RRB Technician Grade III 2024, 2025 PYQs

Concept: A pure recurring decimal $0.\overline{ab}=\frac{ab}{99}$.

1. Let $x=0.\overline{52}$, then $100x=52.\overline{52}$.
2. $100x-x=52\Rightarrow 99x=52$.
3. $x=\frac{52}{99}$.

Yaad rakhein: Repeating digits over as many 9s as there are repeating digits.

Answer: (D) $\frac{52}{99}$

Concept: A cylinder circumscribing a sphere of radius $r$ has radius $r$ and height $2r$.

1. $V_{\text{sphere}}=\frac{4}{3}\pi r^3$.
2. $V_{\text{cylinder}}=\pi r^2\times 2r=2\pi r^3$.
3. Ratio $=\frac{4}{3}:2=4:6=2:3$.

Yaad rakhein: Archimedes' result: sphere $=\frac{2}{3}$ of its circumscribing cylinder (volume and surface).

Answer: (D) 2 : 3

Concept: Express all ages in terms of B's age.

1. Let B $=x$, then A $=2x$ and C $=3x$.
2. $x+2x+3x=42\Rightarrow x=7$.
3. A + B $=2x+x=21$ years.

Trick: Ratio A : B : C $=2:1:3$; A + B is $\frac{3}{6}$ of $42=21$.

Answer: (B) 21 years

Concept: Do the multiplications first, then subtract, then divide (BODMAS).

1. $0.5\times 10.6=5.3$ and $0.2\times 1.4=0.28$.
2. $5.3-0.28=5.02$.
3. $\frac{5.02}{0.4}=12.55$.

Trick: Dividing by $0.4$ is the same as multiplying by $2.5$: $5.02\times 2.5=12.55$.

Answer: (A) 12.55

Concept: $\text{Savings}=\text{Income}-\text{Expenditure}$; take income $=100$.

1. Old: income $100$, expenditure $75$, savings $25$.
2. New: income $120$, expenditure $75\times 1.1=82.5$.
3. New savings $=120-82.5=37.5$.
4. Increase $=\frac{37.5-25}{25}\times 100=50\%$.

Trick: Assume income $=100$ in every income–expenditure–savings question.

Answer: (A) 50%

Concept: For equal distances, average speed $=\frac{2xy}{x+y}$.

1. $x=30$, $y=60$ km/h.
2. Average speed $=\frac{2\times 30\times 60}{30+60}=\frac{3600}{90}=40$ km/h.

Trick: Check: total distance $180$ km, time $3+1.5=4.5$ h, $\frac{180}{4.5}=40$.

Answer: (C) 40 km/h

Concept: List the primes between 61 and 100.

1. Primes: $61,67,71,73,79,83,89,97$.
2. Second smallest $=67$, second greatest $=89$.
3. Sum $=67+89=156$.

Yaad rakhein: Primes from 61 to 100 are 8 in number: 61, 67, 71, 73, 79, 83, 89, 97.

Answer: (D) 156

Concept: If $a,b,c,d$ are in proportion, $a\times d=b\times c$.

1. $12\times y=24\times 45$.
2. $y=\frac{1080}{12}=90$.

Trick: $\frac{24}{12}=2$, so $y=2\times 45=90$.

Answer: (D) 90

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

1. $12+15-\frac{12\times 15}{100}$.
2. $=27-1.8=25.2\%$.

Trick: Price after both $=0.88\times 0.85=0.748$, so discount $=25.2\%$.

Answer: (D) 25.2%

Concept: Convert workers to one unit by comparing the two groups (same work, so total work is equal).

1. $(6M+8B)\times 10=(26M+48B)\times 2$
2. $60M+80B=52M+96B\Rightarrow 8M=16B\Rightarrow M=2B$
3. Group 1 $=6(2B)+8B=20B$, takes 10 days, so work $=20\times 10=200$ boy-days
4. 25 boys need $\frac{200}{25}=8$ days

Trick: Work = workers $\times$ days. Turn everything into boy-days first.

Answer: (D) 8 days

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

1. Two 20% discounts: factor $=0.8\times 0.8=0.64$
2. $\text{MRP}\times 0.64=16000$
3. $\text{MRP}=\frac{16000}{0.64}=25000$

Trick: Two successive 20% discounts = single discount of $20+20-\frac{20\times 20}{100}=36\%$, so 64% of MRP is paid.

Answer: (A) ₹25,000

Concept: Use Pythagoras to get the missing leg; area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Other leg $=\sqrt{{10}^2-8^2}=\sqrt{100-64}=\sqrt{36}=6$
2. Legs 8 and 6 are perpendicular
3. Area $=\frac{1}{2}\times 8\times 6=24$ sq. units

Trick: $(6,8,10)$ is the triplet $(3,4,5)\times 2$ — no calculation needed.

Answer: (D) 24 sq. units

Concept: In simple interest, the amount grows by the same interest every year.

1. Interest for $5-2=3$ years $=2000-1250=750$, so 1 year $=250$
2. Principal $=1250-2\times 250=750$
3. Rate $=\frac{250}{750}\times 100=33.33\%$

Trick: Difference of amounts ÷ difference of years = yearly SI.

Answer: (A) 33.33%

Concept: A prime has exactly two factors, 1 and itself.

1. Primes below 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
2. Count them: 10

Yaad rakhein: There are 4 primes below 10, 8 below 20, 10 below 30 and 25 below 100.

Answer: (A) 10

Concept: $a^3-b^3=(a-b{)}^3+3ab(a-b)$.

1. Put $a-b=1$: $61=1^3+3ab(1)$
2. $3ab=60$
3. $ab=20$

Trick: Check: $a=5,b=4$ gives $125-64=61$.

Answer: (A) 20

Concept: Negative power flips the fraction: ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$; even powers make the sign positive.

1. ${\left(\frac{-2}{5}\right)}^{-2}={\left(\frac{5}{-2}\right)}^2=\frac{25}{4}$
2. ${\left(\frac{-4}{5}\right)}^4=\frac{256}{625}$
3. $\frac{25}{4}\times \frac{256}{625}=\frac{64}{25}$

Trick: Even power → answer positive, so only positive options remain.

Answer: (C) $\frac{64}{25}$

Concept: Value of a sector $=\frac{\text{sector }\%}{100}\times \text{total}$.

1. Mafindre share from the chart $=18\%$
2. Cars $=\frac{18}{100}\times 1500000=270000$

Trick: 1% of 15,00,000 is 15,000; so $18\times 15000=270000$.

Answer: (D) 2,70,000

Concept: For equal distances at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$.

1. $x=4,y=20$
2. Average $=\frac{2\times 4\times 20}{4+20}=\frac{160}{24}$
3. $=\frac{20}{3}$ km/h

Trick: Average speed is never the simple average $\frac{4+20}{2}=12$; it is closer to the slower speed.

Answer: (C) $\frac{20}{3}$ km/h

Concept: With the same money, the saving due to price cut buys the extra quantity.

1. 20% of ₹100 = ₹20 is saved
2. ₹20 buys 10 kg extra, so new price $=\frac{20}{10}=2$ per kg
3. Original price $=\frac{2}{0.8}=2.50$ per kg

Trick: Reduced price $=\frac{\%\times \text{money}}{100\times \text{extra qty}}$.

Answer: (A) ₹2.50

Concept: Perimeter $=2(l+b)$, Area $=l\times b$.

1. $l=3b$, so $2(3b+b)=8b=48\Rightarrow b=6$
2. $l=18$
3. Area $=18\times 6=108{\text{cm}}^2$

Trick: Perimeter $÷$ 8 directly gives the breadth when $l=3b$.

Answer: (C) 108