RRB Technician Grade III 2024, 2025 PYQs

Concept: Let the numbers be $7k$ and $5k$.

1. Difference $=7k-5k=2k=360$
2. $k=180$
3. Greater number $=7\times 180=1260$

Trick: $2$ parts $=360\Rightarrow 1$ part $=180$; greater $=7$ parts.

Answer: (A) 1260

Concept: Division and multiplication are done left to right before addition/subtraction.

1. $3÷3=1$, then $1\times 3=3$
2. $3+3-3=3$

Yaad rakhein: Same-level operations ($÷,\times$) go strictly left to right.

Answer: (D) 3

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; difference depends only on difference of rates.

1. Rate difference $=6.5\%-5.5\%=1\%$
2. Difference $=\frac{300000\times 1\times 2}{100}=6000$

Trick: Same P and T: $\text{diff}=\frac{P\times (R_1-R_2)\times T}{100}$

Answer: (C) ₹6,000

Concept: Form the equation $\frac{5}{4}x-\frac{3}{4}x=7$.

1. $\frac{2}{4}x=7\Rightarrow \frac{x}{2}=7$
2. $x=14$

Trick: Difference of fractions $\frac{5}{4}-\frac{3}{4}=\frac{1}{2}$, so number $=2\times 7$

Answer: (B) 14

Concept: Find each pipe's rate in litres per hour, add, and divide the new capacity.

1. A $=\frac{1000}{5}=200$ L/h, B $=\frac{1000}{6}=\frac{500}{3}$ L/h
2. Together $=200+\frac{500}{3}=\frac{1100}{3}$ L/h
3. Time $=\frac{1800}{1100/3}=\frac{5400}{1100}=\frac{54}{11}$ h

Trick: For 1000 L together: $\frac{5\times 6}{11}=\frac{30}{11}$ h; for 1800 L multiply by $1.8$: $\frac{54}{11}$

Answer: (A) $\frac{54}{11}$

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

1. $\text{SI}=\frac{65000\times 7.5\times 6}{100}=29250$
2. Amount $=65000+29250=94250$

Trick: Options ₹$28750$/₹$29250$ are only interest — question asks for the amount.

Answer: (B) ₹94,250

Concept: Exterior angle $={180}^{\circ }-$ interior angle $=$ sum of the two opposite interior angles.

1. $\angle C={180}^{\circ }-{50}^{\circ }-{70}^{\circ }={60}^{\circ }$
2. Exterior $\angle A={180}^{\circ }-{50}^{\circ }={130}^{\circ }$
3. Check: $\angle B+\angle C={70}^{\circ }+{60}^{\circ }={130}^{\circ }$ ✓

Yaad rakhein: Exterior angle at a vertex $={180}^{\circ }-$ that vertex's angle.

Answer: (A) ${130}^{\circ }$

Concept: For $n=10$ (even) values, median $=$ average of the $5^{th}$ and $6^{th}$ values.

1. $5^{th}=x$, $6^{th}=x+2$
2. $\frac{x+x+2}{2}=63\Rightarrow x+1=63$
3. $x=62$

Trick: Two consecutive middle terms differing by $2$: median $=x+1$.

Answer: (D) 62

Concept: Write each decimal as a power of 10 and add the exponents.

1. $0.1={10}^{-1}$, $0.001={10}^{-3}$, $0.0001={10}^{-4}$
2. Product $={10}^{-1-3-4+6}={10}^{-2}$
3. $=\frac{1}{100}$

Trick: Count decimal places: $1+3+4=8$; ${10}^6$ cancels $6$, leaving ${10}^{-2}$

Answer: (D) $\frac{1}{100}$

Concept: Translate: $75\%$ of $x=15\%$ of $1500-15$.

1. $15\%$ of $1500=225$
2. $0.75x=225-15=210$
3. $x=\frac{210}{0.75}=280$

Trick: $75\%=\frac{3}{4}$, so $x=210\times \frac{4}{3}=280$

Answer: (A) 280

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

1. $28400=\text{MP}\times \frac{80}{100}$
2. $\text{MP}=\frac{28400\times 100}{80}=35500$

Trick: $80\%\to 28400\Rightarrow 1\%\to 355\Rightarrow 100\%\to 35500$

Answer: (D) ₹35,500

Concept: Sum $=$ average $\times$ count.

1. Sum of 8 numbers $=8\times 14=112$
2. Sum of 6 numbers $=6\times 16=96$
3. Remaining two $=112-96=16$; average $=\frac{16}{2}=8$

Trick: Six numbers are $2$ above average each: excess $=12$, so the other two are $12$ below in total: $14-\frac{12}{2}=8$

Answer: (C) 8

Concept: $\sin \theta =\cos ({90}^{\circ }-\theta )$ and $\cos \theta =\sin ({90}^{\circ }-\theta )$.

1. $\sin {54}^{\circ }=\cos ({90}^{\circ }-{54}^{\circ })=\cos {36}^{\circ }$
2. $\cos {72}^{\circ }=\sin ({90}^{\circ }-{72}^{\circ })=\sin {18}^{\circ }$
3. Sum $=\cos {36}^{\circ }+\sin {18}^{\circ }$

Yaad rakhein: Complementary swap: $\sin \leftrightarrow \cos$, angle $\to {90}^{\circ }-$ angle.

Answer: (D) $\cos {36}^{\circ }+\sin {18}^{\circ }$

Concept: Time taken is inversely proportional to efficiency.

1. Efficiency ratio Nayan : Tarun $=100:125=4:5$.
2. So time ratio Nayan : Tarun $=5:4$.
3. Tarun's time $=25\times \frac{4}{5}=20$ days.

Trick: $x\%$ more efficient → time becomes $\frac{100}{100+x}$ of original: $25\times \frac{100}{125}=20$.

Answer: (D) 20

Concept: In a regular polygon, each exterior angle $={180}^{\circ }-\text{interior angle}$ and $n=\frac{{360}^{\circ }}{\text{exterior angle}}$.

1. Exterior angle $={180}^{\circ }-{165}^{\circ }={15}^{\circ }$.
2. Number of sides $n=\frac{{360}^{\circ }}{{15}^{\circ }}=24$.

Yaad rakhein: Always go via the exterior angle – the exterior angles of any polygon add up to ${360}^{\circ }$.

Answer: (A) 24

Concept: Convert the given difference into ratio parts to find the value of one part.

1. C − B $=9-7=2$ parts $=$ ₹800.
2. 1 part $=\frac{800}{2}=$ ₹400.
3. D $=3$ parts $=3\times 400=$ ₹1,200.

Trick: Difference of parts ↔ difference of money; then multiply by D's parts directly: $\frac{800}{2}\times 3=1200$.

Answer: (C) ₹1,200

Concept: Find the CP and SP of one pencil, then $\text{Gain}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. CP of 1 pencil $=$ ₹$\frac{5}{6}$; SP of 1 pencil $=$ ₹$\frac{6}{5}$.
2. $\frac{\text{SP}}{\text{CP}}=\frac{6}{5}\times \frac{6}{5}=\frac{36}{25}$.
3. Gain $\%=\frac{36-25}{25}\times 100=44\%$.

Trick: For "$a$ for ₹$b$, sold $b$ for ₹$a$" type: gain $\%=\frac{a^2-b^2}{b^2}\times 100=\frac{36-25}{25}\times 100$.

Answer: (D) 44%

Concept: Leak rate $=$ (rate of X + Y) $-$ (actual filling rate with leak).

1. X + Y fill per hour $=\frac{1}{21}+\frac{1}{24}=\frac{15}{168}=\frac{5}{56}$ → time $=\frac{56}{5}=11.2$ h $=11$ h 12 min.
2. With leak: $11$ h 12 min $+48$ min $=12$ h, rate $=\frac{1}{12}$.
3. Leak rate $=\frac{5}{56}-\frac{1}{12}=\frac{15-14}{168}=\frac{1}{168}$.
4. Leak empties the full tank in $168$ hours.

Trick: Leak time $=\frac{t_1\times t_2}{t_2-t_1}=\frac{11.2\times 12}{0.8}=168$ h.

Answer: (C) 168 hours

Concept: For $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$: if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, the lines are parallel → no solution.

1. $\frac{a_1}{a_2}=\frac{3}{-6}=-\frac{1}{2}$, $\frac{b_1}{b_2}=\frac{-5}{10}=-\frac{1}{2}$.
2. $\frac{c_1}{c_2}=\frac{7}{7}=1\ne -\frac{1}{2}$.
3. So the system is inconsistent: no solution.

Trick: Multiply the first equation by $-2$: $-6x+10y=-14$, which can never equal $7$ – parallel lines.

Answer: (C) no solution

Concept: Use the perimeter to get the sides, then equate areas.

1. $l+b=\frac{50}{2}=25$; ratio $1:4$ → $b=5$ cm, $l=20$ cm.
2. Area $=20\times 5=100{\text{cm}}^2$.
3. Square side $=\sqrt{100}=10$ cm, perimeter $=4\times 10=40$ cm.

Trick: Area $=k\times 4k=4k^2$ and side of square $=2k$; perimeter $=8k=8\times 5=40$.

Answer: (B) 40 cm