RRB Technician Grade III 2024, 2025 PYQs

Concept: For a cube of side $a$: $V=a^3$ and main (body) diagonal $d=a\sqrt{3}$.

1. $a^3=729\Rightarrow a=9$ cm.
2. $d=9\sqrt{3}$ cm.

Yaad rakhein: Face diagonal $=a\sqrt{2}$, body diagonal $=a\sqrt{3}$.

Answer: (D) $9\sqrt{3}$ cm

Concept: Work done = one-day work × days; the remaining work is finished by B alone.

1. $(A+B)$ one-day work $=\frac{1}{12}+\frac{1}{16}=\frac{7}{48}$.
2. In 3 days: $\frac{21}{48}=\frac{7}{16}$; remaining $=\frac{9}{16}$.
3. B's time $=\frac{9}{16}\times 16=9$ days.

Trick: LCM method: total work $=48$ units, A $=4$, B $=3$ per day; 3 days $=21$ units; left $27÷3=9$.

Answer: (D) 9 days

Concept: A pure recurring decimal = (repeating block) $÷$ (as many 9s as digits in the block).

1. Let $x=0.\overline{33}$; the block "33" has 2 digits.
2. $100x-x=33\Rightarrow 99x=33$
3. $x=\frac{33}{99}=\frac{1}{3}$

Trick: $0.\bar{3}=0.\overline{33}=\frac{1}{3}$ — both are the same number 0.3333…

Answer: (B) $\frac{1}{3}$

Concept: $\text{SP}=\text{CP}\times (1+\frac{\text{markup}}{100})\times (1-\frac{\text{discount}}{100})$

1. Let CP = ₹$x$. List price $=1.22x$.
2. SP $=1.22x\times 0.85=1.037x$
3. $1.037x=5185\Rightarrow x=\frac{5185}{1.037}=5000$

Trick: Net effect $=22-15-\frac{22\times 15}{100}=3.7\%$ gain, so $\text{CP}=\frac{5185}{1.037}$.

Answer: (B) 5,000

Concept: Average = sum of the required values $÷$ number of values. Read only the four asked bars.

1. Indian = 6000, Chinese = 5500, Thai = 3500, Miscellaneous = 6500
2. Sum $=6000+5500+3500+6500=21500$
3. Average $=\frac{21500}{4}=5375$

Trick: Answer must lie between the smallest (3500) and largest (6500) value; check the last digits: sum ends in 500, so average ends in 75.

Answer: (D) 5375

Concept: $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$

1. Distance = 150 km, Time = 3 h
2. Speed $=\frac{150}{3}=50$ km/h

Yaad rakhein: D = S $\times$ T; cover S to get $\frac{D}{T}$.

Answer: (B) 50 km/h

Concept: Turn each condition into an equation and solve the pair.

1. Case 1: $B+5=2(A-5)$
2. Case 2: $A+1=B-1\Rightarrow A=B-2$
3. $B+5=2(B-7)=2B-14\Rightarrow B=19$
4. Then $A=17$. Check: $A-5=12$, $B+5=24=2\times 12$ ✓

Trick: Equalising by sending 1 student means the difference $B-A=2$; test options: $B=19\Rightarrow A=17$ works.

Answer: (C) 19

Concept: One-day work adds up: $\frac{1}{A}+\frac{1}{B}=\frac{1}{A+B\text{together}}$.

1. B’s one-day work $=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
2. So B alone does the work in 4 days.
3. 19 times the work $=19\times 4=76$ days

Trick: Time for B $=\frac{2\times 4}{4-2}=4$ days (product $÷$ difference).

Answer: (C) 76

Concept: $\text{CI}=P[{(1+\frac{R}{100})}^n-1]$

1. ${\left(1.1\right)}^3=1.331$
2. CI $=12000\times (1.331-1)=12000\times 0.331$
3. CI $=3972$

Trick: At 10% for 3 years, CI = 33.1% of principal: $0.331\times 12000=3972$.

Answer: (D) ₹3,972

Concept: Expenditure = Price $\times$ Consumption. If price falls by $r\%$, consumption must rise by $\frac{r}{100-r}\times 100\%$.

1. $r=20$
2. Increase $=\frac{20}{80}\times 100=25\%$
3. Check: $0.8\times 1.25=1$ ✓

Trick: Price 5 → 4 (fall of $\frac{1}{5}$); consumption 4 → 5 (rise of $\frac{1}{4}$ = 25%).

Answer: (C) 25%

Concept: For an equilateral triangle of side $a$: altitude $h=\frac{\sqrt{3}}{2}a$ and area $=\frac{\sqrt{3}}{4}{a}^2=\frac{h^2}{\sqrt{3}}$.

1. $9=\frac{\sqrt{3}}{2}a\Rightarrow a=\frac{18}{\sqrt{3}}=6\sqrt{3}$ cm
2. Area $=\frac{1}{2}\times a\times h=\frac{1}{2}\times 6\sqrt{3}\times 9$
3. Area $=27\sqrt{3}{\text{cm}}^2$

Trick: Directly: Area $=\frac{h^2}{\sqrt{3}}=\frac{81}{\sqrt{3}}=27\sqrt{3}$.

Answer: (D) $27\sqrt{3}$

Concept: $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$, so Sum = Mean $\times$ n.

1. 7 values, mean 30 ⇒ sum $=30\times 7=210$
2. Known values: $22.5+56+42.5+36=157$
3. Unknown: $(2x+1)+(x-2)+3x=6x-1$
4. $157+6x-1=210\Rightarrow 6x=54\Rightarrow x=9$

Trick: Put option $x=9$: $19+7+27=53$, and $157+53=210$ ✓

Answer: (A) 9

Concept: A sum doubles when SI = Principal, i.e. $\frac{P\times R\times T}{100}=P\Rightarrow T=\frac{100}{R}$.

1. $R=5\%$
2. $T=\frac{100}{5}=20$ years

Trick: Doubling time at SI $=\frac{100}{R}$; tripling $=\frac{200}{R}$.

Answer: (D) 20 years

Concept: When two items are sold at the same price, one at $x\%$ gain and other at $x\%$ loss, there is always a loss of $\frac{x^2}{100}\%$.

1. $x=5$
2. Loss% $=\frac{5^2}{100}=\frac{25}{100}=0.25\%$

Trick: Same SP + same % gain and loss ⇒ always LOSS of ${\left(\frac{x}{10}\right)}^2\%$.

Answer: (A) 0.25% loss

Concept: For the same time period, profit is shared in the ratio of investments.

1. Ratio $=48000:72000=2:3$
2. Ram’s share $=\frac{3}{5}\times 6530$
3. $=3\times 1306=3918$

Trick: Ram gets more than half ($>3265$) — only ₹3,918 fits.

Answer: (D) ₹3,918

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; each exterior angle of a regular $n$-gon $=\frac{{360}^{\circ }}{n}$.

1. Decagon ⇒ $n=10$
2. Each exterior angle $=\frac{{360}^{\circ }}{10}={36}^{\circ }$

Yaad rakhein: Interior angle $={180}^{\circ }-{36}^{\circ }={144}^{\circ }$ for a decagon.

Answer: (C) ${36}^{\circ }$

Concept: Follow BODMAS: division and multiplication before addition; $(-)\times (-)=(+)$.

1. $(-20)\times (-1)=20$
2. $14÷7=2$
3. $20+2=22$

Yaad rakhein: Negative $\times$ negative = positive.

Answer: (C) 22

Concept: For cubes: volume ratio $=a_1^3:a_2^3$ and surface area ratio $=a_1^2:a_2^2$.

1. $a_1^3:a_2^3=729:1331=9^3:{11}^3$
2. So $a_1:a_2=9:11$
3. TSA ratio $=9^2:{11}^2=81:121$

Trick: Volume ratio → take cube root → square it.

Answer: (C) 81 : 121

Concept: $x^{\frac{1}{2}}=\sqrt{x}$, $Y^{-1}=\frac{1}{Y}$, $z^{\frac{2}{3}}={\left(\sqrt[3]{z}\right)}^2$.

1. $9^{\frac{1}{2}}=3$
2. $3^{-1}=\frac{1}{3}$
3. $8^{\frac{2}{3}}=2^2=4$
4. Product $=3\times \frac{1}{3}\times 4=4$

Yaad rakhein: For $a^{\frac{m}{n}}$, take the $n$th root first, then the power $m$.

Answer: (C) 4

Concept: Write present ages using the ratio multiplier and use the past/future condition.

1. Amit $=7k$, Amita $=5k$
2. $7k-5=5k+5\Rightarrow k=5$; ages 35 and 25
3. Present sum $=60$; needed 72 ⇒ increase 12
4. Each ages 1 year per year ⇒ sum rises by 2 per year ⇒ $\frac{12}{2}=6$ years

Trick: Sum of two ages grows by 2 every year.

Answer: (B) 6 years