RRB Technician Grade III 2024, 2025 PYQs

Concept: Open brackets from innermost: $()$, then $\{\}$, then $[]$.

1. $(3-3-6)=-6$.
2. $\{3-(-6)\}=9$.
3. $[4-9]=-5$; $5-(-5)=10$.

Yaad rakhein: Minus before a bracket flips signs — watch $-(-6)=+6$.

Answer: (A) 10

Concept: Volume of sphere $=\frac{4}{3}\pi r^3$, with $r=\frac{d}{2}$.

1. $r=\frac{6}{2}=3$ cm.
2. $V=\frac{4}{3}\pi \times 3^3=\frac{4}{3}\pi \times 27=36\pi {\text{cm}}^3$.

Trick: For diameter $d$: $V=\frac{\pi d^3}{6}=\frac{216\pi }{6}=36\pi$.

Answer: (B) $36\pi c{m}^3$

Concept: For $n$ (odd) values in order, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Sorted: $33,35,41,46,55,58,64,77,87,90,92$; $n=11$, median $=6^{\text{th}}=58$.
2. New data: $33,35,43,46,55,58,64,77,87,90,99$; median $=58$.
3. Difference $=58-58=0$.

Trick: Changing values that stay on the same side of the median (92→99 above, 41→43 below) does not change the median.

Answer: (D) 0

Concept: Same material ⇒ weight $\propto$ volume. Hollow sphere volume $\propto R^3-r^3$; we can use diameters directly.

1. Solid: $6^3=216$ units → $6$ kg.
2. Hollow: ${18}^3-{12}^3=5832-1728=4104$ units.
3. Weight $=\frac{4104}{216}\times 6=19\times 6=114$ kg.

Trick: Divide by $6$ first: $3^3-2^3=27-8=19$ times the solid sphere ⇒ $19\times 6=114$ kg.

Answer: (D) 114 kg

Concept: Same direction: relative speed $=$ difference; distance $=$ sum of lengths. $1\text{km/h}=\frac{5}{18}\text{m/s}$.

1. Relative speed $=72-36=36$ km/h $=36\times \frac{5}{18}=10$ m/s.
2. Distance $=200+300=500$ m.
3. Time $=\frac{500}{10}=50$ s.

Yaad rakhein: Opposite direction → add speeds; same direction → subtract.

Answer: (C) 50 sec

Concept: Total cost of both lots is compared with the total selling price of the whole mixture.

1. Cost price $=25\times 4+35\times 4.50=100+157.50=$ ₹257.50
2. Total quantity $=25+35=60$ kg
3. Selling price $=60\times 4.25=$ ₹255
4. Loss $=257.50-255=$ ₹2.50

Trick: Average cost per kg $=\frac{257.5}{60}\approx 4.29>4.25$, so it must be a loss.

Answer: (C) ₹2.50, loss

Concept: For a pure recurring part, $0.\overline{ab}=\frac{ab}{99}$.

1. $8.\overline{46}=8+0.\overline{46}=8+\frac{46}{99}$
2. $=\frac{8\times 99+46}{99}=\frac{792+46}{99}$
3. $=\frac{838}{99}$

Trick: Whole number with bar: $\frac{846-8}{99}=\frac{838}{99}$ (number minus non-repeating part, over $99$).

Answer: (A) $\frac{838}{99}$

Concept: Take the numbers as $4x$ and $x$; their difference is given.

1. $4x-x=3x=54$
2. $x=18$
3. Greater number $=4x=4\times 18=72$

Trick: Difference covers $3$ parts → $1$ part $=18$, greater $=4$ parts.

Answer: (A) 72

Concept: Net filling rate with leak $=$ pipes' rate $-$ leak's rate.

1. Pipes' rate $=\frac{1}{20}$ tank/min; with leak they take $30$ min, rate $=\frac{1}{30}$
2. Leak rate $=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}=\frac{1}{60}$
3. Leak empties full tank in $60$ min

Trick: Leak time $=\frac{a\times b}{b-a}=\frac{20\times 30}{10}=60$ min.

Answer: (B) 60 min

Concept: Given sum and difference: larger $=\frac{\text{sum}+\text{difference}}{2}$.

1. Elder $+$ younger $=21$, elder $-$ younger $=7$
2. Adding: $2\times \text{elder}=28$
3. Elder $=14$ years (younger $=7$)

Trick: Check options: $14-7=7$ and $14+7=21$ ✓.

Answer: (D) 14 years

Concept: A prime number has exactly two factors, $1$ and itself.

1. $1$–$10$: $2,3,5,7$ → $4$
2. $11$–$20$: $11,13,17,19$ → $4$
3. $21$–$30$: $23,29$ → $2$
4. $31$–$40$: $31,37$; $41$–$50$: $41,43,47$ → $5$
5. Total $=4+4+2+5=15$

Yaad rakhein: Primes up to $50$ = $15$, up to $100$ = $25$.

Answer: (D) 15

Concept: $1\text{ kg}=1000\text{ g}$; add grams first and carry over.

1. Grams: $589+678=1267\text{ g}=1\text{ kg }267\text{ g}$
2. Kilograms: $22+19+1=42$ kg
3. Total $=42$ kg $267$ g

Trick: $22.589+19.678=42.267$ kg directly.

Answer: (C) 42 kg 267 g

Concept: BODMAS: Brackets first; inside the bracket division before subtraction.

1. $6÷3=2$
2. $5-2=3$
3. $3-3=0$

Yaad rakhein: Order: Bracket → Of → Division → Multiplication → Addition → Subtraction.

Answer: (D) 0

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

1. SI $=2040-1500=$ ₹540
2. $T=\frac{540\times 100}{1500\times 6}=\frac{54000}{9000}$
3. $T=6$ years

Trick: Interest per year $=6\%$ of $1500=90$; $540÷90=6$ years.

Answer: (A) 6 years

Concept: Pass marks $=$ marks obtained $+$ marks by which he failed.

1. Pass marks $=178+22=200$
2. $40\%$ of maximum $=200$
3. Maximum $=\frac{200\times 100}{40}=500$

Trick: $40\%\to 200$ ⇒ $100\%\to 500$.

Answer: (B) 500

Concept: $a^m\times a^n=a^{m+n}$; equal bases ⇒ equal powers.

1. $4\times 4^{n+5}=4^{n+6}$
2. $256=4^4$, so $n+6=4$ ⇒ $n=-2$
3. $5^{n+4}=5^2=25$

Trick: Note $n+4=(n+6)-2=4-2=2$ directly.

Answer: (B) 25

Concept: ${\sin }^2\theta +{\cos }^2\theta =1$, ${\sec }^2\theta -{\tan }^2\theta =1$, $\cot \theta =\frac{\cos \theta }{\sin \theta }$.

1. ${\sin }^2\theta +{\cos }^2\theta -({\sec }^2\theta -{\tan }^2\theta )=1-1=0$
2. $\cot \theta \sin \theta =\frac{\cos \theta }{\sin \theta }\times \sin \theta =\cos \theta$
3. Expression $=0+\cos \theta -\cos \theta =0$

Trick: Put $\theta ={45}^{\circ }$: $1-1+1\times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0$.

Answer: (B) 0

Concept: Take CP $=100$; SP $=$ MP after discount.

1. MP $=100+25\%=125$
2. Discount $10\%$ of $125=12.5$
3. SP $=112.5$ ⇒ gain $=12.5\%$

Trick: Successive change: $25-10-\frac{25\times 10}{100}=12.5\%$.

Answer: (A) 12.5%

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with rate and time in the same unit (months here).

1. $P=700$, $R=9\%$ per month, $T=7$ months
2. $\text{SI}=\frac{700\times 9\times 7}{100}=7\times 63$
3. $\text{SI}=$ ₹441

Trick: $1\%$ of $700=7$ ⇒ per month $63$; $63\times 7=441$.

Answer: (C) ₹441

Concept: Read only the 'males' column for shops A and B.

1. Shop A to males $=21$
2. Shop B to males $=39$
3. Difference $=39-21=18$

Trick: Ignore the females column – it is a distractor here.

Answer: (B) 18