RRB Technician Grade III 2024, 2025 PYQs

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. 86325k6 from the right: odd places $6+5+3+8=22$
2. Even places $k+2+6=k+8$
3. $22-(k+8)=14-k$ must be 0 or 11 $\Rightarrow k=3$

Trick: Since $k\le 9$, the difference $14-k$ can only be 11, so $k=3$ (8632536 is divisible by 11).

Answer: (B) 3

Concept: Take tank capacity = LCM of 7, 10, 14 = 70 units.

1. Rates: A $=+10$, B $=+7$, C $=-5$ units/h
2. All three for 4.5 h: $(10+7-5)\times 4.5=54$ units
3. Remaining $=70-54=16$ units
4. B and C together $=7-5=2$ units/h, time $=\frac{16}{2}=8$ h

Trick: LCM-unit method avoids fractions completely.

Answer: (D) 8

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \%}{100}$.

1. At 30% loss: $\text{CP}=\frac{1148}{0.70}=1640$
2. For 5% gain: $\text{SP}=1640\times 1.05=1722$

Trick: Direct: $\text{New SP}=1148\times \frac{105}{70}=1722$.

Answer: (D) ₹1,722

Concept: Arrange data in order; for an even count, median = average of the two middle values.

1. Ordered: 4, 4, 6, 7, 8, 9, 9, 11, 12, 14 ($n=10$)
2. Middle terms: 5th $=8$ and 6th $=9$
3. Median $=\frac{8+9}{2}=8.5$

Yaad rakhein: For even $n$, the median is the mean of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th terms.

Answer: (C) 8.5

Concept: $x$ as a percentage of $y$ $=\frac{x}{y}\times 100$.

1. B as % of A $=\frac{60}{75}\times 100$
2. $=\frac{4}{5}\times 100=80\%$

Trick: $60:75=4:5$, and $\frac{4}{5}=80\%$.

Answer: (C) 80%

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

1. B $-$ A $=7-4=3$ parts $=600$, so 1 part $=200$
2. C $-$ D $=9-3=6$ parts
3. $6\times 200=1200$

Trick: Difference of parts × value of 1 part — no need to find the total.

Answer: (C) ₹1,200

Concept: BODMAS: innermost bracket first; inside it, ÷ and × from left to right.

1. $3\times 10÷2=30÷2=15$
2. $12+15=27$
3. $36-27=9$

Trick: Solve ( ) → { } → [ ] in order.

Answer: (D) 9

Concept: In an isosceles triangle the altitude to the base bisects the base.

1. Sides $3k,3k,4k$: $10k=20\Rightarrow k=2$, so sides 6, 6, 8
2. Half base $=4$
3. Altitude $h=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}$ cm

Trick: Altitude to base $=\sqrt{a^2-\frac{b^2}{4}}$ for equal side $a$, base $b$.

Answer: (A) $2\sqrt{5}\text{ cm}$

Concept: If $a,b,c,d$ are in proportion, $a:b=c:d$, i.e. $ad=bc$.

1. $7\times k=12\times 21$
2. $k=\frac{252}{7}=36$

Yaad rakhein: Product of extremes = product of means.

Answer: (B) 36

Concept: A number divisible by all of them must be a multiple of their LCM.

1. $\text{LCM}(8,10,12,16)=2^4\times 3\times 5=240$
2. $100000÷240=416$ remainder 160
3. Next multiple: $100000+(240-160)=100080$

Trick: Smallest 6-digit multiple $=100000+(\text{LCM}-\text{remainder})$.

Answer: (D) 100080

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with $R$ and $T$ in the same time unit.

1. Rate is per month, time 9 months
2. $\text{SI}=\frac{700\times 7\times 9}{100}=7\times 63=441$

Trick: Keep rate and time in the same unit — here both monthly.

Answer: (D) ₹441

Concept: Odd primes exclude 2.

1. First 10 odd primes: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
2. Sum $=158$
3. Average $=\frac{158}{10}=15.8$

Yaad rakhein: 2 is the only even prime — drop it when "odd primes" are asked.

Answer: (D) 15.8

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

1. Sum $=\frac{{\cos }^2{50}^{\circ }+(1+\sin {50}^{\circ }{)}^2}{\cos {50}^{\circ }(1+\sin {50}^{\circ })}$
2. Numerator $={\cos }^2+{\sin }^2+1+2\sin {50}^{\circ }=2(1+\sin {50}^{\circ })$
3. Result $=\frac{2}{\cos {50}^{\circ }}=2\sec {50}^{\circ }$

Trick: $\frac{\cos \theta }{1+\sin \theta }+\frac{1+\sin \theta }{\cos \theta }=2\sec \theta$ for any $\theta$.

Answer: (B) $2\sec {50}^{\circ }$

Concept: Difference of ratio parts equals the actual age difference.

1. $8-7=1$ part $=3$ years
2. Kishan $=8\times 3=24$ years

Trick: Check: Ravi $=21$, and $21:24=7:8$ ✓.

Answer: (A) 24 years

Concept: Curved surface area $=2\pi rh$, so it varies as $r\times h$.

1. New $r\times h=1.2\times 0.7=0.84$ of old
2. Decrease $=1-0.84=0.16=16\%$

Trick: Net % $=a+b+\frac{ab}{100}=20-30-\frac{600}{100}=-16\%$.

Answer: (B) 16%

Concept: $A=P{(1+\frac{R}{100})}^n$, CI $=A-P$.

1. $A=13000\times (1.04{)}^2=13000\times 1.0816$.
2. $A=14060.80$.
3. CI $=14060.80-13000=1060.80$.

Trick: For 2 years, effective rate $=4+4+\frac{4\times 4}{100}=8.16\%$; $8.16\%$ of 13000 $=1060.80$.

Answer: (A) ₹1,060.80

Concept: Sum of values = average × number of values.

1. Mon + Tue + Wed + Thu $=4\times 48=192$.
2. Tue + Wed + Thu + Fri $=4\times 46=184$.
3. Subtract: Mon − Fri $=192-184=8$.
4. Fri $=42-8={34}^{\circ }\text{C}$.

Trick: Difference of averages × 4 = Mon − Fri: $2\times 4=8$, so Fri $=42-8=34$.

Answer: (A) 34°C

Concept: Votes of the losing candidate are taken from valid votes only.

1. Valid votes $=80\%$ of $7500=6000$.
2. Other candidate got $100\%-55\%=45\%$ of valid votes.
3. $45\%$ of $6000=2700$.

Trick: Directly: $7500\times 0.8\times 0.45=2700$.

Answer: (B) 2700

Concept: Mode is the observation that occurs most often.

1. Count each value: 20 appears 5 times.
2. Every other value (23, 25, 30, 31, 32, 35, 41, 42, 43) appears only once.
3. So the most frequent value is 20.

Yaad rakhein: Just scan for repeated numbers – the most repeated one is the mode.

Answer: (B) 20

Concept: One-day work adds up: (A+B) − A = B.

1. B's one-day work $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$.
2. So B alone finishes 1 work in 4 days.
3. 21 times the work $=21\times 4=84$ days.

Trick: B's time $=\frac{2\times 4}{4-2}=4$ days, then $\times 21$.

Answer: (B) 84