RRB Technician Grade III 2024, 2025 PYQs

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100+\text{Profit}\%}$.

1. $\text{CP}=\frac{19.50\times 100}{130}=15$.
2. New SP for $40\%$ profit $=15\times 1.4=21$.
3. Increase $=21-19.50=1.50$.

Trick: Extra $10\%$ profit is on CP: $10\%$ of 15 $=$ ₹1.50.

Answer: (D) ₹1.50

Concept: $x^2+y^2=(x+y{)}^2-2xy$.

1. $(x+y{)}^2={20}^2=400$.
2. $x^2+y^2=400-2\times 84=400-168=232$.

Trick: Here $x,y=14,6$: ${14}^2+6^2=196+36=232$ ✓.

Answer: (C) 232

Concept: If $a:b::b:x$, then $x=\frac{b^2}{a}$ (third proportional).

1. $1.6:0.6::0.6:x$ ⇒ $1.6x=0.6\times 0.6=0.36$.
2. $x=\frac{0.36}{1.6}=0.225$.

Trick: Product of extremes $=$ product of means: $1.6\times x=0.6\times 0.6$.

Answer: (B) 0.225

Concept: Let Sarala $=S$, Kishore $=26-S$. Three years ago: $K-3=\frac{3}{2}(S-3)$.

1. $23-S=\frac{3}{2}(S-3)$ ⇒ $46-2S=3S-9$.
2. $5S=55$ ⇒ $S=11$.

Trick: 3 years ago their sum was $26-6=20$, split $2:3$ ⇒ Sarala $8$, now $8+3=11$.

Answer: (D) 11 years

Concept: $a^n\times b^n=(ab{)}^n$ and $a^m\times a^n=a^{m+n}$.

1. $2^3\times 3^3=(2\times 3{)}^3=6^3$.
2. $6^3\times 6^3=6^6$ ⇒ $M=6$.

Trick: Same power 3 on all: $(2\times 3\times 6{)}^3={36}^3=6^6$.

Answer: (A) 6

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

1. After $20\%$: $2000\times 0.8=1600$.
2. After $15\%$: $1600\times 0.85=1360$.

Trick: Net discount $=20+15-\frac{20\times 15}{100}=32\%$ ⇒ SP $=68\%$ of 2000 $=$ ₹1,360.

Answer: (A) ₹1,360

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

1. SI $=1860-1500=$ ₹360
2. Interest per year $=\frac{1500\times 8}{100}=$ ₹120
3. $T=\frac{360}{120}=3$ years

Trick: Find one year's interest first; total interest ÷ yearly interest = years.

Answer: (C) 3 years

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

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

Trick: Cancel the two identities first ($1-1=0$); $\cot \theta \sin \theta$ simply becomes $\cos \theta$.

Answer: (A) $2\cos \theta$

Concept: A path of width $w$ inside the field reduces each side of the lawn by $2w$. Path area = field area − lawn area.

1. Field area $=50\times 42=2100{\text{ m}}^2$
2. Lawn $=(50-12)\times (42-12)=38\times 30=1140{\text{ m}}^2$
3. Path area $=2100-1140=960{\text{ m}}^2$

Trick: Inner path area $=2w(l+b-2w)=2\times 6\times (50+42-12)=12\times 80=960$.

Answer: (D) 960 m²

Concept: Required number = target − given number.

1. Required $=\frac{-5}{6}-\left(\frac{-5}{9}\right)=\frac{-5}{6}+\frac{5}{9}$
2. LCM of $6,9$ is $18$: $\frac{-15}{18}+\frac{10}{18}=\frac{-5}{18}$

Trick: Check: $\frac{-5}{9}+\frac{-5}{18}=\frac{-10-5}{18}=\frac{-15}{18}=\frac{-5}{6}$ ✓

Answer: (A) $\frac{-5}{18}$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Distance $=90\times 1+60\times 1=150$ km
2. Time $=2$ h
3. Average speed $=\frac{150}{2}=75$ km/h

Trick: For equal time intervals, average speed is the simple mean of the speeds: $\frac{90+60}{2}=75$.

Answer: (B) 75 km/h

Concept: $a^m\times a^n=a^{m+n}$; equal bases ⇒ equal powers.

1. $3^{K-2}\times 3^K=3^{2K-2}$
2. $729=3^6$
3. $2K-2=6\Rightarrow K=4$

Yaad rakhein: Remember $3^6=729$ (and $9^3=729$).

Answer: (B) 4

Concept: Single equivalent discount $=a+b-\frac{ab}{100}$.

1. $a=2,b=5$
2. Equivalent $=2+5-\frac{2\times 5}{100}=7-0.1=6.9\%$

Trick: Successive discounts are always a little less than their sum.

Answer: (C) 6.9%

Concept: $a^2+\frac{1}{a^2}={(a+\frac{1}{a})}^2-2$.

1. ${(a+\frac{1}{a})}^2=8^2=64$
2. $a^2+\frac{1}{a^2}=64-2=62$

Yaad rakhein: If $a+\frac{1}{a}=k$, then $a^2+\frac{1}{a^2}=k^2-2$.

Answer: (D) 62

Concept: Share of a part $=\frac{\text{its ratio term}}{\text{sum of terms}}\times \text{total}$.

1. Sum of ratio terms $=3+2=5$
2. Milk $=\frac{3}{5}\times 10=6$ litres

Trick: $5$ parts $=10$ L, so $1$ part $=2$ L; milk $=3\times 2=6$ L.

Answer: (B) 6 litres

Concept: If $a,b,c,d$ are in proportion, $a:b=c:d$, i.e. $ad=bc$.

1. $6\times x=18\times 39$
2. $x=\frac{18\times 39}{6}=3\times 39=117$

Trick: $18$ is $3$ times $6$, so $x$ is $3$ times $39$.

Answer: (A) 117

Concept: A system is consistent when it has at least one solution.

1. Intersecting lines → exactly one solution (consistent)
2. Coincident lines → infinitely many solutions (consistent)
3. Parallel lines → no solution (inconsistent)

Yaad rakhein: Consistent = intersecting or coincident; inconsistent = parallel.

Answer: (A) intersecting or coincident

Concept: Convert everything to grams: $1$ kg $=1000$ g.

1. Box with chocolates $=12.650$ kg $=12650$ g
2. Chocolates $=6$ kg $350$ g $=6350$ g
3. Empty box $=12650-6350=6300$ g

Trick: Subtract kg and g separately: $12-6=6$ kg, $650-350=300$ g.

Answer: (C) 6300 g

Concept: Area of 4 walls $=2h(l+b)$; ceiling $=l\times b$.

1. Walls $=2\times 21\times (15+17)=1344{\text{ m}}^2$
2. Ceiling $=15\times 17=255{\text{ m}}^2$
3. Total $=1344+255=1599{\text{ m}}^2$
4. Cost $=1599\times 42=$ ₹67,158

Trick: The unit digit of $1599\times 42$ is $8$ – only ₹67,158 and ₹68,918 qualify; estimate $1600\times 42=67200$.

Answer: (D) ₹67,158

Concept: Test the options: the pair must add to $35$ and have LCM $306$.

1. All options add to $35$
2. $17$ and $18$ are co-prime, so LCM $=17\times 18=306$ ✓
3. $14,21$: LCM $=42$; $16,19$: $304$; $13,22$: $286$ ✗

Trick: $306=2\times 3^2\times 17$ – the pair must contain $17$.

Answer: (A) 17, 18