RRB Technician Grade III 2024, 2025 PYQs

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

Concept: Equivalent discount of $a\%$ and $b\%$ $=a+b-\frac{ab}{100}$.

1. $40+20-\frac{40\times 20}{100}$
2. $=60-8=52\%$

Trick: On ₹100: $100\to 60\to 48$, so total discount $=52\%$.

Answer: (C) 52%

Concept: Rise in level $=\frac{\text{Volume displaced}}{\text{Base area of tank}}$.

1. Volume displaced $=50\times 4=200$ m${}^3$.
2. Base area $=40\times 20=800$ m${}^2$.
3. Rise $=\frac{200}{800}=0.25$ m $=25$ cm.

Trick: Always convert the final metres to cm: $\times 100$.

Answer: (D) 25 cm

Concept: If $\frac{A}{2}=\frac{B}{3}$, then $A:B=2:3$.

1. Total parts $=2+3=5$.
2. B's part $=\frac{3}{5}\times 2000=1200$.

Trick: Check: A $=800$, $\frac{800}{2}=400=\frac{1200}{3}$ ✓.

Answer: (A) 1,200

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

1. $240=2^4\times 3\times 5$, $360=2^3\times 3^2\times 5$, $480=2^5\times 3\times 5$.
2. HCF $=2^3\times 3\times 5=120$.

Trick: $240=2\times 120$, $360=3\times 120$, $480=4\times 120$, and 2, 3, 4 have HCF 1.

Answer: (D) 120

Concept: $a^m÷a^n=a^{m-n}$.

1. ${10}^{150}÷{10}^{146}={10}^{150-146}$
2. $={10}^4=10000$

Yaad rakhein: Same base in division ⇒ subtract the powers.

Answer: (A) 10000

Concept: After successive discounts, $\text{SP}=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. $\text{SP}=\text{MP}\times 0.80\times 0.85=0.68\text{MP}$.
2. $0.68\text{MP}=571.20$.
3. $\text{MP}=\frac{571.20}{0.68}=840$.

Trick: Net discount of 20% and 15% $=20+15-\frac{20\times 15}{100}=32\%$, so SP is $68\%$ of MP.

Answer: (C) 840