RRB Technician Grade III 2024, 2025 PYQs

Concept: If A is $r\%$ more than B, then B is less than A by $\frac{r}{100+r}\times 100\%$.

1. $33\frac{1}{3}\%=\frac{1}{3}$: B $=3$, A $=4$.
2. B is less by $\frac{1}{4}\times 100=25\%$.

Trick: Fraction method: $\frac{1}{3}$ more $\Rightarrow \frac{1}{3+1}=\frac{1}{4}$ less $=25\%$.

Answer: (A) 25%

Concept: Fourth proportional $x$ of $a,b,c$: $a:b=c:x\Rightarrow x=\frac{b\times c}{a}$.

1. $24:8=9:x$.
2. $x=\frac{8\times 9}{24}=3$.

Yaad rakhein: Product of extremes = product of means: $24\times x=8\times 9$.

Answer: (C) 3

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with rate and time in the same unit (here per month and months).

1. $P=800$, $R=7\%$ per month, $T=9$ months.
2. $\text{SI}=\frac{800\times 7\times 9}{100}=8\times 63=504$.

Trick: $1\%$ of 800 $=8$; total rate $=7\times 9=63\%$, so SI $=8\times 63=504$.

Answer: (D) ₹504

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{gain}\%}{100}$ and $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$.

1. $\text{SP}=136\times \frac{120}{100}=163.2$.
2. $\text{MP}\times \frac{85}{100}=163.2\Rightarrow \text{MP}=\frac{163.2\times 100}{85}=192$.

Trick: $\text{MP}=136\times \frac{120}{85}=136\times \frac{24}{17}=8\times 24=192$.

Answer: (D) ₹192

Concept: Mode is the observation that occurs the maximum number of times.

1. Count: 6 appears 4 times; every other value appears only once.
2. So mode $=6$.

Yaad rakhein: Mode = most frequent value; mean = average; median = middle value.

Answer: (A) 6

Concept: Items of a product $=\frac{\text{its }\%}{100}\times \text{total items}$.

1. From the chart: Raincoats $=13\%$, Jackets $=8\%$.
2. Together $=21\%$ of 1000 $=210$.

Trick: Total is 1000, so $1\%=10$ items; $21\%=210$.

Answer: (D) 210

Concept: Sum of the four angles of a quadrilateral $={360}^{\circ }$.

1. $3x+5x+4x+6x={360}^{\circ }\Rightarrow 18x={360}^{\circ }\Rightarrow x={20}^{\circ }$.
2. $A={60}^{\circ }$, $B={100}^{\circ }$.
3. $3A+2B={180}^{\circ }+{200}^{\circ }={380}^{\circ }$.

Trick: $3A+2B=(3\times 3+2\times 5)x=19x=19\times {20}^{\circ }={380}^{\circ }$.

Answer: (C) ${380}^{\circ }$

Concept: If the sum of two numbers is known, other number $=\text{sum}-\text{given number}$.

1. Other $=-4-\left(\frac{-13}{25}\right)=-4+\frac{13}{25}$
2. $-4=\frac{-100}{25}$, so other $=\frac{-100+13}{25}=\frac{-87}{25}$

Trick: Check: $\frac{-87}{25}+\frac{-13}{25}=\frac{-100}{25}=-4$ ✓

Answer: (A) $\frac{-87}{25}$

Concept: $V=\frac{4}{3}\pi r^3$, so volume changes as the cube of the radius.

1. New radius $=0.9r$
2. New volume $=(0.9{)}^{3}V=0.729V$
3. Decrease $=1-0.729=0.271=27.1\%$

Trick: For a change of $x\%$ in all three dimensions: $\text{net}=3x+\frac{3x^2}{100}+\frac{x^3}{{100}^2}$ with $x=-10$: $-30+3-0.1=-27.1\%$

Answer: (D) 27.1%

Concept: Father's age $=3\times$ son's age.

1. Son $=15$ years, father $=3\times 15=45$ years
2. Sum $=45+15=60$ years

Trick: Sum $=(3+1)\times \text{son}=4\times 15=60$

Answer: (C) 60 years

Concept: Distance lost due to stoppages per hour = difference of speeds; time stopped $=\frac{\text{lost distance}}{\text{speed without stoppage}}$.

1. Lost distance per hour $=45-36=9$ km
2. Time $=\frac{9}{45}$ h $=\frac{9}{45}\times 60=12$ min

Trick: Stoppage time $=\frac{\text{difference}}{\text{faster speed}}\times 60$ minutes

Answer: (D) 12 min

Concept: Read the values of Company X (blue line) for the required years and subtract.

1. Imports of X in 2019 $=120$ crore
2. Imports of X in 2021 $=80$ crore
3. Difference $=120-80=40$ crore

Trick: Take the label on the correct line only — Company Y (orange) values are a trap here.

Answer: (C) 40 crore

Concept: Shortfall in marks = shortfall in percent of the maximum marks.

1. Shortfall $=975-870=105$ marks
2. $7\%$ of maximum $=105$
3. Maximum $=\frac{105\times 100}{7}=1500$

Trick: Check: pass marks $975=65\%$ of $1500$, Priya $870=58\%$; gap $=7\%$ ✓

Answer: (C) 1500

Concept: $\text{Gain}\%=\frac{\text{Gain}}{\text{CP}}\times 100$

1. A: gain $=480-384=96$, $\frac{96}{384}\times 100=25\%$
2. B: gain $=1672-1254=418$, $\frac{418}{1254}\times 100=33\frac{1}{3}\%$
3. Ratio $=\frac{25}{100/3}=\frac{75}{100}=\frac{3}{4}$

Trick: Use fractions: $\frac{96}{384}=\frac{1}{4}$, $\frac{418}{1254}=\frac{1}{3}$; ratio $=\frac{1/4}{1/3}=\frac{3}{4}$

Answer: (B) $\frac{3}{4}$

Concept: Volume of cylinder $V=\pi r^{2}h$.

1. Let $r=3k$, $h=4k$
2. $\frac{22}{7}\times 9k^2\times 4k=38808\Rightarrow k^3=\frac{38808\times 7}{792}=343$
3. $k=7$, $r=21$ cm, diameter $=2r=42$ cm

Trick: Check options: $d=42\Rightarrow r=21,h=28$; $\frac{22}{7}\times 441\times 28=38808$ ✓

Answer: (D) 42 cm

Concept: $1\text{ cm}=10\text{ mm}$; add mm and cm separately and carry over.

1. mm: $7+6=13$ mm $=1$ cm $3$ mm
2. cm: $5+9+1=15$ cm
3. Total $=15$ cm $3$ mm

Trick: Convert to mm: $57+96=153$ mm $=15.3$ cm

Answer: (D) 15 cm 3 mm

Concept: $0.\overline{abc}=\frac{abc}{999}$ (as many 9s as repeating digits).

1. Let $x=0.512512\dots$
2. $1000x=512.512\dots$
3. $999x=512\Rightarrow x=\frac{512}{999}$

Trick: Pure recurring: repeating block over the same number of 9s.

Answer: (A) $\frac{512}{999}$

Concept: If A takes $a$ hours and B takes $a+6$ hours, then $\frac{1}{a}+\frac{1}{a+6}=\frac{1}{4}$.

1. $4(2a+6)=a(a+6)\Rightarrow a^2-2a-24=0$
2. $(a-6)(a+4)=0\Rightarrow a=6$ hours
3. Check: $\frac{1}{6}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}$ ✓

Trick: Try options: $a=6$, $b=12$; together $\frac{6\times 12}{6+12}=4$ ✓

Answer: (B) 6 hours

Concept: Volume $=l\times b\times h$, so factorise the expression into three factors.

1. $3x^2-27=3(x^2-9)$
2. $x^2-9=(x-3)(x+3)$
3. So dimensions: $3,(x-3),(x+3)$

Yaad rakhein: $a^2-b^2=(a-b)(a+b)$ after taking out the common factor.

Answer: (C) $3,x-3,x+3$

Concept: Fourth proportional of $a,b,c$ is $d=\frac{b\times c}{a}$ (since $a:b=c:d$).

1. $d=\frac{10\times 12}{15}$
2. $d=\frac{120}{15}=8$

Trick: $15:10=12:d\Rightarrow$ ratio $\frac{2}{3}$, so $d=12\times \frac{2}{3}=8$

Answer: (D) 8