Kindly solve

Page no : 2

(Guest)
Originally posted by : Manjul Soni.




Originally posted by : !..SaNKeT..!






See total Age of 10 Students = 10 Students * 20 Years = 200

New Average = 20 Years + 2 Years = 22 Years

Total Age of 12 Students = 12(10+2) Students * 22 Years = 264

Total age of the 2 new students = 264 - 200 = 64


Avg age of the new students = 64 / 2 = 32 years






 


K.K.K. (Professional) (131 Points)
Replied 18 March 2012

Originally posted by : swati ghosh


if x is inversely related with square of y for x =1 & y=2 then find the value of x when y=6 :

a).3                                                                                                                                                                  b).9

c).1/3                                                                                                                                                                d).1/9


 

x=k * 1/y2  (where k is a constant)

(note: constant is used because the question uses term "related" and the word "equal")

put  x=1  & y=2, it will give the value of k as 4.

Let y=6, & k being constant so having value as 4 as above, then putting these vaues:

x=4* 1/6squire

x=4*1/36

x=1/9 (it is the answer)


vikash (Chartered Accountant) (279 Points)
Replied 20 March 2012

C is the correct answer


vikash (Chartered Accountant) (279 Points)
Replied 20 March 2012

The Total Age of 10 Students are 10 Students * 20 Years = 200

New Average becomes 20 Years + 2 Years = 22 Years

Total Age of 12 Students = 12(10+2) Students * 22 Years = 264

Total age of the 2 new students = 264 - 200 = 64


Average age of the new students = 64 / 2 = 32 years.


Fahad Ali Sayed (Praciticing CA) (40 Points)
Replied 21 March 2012

Ans is 32.

 

Existing Students  10*20=200

New Students           2*32 =64

Total Age of   All Students 264

Divide 264 by 12 u get 22.




Leave a reply

Your are not logged in . Please login to post replies

Click here to Login / Register  

Join CCI Pro


Subscribe to the latest topics :

Search Forum: