RRB Technician Grade III 2024, 2025 PYQs

Concept: Final = Original $\times (1+r_1)(1-r_2)(1+r_3)$.

1. $1,20,000\times 1.20=1,44,000$
2. $1,44,000\times 0.90=1,29,600$
3. $1,29,600\times 1.15=$ ₹1,49,040

Trick: Combine multipliers: $1.2\times 0.9\times 1.15=1.242$ → $1,20,000\times 1.242=1,49,040$.

Answer: (A) ₹1,49,040

Concept: Do × and ÷ first, then − and + (BODMAS).

1. $3.6\times 2.5=9$
2. $4.8÷1.6=3$
3. $3.2\times 0.5=1.6$
4. $9-3+1.6=7.6$

Trick: $\times 2.5=\times 10÷4$ → $36÷4=9$; $\times 0.5$ = half → 1.6.

Answer: (C) 7.6

Concept: Amount = $P(1+\frac{RT}{100})$ for simple interest.

1. $RT=15\times 3=45\%$ → $A=1.45P$
2. $1.45P=17,400$ → $P=$ ₹12,000
3. $x=\frac{12,000}{80,000}\times 100=15\%$

Trick: Check options: 15% of 80,000 = 12,000; $12,000\times 1.45=17,400$ ✓.

Answer: (D) 15

Concept: Mode = the value that occurs most often.

1. Count: 40 → 4 times; 45 → 3 times
2. 43, 46, 47 → 2 times each; others once
3. Highest frequency = 40

Trick: Only tally the option values (40, 43, 46, 45) — saves time.

Answer: (A) 40

Concept: Third proportional of a and b = $\frac{b^2}{a}$; also $a^3-b^3=(a-b)(a^2+ab+b^2)$.

1. Third proportional = $\frac{(a^2+ab+b^2{)}^2}{a^3-b^3}=\frac{a^2+ab+b^2}{a-b}$
2. $a^2+ab+b^2=256+192+144=592$
3. $a-b=4$ → $\frac{592}{4}=148$

Trick: Cancel first using the identity — then only $\frac{a^2+ab+b^2}{a-b}$ is needed.

Answer: (C) 148

Concept: Empirical relation: Mode = 3 Median − 2 Mean.

1. Mode = $3\times 61-2\times 47$
2. $=183-94=89$

Trick: Mode = Mean − 3(Mean − Median) = $47-3(47-61)=47+42=89$.

Answer: (D) 89

Concept: Solve brackets first, then ÷ and × from left to right, then + and −.

1. Numerator: $(120-12)÷(24÷2)-12+7=108÷12-12+7=9-12+7=4$
2. Denominator: $5\times 12÷10-(6\times 4)÷12=60÷10-24÷12=6-2=4$
3. Value = $\frac{4}{4}=1$

Trick: Numerator and denominator both come out as 4, so the answer is 1 without any further work.

Answer: (D) $1$

Concept: $(a+b+c)(ab+bc+ca)=ab(a+b)+bc(b+c)+ca(c+a)+3abc$.

1. The given expression is exactly $(a+b+c)(ab+bc+ca)$
2. $=13\times 4=52$

Trick: Check with $a=b=c=1$: LHS = $2+2+2+3=9$ and $(3)(3)=9$. The identity holds, so just multiply the two given values.

Answer: (D) 52

Concept: Ages 5 years ago: their sum and product are known, so find two numbers.

1. 5 years ago, sum of ages = $60-10=50$, product = 525
2. Find two numbers with sum 50 and product 525: $35\times 15=525$ and $35+15=50$
3. Father then = 35, now = $35+5=40$ years

Trick: Option check: if the father is 40, the son is 20. 5 years ago they were $35\times 15=525$. ✓

Answer: (B) 40

Concept: $55=5\times 11$. Last digit must be 0 or 5, and the alternate-digit difference must be a multiple of 11.

1. Divisibility by 11: $(9+7)-(A+B)=16-(A+B)$ must be 0 or 11, so $A+B=16$ or 5
2. By 5: $B=0$ or 5. If $A+B=16$, then $A=16$ or 11, which is not a digit
3. So $A+B=5$ (e.g. 9570 or 9075, both ÷ 55 exactly)

Trick: $A+B=16$ needs $B\ge 7$, but B can only be 0 or 5. So reject 16 at once.

Answer: (D) 5

Concept: Mode = the observation that occurs most often.

1. Count: 44 occurs 3 times
2. 43, 48, 53 and 55 occur only 2 times each; the rest occur once
3. Mode = 44

Trick: Only count the values given in the options: 55 (2), 43 (2), 53 (2), 44 (3).

Answer: (D) 44

Concept: Divisible by both 6 and 9 means divisible by LCM 18, so the number must be even and its digit sum must be divisible by 9.

1. 45 is odd, so it is not divisible by 6 ($45÷6=7.5$)
2. $72=18\times 4$, $54=18\times 3$, $90=18\times 5$, so all three are divisible by both

Trick: All options have a digit sum of 9, so just look for the odd number: 45.

Answer: (A) 45

Concept: $\cot \theta =\frac{1}{\sqrt{3}}$ gives $\theta ={60}^{\circ }$. Substitute the standard values.

1. ${\sin }^2{60}^{\circ }=\frac{3}{4}$, ${\cos }^2{60}^{\circ }=\frac{1}{4}$, ${\csc }^2{60}^{\circ }=\frac{4}{3}$, ${\sec }^2{60}^{\circ }=4$
2. $\frac{2-\frac{3}{4}}{1+\frac{1}{4}}=\frac{5/4}{5/4}=1$
3. Value = $1+\frac{4}{3}-4=-\frac{5}{3}$

Trick: ${\sec }^2$ is large (4), so the result must be negative. That leaves only $-\frac{2}{3}$ and $-\frac{5}{3}$; compute once.

Answer: (C) $-\frac{5}{3}$

Concept: Diagonal² = $l^2+b^2$ (Pythagoras); Perimeter = $2(l+b)$.

1. Other side = $\sqrt{{25}^2-{15}^2}=\sqrt{625-225}=\sqrt{400}=20$ m
2. Perimeter = $2(15+20)=70$ m

Trick: 15 : 20 : 25 is the triplet 3 : 4 : 5 × 5, so the other side = $4\times 5=20$ directly.

Answer: (A) 70 meters

Concept: Form a linear equation and solve it.

1. $\frac{x}{3}+25=100\Rightarrow \frac{x}{3}=75$
2. $x=225$
3. Sum of digits = $2+2+5=9$

Trick: Work backwards: $100-25=75$, then $75\times 3=225$.

Answer: (C) 9

Concept: Net filling rate = inlet rate − outlet rate.

1. Net rate = $\frac{1}{14}-\frac{1}{63}=\frac{9-2}{126}=\frac{7}{126}=\frac{1}{18}$ tank per min
2. Full tank takes 18 min
3. One-third tank = $\frac{18}{3}=6$ min

Trick: LCM method: tank = 126 units; inlet +9, outlet −2, net 7 per min. $\frac{1}{3}$ tank = 42 units ÷ 7 = 6 min.

Answer: (A) 6

Concept: If the total time is t, then pipe Z works only $(t-3)$ hours.

1. Tank = $\text{LCM}(12,15,20)=60$ units; rates $X=5$, $Y=4$, $Z=3$ units/h
2. $5t+4t+3(t-3)=60\Rightarrow 12t-9=60\Rightarrow 12t=69$
3. $t=5.75$ h = 5 hours 45 minutes

Trick: Assume Z also ran for the last 3 h: it would add 9 more units, so the total becomes 69 units at 12 units/h, which is 5.75 h.

Answer: (C) 5 hours 45 minutes

Concept: Area of an equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$; Perimeter = $3a$. (Area is 900 m².)

1. $\frac{\sqrt{3}}{4}{a}^2=900\Rightarrow a^2=\frac{3600}{\sqrt{3}}$
2. $a=\frac{60}{3^{1/4}}=\frac{60\times 3^{3/4}}{3}=20\times 3^{3/4}$
3. Perimeter = $3a=60\times 3^{3/4}=60\sqrt[4]{27}$ m

Trick: $a^2=\frac{3600}{\sqrt{3}}$, so $a=60÷\sqrt[4]{3}$. Rationalise: $\frac{1}{\sqrt[4]{3}}=\frac{\sqrt[4]{27}}{3}$. Then $3a=60\sqrt[4]{27}$.

Answer: (A) $60\sqrt[4]{27}$ m

Concept: Solve the innermost bracket first: ( ), then { }, then [ ].

1. $24÷3=8$; $76\times 8=608$; $2\times 18=36$
2. $\{608+36\}=644$; $644÷644=1$
3. $1\times \frac{20}{2}=10$

Trick: When the curly bracket equals the divisor (644), the square bracket is 1, so the answer is just $\frac{20}{2}$.

Answer: (B) 10

Concept: $\text{Smaller number}=\frac{\text{Sum}-\text{Difference}}{2}$.

1. 5 years ago sum = 83 ⇒ present sum = $83+10=93$
2. $A-N=21$
3. Nidhi = $\frac{93-21}{2}=36$ years

Trick: Remember to add 5 years for each person (+10 in total) when going from past sum to present sum.

Answer: (B) 36