Jump to content
Urch Forums

Law of averages


jatinet

Recommended Posts

The average height for a group of students is 164. If the average height of the female students is 160, is the number of the female greater than the number of the males?

1) The average of the males is 170

2) There are 20 female students.

Link to comment
Share on other sites

dsaqwert, could you please explain further.

 

My answer is A

My approach:

 

Females/x = 160, where x is the number of females -> females' height is 160x

males/y= ?, where y is the number of males

females+males/x+y=164

 

from statement 1: males/y=170 -> males' height=170y

 

160x+170y/x+y =164

160x+170y=164x+164y

4x=6y

x=3/2 *y

 

Consequently, x>y, i.e. number of females is bigger than that of males -> statement 1 is enough

 

Statement 2 is not enough as we don't know the combined height or the average or the number of males.

Link to comment
Share on other sites

My ans is A similar approach to alex.

 

dsaqwert could you explain your approach or if their is a shorter logic to get to the ans.

 

ok, let;s make it more clear;

 

assume number of boys and girls are equal, and knowing that the average of girls is 160 and the average of boys is 170, what overall average would you expect.... yes 165 ([160x + 170x]/2x=165) however the overall average is 164, in other words number of girls should be more than number of boys to ensure that the overall average is closer to the girls' average

 

in other words (mathematically);

x=number of girls

y=number of boys

160x+170y=164(x+y)

6y=4x

x=1.5y

x>y

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...