RRB Technician Grade III 2024, 2025 PYQs

Concept: Let age at marriage $=x$; present age $=x+6$.

1. $x+6=\frac{5}{4}x\Rightarrow \frac{x}{4}=6\Rightarrow x=24$.
2. Namya now $=30$ years.
3. Son $=\frac{1}{5}\times 30=6$ years.

Trick: $1\frac{1}{4}$ times means the extra $\frac{1}{4}$ part $=6$ years, so $x=24$.

Answer: (C) 6 years

Concept: $a$ is what % of $b$ $=\frac{a}{b}\times 100$.

1. $\frac{3}{20}÷\frac{12}{25}=\frac{3}{20}\times \frac{25}{12}=\frac{75}{240}=\frac{5}{16}$.
2. $\frac{5}{16}\times 100=31.25\%$.

Yaad rakhein: $\frac{1}{16}=6.25\%$, so $\frac{5}{16}=31.25\%$.

Answer: (A) 31.25%

Concept: Leak rate $=$ (filling rate without leak) $-$ (net rate with leak).

1. Without leak: 7 h; with leak: 8 h.
2. Leak rate $=\frac{1}{7}-\frac{1}{8}=\frac{1}{56}$.
3. Leak empties the full cistern in $56$ hours.

Trick: Leak time $=\frac{a\times b}{b-a}=\frac{7\times 8}{8-7}=56$ h.

Answer: (A) 56 hours

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=80-68=$ ₹12.
2. Rate $=\frac{12}{80}\times 100=15\%$.

Yaad rakhein: Discount % is always on the marked price, not the selling price.

Answer: (D) 15%

Concept: Equal profit % and loss % means equal profit and loss amounts, so CP is the average of the two SPs.

1. $\text{CP}=\frac{1920+1500}{2}=$ ₹1710.
2. For 10% profit: $\text{SP}=1710\times \frac{110}{100}=$ ₹1881.

Trick: Same CP, equal % gain and loss → $\text{CP}=\frac{{\text{SP}}_1+{\text{SP}}_2}{2}$.

Answer: (B) ₹1,881

Concept: BODMAS: division and multiplication first, then addition/subtraction from left to right.

1. $36÷4=9$ and $2\times 5=10$.
2. $4-9+10=5$.

Yaad rakhein: Never do $4-36$ first — that breaks the order of operations.

Answer: (D) 5

Concept: $\text{CSA}=2\pi rh$, $V=\pi r^{2}h$.

1. $2\times \frac{22}{7}\times r\times 7=484\Rightarrow 44r=484\Rightarrow r=11$ cm.
2. $V=\frac{22}{7}\times {11}^2\times 7=22\times 121=2662$ cm${}^3$.

Trick: $V=\frac{\text{CSA}\times r}{2}=\frac{484\times 11}{2}=2662$.

Answer: (A) 2662

Concept: List the primes after 20, then average $=\frac{\text{sum}}{\text{count}}$.

1. First five primes above 20: $23,29,31,37,41$.
2. Sum $=161$.
3. Average $=\frac{161}{5}=32.2$.

Trick: Skip 21, 25, 27, 33, 35, 39 — all composite.

Answer: (B) 32.2

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. $\text{SI}=\frac{600\times 7\times 9}{100}=6\times 63=$ ₹378.

Yaad rakhein: Match units: monthly rate → time in months.

Answer: (D) ₹378

Concept: Share of A $=\frac{a}{a+b}\times \text{total}$.

1. Total parts $=4+5=9$.
2. A $=\frac{4}{9}\times 63=28$.

Trick: 1 part $=\frac{63}{9}=7$, so A $=4\times 7=28$.

Answer: (A) 28

Concept: Convert averages to totals; B is common to both pairs.

1. $A+B+C=3\times 45=135$.
2. $A+B=80$, $B+C=86$.
3. $B=(A+B)+(B+C)-(A+B+C)=80+86-135=31$ kg.

Trick: Middle common term $=$ sum of pair totals $-$ grand total.

Answer: (B) 31 kg

Concept: Divide each term by ${\cos }^2\theta$ and use ${\sec }^2\theta =1+{\tan }^2\theta$.

1. $\frac{3-4{\sin }^2\theta }{{\cos }^2\theta }=3{\sec }^2\theta -4{\tan }^2\theta$.
2. $=3+3{\tan }^2\theta -4{\tan }^2\theta =3-{\tan }^2\theta$.
3. Add $2{\tan }^2\theta$: $3-{\tan }^2\theta +2{\tan }^2\theta =3+{\tan }^2\theta$.

Trick: Put $\theta ={45}^{\circ }$: $\frac{3-2}{\frac{1}{2}}+2=4$ and $3+{\tan }^2{45}^{\circ }=4$ ✓.

Answer: (C) $3+{\tan }^2\theta$

Concept: If $x+y=S$ and $x-y=D$, then $x=\frac{S+D}{2}$, $y=\frac{S-D}{2}$.

1. $x=\frac{55+19}{2}=37$.
2. $y=\frac{55-19}{2}=18$.

Trick: Check options: $37+18=55$ and $37-18=19$ ✓.

Answer: (C) 37, 18

Concept: Read only the "females" column for both shops.

1. Shop A to females $=78$.
2. Shop B to females $=40$.
3. Difference $=78-40=38$.

Trick: Ignore the males column — it is only a distractor here.

Answer: (A) 38

Concept: Convert everything to millilitres: $1$ L $=1000$ mL.

1. $5.5$ L $=5500$ mL; $3$ L $925$ mL $=3925$ mL.
2. Left $=5500-3925=1575$ mL $=1$ L $575$ mL.

Trick: Work in mL, then split back into L and mL.

Answer: (A) 1 ltr 575 ml

Concept: Amount $=\frac{8}{5}P$ means $\text{SI}=\frac{3}{5}P$; $R=\frac{\text{SI}\times 100}{P\times T}$.

1. $\text{SI}=\frac{8}{5}P-P=\frac{3}{5}P=60\%$ of $P$.
2. In 4 years: $R=\frac{60}{4}=15\%$ p.a.

Trick: $\frac{\text{Total SI}\%}{\text{years}}=\text{rate}$: $60\%÷4=15\%$.

Answer: (A) 15%

Concept: Remainder of a product $=$ product of individual remainders (then reduce again).

1. $171=170+1$ → remainder $1$.
2. $173=170+3$ → remainder $3$.
3. $1\times 3=3$, which is less than 17.

Trick: $170=17\times 10$ — subtract the nearest multiple of 17 first.

Answer: (B) 3

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

1. $P=12000$, $R=10\%$, $n=2$.
2. $A=12000\times {\left(\frac{11}{10}\right)}^2=12000\times \frac{121}{100}$.
3. $A=14520$, so Swarna pays ₹14,520.

Trick: At $10\%$ for 2 years, CI adds $21\%$: $12000+2520=14520$.

Answer: (A) ₹14,520

Concept: Find the square root using the unit digit and the nearest known squares.

1. $6084$ ends in 4, so the root ends in 2 or 8.
2. ${70}^2=4900$ and ${80}^2=6400$, so the root lies between 70 and 80.
3. ${78}^2=6084$ ✓ (since ${72}^2=5184$ is too small).

Trick: Root near 80: ${80}^2-{78}^2=(80-78)(80+78)=316$ and $6400-316=6084$.

Answer: (C) 78

Concept: In simple interest, the interest is the same every year, so the difference between two amounts is the interest for the extra years.

1. SI for $5-2=3$ years $=2000-1400=600$, so SI per year $=200$.
2. Principal $=1400-2\times 200=1000$.
3. $R=\frac{200}{1000}\times 100=20\%$.

Trick: $R=\frac{100\times (A_2-A_1)}{A_1{t}_2-A_2{t}_1}=\frac{100\times 600}{1400\times 5-2000\times 2}=\frac{60000}{3000}=20\%$.

Answer: (D) 20%