RRB Technician Grade III 2024, 2025 PYQs

Concept: For $n=10$ (even) values, median $=$ average of the $5^{th}$ and $6^{th}$ values.

1. $5^{th}=x$, $6^{th}=x+2$
2. $\frac{x+x+2}{2}=63\Rightarrow x+1=63$
3. $x=62$

Trick: Two consecutive middle terms differing by $2$: median $=x+1$.

Answer: (D) 62

Concept: Write each decimal as a power of 10 and add the exponents.

1. $0.1={10}^{-1}$, $0.001={10}^{-3}$, $0.0001={10}^{-4}$
2. Product $={10}^{-1-3-4+6}={10}^{-2}$
3. $=\frac{1}{100}$

Trick: Count decimal places: $1+3+4=8$; ${10}^6$ cancels $6$, leaving ${10}^{-2}$

Answer: (D) $\frac{1}{100}$

Concept: Translate: $75\%$ of $x=15\%$ of $1500-15$.

1. $15\%$ of $1500=225$
2. $0.75x=225-15=210$
3. $x=\frac{210}{0.75}=280$

Trick: $75\%=\frac{3}{4}$, so $x=210\times \frac{4}{3}=280$

Answer: (A) 280

Concept: $\text{SP}=\text{MP}\times \frac{100-\text{discount}\%}{100}$

1. $28400=\text{MP}\times \frac{80}{100}$
2. $\text{MP}=\frac{28400\times 100}{80}=35500$

Trick: $80\%\to 28400\Rightarrow 1\%\to 355\Rightarrow 100\%\to 35500$

Answer: (D) ₹35,500

Concept: Sum $=$ average $\times$ count.

1. Sum of 8 numbers $=8\times 14=112$
2. Sum of 6 numbers $=6\times 16=96$
3. Remaining two $=112-96=16$; average $=\frac{16}{2}=8$

Trick: Six numbers are $2$ above average each: excess $=12$, so the other two are $12$ below in total: $14-\frac{12}{2}=8$

Answer: (C) 8

Concept: $\sin \theta =\cos ({90}^{\circ }-\theta )$ and $\cos \theta =\sin ({90}^{\circ }-\theta )$.

1. $\sin {54}^{\circ }=\cos ({90}^{\circ }-{54}^{\circ })=\cos {36}^{\circ }$
2. $\cos {72}^{\circ }=\sin ({90}^{\circ }-{72}^{\circ })=\sin {18}^{\circ }$
3. Sum $=\cos {36}^{\circ }+\sin {18}^{\circ }$

Yaad rakhein: Complementary swap: $\sin \leftrightarrow \cos$, angle $\to {90}^{\circ }-$ angle.

Answer: (D) $\cos {36}^{\circ }+\sin {18}^{\circ }$

Concept: Facing the centre, a person's left is clockwise and right is anticlockwise. Fix Kate and place the others.

1. Kate at the top; John is second to her left (clockwise), Neil is immediately to her right.
2. Lisa is third to Kate's right, i.e. directly opposite; Mary is immediately to Lisa's left.
3. The only empty seat (between Kate and John) goes to Omar.
4. From Omar towards his right: Kate, Neil, then Mary → $2$ people between.

Trick: Facing centre: left = clockwise. In a 6-seat table, 'third to right/left' is always the opposite seat.

Answer: (D) Two

Concept: Draw the Venn diagram: Oranges circle inside Mangoes; Bananas completely separate from Mangoes.

1. I: 'No mango is a banana' ⇒ 'No banana is a mango' (an E-statement converts to itself) → follows.
2. II: 'All oranges are mangoes' only means Oranges ⊂ Mangoes; mangoes outside oranges are possible → does not follow.

Yaad rakhein: 'All A are B' never gives 'All B are A' — it only gives 'Some B are A'. 'No A is B' = 'No B is A'.

Answer: (C) Only conclusion I follows

Concept: Facing the centre: left = clockwise, right = anticlockwise. Start with the S–U–T chain.

1. S is immediately left of U; going left of S, one person is skipped and then T sits.
2. P is a neighbour of T and of Q → P is next to T on the far side, then Q.
3. Two seats remain between S and Q; R and V must have $2$ people between them counting from R's right → R next to S (left side), V next to Q.
4. Order clockwise: S, R, T, P, Q, V, U. From Q to the left: V, U, S, then R → $3$ between.

Trick: Fix one person (S) and convert every 'left/right' into clockwise/anticlockwise before placing.

Answer: (B) 3

Concept: Compare letter positions of NJQO and HDKI.

1. N(14)→H(8), J(10)→D(4), Q(17)→K(11), O(15)→I(9): each $-6$.
2. Check SOVT→MIPN: each $-6$ ✓.
3. PLSQ $-6$: P→J, L→F, S→M, Q→K = JFMK.

Trick: Same shift on all four letters — just move each letter back $6$.

Answer: (D) JFMK

Concept: Track the path with coordinates (East $+x$, North $+y$), starting at A $(0,0)$.

1. $30$ km East → $(30,0)$; right = South $18$ → $(30,-18)$.
2. Right = West $17$ → $(13,-18)$; right = North $9$ → $(13,-9)$.
3. Left = West $13$ → P $(0,-9)$.
4. A is $9$ km straight North of P.

Trick: East–west: $30-17-13=0$, North–south: $-18+9=-9$ → only a $9$ km gap, P is south of A.

Answer: (B) 9 km towards North

Concept: Check the position gaps inside each triad.

1. JE-KF-LM: J, E $=$ J$-5$; K, F (each $+1$); L $=$ K$+1$, M $=$ L$+1$.
2. NI-OJ-PQ follows the same: N, I; O, J; P, Q.
3. LG-MH-NO: L, G $=$ L$-5$; M, H; N, O $=$ N$+1$ ✓.
4. Others fail in the last pair (QS, SU, RT jump by $2$).

Trick: Look at the 3rd pair first — its two letters must be consecutive.

Answer: (C) LG-MH-NO

Concept: Number the positions 1 (bottom) to 7 (top) and place the boxes with fixed positions first.

1. Only three boxes below L → L is at position $4$.
2. Only A is above B → B is at $6$ and A at $7$ (the top).
3. Two boxes between L and Z → Z is at $1$ or $7$; position $7$ is A, so Z is at $1$ (the bottom).
4. Remaining positions $2,3,5$ for Y, X, C with Y above X above C → C $=2$, X $=3$, Y $=5$.
5. Order from bottom: Z, C, X, L, Y, B, A. Second from the bottom = C.

Trick: Fill 'only $n$ below/above' clues first — they fix exact positions.

Answer: (D) C

Concept: Words common to two sentences share the common codes.

1. 'breeze rustles' is common → codes 'eg lz'.
2. Sentence 2 left with 'softly' and code 'cw'.
3. So softly = cw.

Trick: Remove common words and common codes; what remains pairs up.

Answer: (A) cw

Concept: In a series of $n$ terms, $k^{th}$ from right $=(n-k+1{)}^{th}$ from left.

1. There are $30$ digits; 1st from left $=1$.
2. 12th from right $=30-12+1={19}^{th}$ from left $=2$.
3. Sum $=1+2=3$.

Trick: Count backwards from the right end directly: …$240700492990$ → the ${12}^{th}$ is $2$.

Answer: (D) 3

Concept: Decode each symbol step by step and build the family tree.

1. P × Q: P is mother of Q. Q ÷ R: Q is brother of R → P is mother of R too.
2. R – S: R is husband of S. S × T: S is mother of T → R is father of T.
3. P is the mother of T's father.

Trick: Work from the target (T) upward: T → father R → R's mother P.

Answer: (B) Father's mother

Concept: Find the position gaps within and between the two clusters.

1. RY: R$+7=$Y; next A $=$ Y$+2$; S $=$ A$-8$ (i.e. $+18$).
2. OV: $+7$; X $=$ V$+2$; P $=$ X$-8$ ✓. NU: $+7$; W $=$ U$+2$; O $=$ W$-8$ ✓.
3. DI: D$+5=$I ✗ (not $+7$) → odd one.

Trick: Check only the first cluster's gap ($+7$) — the pair that breaks it is odd.

Answer: (D) DI – PW

Concept: Swap $÷$ and $\times$, then apply BODMAS.

1. New expression: $54÷9\times 8+22-33$
2. $54÷9=6$, $6\times 8=48$
3. $48+22-33=37$

Trick: Rewrite the full expression first, then solve — never solve and swap together.

Answer: (B) 37

Concept: Compare each letter with the same letter of the next term.

1. B→K, E→N, H→Q: each $+9$.
2. KNQ→TWZ ($+9$), TWZ→CFI ($+9$, wrapping after Z).
3. CFI $+9$ → LOR.

Trick: Inside each term letters jump by $3$ (BEH); so only the first letter matters: C$+9=$L → LOR.

Answer: (A) LOR

Concept: Replace the letters by operators, then apply BODMAS.

1. $5\times 12-56+21÷7$
2. $=60-56+3$
3. $=7$

Trick: Do $÷$ and $\times$ first: $60$ and $3$, then $60-56+3$.

Answer: (D) 7