $p$. Then $p-s=7000$ and $p=2s$.
2. $2s-s=7000\Rightarrow s=7000$.
3. Check: printer = $14000$, difference = $7000$ ✓.

Trick: If one thing is double the other, their difference equals the smaller one.

Answer: (C) ₹7,000