RRB Technician Grade III 2024, 2025 PYQs

Concept: Take present ages as multiples of the ratio and use the second condition.

1. Amit $=7x$, Amita $=5x$.
2. $7x-5=5x+5\Rightarrow 2x=10\Rightarrow x=5$; ages $35$ and $25$.
3. Amit after 5 years $=40$, Amita 5 years ago $=20$.
4. Ratio $=40:20=2:1$.

Trick: A 1 : 1 ratio means the two ages are equal – set them equal directly.

Answer: (B) 2 : 1

Concept: The sum of deviations of all values from their arithmetic mean is always zero: $\sum (x_i-\overline{x})=0$.

1. Mean $\overline{x}=\frac{5+6+7+9+10+12+14}{7}=\frac{63}{7}=9$.
2. Deviations: $-4,-3,-2,0,1,3,5$.
3. Sum $=-9+9=0$.

Yaad rakhein: Sum of deviations from the mean is 0 for every data set – no calculation needed.

Answer: (B) 0

Concept: Write decimals as fractions: $\text{LCM}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$.

1. $5=\frac{5}{1},12.5=\frac{25}{2},0.05=\frac{1}{20}$.
2. LCM of $5,25,1=25$; HCF of $1,2,20=1$.
3. LCM $=\frac{25}{1}=25$.

Trick: Check: $25÷5=5$, $25÷12.5=2$, $25÷0.05=500$ – all whole numbers.

Answer: (D) 25

Concept: In $a:b::b:c$, $b$ is the mean proportional: $b^2=ac$.

1. ${12}^2=9\times k\Rightarrow 144=9k$.
2. $k=16$.
3. $k+1=17$.

Trick: Third proportional $=\frac{b^2}{a}=\frac{144}{9}=16$.

Answer: (B) 17

Concept: Net change for $a\%$ and $b\%$: $a+b+\frac{ab}{100}$ (decrease taken negative).

1. $a=-16,b=+25$.
2. Net $=-16+25+\frac{(-16)(25)}{100}=9-4=5\%$.
3. Positive, so it is a 5% increase.

Trick: Multiplier method: $0.84\times 1.25=1.05\Rightarrow +5\%$.

Answer: (D) 5% increase

Concept: One-day work: $\frac{1}{B}=\frac{1}{A+B\text{ time}}-\frac{1}{A}$.

1. B's one-day work $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$, so B alone takes $4$ days.
2. 17 times the work $=17\times 4=68$ days.

Trick: $\text{B}=\frac{2\times 4}{4-2}=4$ days directly.

Answer: (B) 68

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

1. $\text{SP}=1200\times \frac{85}{100}$.
2. $=1020$.

Trick: 15% of 1200 = 180; $1200-180=1020$.

Answer: (B) ₹1,020

Concept: At the same pace, time is directly proportional to the number of litchis.

1. Time for 1 litchi $=\frac{60}{21}$ min.
2. Time for 35 $=\frac{60}{21}\times 35=\frac{60\times 5}{3}=100$ min.

Trick: $21:35=3:5$, so time $=60\times \frac{5}{3}=100$.

Answer: (D) 100 minutes

Concept: Write in $\sin ,\cos$ and use $1-{\sin }^{2}A={\cos }^{2}A$.

1. $\sec A+\tan A=\frac{1+\sin A}{\cos A}$.
2. Product $=\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-{\sin }^{2}A}{\cos A}$.
3. $=\frac{{\cos }^{2}A}{\cos A}=\cos A$.

Trick: Put $A=0^{\circ }$: $(1+0)(1-0)=1=\cos 0^{\circ }$; only option (D) gives 1.

Answer: (D) $\cos A$

Concept: Repeatedly use $(a+b)(a-b)=a^2-b^2$.

1. $(x+\frac{1}{x})(x-\frac{1}{x})=x^2-\frac{1}{x^2}$.
2. $(x^2-\frac{1}{x^2})(x^2+\frac{1}{x^2})=x^4-\frac{1}{x^4}$.
3. $(x^4-\frac{1}{x^4})(x^4+\frac{1}{x^4})=x^8-\frac{1}{x^8}$.

Trick: Each extra bracket doubles the power: $2\to 4\to 8$.

Answer: (A) $x^8-\frac{1}{x^8}$

Concept: $A=P+\frac{P\times R\times T}{100}$.

1. SI for 2 years $=2\times 12\frac{1}{2}\%=25\%$ of $P$.
2. $1.25P=8250\Rightarrow P=6600$.
3. SI for 1 year $=6600\times \frac{12.5}{100}=825$.

Trick: $12\frac{1}{2}\%=\frac{1}{8}$: amount $=\frac{10}{8}P$, yearly SI $=\frac{P}{8}=\frac{8250}{10}=825$.

Answer: (C) ₹825

Concept: Find the next perfect square above the number and take the difference.

1. $\sqrt{6072}$ lies between $77$ and $78$ (${77}^2=5929$).
2. Next square: ${78}^2=6084$.
3. Number to add $=6084-6072=12$.

Trick: ${78}^2=(80-2{)}^2=6400-320+4=6084$.

Answer: (A) 12

Concept: Opposite directions: relative speed = sum of speeds; distance = sum of lengths.

1. Relative speed $=42+30=72$ km/h $=72\times \frac{5}{18}=20$ m/s.
2. Distance $=220+280=500$ m.
3. Time $=\frac{500}{20}=25$ s.

Trick: km/h → m/s: multiply by $\frac{5}{18}$; 72 km/h = 20 m/s.

Answer: (B) 25 sec

Concept: HCF = product of common prime factors with the lowest powers.

1. $45=3^2\times 5$, $78=2\times 3\times 13$, $117=3^2\times 13$.
2. Only common prime: $3$ (lowest power $3^1$).
3. HCF $=3$.

Trick: 78 is not divisible by 9, so 9 is ruled out; all three are divisible by 3.

Answer: (C) 3

Concept: $V\propto r^3$ and $S\propto r^2$.

1. $\frac{r_1^3}{r_2^3}=\frac{1}{8}\Rightarrow \frac{r_1}{r_2}=\frac{1}{2}$.
2. $\frac{S_1}{S_2}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$.

Trick: Area ratio $=(\text{volume ratio}{)}^{\frac{2}{3}}$: $8^{\frac{2}{3}}=4$.

Answer: (B) 1:4

Concept: Convert averages of pairs into sums, then add all three.

1. $P+Q=10100$, $Q+R=12500$, $P+R=10400$.
2. Adding: $2(P+Q+R)=33000\Rightarrow P+Q+R=16500$.
3. $P=16500-12500=4000$.

Trick: $P=\frac{(P+Q)+(P+R)-(Q+R)}{2}=\frac{10100+10400-12500}{2}=4000$.

Answer: (A) 4,000

Concept: $\frac{a^2-b^2}{a-b}=a+b$.

1. $a=2.46,b=1.46$.
2. Value $=2.46+1.46=3.92$.
3. $3.92=\frac{392}{100}$.

Trick: Spot $a\times a-b\times b$ over $a-b$ – the answer is simply $a+b$.

Answer: (A) $\frac{392}{100}$

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{Gain}\%}{100}$ and $\text{MRP}=\frac{\text{SP}\times 100}{100-\text{Discount}\%}$.

1. $\text{SP}=210\times \frac{190}{100}=399$.
2. $\text{MRP}=\frac{399\times 100}{95}=420$.

Trick: One line: $\text{MRP}=210\times \frac{190}{95}=210\times 2=420$.

Answer: (C) ₹420

Concept: If B lies on segment AC (between A and C), then $AB+BC=AC$.

1. $(k+3)+2k=5k-5$.
2. $3k+3=5k-5\Rightarrow 2k=8\Rightarrow k=4$.
3. Check: $AB=7,BC=8,AC=15$ ✓.

Yaad rakhein: Betweenness rule: the two parts must add up to the whole.

Answer: (B) 4

Concept: A number is divisible by 11 if (sum of odd-place digits) − (sum of even-place digits) is $0$ or a multiple of 11.

1. For $2y5$: $(2+5)-y=7-y$.
2. $7-y=0\Rightarrow y=7$ (a digit cannot make $7-y=\pm 11$).
3. Check: $275=11\times 25$ ✓.

Trick: For a 3-digit number $abc$, usually middle digit $b=a+c$ (if $a+c<10$).

Answer: (A) 7