RRB Technician Grade III 2024, 2025 PYQs

Concept: Inverse ratio of $a:b:c$ is $\frac{1}{a}:\frac{1}{b}:\frac{1}{c}$.

1. Inverse of $1:2:3$ $=1:\frac{1}{2}:\frac{1}{3}$.
2. Multiply by LCM 6: $6:3:2$.

Trick: Multiply LCM of terms by each reciprocal: $\frac{6}{1}:\frac{6}{2}:\frac{6}{3}=6:3:2$.

Answer: (C) 6 : 3 : 2

Concept: If adjacent face areas are $lb,bh,lh$, then $V=\sqrt{lb\times bh\times lh}$.

1. $V^2=66\times 192\times 198$.
2. $66\times 198=13068$, and $13068\times 192=2509056$.
3. $V=\sqrt{2509056}=1584{\text{cm}}^3$.

Trick: Factorise: $66\times 192\times 198=(2^4\times 3^2\times 11{)}^2$, so $V=16\times 9\times 11=1584$.

Answer: (B) 1584

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100-\text{Loss}\%}$, $\text{SP}=\frac{\text{CP}\times (100+\text{Gain}\%)}{100}$.

1. SP of 1 orange $=\frac{40}{8}=5$ (₹).
2. CP $=\frac{5\times 100}{80}=6.25$ (₹).
3. Required SP $=6.25\times 1.2=7.5$ (₹).
4. Number for ₹30 $=\frac{30}{7.5}=4$.

Trick: SP ratio $=\frac{120}{80}=\frac{3}{2}$: price per orange goes from 5 to 7.5, so ₹30 buys $\frac{30}{7.5}=4$.

Answer: (A) 4

Concept: Sphere surface $=4\pi r^2$; cylinder CSA $=2\pi Rh$.

1. Cylinder: $R=6$, $h=12$, CSA $=2\pi \times 6\times 12=144\pi$.
2. $4\pi r^2=144\pi \Rightarrow r^2=36$.
3. $r=6$ cm.

Yaad rakhein: When cylinder height = its diameter, its CSA equals the surface of a sphere of the same radius.

Answer: (A) 6 cm

Concept: $(a^m\times b^n{)}^k=a^{mk}\times b^{nk}$.

1. $x^4=(3^2\times 2^3{)}^4$.
2. $=3^{2\times 4}\times 2^{3\times 4}$.
3. $=3^8\times 2^{12}$.

Trick: Multiply every exponent by 4: $2\to 8$, $3\to 12$.

Answer: (D) $3^8\times 2^{12}$

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

1. Sum $=4+5+1+k+6+0+3=19+k$.
2. Next multiple of 9 after 19 is 27.
3. $19+k=27\Rightarrow k=8$.

Trick: Single digit k only: 27 is the only reachable multiple of 9.

Answer: (B) 8

Concept: $\tan \theta ·\cot \theta =1$, so ${\tan }^2\theta +{\cot }^2\theta =(\tan \theta +\cot \theta {)}^2-2$.

1. ${\tan }^2\theta +{\cot }^2\theta ={13}^2-2=167$.
2. $8({\tan }^2\theta +{\cot }^2\theta )=8\times 167=1336$.

Trick: If $x+\frac{1}{x}=a$, then $x^2+\frac{1}{x^2}=a^2-2$.

Answer: (A) 1336

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

1. $P=\frac{960\times 100}{16\times 3}$.
2. $=\frac{96000}{48}=2000$.

Trick: Total rate $=16\times 3=48\%$; $48\%$ of P $=960\Rightarrow P=2000$.

Answer: (B) ₹2,000

Concept: Solve the bracket first, then multiply.

1. $24÷3=8$.
2. $8\times 2=16$.

Yaad rakhein: BODMAS: Brackets → Of → Division → Multiplication → Addition → Subtraction.

Answer: (A) 16

Concept: Make the divisor a whole number by shifting the decimal in both numbers.

1. $\frac{0.0488}{0.002}=\frac{48.8}{2}$ (multiply both by 1000).
2. $\frac{48.8}{2}=24.4$.

Trick: Ignore decimals: $488÷2=244$; decimal places $4-3=1$ → 24.4.

Answer: (D) 24.4

Concept: Take ages one year ago as $3x$ and $4x$; the gap between the two times is 5 years.

1. Four years hence: Sara $=3x+5$, Dipti $=4x+5$.
2. $\frac{3x+5}{4x+5}=\frac{4}{5}\Rightarrow 15x+25=16x+20$.
3. $x=5$; Sara one year ago $=15$.
4. Present age $=15+1=16$ years.

Trick: Check: one year ago $15:20=3:4$; four years hence $20:25=4:5$ ✓.

Answer: (C) 16 years

Concept: $P_n=P{(1+\frac{R}{100})}^n$.

1. $P_2=176400\times {\left(\frac{21}{20}\right)}^2$.
2. $=176400\times \frac{441}{400}=441\times 441$.
3. $=194481$.

Trick: $176400={420}^2$, so answer $=(420\times 1.05{)}^2={441}^2=194481$.

Answer: (D) 1,94,481

Concept: If $\angle Q=\angle R$, then $PQ=PR$ and the altitude from P bisects QR.

1. $PQ=PR=10$ cm, $QS=SR=\frac{12}{2}=6$ cm.
2. In right $\Delta PSR$: $P{S}^2=P{R}^2-S{R}^2=100-36=64$.
3. $PS=8$ cm.

Trick: 6-8-10 is a Pythagorean triplet (3-4-5 doubled).

Answer: (C) 8 cm

Concept: Form two equations from the two time points.

1. $M-6=6(D-6)\Rightarrow M=6D-30$.
2. $M+3=3(D+3)\Rightarrow M=3D+6$.
3. $6D-30=3D+6\Rightarrow D=12$, $M=42$.
4. Difference $=42-12=30$ years.

Trick: Age difference never changes: $5(D-6)=2(D+3)$ gives $D-6=6$, gap $=5\times 6=30$.

Answer: (D) 30 years

Concept: In $a:b::c:d$, product of extremes = product of means.

1. $3\times a=6\times 6$.
2. $a=\frac{36}{3}=12$.

Trick: Here 6 is the mean proportional: $a=\frac{6^2}{3}=12$.

Answer: (C) 12

Concept: Do the multiplication first, then the bracket, then add.

1. $(-\frac{5}{3})\times \frac{9}{25}=-\frac{3}{5}$.
2. $\frac{1}{5}-3=-\frac{14}{5}$.
3. $-\frac{3}{5}-\frac{14}{5}=-\frac{17}{5}$.

Trick: Use denominator 5 throughout: $-\frac{3}{5}+\frac{1}{5}-\frac{15}{5}=-\frac{17}{5}$.

Answer: (B) $-\frac{17}{5}$

Concept: An identity is true for all values of the variables.

1. $(a+b{)}^2=a^2+b^2+2ab$, so $(a+b{)}^2=a^2+b^2$ misses $2ab$.
2. Test $a=b=1$: LHS $=4$, RHS $=2$ – false.
3. Options (B), (C), (D) are standard identities.

Trick: Put $a=b=1$ in each option; the one that fails is not an identity.

Answer: (A) $(a+b{)}^2=a^2+b^2$

Concept: Equal work: (workers × days) must be equal in both cases.

1. $10(6m+8b)=2(26m+48b)\Rightarrow 60m+80b=52m+96b$.
2. $8m=16b\Rightarrow m=2b$.
3. Total work $=10(12b+8b)=200$ boy-days.
4. $40m+20b=100b$; time $=\frac{200}{100}=2$ days.

Trick: Convert all to boys: $1\text{man}=2\text{boys}$, then days $=\frac{\text{total work}}{\text{boys}}$.

Answer: (C) 2 days

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{P}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{D}\%}{100}$.

1. SP $=120\times 1.1=132$ (₹).
2. MP $\times 0.96=132$.
3. MP $=\frac{132}{0.96}=137.50$ (₹).

Trick: $\text{MP}=\text{CP}\times \frac{100+P}{100-D}=120\times \frac{110}{96}=137.5$.

Answer: (D) ₹137.50

Concept: To cross a bridge, the train covers (train length + bridge length).

1. Train length $=(24+1)\times 60=1500$ m.
2. Distance $=1500+1500=3000$ m $=3$ km.
3. Time $=\frac{3}{50}$ h $=\frac{3}{50}\times 60=3.6$ min.

Trick: Keep km and hours: $\frac{3\text{km}}{50\text{km/h}}\times 60$ gives minutes directly.

Answer: (D) 3.6 min