RRB Technician Grade III 2024, 2025 PYQs

Concept: Convert workers to one unit by comparing the two groups (same work, so total work is equal).

1. $(6M+8B)\times 10=(26M+48B)\times 2$
2. $60M+80B=52M+96B\Rightarrow 8M=16B\Rightarrow M=2B$
3. Group 1 $=6(2B)+8B=20B$, takes 10 days, so work $=20\times 10=200$ boy-days
4. 25 boys need $\frac{200}{25}=8$ days

Trick: Work = workers $\times$ days. Turn everything into boy-days first.

Answer: (D) 8 days

Concept: After successive discounts, $\text{SP}=\text{MRP}\times (1-d_1)(1-d_2)$.

1. Two 20% discounts: factor $=0.8\times 0.8=0.64$
2. $\text{MRP}\times 0.64=16000$
3. $\text{MRP}=\frac{16000}{0.64}=25000$

Trick: Two successive 20% discounts = single discount of $20+20-\frac{20\times 20}{100}=36\%$, so 64% of MRP is paid.

Answer: (A) ₹25,000

Concept: Use Pythagoras to get the missing leg; area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Other leg $=\sqrt{{10}^2-8^2}=\sqrt{100-64}=\sqrt{36}=6$
2. Legs 8 and 6 are perpendicular
3. Area $=\frac{1}{2}\times 8\times 6=24$ sq. units

Trick: $(6,8,10)$ is the triplet $(3,4,5)\times 2$ — no calculation needed.

Answer: (D) 24 sq. units

Concept: In simple interest, the amount grows by the same interest every year.

1. Interest for $5-2=3$ years $=2000-1250=750$, so 1 year $=250$
2. Principal $=1250-2\times 250=750$
3. Rate $=\frac{250}{750}\times 100=33.33\%$

Trick: Difference of amounts ÷ difference of years = yearly SI.

Answer: (A) 33.33%

Concept: A prime has exactly two factors, 1 and itself.

1. Primes below 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
2. Count them: 10

Yaad rakhein: There are 4 primes below 10, 8 below 20, 10 below 30 and 25 below 100.

Answer: (A) 10

Concept: $a^3-b^3=(a-b{)}^3+3ab(a-b)$.

1. Put $a-b=1$: $61=1^3+3ab(1)$
2. $3ab=60$
3. $ab=20$

Trick: Check: $a=5,b=4$ gives $125-64=61$.

Answer: (A) 20

Concept: Negative power flips the fraction: ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$; even powers make the sign positive.

1. ${\left(\frac{-2}{5}\right)}^{-2}={\left(\frac{5}{-2}\right)}^2=\frac{25}{4}$
2. ${\left(\frac{-4}{5}\right)}^4=\frac{256}{625}$
3. $\frac{25}{4}\times \frac{256}{625}=\frac{64}{25}$

Trick: Even power → answer positive, so only positive options remain.

Answer: (C) $\frac{64}{25}$

Concept: Value of a sector $=\frac{\text{sector }\%}{100}\times \text{total}$.

1. Mafindre share from the chart $=18\%$
2. Cars $=\frac{18}{100}\times 1500000=270000$

Trick: 1% of 15,00,000 is 15,000; so $18\times 15000=270000$.

Answer: (D) 2,70,000

Concept: For equal distances at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$.

1. $x=4,y=20$
2. Average $=\frac{2\times 4\times 20}{4+20}=\frac{160}{24}$
3. $=\frac{20}{3}$ km/h

Trick: Average speed is never the simple average $\frac{4+20}{2}=12$; it is closer to the slower speed.

Answer: (C) $\frac{20}{3}$ km/h

Concept: With the same money, the saving due to price cut buys the extra quantity.

1. 20% of ₹100 = ₹20 is saved
2. ₹20 buys 10 kg extra, so new price $=\frac{20}{10}=2$ per kg
3. Original price $=\frac{2}{0.8}=2.50$ per kg

Trick: Reduced price $=\frac{\%\times \text{money}}{100\times \text{extra qty}}$.

Answer: (A) ₹2.50

Concept: Perimeter $=2(l+b)$, Area $=l\times b$.

1. $l=3b$, so $2(3b+b)=8b=48\Rightarrow b=6$
2. $l=18$
3. Area $=18\times 6=108{\text{cm}}^2$

Trick: Perimeter $÷$ 8 directly gives the breadth when $l=3b$.

Answer: (C) 108

Concept: A number is divisible by 11 if (sum of odd-place digits) $-$ (sum of even-place digits) is 0 or a multiple of 11.

1. 86325k6 from the right: odd places $6+5+3+8=22$
2. Even places $k+2+6=k+8$
3. $22-(k+8)=14-k$ must be 0 or 11 $\Rightarrow k=3$

Trick: Since $k\le 9$, the difference $14-k$ can only be 11, so $k=3$ (8632536 is divisible by 11).

Answer: (B) 3

Concept: Take tank capacity = LCM of 7, 10, 14 = 70 units.

1. Rates: A $=+10$, B $=+7$, C $=-5$ units/h
2. All three for 4.5 h: $(10+7-5)\times 4.5=54$ units
3. Remaining $=70-54=16$ units
4. B and C together $=7-5=2$ units/h, time $=\frac{16}{2}=8$ h

Trick: LCM-unit method avoids fractions completely.

Answer: (D) 8

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

1. At 30% loss: $\text{CP}=\frac{1148}{0.70}=1640$
2. For 5% gain: $\text{SP}=1640\times 1.05=1722$

Trick: Direct: $\text{New SP}=1148\times \frac{105}{70}=1722$.

Answer: (D) ₹1,722

Concept: Arrange data in order; for an even count, median = average of the two middle values.

1. Ordered: 4, 4, 6, 7, 8, 9, 9, 11, 12, 14 ($n=10$)
2. Middle terms: 5th $=8$ and 6th $=9$
3. Median $=\frac{8+9}{2}=8.5$

Yaad rakhein: For even $n$, the median is the mean of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th terms.

Answer: (C) 8.5

Concept: $x$ as a percentage of $y$ $=\frac{x}{y}\times 100$.

1. B as % of A $=\frac{60}{75}\times 100$
2. $=\frac{4}{5}\times 100=80\%$

Trick: $60:75=4:5$, and $\frac{4}{5}=80\%$.

Answer: (C) 80%

Concept: Find the value of one ratio part from the given difference.

1. B $-$ A $=7-4=3$ parts $=600$, so 1 part $=200$
2. C $-$ D $=9-3=6$ parts
3. $6\times 200=1200$

Trick: Difference of parts × value of 1 part — no need to find the total.

Answer: (C) ₹1,200

Concept: BODMAS: innermost bracket first; inside it, ÷ and × from left to right.

1. $3\times 10÷2=30÷2=15$
2. $12+15=27$
3. $36-27=9$

Trick: Solve ( ) → { } → [ ] in order.

Answer: (D) 9

Concept: In an isosceles triangle the altitude to the base bisects the base.

1. Sides $3k,3k,4k$: $10k=20\Rightarrow k=2$, so sides 6, 6, 8
2. Half base $=4$
3. Altitude $h=\sqrt{6^2-4^2}=\sqrt{20}=2\sqrt{5}$ cm

Trick: Altitude to base $=\sqrt{a^2-\frac{b^2}{4}}$ for equal side $a$, base $b$.

Answer: (A) $2\sqrt{5}\text{ cm}$

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

1. $7\times k=12\times 21$
2. $k=\frac{252}{7}=36$

Yaad rakhein: Product of extremes = product of means.

Answer: (B) 36