RRB Technician Grade III 2024, 2025 PYQs

Concept: Mode is the observation that occurs the maximum number of times.

1. Count: 66 → 3, 69 → 4, 71 → 3, 73 → 1, 74 → 3, 83 → 5, 84 → 3, 90 → 2
2. Highest frequency = 5, for 83

Trick: Count only the option values (90, 83, 74, 84) to save time.

Answer: (B) 83

Concept: If A is $r\%$ less than B, then $A=(100-r)\%$ of B.

1. Let B = 100
2. A $=100-40=60$
3. $\frac{A}{B}\times 100=60\%$

Trick: Take the base as 100 for percentage questions.

Answer: (B) 60%

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

1. $9993=10000-7$, $10007=10000+7$
2. Product $={10000}^2-7^2$
3. $=100000000-49=99999951$

Trick: Numbers equally spaced around a round number: (round number)${}^2$ − (gap)${}^2$.

Answer: (C) 99999951

Concept: Find the value of one ratio part from the given difference.

1. C − D $=9-3=6$ parts $=720$ ⇒ 1 part $=120$
2. C − B $=9-7=2$ parts
3. $=2\times 120=240$

Trick: Difference of shares = difference of ratio terms × value of 1 part.

Answer: (A) ₹240

Concept: Solve the bracket first, then multiplication, then addition/subtraction from left to right.

1. $\frac{2}{3}+\frac{1}{3}=1$
2. $1\times 23=23$
3. $23-13+9=19$

Yaad rakhein: Bracket → Of → Division/Multiplication → Addition/Subtraction.

Answer: (D) 19

Concept: A prime number is a natural number greater than 1 that is divisible only by 1 and itself.

1. So every prime has exactly two factors: 1 and the number.
2. Example: 7 → factors 1, 7; 13 → factors 1, 13.

Yaad rakhein: 1 has only one factor, so it is neither prime nor composite; 2 is the only even prime.

Answer: (B) 2

Concept: Use $1-{\sin }^2\theta ={\cos }^2\theta$ to simplify the denominator.

1. Denominator $=1-{\sin }^2\theta +\cos \theta ={\cos }^2\theta +\cos \theta$
2. $=\cos \theta (\cos \theta +1)$
3. Expression $=\frac{\sin \theta (1+\cos \theta )}{\cos \theta (1+\cos \theta )}=\frac{\sin \theta }{\cos \theta }=\tan \theta$

Trick: Put $\theta ={45}^{\circ }$: $\frac{0.707\times 1.707}{1.707-0.5}\approx 1=\tan {45}^{\circ }$.

Answer: (C) $\tan \theta$

Concept: Take total work as LCM of the days and work in repeating 3-day cycles.

1. Total work = LCM(15, 10, 30) = 30 units; A = 2, B = 3, C = 1 unit/day
2. One 3-day cycle: A alone 2 days + (A+B+C) on 3rd day $=2+2+6=10$ units
3. 30 units need $\frac{30}{10}=3$ cycles $=3\times 3=9$ days

Trick: Cycle questions: work done per cycle → number of full cycles → check if work ends exactly at a cycle end.

Answer: (B) 9 days

Concept: Net rate = filling rate − emptying rate (work per hour).

1. Filling pipe: $\frac{1}{9}$ tank/h; emptying pipe: $\frac{1}{36}$ tank/h.
2. Net rate $=\frac{1}{9}-\frac{1}{36}=\frac{4-1}{36}=\frac{1}{12}$ tank/h, so the full tank needs 12 h.
3. One-third of the tank: $\frac{1}{3}\times 12=4$ h.

Trick: Take LCM 36 as tank capacity: inlet $+4$, outlet $-1$, net $3$ units/h; $\frac{1}{3}$ tank $=12$ units $\Rightarrow 4$ h.

Answer: (A) 4

Concept: Average = sum of values ÷ number of values.

1. Values (2017–2022): 4300, 4500, 5000, 5000, 6000, 5500.
2. Sum $=4300+4500+5000+5000+6000+5500=30300$.
3. Average $=\frac{30300}{6}=5050$.

Trick: Take 5000 as base: deviations $-700,-500,0,0,+1000,+500$ add to $+300$; $5000+\frac{300}{6}=5050$.

Answer: (C) 5050

Concept: Winner's share $\%=\frac{\text{winner's votes}}{\text{total votes}}\times 100$.

1. Total votes $=1854+618=2472$.
2. Notice $1854=3\times 618$, so the ratio winner : loser $=3:1$.
3. Share $=\frac{3}{4}\times 100=75\%$.

Trick: Whenever one number is a multiple of the other, convert to a ratio first — no long division needed.

Answer: (B) 75%

Concept: Wet surface of an open cistern = floor area + area of the four walls up to the water level: $lb+2(l+b)h$.

1. $l=6$ m, $b=4$ m, $h=1\text{ m }25\text{ cm}=1.25$ m.
2. Floor $=6\times 4=24{\text{ m}}^2$.
3. Walls $=2(6+4)\times 1.25=25{\text{ m}}^2$.
4. Total $=24+25=49{\text{ m}}^2$.

Yaad rakhein: The top surface is water, not a wall — never add a second $lb$.

Answer: (D) 49 $m^2$

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

1. $63=3\text{Median}-2\times 36=3\text{Median}-72$.
2. $3\text{Median}=135$.
3. $\text{Median}=45$.

Trick: Direct form: $\text{Median}=\frac{\text{Mode}+2\text{Mean}}{3}=\frac{63+72}{3}=45$.

Answer: (B) 45

Concept: Let present ages be $7x$ and $6x$; use the future ratio to find $x$, then the past ratio to find the years.

1. $\frac{7x+6}{6x+6}=\frac{8}{7}\Rightarrow 49x+42=48x+48\Rightarrow x=6$.
2. Present ages: Manmohan $42$, Manpreet $36$.
3. $t$ years ago: $\frac{42-t}{36-t}=\frac{4}{3}\Rightarrow 126-3t=144-4t\Rightarrow t=18$.

Trick: Age difference stays constant: $7-6=1$ part $=6$ yrs; in $4:3$ also 1 part $=6$, so ages were $24$ and $18$; $42-24=18$.

Answer: (B) 18 years

Concept: Three consecutive multiples of 6: $a,a+6,a+12$.

1. $3a+18=3240$.
2. $3a=3222\Rightarrow a=1074$.
3. Check: $1074+1080+1086=3240$.

Trick: Middle term $=\frac{3240}{3}=1080$; smallest $=1080-6=1074$.

Answer: (B) 1074

Concept: Time is inversely proportional to efficiency.

1. Efficiency A : B $=1:3$, so time A : B $=3x:x$.
2. Difference $3x-x=60\Rightarrow x=30$: A takes 90 days, B takes 30 days.
3. Together: $\frac{1}{90}+\frac{1}{30}=\frac{4}{90}$ per day $\Rightarrow \frac{90}{4}=22\frac{1}{2}$ days.

Trick: Total work $=90$ units, efficiencies $1$ and $3$; together $\frac{90}{1+3}=22.5$ days.

Answer: (A) $22\frac{1}{2}$ days

Concept: Selling price after discount $=\text{List price}\times (1-\frac{d}{100})$; this is the retailer's cost price.

1. List price $=$ ₹13,200, discount $=20\%$.
2. Price $=13200\times \frac{80}{100}=10560$.
3. So the retailer pays ₹10,560.

Trick: $20\%$ of 13200 is 2640; $13200-2640=10560$.

Answer: (A) ₹10,560

Concept: The unit digit of a square depends only on the unit digit of the number.

1. Squares of 0–9 end in: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1.
2. Possible unit digits: 0, 1, 4, 5, 6, 9.
3. Digits 2, 3, 7, 8 never occur; among the options only 8 is impossible.

Yaad rakhein: A perfect square never ends in 2, 3, 7 or 8.

Answer: (D) 8

Concept: Money doubles when SI $=$ Principal, i.e. $\frac{P\times R\times T}{100}=P$.

1. $R\times T=100$.
2. $R\times 8=100$.
3. $R=\frac{100}{8}=12.5\%$.

Trick: Doubling: $R=\frac{100}{T}$; becomes $n$ times: $R=\frac{100(n-1)}{T}$.

Answer: (B) 12.5%

Concept: Convert the length to one unit, then divide equally.

1. $8\text{ m }54\text{ cm}=854$ cm.
2. Each piece $=\frac{854}{7}=122$ cm.
3. $122\text{ cm}=1.22$ m.

Trick: Check by multiplication: $1.22\times 7=8.54$ m.

Answer: (A) 1.22 m