Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

Hence, Nia will go on the trip on 16^th September.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2. Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Dia goes on the trip exactly before last.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Except Mia - Pia all are born in the same month.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

 

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

Hence, Only Dia and Ria goes on the trip after Gia.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Tia goes on the 27^th february.

The least possible diagram for the given statements is as follows

Conclusions:

I. All balls are tablets → True (All balls are dolls and all dolls are mobile and all mobiles are tablets so all balls are also tablets)

II. Some mobiles are dolls → True (all dolls are mobile so some mobiles are dolls)

Conclusion I and II both follow.

The least possible diagram for the given statements is as follows

Conclusions:

I. No potato is orange → True (Because all potatoes are lime and no lime is orange so no potato is orange)

II. Some apples are limes → False (Because no orange are limes and all apples are orange so some apples are limes it is not possible)

Only conclusion I follow.

Solution:

The least possible Venn diagram for the given statements is as follows,

Conclusions:

I. No bucky is captain. →  False (it is possible but not definite, because no direct relation between bucky and captain is given)

II. Some america is bucky. →  False (because only captain can be america)

Hence, Neither I nor II follows is the correct answer.

Additional Information:

Only captain is america means all america is captain and no other element can be america.

Only a few captain are hydra means some captains are hydra and some are not.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Now, consonants which are immediately preceded by a symbol and immediately followed by a number: Symbol → Consonant → Number

Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £

Hence, there are two consonants N and Z which are immediately preceded by a symbol and immediately followed by a number → @ N 2 and ¥ Z 8.

Note: A, E, I, O, U are the vowels in English alphabet and all else are called consonants.

Mistake Points 

Please note that U is not a consonant.

According to the question, the pattern should be

Symbol → Consonant → Number

Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £

Hence, there are two consonants N and Z which are immediately preceded by a symbol and immediately followed by a number → @ N 2 and ¥ Z 8.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Here, each element of each term moves 6 steps forward to give the corresponding element of the next term.

@ + 6 = B; B + 6 = 3; 3 + 6 = I

2 + 6 = 6; 6 + 6 = A; A + 6 = Z

U + 6 = S; S + 6 = 7; 7 + 6 = £

Hence, according to this the missing term will be 3 A 7.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

When all the vowels are dropped, the new sequence is;

Y 5 @ N 2 # 9 B 6 % S © 3 Q & 7 W ¥ Z 8 £

Now, the 8^th term from the right in the new sequence:

(Y 5 @ N 2 # 9 B 6 % S © 3 Q & 7 W ¥ Z 8 £) → Q.

Hence, “Q” is the correct answer. 

Note: A, E, I, O, U are the vowels in English alphabet and all else are called consonants.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

The new sequence after reversing the last 12 elements;

Y 5 @ N 2 # U 9 B E 6 % S £ 8 Z ¥ I W 7 & A Q 3 ©

As, Left – Left = Left

21^st from the left – 6^th from the left = 15^th from the left

Hence, 15^th from the left is 8.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Here, the group is formed in which the first and second elements move 2 steps forward and 4 steps backward respectively to give the second and third elements.

1. + 2 = 9; 9 - 4 = N

2) 6 + 2 = S; S + 2 = 3

3. 7 + 2 = I; I - 4 = A

4. ¥ + 2 = 8; 8 - 4 = W

5. @ + 2 = 2; 2 - 4 = Y

Hence, 6 S 3 does not belong to that group.

**Given

Statements:

** T = V < U ≥ X; V ≤ S; H > U

On Combining : T = V < U ≥ X; V ≤ S; H > U

Conclusions:

I. S ≥ T → True  (As V ≤ S and T = V → S ≥ T)

II. X < H → True (As H > U and U  ≥  X → X < H )

Therefore Both I and II are True

Given Statements : L < O < M ; Q> N ≥ P > M 

On Combining : Q > N ≥ P > M > O > L

Conclusions :

I. P > L → True ( As P > M > O > L → thus clear relation P and L can be determined )

II. M > N → False ( As N ≥ P > M → thus we conclude that M is less than N not greater )

Therefore Only I is true.

**Given

Statements:

**  X ≤ U < V > T; U < S; H > V

On combining:X ≤ U < V > T; U < S; H > V

Conclusions:

I. X < H → True (H > V and  V > U ≥ X → X < H)

II. T ≤ S → False (U <  S and U < V > T  → thus clear relation between T and  S cannot be determined)

Therfore Only conclusion I is True.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, two boxes are placed above box U.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, box V is placed second from the bottom.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, five boxes are placed between box T and box S.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Here, box S is placed exactly above box P, and box R is placed exactly above box Q, in the same way, box V is placed exactly above box T.

Clearly, box V is related to box T.

The word 'at least' means the minimum number, while 'at most' means the maximum number.