RRB Technician Grade III 2024, 2025 PYQs

Concept: The difference of ratio parts equals the actual age difference.

1. Ratio $5:8$, difference $=8-5=3$ parts.
2. $3$ parts $=3$ years, so $1$ part $=1$ year.
3. Surya $=8\times 1=8$ years.

Trick: Here 1 part = 1 year, so the ratio terms are the ages themselves (5 and 8).

Answer: (C) 8 years

Concept: Successive increase of $r\%$ for $n$ years: $\text{Final}=P{(1+\frac{r}{100})}^n$.

1. 2019 to 2021 is 2 years.
2. Income $=21000\times {\left(\frac{11}{10}\right)}^2=21000\times 1.21$.
3. $=25410$.

Trick: Two $10\%$ increases $=21\%$: $21000+4410=25410$.

Answer: (C) ₹25,410

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100\pm \text{rate}}$ (+ for profit, − for loss).

1. Shirt: $\text{CP}=\frac{960\times 100}{96}=1000$, loss $=40$.
2. Sweater: $\text{CP}=\frac{840\times 100}{120}=700$, profit $=140$.
3. Net $=140-40=100$ profit.

Trick: Compare totals: CP $=1700$, SP $=1800$, so profit ₹100.

Answer: (D) Profit, ₹100

Concept: Total cost includes transport. $\text{SP per item}=\text{CP per item}+\text{profit per item}$.

1. Total cost $=4800+800=$ ₹5,600
2. CP per apple $=\frac{5600}{800}=$ ₹7
3. SP per apple $=7+10=$ ₹17

Trick: Always add overheads (transport, repair) to the cost price before dividing.

Answer: (D) ₹17

Concept: When a transversal cuts two parallel lines, the interior angles on the same side (co-interior angles) are supplementary: $\angle 1+\angle 2={180}^{\circ }$.

1. $\angle 1+\angle 2={180}^{\circ }$
2. ${65}^{\circ }+\angle 2={180}^{\circ }$
3. $\angle 2={180}^{\circ }-{65}^{\circ }={115}^{\circ }$

Yaad rakhein: Co-interior = sum ${180}^{\circ }$; alternate and corresponding angles = equal.

Answer: (C) ${115}^{\circ }$

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

1. $\text{SI}=\frac{1232\times 6\times 3}{100}=\frac{22176}{100}=221.76$
2. $\text{Amount}=1232+221.76=1453.76$

Trick: Total rate $=6\times 3=18\%$, so amount $=1232\times 1.18=1453.76$ in one step.

Answer: (B) ₹1,453.76

Concept: With simple interest, $A=P(1+\frac{RT}{100})$.

1. $RT=4\times 6=24\%$
2. $A=P\times \frac{124}{100}$
3. $P=\frac{1116\times 100}{124}=900$

Trick: Amount = $124\%$ of principal; $1\%=\frac{1116}{124}=9$, so $100\%=900$.

Answer: (B) ₹900

Concept: A prime number has exactly two factors, 1 and itself. 1 is not prime; 2 is the only even prime.

1. First five primes: $2,3,5,7,11$
2. Sum $=2+3+5+7+11=28$

Yaad rakhein: Primes up to 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 (ten primes).

Answer: (C) 28

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

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

Trick: Put $A={45}^{\circ }$: $2+1-2+1=2$ and $2{\cot }^2{45}^{\circ }=2$ – matches option (A).

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

Concept: Read only the column asked for (females) and subtract.

1. Apples sold to females: A $=26$, B $=73$
2. Difference $=73-26=47$

Trick: Mark the needed column first so you don't mix up male and female figures.

Answer: (A) 47

Concept: If A is $x\%$ more than B, then B is $\frac{x}{100+x}\times 100\%$ less than A.

1. Let water $=100$, milk $=115$
2. Water is less by $\frac{15}{115}\times 100\%$
3. $=13.04\%$

Trick: $x\%$ more $\to$ $\frac{x}{100+x}$ less: $\frac{15}{115}=\frac{3}{23}\approx 13.04\%$.

Answer: (D) 13.04%

Concept: Diagonals of a rhombus bisect each other at right angles, so ${\text{side}}^2={\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2$; $\text{Area}=\frac{1}{2}{d}_1{d}_2$.

1. Side $=\frac{148}{4}=37$ cm
2. $\frac{d_2}{2}=\sqrt{{37}^2-{12}^2}=\sqrt{1369-144}=\sqrt{1225}=35$
3. $d_2=70$ cm
4. Area $=\frac{1}{2}\times 24\times 70=840$

Trick: $(12,35,37)$ is a Pythagorean triplet – no square roots needed.

Answer: (B) 840

Concept: Evaluate powers first, then division, then addition.

1. ${36}^2÷9^2={\left(\frac{36}{9}\right)}^2=4^2=16$
2. $16+4^2=16+16=32$

Trick: $a^2÷b^2={\left(\frac{a}{b}\right)}^2$ – saves big multiplications.

Answer: (D) 32

Concept: Successive discounts are applied one after another: $\text{SP}=\text{MP}\times \frac{100-d_1}{100}\times \frac{100-d_2}{100}$.

1. After $10\%$: $28000\times 0.9=25200$
2. After $15\%$: $25200\times 0.85=21420$

Trick: Net discount $=10+15-\frac{10\times 15}{100}=23.5\%$; $28000\times 0.765=21420$.

Answer: (C) ₹21,420

Concept: Take total work = LCM of 9, 12, 18 $=36$ units. Efficiencies: A $=4$, B $=3$, C $=2$ units/day.

1. First 3 days (A+B+C): $3\times 9=27$ units
2. Remaining $=36-27=9$ units
3. B+C do $5$ units/day: $\frac{9}{5}$ days
4. Total $=3+\frac{9}{5}=\frac{24}{5}$ days

Trick: LCM-unit method avoids fractions; add the days before and after the person leaves.

Answer: (B) $\frac{24}{5}$

Concept: After $k$ years both ages increase by $k$; set up the ratio equation.

1. $\frac{46+k}{60+k}=\frac{4}{5}$
2. $230+5k=240+4k$
3. $k=10$

Trick: Age difference $14$ stays constant; ratio $4:5$ differs by 1 part, so 1 part $=14$, ages $56,70$ $\Rightarrow k=10$.

Answer: (B) 10

Concept: Difference in shares equals difference in ratio parts.

1. C − D $=9-3=6$ parts $=$ ₹720
2. 1 part $=\frac{720}{6}=$ ₹120
3. D's share $=3\times 120=$ ₹360

Trick: Find value of 1 part from the given difference, then multiply.

Answer: (A) ₹360

Concept: Moving towards each other, relative speed $=$ sum of speeds. To be 2 km apart, they must cover $56-2=54$ km together.

1. Relative speed $=13+14=27$ km/h
2. Distance to close $=54$ km
3. Time $=\frac{54}{27}=2$ hours

Trick: (After crossing they would again be 2 km apart at $\frac{58}{27}$ h, which is not an option.)

Answer: (A) 2 hours

Concept: For a circle through all four corners of a square, the diagonal of the square is the diameter: $d=a\sqrt{2}$.

1. Diameter $=7\sqrt{2}$ cm
2. Circumference $=\pi d=\frac{22}{7}\times 7\sqrt{2}$
3. $=22\sqrt{2}$ cm

Yaad rakhein: Circumscribed circle $\to$ diameter = diagonal; inscribed circle $\to$ diameter = side.

Answer: (B) $22\sqrt{2}$ cm

Concept: Take tank $=30$ units (LCM of 15, 10, 30). P $=+2$, Q $=+3$, R $=-1$ unit/hour.

1. Normal time with P+Q $=\frac{30}{5}=6$ h
2. First 2 h all open: $2\times (2+3-1)=8$ units
3. Remaining $22$ units by P+Q: $\frac{22}{5}=4.4$ h
4. Total $=6.4$ h $\Rightarrow$ extra $=0.4$ h $=24$ min

Trick: Extra time $=\frac{\text{water lost to R}}{\text{P+Q rate}}=\frac{2\times 1}{5}$ h $=24$ min.

Answer: (B) 24 min