Puzzle Test Questions and Answers

Answer: C

Given F is a psychologist.
B is the mother of F and E is the brother or sister of F.
There are only two married couples in the family. Since, D is married to A, so C, the jeweller, who is married to a lawyer, will be married to B.
Again, the Manager D is married to A means A is the doctor and Grandfather of F and E.
Also, no one else is an Engineer. So, E must be an Engineer.
Clearly, D, the manager is married to A. So, AD is one of the couples in the family.
Hence, the answer is (c).

Answer: B

In a term of six if five boys are to be selected then both A and D together are selected.
If C is selected, a girl Q will be selected.
From B and E, one has to be selected. So, we select E because B will be accompanied by a girl.
The fifth boy will be F. So, the only girl will be Q in the team A C D E F G.
Hence, the answer is (b).

Answer: A

If P is included, D and hence A cannot be included.
If Q is selected, C has to be selected and so S cannot be selected.
T goes with S. So, T is also not selected.
The third girl can be R. With R, B will be selected but with B, E cannot be selected. So, the sixth member can be F only. i.e, the team become P Q C R B F. So, the answer is (a).

Answer: B

If team contains C, Q will be include. If another girl include is R, B will be there and hence E cannot be there.
A and D have to be together. So, they are also included and only F can be excluded. Thus, the team is C Q R B A D.
So, the answer is (b).

Answer: B

If E is included, B cannot be included. A and D have to be together. So, they are both included. Without B, R will not be there. With D, P cannot be there. So, two girls together can be only S and T. If S is there, C cannot be there. So the fourth boy can be F alone.
Thus, the team becomes E A D S T F. So, the answer is (b).

Answer: B

n four girls, S and I are taken together. With S, C cannot be there. So, Q will not be there.
If P is included, D and hence A cannot be there.
If R is included, B will be there and hence E cannot be there.
So, only F can be there.
Thus, the team is S T P R B F. So, the answer is (b).

Answer: D

We first form the sequence of visit using (iv), (v) and (vi).
From (iv), we know that Q visits first and U visits last.
From (v), we know that W visits first after break and is followed by R.
From (vi), we know that P visits after break.
Thus, the sequence of visit after break becomes W R P U.
Also, S has the same position in morning session as P in afternoon session. So, sequence of visit before break is Q, T / V, S, V / T.
Clearly, P visits between R and U

Answer: C

We first form the sequence of visit using (iv), (v) and (vi).
From (iv), we know that Q visits first and U visits last.
From (v), we know that W visits first after break and is followed by R.
From (vi), we know that P visits after break.
Thus, the sequence of visit after break becomes W R P U.
Also, S has the same position in morning session as P in afternoon session. So, sequence of visit before break is Q, T / V, S, V / T.
The time of visit of W is 2 p.m. to 3 p.m., that of doctor R is 3 p.m. to 4 p.m.
So, the visit of doctor R is over at 4 p.m.

Answer: C

We first form the sequence of visit using (iv), (v) and (vi).
From (iv), we know that Q visits first and U visits last.
From (v), we know that W visits first after break and is followed by R.
From (vi), we know that P visits after break.
Thus, the sequence of visit after break becomes W R P U.
Also, S has the same position in morning session as P in afternoon session. So, sequence of visit before break is Q, T / V, S, V / T.
Clearly, T visits either second or fourth. So, the time of visit on Saturday will be either 9.30 a.m. or 10.30 a.m.
Thus, T's visit will be over at either 10 a.m. or 11 a.m.

Answer: B

We first form the sequence of visit using (iv), (v) and (vi).
From (iv), we know that Q visits first and U visits last.
From (v), we know that W visits first after break and is followed by R.
From (vi), we know that P visits after break.
Thus, the sequence of visit after break becomes W R P U.
Also, S has the same position in morning session as P in afternoon session. So, sequence of visit before break is Q, T / V, S, V / T.
Clearly, as mentioned, lunch break will be over and doctor W will visit at 1.45 p.m., doctor R will visit at 2.30 p.m., doctor P will visit at 3.15 p.m. and U will visit at 4 p.m.