Logical Reasoning & Puzzles

Opposite Alphabet Pairs:
R ↔ I, O ↔ L, S ↔ H, E ↔ V.
Similarly:
H ↔ S
A ↔ Z
N ↔ M
D ↔ W
Therefore, HAND = SZMW.

This is an alte ating addition and subtraction series: 7 (+3) = 10, 10 (-2) = 8, 8 (+3) = 11, 11 (-2) = 9, 9 (+3) = 12.

The next operation is subtracting 2: $12 - 2 = 10$.

'The only son of my mother' refers to Suresh himself (since his mother has only one son).
Therefore, the boy in the photo is Suresh's son.
Hence, Suresh is the Father of that boy.

All mangoes belong to the set of golden things. No golden things belong to the set of cheap things.
Therefore, no mango can be cheap.
Conclusion I ('All mangoes are cheap') is false.
Conclusion II ('Golden-colored mangoes are not cheap') is true, since no golden thing is cheap.

Let us analyze the transformation from COMPUTER to RFUVQNPC:
1. Reverse COMPUTER → RETUPMOC.
2. First and last letters stay unchanged: R remains R, C remains C.
3. All middle letters are shifted forward by +1:
E → F
T → U
U → V
P → Q
M → N
O → P
Result: R F U V Q N P C.

Applying this rule to MEDICINE:
1. Reverse → ENICIDEM.
2. First (E) and last (M) letters stay intact.
3. Middle letters shifted by +1:
N → O
I → J
C → D
I → J
D → E
E → F
Result: E O J D J E F M.

1. Starting at origin $(0,0)$, walking 5 km South reaches $(0, -5)$.
2. Facing South, tu ing right means facing West. Walking 3 km reaches $(-3, -5)$.
3. Facing West, tu ing left means facing South. Walking 5 km South reaches $(-3, -10)$.
The final coordinates $(-3, -10)$ are South and West of origin → South-West.

At 4 o'clock, the minute hand is 20 minute spaces behind the hour hand. For hands to be at right angle ($90^\circ$), they must be 15 minute spaces apart.

Case 1: When minute hand is 15 minutes behind the hour hand → $(20 - 15) = 5\text{ minutes spaces to gain}$.
Time = $5 \times \frac{12}{11} = \frac{60}{11} = 5 \frac{5}{11}\text{ min past 4}$.

Case 2: When minute hand is 15 minutes ahead of the hour hand → $(20 + 15) = 35\text{ minutes spaces to gain}$.
Time = $35 \times \frac{12}{11} = \frac{420}{11} = 38 \frac{2}{11}\text{ min past 4}$.

Therefore, both options are valid times where a right angle is formed.

1. Seema is to the right of Bindu → (Bindu, Seema).
2. Seema is to the left of Rani → (Bindu, Seema, Rani).
3. Reeta is between Rani and Mary, and Mary is to the right of Rani → (Bindu, Seema, Rani, Reeta, Mary).
Looking at the ordered sequence from left to right: Bindu → Seema → Rani → Reeta → Mary.
Immediately to the right of Reeta is Mary.

Each of the numbers 3, 5, 11, 17, and 21 are odd numbers, whereas 14 is an even number. Furthermore, 14 is the only composite even number in the prime-dense sequence.

In expression P - R + Q:

• P - R means P is the sister of R.
• R + Q means R is the father of Q.
• The sister of someone's father is their pate al aunt (Bua / Chachi). Hence, P is the pate al aunt of Q.

Each letter is incremented by +1 alphabetical position:

• L (+1) → M, I (+1) → J, G (+1) → H, H (+1) → I, T (+1) → U

Applying the same logic to SPARK:

• S (+1) → T, P (+1) → Q, A (+1) → B, R (+1) → S, K (+1) → L → TQBSL

Cleaning is undertaken because drains are currently clogged with silt and debris (Assumption I is valid). The purpose of undertaking this municipal effort before monsoons is to mitigate urban waterlogging and overflow (Assumption II is valid). Therefore, both I and II are implicit.

Differences between consecutive terms:

• $6 - 2 = 4$
• $12 - 6 = 6$
• $20 - 12 = 8$
• $30 - 20 = 10$
• $42 - 30 = 12$

The difference sequence increases by +2 each step. Next difference = $12 + 2 = 14$.

Next term = $42 + 14 = 56$ (Alte atively, $n(n+1)$ where $n=7 \implies 7 \times 8 = 56$).

Term 1 = 5; Term 2 = 2(5) + 1 = 11; Term 3 = 2(11) + 1 = 23; Term 4 = 2(23) + 1 = 47.

Z + 2 = B; E + 2 = G; B + 2 = D; R + 2 = T; A + 2 = C. Therefore, ZEBRA becomes BGDTC.