Latest TCS Aptitude Question SOLUTION: In a family of five person A,B,C,D and E every one loves one another. there birthdays are in different month and on diff dates . A bday is between 25th and 30th, B is betwn 20th an

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In a family of five person A,B,C,D and E every one loves one another. there birthdays are in different month and on diff dates . A bday is between 25th and 30th, B is betwn 20th and 25th,C is betwn 10th and 20th, D is betwn 5th and 10th and E is betwn 1st to 5th of the month.the sum of the of birth is definrd as the addition of date and month,for example 12th jan will be written as 12/1 and will add to a sum of the date of 13.(between 25th and 30th includes both 25 and 30). What may be the max average of their sum of the dates of birth ?

Asked In TCS Kiran Lokhande (10 years ago)

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Read Solution (4) :

Vaibhav Mishra  (10 years ago)

Here may be different combinations of date of birth.
As per question, to find max average of their sum of DOB,we can't take starting months
we have combinations
30/8,25/9,20/10,10/11,5/12
here average = (38+34+30+21+17)/5 =140/5 = 28
30/8,25/12,20/11,5/9,10/10
Avg = 28
30/11,25/12,20/810/9,5/10
Avg = 28
30/12,25/11,20/10,10/9,5/8
Avg = 27.6
So maximum average of sum of their Dob = 28 Ans

Is this solution Helpfull? Yes (20) | No   

babul  (10 years ago)

moths will defer => we have to take maximum value of their sum of DOB => so takes dates and moth max possible as shown below:
range date month
25- 30(A) 30 12
20-25(B) 25 11
10-20(C) 20 10
5-10(D) 10 9
1-5(E) 5 8

now take the sum (30+12)+(25+11)+(20+10)+(10+9)+(5+8) / 5 = 28 (ans)

Is this solution Helpfull? Yes (6) | No (2)  

stutisahoo  (10 years ago)

for max avg if we take,
30/12,25/11,20/10,10/9,5/8,1/7 then avg is 37.6

Is this solution Helpfull? Yes (2) | No (5)  

komal  (10 years ago)

given,
25

Is this solution Helpfull? Yes | No (8)  

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