RRB Technician Grade III 2024, 2025 PYQs

Concept: Order: Brackets → Division/Multiplication → Addition/Subtraction.

1. Bracket: $4\times 6=24$
2. Division: $48÷24=2$
3. $10+8+6-2=22$

Yaad rakhein: Always solve the bracket first, then the division.

Answer: (B) 22

Concept: ${\sin }^2\theta +{\cos }^2\theta =1$, ${\sec }^2\theta -{\tan }^2\theta =1$, $\tan \theta =\frac{\sin \theta }{\cos \theta }$.

1. ${\sin }^2\theta +{\cos }^2\theta -({\sec }^2\theta -{\tan }^2\theta )=1-1=0$
2. $\tan \theta \cos \theta =\frac{\sin \theta }{\cos \theta }\times \cos \theta =\sin \theta$
3. Total $=0+\sin \theta +\sin \theta =2\sin \theta$

Trick: Put $\theta ={30}^{\circ }$: $0+\frac{1}{2}+\frac{1}{2}=1=2\sin {30}^{\circ }$.

Answer: (B) $2\sin \theta$

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

1. $\text{SI}=\frac{200000\times 11\times 2}{100}$
2. $=2000\times 22=44000$

Trick: 11% of 2 lakh $=22000$ per year; 2 years → 44000.

Answer: (A) ₹44,000

Concept: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$

1. After the 2nd discount: $x\times \frac{7}{8}=105\Rightarrow x=120$
2. First discount $=150-120=30$
3. $\frac{30}{150}\times 100=20\%$

Trick: $12.5\%=\frac{1}{8}$: price before it $=105\times \frac{8}{7}=120$.

Answer: (B) 20%

Concept: Do multiplication and division before addition and subtraction (left to right).

1. $5\times 3=15$ and $12÷4=3$
2. $15-3+8=20$

Yaad rakhein: Never subtract before dividing: $12÷4$ comes first.

Answer: (C) 20

Concept: $1\text{km/h}=\frac{5}{18}\text{m/s}$; $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.

1. $10\text{km/h}=10\times \frac{5}{18}=\frac{25}{9}\text{m/s}$
2. Time $=200÷\frac{25}{9}=\frac{200\times 9}{25}=72$ s

Trick: 10 km/h = 10,000 m per 3600 s → 200 m takes $\frac{200}{10000}\times 3600=72$ s.

Answer: (D) 72 seconds

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

1. $25+8-\frac{25\times 8}{100}$
2. $=33-2=31\%$

Trick: On ₹100: after 25% → 75; 8% of 75 = 6 → pays 69; discount 31%.

Answer: (D) 31%

Concept: Form an equation linking the present age and the age at marriage.

1. Age at marriage $=x$; present $=x+10=\frac{6}{5}x$
2. $\frac{x}{5}=10\Rightarrow x=50$; present age $=60$
3. Son $=\frac{60}{10}=6$ years

Trick: The extra $\frac{1}{5}$ part equals the 10 years → $x=50$.

Answer: (A) 6 years

Concept: Percent less $=\frac{\text{difference}}{\text{base}}\times 100$, base = the number compared with (first).

1. Third $=100$: first $=70$, second $=63$
2. Difference $=7$
3. $\frac{7}{70}\times 100=10\%$

Trick: Base is the first number (70), not 100 — so not 7%.

Answer: (D) 10%

Concept: Read the data labels on the chart for each company and month.

1. Company X, February $=35$ crore
2. Company Y, April $=55$ crore
3. Difference $=55-35=20$ crore

Trick: Use the printed labels, not the axis height — the Y line sits above X on the scale.

Answer: (B) 20 crore

Concept: Find the work done in one 2-hour cycle, then finish the remainder.

1. C: $\frac{1}{6}$/h, D: $\frac{1}{9}$/h; one cycle $=\frac{1}{6}+\frac{1}{9}=\frac{5}{18}$
2. 3 cycles (6 h) $=\frac{15}{18}$; left $=\frac{3}{18}=\frac{1}{6}$
3. C (7th hour) fills $\frac{1}{6}$ exactly → total 7 h

Trick: LCM 18 units: C = 3, D = 2 per hour; 15 in 6 h, C adds the last 3 in hour 7.

Answer: (D) 7 hours

Concept: For an odd count of consecutive numbers, the average is the middle number.

1. 7 numbers → the $4^{\text{th}}$ number is the middle one $=33$.
2. Numbers: 30, 31, 32, 33, 34, 35, 36.
3. Greatest $=33+3=36$.

Trick: Greatest $=\text{average}+\frac{n-1}{2}=33+\frac{7-1}{2}=36$.

Answer: (C) 36

Concept: A number divisible by every number from 1 to 10 must be a multiple of their LCM.

1. $\text{LCM}(1,2,\dots ,10)=2^3\times 3^2\times 5\times 7$
2. $=8\times 9\times 5\times 7=2520$
3. 2520 is in the options; 1000, 504 and 608 fail (e.g. 1000 is not divisible by 3, 504 not by 5).

Yaad rakhein: 2520 is the smallest number divisible by all of 1 to 10 — a classic fact worth memorising.

Answer: (B) 2520

Concept: Volume of a cube $=a^3$; cube every factor separately.

1. $a=6xy{z}^2$
2. $a^3=6^3\times x^3\times y^3\times (z^2{)}^3$
3. $=216x^3{y}^3{z}^6$

Trick: $(z^2{)}^3=z^{2\times 3}=z^6$ — multiply the powers.

Answer: (D) $216x^3{y}^3{z}^6$

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

1. $P=5000,R=4\%,T=5$
2. $\text{SI}=\frac{5000\times 4\times 5}{100}$
3. $=1000$

Trick: $4\%\times 5=20\%$ of 5000 $=1000$.

Answer: (D) ₹1,000

Concept: Selling price $=\text{total cost}\times (1+\frac{\text{profit}\%}{100})$

1. Transport $=10\%$ of 2000 $=200$; total cost $=2200$.
2. SP $=2200\times 1.15$
3. $=2530$

Trick: Chain multiply: $2000\times 1.1\times 1.15=2530$.

Answer: (D) ₹2,530

Concept: One day's work = $\frac{1}{\text{days}}$; works of A, B, C add up.

1. $(A+B+C)$ one day $=\frac{1}{2}$
2. $A+B=\frac{1}{4}+\frac{1}{20}=\frac{5+1}{20}=\frac{3}{10}$
3. $C=\frac{1}{2}-\frac{3}{10}=\frac{2}{10}=\frac{1}{5}$ → 5 days

Trick: Total work = LCM$(4,20,2)=20$ units: A = 5, B = 1, all = 10 per day → C = 4 per day → $\frac{20}{4}=5$ days.

Answer: (C) 5

Concept: $\text{Distance}=\text{Speed}\times \text{Time}$ (time in hours when speed is in km/h).

1. $10\text{min}=\frac{10}{60}=\frac{1}{6}$ h
2. $\text{Distance}=90\times \frac{1}{6}$
3. $=15$ km

Trick: 90 km/h = 1.5 km per minute; $1.5\times 10=15$ km.

Answer: (C) 15 km

Concept: Convert all fractions to a common denominator (LCM of 7 and 14 = 14).

1. $\frac{3}{7}+\frac{2}{7}=\frac{5}{7}=\frac{10}{14}$
2. $\frac{10}{14}+\frac{1}{14}=\frac{11}{14}$
3. $\frac{11}{14}-3=\frac{11-42}{14}=-\frac{31}{14}$

Trick: Sum of fractions $<1$, so the answer must be negative and lie between $-3$ and $-2$: only $-\frac{31}{14}\approx -2.2$ fits.

Answer: (C) $-\frac{31}{14}$

Concept: Base area $=\pi r^2$, CSA $=2\pi rh$, Volume $=\pi r^{2}h$.

1. $\frac{22}{7}{r}^2=38.5\Rightarrow r^2=12.25\Rightarrow r=3.5$ cm
2. $2\times \frac{22}{7}\times 3.5\times h=616\Rightarrow 22h=616\Rightarrow h=28$ cm
3. $V=\text{base area}\times h=38.5\times 28=1078{\text{cm}}^3$

Trick: $V=\frac{\text{CSA}\times r}{2}=\frac{616\times 3.5}{2}=1078$.

Answer: (A) 1078