my love for GRE....
The average ( arithmetic mean) of five number is 25. After one of the numbers is removed, the average (arithmetic mean) of them remaining numbers is 31.
What number has been removed?
(A) 1
(B) 6
(C) 11
(D) 24
(E) It cannot be determined from the information given.
posted by *fumiko* @ 6:25 AM
4 Comments
4 Comments:
a+b+c+d+e=25*5=125
say, 'e' is removed,
a+b+c+d=31*4=124
so, 125-124=1=e
therefore, 1 is the number that has been removed.
http://www.ets.org/portal/site/ets/menuitem.1488512ecfd5b8849a77b13bc3921509/?vgnextoid=302b66f22c6a5010VgnVCM10000022f95190RCRD&vgnextchannel=d687e3b5f64f4010VgnVCM10000022f95190RCRD
http://www.petersons.com/default.asp#practice
http://number2.com/
these are good GRE preparation websites! have fun!
What i hate about questions like it's not that hard to solve it.
But somehow when time is crunched, i panic and think it's super duper difficult question!!!
I am just Baka!
this question sounds tricky because it seems like you have 4 or 5 unknown numbers. but actually you only have one(well two if you add up the four numbers) unknown number that you need to get the value of, and have two equation for it.
so you are sure that you can get the value of the one unknown number from the question if you read the question carefully.
I'm with you fumiko...my name is Grace and I am baka.
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