RRB Technician Grade III 2024, 2025 PYQs

Concept: By BODMAS, division is done before addition.

1. $6÷3=2$
2. $(-18)+2+5=-18+7$
3. $=-11$

Trick: Do $÷$ and $\times$ first, then add the positives together and the negatives together.

Answer: (D) $-11$

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

1. $P=2,00,000$, $R=7\%$, $T=2$ years.
2. $\text{SI}=\frac{200000\times 7\times 2}{100}$
3. $=2000\times 14=28000$

Trick: 1 year interest $=7\%$ of ₹2,00,000 $=14000$; for 2 years just double it: $28000$.

Answer: (B) ₹28,000

Concept: Take total work as LCM of 9, 12, 18 $=36$ units; efficiency = units per day.

1. Efficiencies: A $=\frac{36}{9}=4$, B $=\frac{36}{12}=3$, C $=\frac{36}{18}=2$.
2. In 3 days A+B+C do $(4+3+2)\times 3=27$ units.
3. Remaining work $=36-27=9$ units.
4. B+C do 5 units/day, so time $=\frac{9}{5}$ days.

Trick: LCM method avoids fractions: work left $÷$ combined efficiency of those still working.

Answer: (D) $\frac{9}{5}$

Concept: Net filling rate = filling rate $-$ emptying rate.

1. Net rate $=\frac{1}{12}-\frac{1}{28}=\frac{7-3}{84}=\frac{4}{84}=\frac{1}{21}$ tank per hour.
2. Full tank takes 21 hours.
3. One-third tank $=\frac{21}{3}=7$ hours.

Trick: Capacity = LCM(12, 28) $=84$ units: $+7-3=4$ units/h; $\frac{1}{3}$ of 84 $=28$ units $\Rightarrow \frac{28}{4}=7$ h.

Answer: (D) 7

Concept: Subtract the two equations to eliminate the tables.

1. $2T+3C=540$ and $2T+C=470$.
2. Subtracting: $2C=70\Rightarrow C=35$.
3. Cost of 35 chairs $=35\times 35=1225$.

Trick: Both bills have the same 2 tables, so the ₹70 difference is the price of 2 chairs.

Answer: (D) ₹1,225

Concept: Co-primes have $\text{HCF}=1$; here both numbers must also be composite.

1. (11, 17): both are prime — rejected.
2. (19, 27): 19 is prime — rejected.
3. (21, 35): both composite but $\text{HCF}=7$ — not co-prime.
4. (8, 25): $8=2^3$, $25=5^2$, no common factor, $\text{HCF}=1$ ✓.

Yaad rakhein: Two composite numbers can still be co-prime if their prime factors are completely different.

Answer: (C) (8, 25)

Concept: Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$.

1. $\frac{\sqrt{3}}{4}{a}^2=16\sqrt{3}$
2. $a^2=16\times 4=64$
3. $a=8$ cm

Trick: Cancel $\sqrt{3}$ on both sides at once: $a^2=4\times 16$.

Answer: (B) 8 cm

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

1. $\text{Mode}=3\times 30-2\times 25$
2. $=90-50=40$

Yaad rakhein: Remember “3 Median minus 2 Mean”.

Answer: (A) 40

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$, $\text{cosec}({90}^{\circ }-\theta )=\sec \theta$, and ${\sin }^2\theta +{\cos }^2\theta =1$, ${\sec }^2\theta -{\tan }^2\theta =1$.

1. $\sin {42}^{\circ }=\cos {48}^{\circ }$, so numerator $={\sin }^2{48}^{\circ }+{\cos }^2{48}^{\circ }=1$.
2. $\text{cosec}{50}^{\circ }=\sec {40}^{\circ }$, so denominator $={\tan }^2{40}^{\circ }-{\sec }^2{40}^{\circ }=-1$.
3. Value $=\frac{1}{-1}=-1$.

Trick: Angles adding to ${90}^{\circ }$ (48+42, 40+50) signal the complementary-angle identities.

Answer: (C) $-1$

Concept: Translate the statement directly into an equation in “60% of the number”.

1. Let 60% of the number $=y$.
2. $y-50=40$
3. $y=90$

Trick: The question asks for 60% itself, so there is no need to find the number ($x=150$).

Answer: (A) 90

Concept: $\text{Qualified}\%=\frac{\text{Qualified}}{\text{Appeared}}\times 100$ for each year; compare.

1. A: $\frac{100}{500}=20\%$ → $\frac{150}{600}=25\%$ (changed).
2. B: $\frac{200}{600}=33.3\%$ → $\frac{250}{800}=31.25\%$ (changed).
3. C: $\frac{50}{400}=12.5\%$ → $\frac{100}{500}=20\%$ (changed).
4. D: $\frac{60}{300}=20\%$ → $\frac{80}{400}=20\%$ (same).

Trick: Check if the ratio stays the same: $\frac{60}{300}=\frac{1}{5}=\frac{80}{400}$.

Answer: (A) D

Concept: Form equations from present and future ages; the combined age rises by 2 every year.

1. Son $=x$, father $=7x$.
2. $7x+3=5(x+3)\Rightarrow 2x=12\Rightarrow x=6$; father $=42$.
3. Present sum $=48$; needed increase $=58-48=10$.
4. Years $=\frac{10}{2}=5$.

Trick: Check options: in 5 years ages are 11 and 47, sum $=58$ ✓.

Answer: (A) 5 years

Concept: $x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$

1. ${(x+\frac{1}{x})}^2=6^2=36$
2. $x^2+\frac{1}{x^2}=36-2=34$

Yaad rakhein: If $x+\frac{1}{x}=k$, then $x^2+\frac{1}{x^2}=k^2-2$.

Answer: (C) 34

Concept: Altitude of an equilateral triangle $h=\frac{\sqrt{3}}{2}a$; area $=\frac{\sqrt{3}}{4}{a}^2=\frac{h^2}{\sqrt{3}}$.

1. $\frac{\sqrt{3}}{2}a=12\Rightarrow a=\frac{24}{\sqrt{3}}=8\sqrt{3}$ cm.
2. $a^2=64\times 3=192$.
3. Area $=\frac{\sqrt{3}}{4}\times 192=48\sqrt{3}$ cm${}^2$.

Trick: Direct: Area $=\frac{h^2}{\sqrt{3}}=\frac{144}{\sqrt{3}}=48\sqrt{3}$.

Answer: (D) $48\sqrt{3}$

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$, i.e. $m^2=ab$.

1. ${72}^2=18\times y$
2. $5184=18y$
3. $y=\frac{5184}{18}=288$

Trick: $y=\frac{72\times 72}{18}=72\times 4=288$ — cancel first.

Answer: (D) 288

Concept: $\text{Sum}=\text{Average}\times \text{Count}$.

1. Sum $=72\times 8=576$.
2. Least number $=\frac{1}{16}\times 576=36$.

Trick: $\frac{8\times 72}{16}=\frac{72}{2}=36$ — half of the average.

Answer: (C) 36

Concept: Test odd numbers (not ending in 5) for divisibility by primes up to $\sqrt{430}\approx 20.7$, i.e. 3, 7, 11, 13, 17, 19.

1. Candidates: 401, 403, 407, 409, 411, 413, 417, 419, 421, 427, 429.
2. Composite: $403=13\times 31$, $407=11\times 37$, $411=3\times 137$, $413=7\times 59$, $417=3\times 139$, $427=7\times 61$, $429=3\times 143$.
3. Primes left: 401, 409, 419, 421.

Trick: First drop multiples of 3 by digit sum, then test only 7, 11, 13, 17, 19.

Answer: (C) 4

Concept: With false weight, $\text{Gain}\%=\frac{\text{Error}}{\text{True value}-\text{Error}}\times 100$.

1. $11\frac{1}{9}\%=\frac{1}{9}$.
2. So error : weight used $=1:9$, i.e. out of 1000 g, error $=\frac{1000}{10}=100$ g.
3. Weight used $=1000-100=900$ g.

Trick: Gain $\frac{1}{n}$ ⇒ he gives $\frac{n}{n+1}$ of the true weight: $\frac{9}{10}\times 1000=900$ g.

Answer: (A) 900 gm weight

Concept: $\text{Present value}=\text{Past value}\times {(1-\frac{r}{100})}^n$.

1. $162000=P\times (0.9{)}^2=0.81P$
2. $P=\frac{162000}{0.81}=200000$

Trick: Check: $200000\to 180000\to 162000$ (10% less each year) ✓.

Answer: (A) ₹2,00,000

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

1. $\frac{A}{P}=\frac{14520}{12000}=1.21$.
2. $(1.1{)}^n=1.21=(1.1{)}^2$.
3. $n=2$ years.

Trick: At 10%: $1.1,1.21,1.331$ — ratio 1.21 means 2 years.

Answer: (B) 2