Given series:

Left Side # 6 B G @ I L 7 3 H A % © D F K E 8 J Q 1 * V T U 2 $ W Right Side

1. Vowels which are immediately followed by a symbol and also immediately preceded by a consonant:

6 B G @ I L 7 3 H A % © D F K E 8 J Q 1 * V T U 2 $ W

Hence, there is 1 Vowel which is immediately followed by a symbol and also immediately preceded by a consonant:

H A %

Given word: ACKNOWLEDGE

2^nd letter = C; 4^th letter = N; 5^th letter = O; 6^th letter = W; 7^th letter = L

Letters given: C N O W L

Possible words: C L O W N (a comic entertainer)

Hence, only one word can be formed and the first letter of the word is “C”.

Seven persons: A, B, C, D, E, F and G

Items: Eraser, Pen, TV, Printer, Mouse, Mug and Phone

1. There is a gap of 4 persons between E and G who purchased a printer.

2. E didn’t come first or at last.

3. A purchased a mouse and there is a gap of 2 persons between A and E. 

4. C purchased eraser immediately before A.

5. D purchased phone immediately before E who purchased a TV. 

6. The many persons came after A are the same as the number of persons came before F who purchased the pen.

7. The mug is purchased by B.

Hence, F purchased the pen.

Seven persons: A, B, C, D, E, F and G

Items: Eraser, Pen, TV, Printer, Mouse, Mug and Phone

1. There is a gap of 4 persons between E and G who purchased a printer.

2. E didn’t come first or at last.

3. A purchased a mouse and there is a gap of 2 persons between A and E. 

4. C purchased eraser immediately before A.

5. D purchased phone immediately before E who purchased a TV. 

6. The many persons came after A are the same as the number of persons came before F who purchased the pen.

7. The mug is purchased by B.

Hence, there are 3 persons between F and G.

Seven persons: A, B, C, D, E, F and G

Items: Eraser, Pen, TV, Printer, Mouse, Mug, and Phone

1. There is a gap of 4 persons between E and G who purchased a printer.

2. E didn’t come first or at last.

3. A purchased a mouse and there is a gap of 2 persons between A and E. 

4. C purchased an eraser immediately before A.

5. D purchased a phone immediately before E who purchased a TV.

6. The many persons who came after A are the same as the number of persons who came before F who purchased the pen.

7. The mug is purchased by B.

Hence, the mouse has been purchased just before mug.

Seven persons: A, B, C, D, E, F and G

Items: Eraser, Pen, TV, Printer, Mouse, Mug and Phone

1. There is a gap of 4 persons between E and G who purchased a printer.

2. E didn’t come first or at last.

3. A purchased a mouse and there is a gap of 2 persons between A and E. 

4. C purchased eraser immediately before A.

5. D purchased phone immediately before E who purchased a TV. 

6. The many persons came after A are the same as the number of persons came before F who purchased the pen.

7. The mug is purchased by B.

The printer was purchased at last.

Seven persons: A, B, C, D, E, F and G

Items: Eraser, Pen, TV, Printer, Mouse, Mug and Phone

1. There is a gap of 4 persons between E and G who purchased a printer.

2. E didn’t come first or at last.

3. A purchased a mouse and there is a gap of 2 persons between A and E. 

4. C purchased eraser immediately before A.

5. D purchased phone immediately before E who purchased a TV. 

6. The many persons came after A are the same as the number of persons came before F who purchased the pen.

7. The mug is purchased by B.

E came just before F to purchase.

**Given

Statements:

** B > C; C ≥ D = E; G ≤ H < E ≥ I = F

On combining: B > C ≥ D = E > H ≥ G; C ≥ D = E ≥ I = F

Conclusions:

I. F ≤ C → True (as F = I ≤ E → F ≤ E and E = D ≤ C → F ≤ C)

II. B > H → True (as B > C ≥ D = E → B > E and H < E → B > H)

Hence, both conclusions I and II follow.

**Given

Statements:

** P = Q; Q ≥ R; R < S; T > P; S ≤ U

On combining: T > P = Q ≥ R < S ≤ U

Conclusions:

I. T > R → True (as T > P = Q ≥ R → T > R)

II. R < U → True (as R < S ≤ U → R < U)

Hence, both conclusions I and II follow.

**Given

Statements:

** V ≥ W; X > Y; V ≥ Z; W ≥ X

On combining: V ≥ W ≥ X > Y; V ≥ Z

Conclusions:

I. X > Z → False (as X ≤ V ≥ Z → thus relationship between X and Z cannot be determined)

II. Z > W → False (as W ≤ V ≥ Z → thus relationship between W and Z cannot be determined)

III. W > Y  → True (W ≥ X > Y)

Therefore, only conclusion III is true and follows the statement.

Odd series = 1, 3, 5, 7,

Even series = 2, 4, 6,

Odd Series = 3, 5, 7,

Even series = 2, 4,

Odd Series = 5, 7,

Even Series = 2,

Odd sereis = 7 =?

Word: DYNAMICS 

on Arranging as per dictionary: ACDIMNSY

Only one alphabet whose positions are unchanged is M

Hence, one is the answer.

Given Word: INEVITABLE 

2^nd letter = N; 3^rd letter = E; 6^th letter = T

Letters have given: N, E, T

Possible words: NET, TEN 

Hence, two words can be made, and the answer is Y.

Given Series: M J % 4 T E K I 9 # O A $ Q 3 8 N 5 U 7 W ⋆ B @ D F © 1 Z 6 H

Question series:

J4E   I#A    Q85   7⋆@   ?

Here, ‘J4E’ have J, J + 2 unit = ‘4’, ‘4’ + 2 unit = E

E + 2 unit = I, I + 2 unit = #, # + 2 unit = A ⇒ I#A is 2^nd element

A + 2 unit = Q, Q + 2 unit = ‘8’, ‘8’ + 2 unit = ‘5’ ⇒ Q85 is 3^rd element

‘5’ + 2 unit = 7, ‘7’ + 2 unit = ⋆, ⋆ + 2 unit = @ ⇒ 7⋆@ is 4^th element

For 5^th element: @ + 2 unit = F, F + 2 unit = ‘1’, ‘1’ + 2 unit = ‘6’.

So, F16 is the next term$.

Given Series: M J % 4 T E K I 9 # O A $ Q 3 8 N 5 U 7 W ⋆ B @ D F © 1 Z 6 H

1. I 9 E: I and 9 are consecutive, E = ‘I’ - 2 unit.

2. 7 W 5: 7 and W are consecutive, 5 = ‘7’ - 2 unit.

3. © 1 D: © and 1 are consecutive, D = ‘©’ - 2 unit.

4. Q A 3: Q and A are not consecutive.

5. E K 4: E and K are consecutive, 4 = ‘E’ - 2 unit.

The pattern followed is first two elements are consecutive to each other, third element is - 2 of 1^st element. Hence, (Q A 3) is odd-one out$.

Given Series: M J % 4 T E K I 9 # O A $ Q 3 8 N 5 U 7 W ⋆ B @ D F © 1 Z 6 H

Removing symbols:

(Left End) M J 4 T E K I 9 O A Q 3 8 N 5 U 7 W B D F 1 Z 6 H (Right End)

12^th from left end = 3

4^th to left of ‘3’ = 9.

Hence answer is 9$.

Given Series: M J % 4 T E K I 9 # O A $ Q 3 8 N 5 U 7 W ⋆ B @ D F © 1 Z 6 H

Removing numbers:

(Left End) M J % T E K I # O A $ Q N U W ⋆ B @ D F © Z H (Right End)

⇒ 13^th from the right end = $

⇒ 6^th to right of ‘$’ = B

Hence answer is ‘B’.

Given Series: M J % 4 T E K I 9 # O A $ Q 3 8 N 5 U 7 W ⋆ B @ D F © 1 Z 6 H

Pattern to look for: Number → Symbols → Vowel

Such Symbols: #.

Hence only one such symbol is there$.

Friends: P, Q, R, S, T, U and V are seven friends.

1. U is shorter than only one person.

__ > U > __ > __ > __ > __ > __

2. P is taller than Q and T, but P is not the tallest.

3. S is taller than only V.

(Now leftover is R therefore R is tallest)

R > U > P > Q/T  > Q/T > S > V

Hence, “P” is 3^rd tallest among them.

Using the following notations, we can draw the family tree,

1. M is the only son of X.

2. C is the wife of X.

3. None of the daughters of X is married.

4. S and Y are the daughters of C.

5. C is the daughter of W and R.

6. Q is the grandson of C.

7. D is the mother of X.

As per family diagram D is Mother-in-law of C.