GRE
Exam
Numerical Ability
Algebra
if x+y+z=21 then find the maximum value of (x+2) (y+2) (z+2)
Read Solution (Total 6)
-
- out of many possibilities, 7+7+7=21 gives the maximum value
9*9*9=729 - 9 years agoHelpfull: Yes(5) No(0)
- for 3 nos. arithmetic mean> geometric mean.
consider nos. x+2,y+2,z+2
their am is (a+b+c)/3
(x+y+z+6)/3=9
g.m. is [(x+2)(y+2)(z+2)]^1/3
am>gm
taking cube both side
729 >(x+2)(y+2)(z+2)
HENCE max. value is 729 - 9 years agoHelpfull: Yes(3) No(0)
- x,y and z will have different values so I guess x=8,y=7,z=6, as 6+7+8=21
so answer will be= 10*9*8=720 - 9 years agoHelpfull: Yes(0) No(1)
- The possible solutions to get 21 are =20,1,0.
.
.
.
9,9,3--(9+2)(9+2)(5)----max - 9 years agoHelpfull: Yes(0) No(1)
- 729
9*9*9=729
- 9 years agoHelpfull: Yes(0) No(0)
- remainder=3
2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256
from above observe last digits 2,4,8,6,2,4,6,8,............... here 2^35 th last digit is 8 ,so 8/5=3. - 8 years agoHelpfull: Yes(0) No(0)
GRE Other Question