RRB Technician Grade III 2024, 2025 PYQs

Concept: For an odd count of consecutive numbers, the average is the middle number.

1. 7 numbers → the $4^{\text{th}}$ number is the middle one $=33$.
2. Numbers: 30, 31, 32, 33, 34, 35, 36.
3. Greatest $=33+3=36$.

Trick: Greatest $=\text{average}+\frac{n-1}{2}=33+\frac{7-1}{2}=36$.

Answer: (C) 36

Concept: A number divisible by every number from 1 to 10 must be a multiple of their LCM.

1. $\text{LCM}(1,2,\dots ,10)=2^3\times 3^2\times 5\times 7$
2. $=8\times 9\times 5\times 7=2520$
3. 2520 is in the options; 1000, 504 and 608 fail (e.g. 1000 is not divisible by 3, 504 not by 5).

Yaad rakhein: 2520 is the smallest number divisible by all of 1 to 10 — a classic fact worth memorising.

Answer: (B) 2520

Concept: Volume of a cube $=a^3$; cube every factor separately.

1. $a=6xy{z}^2$
2. $a^3=6^3\times x^3\times y^3\times (z^2{)}^3$
3. $=216x^3{y}^3{z}^6$

Trick: $(z^2{)}^3=z^{2\times 3}=z^6$ — multiply the powers.

Answer: (D) $216x^3{y}^3{z}^6$

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

1. $P=5000,R=4\%,T=5$
2. $\text{SI}=\frac{5000\times 4\times 5}{100}$
3. $=1000$

Trick: $4\%\times 5=20\%$ of 5000 $=1000$.

Answer: (D) ₹1,000

Concept: Selling price $=\text{total cost}\times (1+\frac{\text{profit}\%}{100})$

1. Transport $=10\%$ of 2000 $=200$; total cost $=2200$.
2. SP $=2200\times 1.15$
3. $=2530$

Trick: Chain multiply: $2000\times 1.1\times 1.15=2530$.

Answer: (D) ₹2,530

Concept: One day's work = $\frac{1}{\text{days}}$; works of A, B, C add up.

1. $(A+B+C)$ one day $=\frac{1}{2}$
2. $A+B=\frac{1}{4}+\frac{1}{20}=\frac{5+1}{20}=\frac{3}{10}$
3. $C=\frac{1}{2}-\frac{3}{10}=\frac{2}{10}=\frac{1}{5}$ → 5 days

Trick: Total work = LCM$(4,20,2)=20$ units: A = 5, B = 1, all = 10 per day → C = 4 per day → $\frac{20}{4}=5$ days.

Answer: (C) 5

Concept: $\text{Distance}=\text{Speed}\times \text{Time}$ (time in hours when speed is in km/h).

1. $10\text{min}=\frac{10}{60}=\frac{1}{6}$ h
2. $\text{Distance}=90\times \frac{1}{6}$
3. $=15$ km

Trick: 90 km/h = 1.5 km per minute; $1.5\times 10=15$ km.

Answer: (C) 15 km

Concept: Convert all fractions to a common denominator (LCM of 7 and 14 = 14).

1. $\frac{3}{7}+\frac{2}{7}=\frac{5}{7}=\frac{10}{14}$
2. $\frac{10}{14}+\frac{1}{14}=\frac{11}{14}$
3. $\frac{11}{14}-3=\frac{11-42}{14}=-\frac{31}{14}$

Trick: Sum of fractions $<1$, so the answer must be negative and lie between $-3$ and $-2$: only $-\frac{31}{14}\approx -2.2$ fits.

Answer: (C) $-\frac{31}{14}$

Concept: Base area $=\pi r^2$, CSA $=2\pi rh$, Volume $=\pi r^{2}h$.

1. $\frac{22}{7}{r}^2=38.5\Rightarrow r^2=12.25\Rightarrow r=3.5$ cm
2. $2\times \frac{22}{7}\times 3.5\times h=616\Rightarrow 22h=616\Rightarrow h=28$ cm
3. $V=\text{base area}\times h=38.5\times 28=1078{\text{cm}}^3$

Trick: $V=\frac{\text{CSA}\times r}{2}=\frac{616\times 3.5}{2}=1078$.

Answer: (A) 1078

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) ${50}^{\circ }$