Examine the information given below and and answer the questions that follow:
P,Q,R,S,T and U are members of a family. There are two married couples in the family. Q is an engineer and the father of T. U is the grandfather of R and is a lawyer. S is the grandmother of T and is a housewife. There is one engineer, one lawyer, one teacher, one housewife and two students in the family.
Question 1
Who is the husband of P?
(A) R
(B) T
(C) Q
(D) S
Question 2
Which of the following are definitely the married couples?
(A) US and QP
(B) US and RT
(C) QP and RT
(D) None of these
Question 3
Which of the following is definitely a group of male members?
(A) Q,U,T
(B) U,T
(C) Q,U,P
(D) Q,U
Question 4
Which of the following is definitely a group of female members?
(A) P,S
(B) S only
(C) P,S,R
(D) P,R
Question 5
How many different family trees can be drawn that satisfy all the conditions given?
(A) 1
(B) 2
(C) 3
(D) More than 3
SOLUTIONS TO DAILY CSAT MISSION # 61
1. (C) 2. (A) 3. (D) 4. (C) 5. (C)
Explanations
1. LCM of 6,7,8,9 is 504. So, they will toll together again after 504 seconds i.e. 8 minutes and 24 seconds after 11:55:00 am.
2.
Start off by filling the innermost circle with one colour. The next circle will be a different colour.
Beyond that, the first colour can again be used for half of the boxes in the next circle. The other half of the boxes will need a third colour (because the second cant be used).
The second colour can again be used to fill the fourth circle completely.
The outermost circle can again be filled by alternate boxes of the first and the third colour respectively.
The figure with colours will be the following. Note that no two boxes with the same colour are adjacent:
3. Let his income were ‘x’ pesos.
Tax in 2009 = 15x/100.
Tax in 2010 = 2000 + (x-10000)*5/100.
Since the two taxes are same, 15x/100 = 2000 + (x-10000)*5/100, which gives x = 10000.
4. The first person can be born on any of the seven days of the week. So, he can be born in 7 ways i.e. either on Sunday or Monday or …… or Saturday.
The second person cannot be born on the same day as the first person. So, there are 6 days in which he can be born i.e. the remaining six days of the week.
The third person cannot be born on either of the two days on which the first or second person were born. So, he can be born in 5 ways.
The total number of ways in which the three people can be born (with these conditions applied) = 7*6*5.
If no condition were there and the people were free to be born on whatever day possible, then the first guy would have seven different options to be born (either Sunday or Monday or …. or Saturday), the second guy would also have all the seven different options (either Sunday or Monday or …. or Saturday), and the third guy would also have all the seven different options (either Sunday or Monday or …. or Saturday). In that case, the total number of possibilities would have been 7*7*7.
The probability with the given conditions is (7*6*5)/(7*7*7) = 30/49
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