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by sanju09 » Thu Jul 29, 2010 4:39 am
If x, y and z are different integers, is x divisible by 11?

1. xyz is divisible by 22 and 33.
2. yz is divisible by 72.

[spoiler]Source: majortests.com[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by selango » Thu Jul 29, 2010 4:44 am
stmt1,

xyz is divisible by 22 and 33

x can be factor of 11.y or z can also be the factor of 11.

3 scenarios are possible.we dont know value of X.

Insuff

stmt2,

yz=72

No info abt X

Insuff

Combining 1 and 2

yz=72

2 and 3 are factors of y and z.

xyz is divisible by 22 and 33.

still we dont know info abt x. x can be factor of 11 or not a factor

Pick E
Last edited by selango on Thu Jul 29, 2010 5:03 am, edited 2 times in total.
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by sanju09 » Thu Jul 29, 2010 4:58 am
selango wrote:stmt1,

xyz is divisible by 22 and 33

xyz=x*2*11=y*3*11

xyz=11*y*2=11*y*3

xyz=11*2*z=11*3*z

3 scenarios are possible.we dont know value of X

Insuff

stmt2,

yz=72

No info abt X

Insuff

Combining 1 and 2

yz=72

xyz is divisble by 22 and 33.

x must be the factor of 11 so that xyz is divisible by 22 and 33

Pick C
Will it mean 2 = 3 there in bold?
The mind is everything. What you think you become. -Lord Buddha



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by selango » Thu Jul 29, 2010 5:04 am
sanju09 wrote:
selango wrote:stmt1,

xyz is divisible by 22 and 33

xyz=x*2*11=y*3*11

xyz=11*y*2=11*y*3

xyz=11*2*z=11*3*z

3 scenarios are possible.we dont know value of X

Insuff

stmt2,

yz=72

No info abt X

Insuff

Combining 1 and 2

yz=72

xyz is divisble by 22 and 33.

x must be the factor of 11 so that xyz is divisible by 22 and 33

Pick C
Will it mean 2 = 3 there in bold?
Edited my post.x,y or z any value can be factors of 11.
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