RRB Technician Grade III 2024, 2025 PYQs

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

Concept: Take LCM and use ${\sin }^2\theta +{\cos }^2\theta =1$.

1. Sum $=\frac{{\cos }^2\theta +(1+\sin \theta {)}^2}{\cos \theta (1+\sin \theta )}$.
2. Numerator $={\cos }^2\theta +1+2\sin \theta +{\sin }^2\theta =2+2\sin \theta =2(1+\sin \theta )$.
3. Cancel $(1+\sin \theta )$: result $=\frac{2}{\cos \theta }=2\sec \theta$.

Trick: Put $\theta =0^{\circ }$: value $2$; at ${60}^{\circ }$ the expression is $4$ and $2\sec {60}^{\circ }=4$ — only (A) fits both.

Answer: (A) $2\sec \theta$

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

1. $a=10001,b=12$.
2. $a^2={10001}^2=100020001$ (since $({10}^4+1{)}^2={10}^8+2\times {10}^4+1$).
3. $b^2=144$; $100020001-144=100019857$.

Trick: Count digits: $\approx {10}^4\times {10}^4={10}^8$, a 9-digit number — only 100019857 has 9 digits.

Answer: (D) 100019857

Concept: Quantity of a part $=\frac{\text{its ratio term}}{\text{sum of terms}}\times \text{total}$.

1. Milk : water $=3:1$, total parts $=4$.
2. Milk $=\frac{3}{4}\times 80=60$ litres.

Trick: 1 part $=\frac{80}{4}=20$ L, milk $=3\times 20=60$ L.

Answer: (D) 60 litres

Concept: Half-yearly compounding: rate becomes $\frac{R}{2}$ and periods become $2T$.

1. Rate per half-year $=5\%$, periods $=2$.
2. $A=16000{(1+\frac{5}{100})}^2=16000\times \frac{21}{20}\times \frac{21}{20}$.
3. $A=16000\times 1.1025=17640$.

Trick: Two periods at $5\%$ = effective $10.25\%$; $10.25\%$ of 16000 $=1640$, so ₹17,640.

Answer: (D) ₹17,640

Concept: $0.\bar{a}=\frac{a}{9}$.

1. $5.2222\dots =5+0.\bar{2}$.
2. $0.\bar{2}=\frac{2}{9}$.
3. $5+\frac{2}{9}=\frac{45+2}{9}=\frac{47}{9}$.

Trick: Shortcut: $\frac{52-5}{9}=\frac{47}{9}$ (whole number with repeating digit minus non-repeating part, over 9).

Answer: (A) $\frac{47}{9}$

Concept: Angle sum of a triangle $={180}^{\circ }$; an exterior and its adjacent interior angle make ${180}^{\circ }$.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{54}^{\circ }={36}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-{34}^{\circ }-{36}^{\circ }={110}^{\circ }$.
3. B, E, C are collinear, so $\angle AEB={180}^{\circ }-{110}^{\circ }={70}^{\circ }$.

Trick: Exterior angle $\angle AEB=\angle CAE+\angle C={34}^{\circ }+{36}^{\circ }={70}^{\circ }$.

Answer: (B) ${70}^{\circ }$

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

1. Digit sum of $42a=4+2+a=6+a$.
2. $6+a$ must be a multiple of 9; since $a\le 9$, only $6+a=9$ is possible (18 would need $a=12$).
3. So $a=3$; check $423=9\times 47$.

Trick: Add the digits and fill the gap to the next multiple of 9.

Answer: (D) 3

Concept: Convert each ratio to a decimal (or common denominator) and compare.

1. $\frac{2}{3}=0.67$, $\frac{5}{4}=1.25$, $\frac{3}{2}=1.5$, $\frac{4}{5}=0.8$.
2. Largest value $=1.5$.

Trick: Ratios with first term larger than second exceed 1; compare only $\frac{5}{4}$ and $\frac{3}{2}$: $5\times 2=10<3\times 4=12$, so $\frac{3}{2}$ is bigger.

Answer: (D) 3 : 2