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## gmat prep question

This topic has 4 member replies
yvonne12 Master | Next Rank: 500 Posts
Joined
07 Feb 2007
Posted:
144 messages

#### gmat prep question

Tue Apr 10, 2007 3:29 pm
If the operation [@] is one of the four arithmetic operations addition, subtraction, multiplication and division is (6@2)@4=6@(2@4)

I. 3@2>3

II. 3@1=3

the answer is I alone, but how can this be if either 3x2 =6 or 3+2 = 5[/code]

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yvonne12 Master | Next Rank: 500 Posts
Joined
07 Feb 2007
Posted:
144 messages
Thu Apr 12, 2007 9:11 am
I get it now, thanks

Cybermusings Legendary Member
Joined
27 Mar 2007
Posted:
559 messages
Followed by:
2 members
5
Thu Apr 12, 2007 8:12 am
If the operation [@] is one of the four arithmetic operations addition, subtraction, multiplication and division is (6@2)@4=6@(2@4)

I. 3@2>3

II. 3@1=3

3@2>3. Thus @ for sure can't be division or subtraction. But it can be either addition or multiplication in in both cases the result would be greater than 3.
Now if @ is addition then (6@2)@4 would be (6+2)+4 = 6+2+4=12; and 6@(2@4)=6+(2+4)=6+2+4=12
If @ is multiplication then (6@2)@4=(6*2)*4=48; and 6@(2@4)=6*(2*4)=48
Hence in both cases (6@2)@4=6@(2@4)
Hence alone sufficient

3@1=3. Here @ can be multiplication or division. If @ we know that the two sides of the equation would be the same. But in case of division
(6/2)/4 is not equal to 6/(2/4)
First value = 3/4
Second value = 12

yvonne12 Master | Next Rank: 500 Posts
Joined
07 Feb 2007
Posted:
144 messages
Thu Apr 12, 2007 9:11 am
I get it now, thanks

Cybermusings Legendary Member
Joined
27 Mar 2007
Posted:
559 messages
Followed by:
2 members
5
Thu Apr 12, 2007 8:12 am
If the operation [@] is one of the four arithmetic operations addition, subtraction, multiplication and division is (6@2)@4=6@(2@4)

I. 3@2>3

II. 3@1=3

3@2>3. Thus @ for sure can't be division or subtraction. But it can be either addition or multiplication in in both cases the result would be greater than 3.
Now if @ is addition then (6@2)@4 would be (6+2)+4 = 6+2+4=12; and 6@(2@4)=6+(2+4)=6+2+4=12
If @ is multiplication then (6@2)@4=(6*2)*4=48; and 6@(2@4)=6*(2*4)=48
Hence in both cases (6@2)@4=6@(2@4)
Hence alone sufficient

3@1=3. Here @ can be multiplication or division. If @ we know that the two sides of the equation would be the same. But in case of division
(6/2)/4 is not equal to 6/(2/4)
First value = 3/4
Second value = 12

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