RRB Technician Grade III 2024, 2025 PYQs

Concept: Work rates add: $\frac{1}{A}+\frac{1}{B}=\frac{1}{\text{together}}$.

1. $\frac{1}{A}=\frac{1}{2.4}-\frac{1}{4}=\frac{5}{12}-\frac{3}{12}=\frac{1}{6}$
2. So A alone fills it in $6$ hours
3. Capacity $=6\times 100=600$ litres

Trick: LCM method: take capacity $12$ units; together $5$/h, B $3$/h, so A $2$/h $=100$ L ⇒ $1$ unit $=50$ L, capacity $=600$ L.

Answer: (B) 600 litres

Concept: Multiply like bases by adding powers; write $8=2^3$.

1. $A\times B=8^{3+5}\times 5^{4+3}=8^8\times 5^7$
2. $8^8=(2^3{)}^8=2^{24}$
3. $A\times B=2^{24}\times 5^7$

Trick: Convert all bases to primes before matching options.

Answer: (A) $2^{24}\times 5^7$

Concept: Pass marks = marks obtained + marks short.

1. Pass marks $=20+40=60$
2. $40\%$ of maximum $=60$
3. Maximum $=\frac{60\times 100}{40}=150$

Trick: $10\%=15$, so $100\%=150$.

Answer: (C) 150

Concept: Diagonals $=\frac{n(n-3)}{2}$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. $\frac{n(n-3)}{2}=20\Rightarrow n(n-3)=40=8\times 5$
2. $n=8$ (octagon)
3. Sum $=(8-2)\times {180}^{\circ }={1080}^{\circ }$

Yaad rakhein: Octagon has $20$ diagonals; hexagon $9$, decagon $35$.

Answer: (A) 1080°

Concept: Net change $=a+b+\frac{ab}{100}$ (decrease negative).

1. $a=-40,b=+50$
2. Net $=-40+50+\frac{(-40)(50)}{100}=10-20=-10\%$
3. Negative ⇒ loss of $10\%$

Trick: Take salary $100$: $100\to 60\to 90$, loss $=10\%$.

Answer: (B) 10%

Concept: Average wage $=\frac{\text{total wages}}{\text{total workers}}$.

1. Total per hour $=25\times 20+15\times 20=500+300=800$
2. Workers $=25+15=40$
3. Average $=\frac{800}{40}=$ ₹20

Trick: If every member gets the same value, the average is that same value.

Answer: (C) ₹20

Concept: Value of a sector $=\frac{\text{its }\%}{100}\times \text{total}$.

1. From the chart, electricity $=5\%$
2. Cost $=\frac{5}{100}\times 100000=5000$

Trick: Total is $100000$, so each $1\%=1000$; $5\%=5000$.

Answer: (B) $5,000

Concept: Difference of ratio terms corresponds to the actual age difference.

1. Ratio $7:8$, difference $=1$ part
2. $1$ part $=3$ years
3. Ishan $=7\times 3=21$ years

Trick: Check: Surya $=24$, $24-21=3$ ✓

Answer: (B) 21 years

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

1. $T=\frac{1050\times 100}{2100\times 10}$
2. $T=\frac{105000}{21000}=5$ years

Trick: SI is half the principal; at $10\%$ per year that takes $50÷10=5$ years.

Answer: (C) 5 years

Concept: Each seller adds his gain on his own cost: $\text{Retail}=C\times 1.10\times 1.15\times 1.25$.

1. $C\times \frac{11}{10}\times \frac{23}{20}\times \frac{5}{4}=1265$
2. $C=\frac{1265\times 10\times 20\times 4}{11\times 23\times 5}$
3. $1265=11\times 23\times 5$, so $C=10\times 20\times 4=800$

Trick: Spot that $1265=5\times 11\times 23$ – it cancels exactly.

Answer: (C) ₹800.00

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})$
2. $24.12=2(\text{Median}-\text{Mean})$
3. $\text{Median}-\text{Mean}=12.06$

Yaad rakhein: Mode–Median gap is always twice the Median–Mean gap.

Answer: (D) 12.06

Concept: Net rate = filling rates − emptying rate.

1. $\frac{1}{20}+\frac{1}{24}=\frac{11}{120}$ tank/min
2. Net rate $=\frac{1}{15}=\frac{8}{120}$
3. Waste pipe empties $\frac{11-8}{120}=\frac{3}{120}=\frac{1}{40}$ tank/min
4. Capacity $=40\times 3=120$ gallons

Trick: Waste pipe alone empties the tank in $40$ min at $3$ gal/min ⇒ $120$ gallons.

Answer: (D) 120 gallons

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=540-405=$ ₹135.
2. $d=\frac{135}{540}\times 100=25$.

Trick: $\frac{405}{540}=\frac{3}{4}$, so SP is $75\%$ of MP; discount $=100-75=25\%$.

Answer: (D) 25

Concept: Read the highest and lowest points of the line graph and subtract.

1. Maximum production: June $=350$; minimum: January $=150$.
2. Difference $=350-150=200$.

Trick: Just scan the labelled peaks and dips – no calculation beyond one subtraction.

Answer: (C) 200

Concept: Subtract the two equations to isolate one variable.

1. $2T+3C=540$ and $2T+C=470$.
2. Subtract: $2C=70\Rightarrow C=35$, so $2T=470-35=435$.
3. Difference $=2T-2C=435-70=365$.

Trick: $2T-2C=(2T+C)-3C=470-105=$ ₹365 directly.

Answer: (C) ₹365

Concept: If price rises by $r\%$, consumption must fall by $\frac{r}{100+r}\times 100\%$ to keep expenditure same.

1. Reduction $=\frac{20}{120}\times 100$.
2. $=\frac{50}{3}=16\frac{2}{3}\%$.

Trick: $20\%=\frac{1}{5}$ increase $\Rightarrow$ $\frac{1}{5+1}=\frac{1}{6}=16\frac{2}{3}\%$ decrease.

Answer: (D) $16\frac{2}{3}\%$

Concept: A number is divisible by 9 if the sum of its digits is divisible by 9.

1. Sum $=5+k+2+1=8+k$.
2. $8+k=9\Rightarrow k=1$ (next, $18$, needs $k=10$ – not a digit).

Trick: Find the gap to the next multiple of 9: $9-8=1$.

Answer: (B) 1

Concept: Make decimal places equal, find LCM of the whole numbers, then put the decimal back.

1. $2.5=2.500$, $0.175$ → $2500$ and $175$ (3 decimal places).
2. $2500=2^2\times 5^4$, $175=5^2\times 7$; LCM $=2^2\times 5^4\times 7=17500$.
3. LCM $=\frac{17500}{1000}=17.5$.

Trick: LCM must be a multiple of both: $17.5÷2.5=7$ and $17.5÷0.175=100$ – both whole.

Answer: (C) 17.5

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

1. $A=2500\times (1.06{)}^2=2500\times 1.1236=2809$.
2. CI $=2809-2500=309$.

Trick: 2-year effective rate $=6+6+\frac{6\times 6}{100}=12.36\%$; $12.36\%$ of $2500=309$.

Answer: (B) ₹309

Concept: Solid hemisphere: TSA $=3\pi r^2$, CSA $=2\pi r^2$.

1. $3\pi r^2=166.32\Rightarrow \pi r^2=55.44$.
2. CSA $=2\pi r^2=2\times 55.44=110.88$ sq cm.

Trick: CSA $=\frac{2}{3}\times$ TSA $=\frac{2}{3}\times 166.32=110.88$.

Answer: (B) 110.88 sq cm