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Tuesday, April 14, 2009

GRE ALGEBRA QUESTION OF THE DAY

If y = 3, then y³(y³-y)=

A. 300
B. 459
C. 648
D. 999
E. 1099

16 comments:

R.Sidharth said...: 648; April 14, 2009 at 8:03 PM
Anonymous said...: C; April 16, 2009 at 12:49 PM
Unknown said...: becoz
27(27-3)
27(24)
multiply last digits
4*7=ends wit 8
therefore 648; April 18, 2009 at 6:54 PM
Unknown said...: option is c; May 28, 2009 at 10:18 AM
Unknown said...: c; June 14, 2009 at 7:58 AM
Anonymous said...: 648 dude; June 20, 2009 at 11:09 AM
Unknown said...: 648; June 23, 2009 at 3:25 AM
Anonymous said...: C:648; June 24, 2009 at 2:39 PM
Unknown said...: 648; June 27, 2009 at 9:58 PM
Anonymous said...: lawra; June 30, 2009 at 4:19 AM
kmk028 said...: C; September 18, 2009 at 8:38 AM
Anonymous said...: c; November 4, 2009 at 12:17 PM
wazza said...: 27(27-3)
Even multiplying 27*27 is a waste of time. Just look at the last digits by distributing it:
= _7 * _7 - _7 * 3
= _ _ 9 - _ 1
= _ _ 8

only number that fits the bill is 648.; December 15, 2009 at 10:31 AM
Aditya said...: 648; March 19, 2010 at 7:12 PM
Anonymous said...: C; March 11, 2011 at 2:24 PM
Anonymous said...: 648; May 11, 2011 at 2:21 PM

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