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06 Feb 2012, 09:11
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C
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Difficulty:
25%
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Question Stats:
76% (01:10) correct
24%
(01:03)
wrong
based on 139
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If m and z are integers, is z divisible by 11 ?
(1) m is a 2 digit number with equal units and tens digits
(2) m - z is divisible by 11
(1) m is a 2 digit number with equal units and tens digits
(2) m - z is divisible by 11
06 Feb 2012, 09:24
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Statement 1: Tells nothing about Z. Not enough info
Statement 2: Lots of number combination will work. Not enough info
1+2= m= [99,88,77,66,55,44,33,22,11,00,-11....] Z could be the same numbers. If Z = 11,22,33 then it is divisible, but if Z was 0 then it's not divisible. Not enough info.
Ans=E
Statement 2: Lots of number combination will work. Not enough info
1+2= m= [99,88,77,66,55,44,33,22,11,00,-11....] Z could be the same numbers. If Z = 11,22,33 then it is divisible, but if Z was 0 then it's not divisible. Not enough info.
Ans=E
06 Feb 2012, 09:37
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If m and z are integers, is z divisible by 11?
(1) m is a 2 digit number with equal units and tens digits --> m is of a type 11, 22, ..., 99 so basically we are told that m is divisible by 11. Though still insufficient, as no info about z.
(2) m - z is divisible by 11. Clearly insufficient.
(1)+(2) {multiple of 11}-z={multiple of 11} --> z={multiple of 11}. Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
(1) m is a 2 digit number with equal units and tens digits --> m is of a type 11, 22, ..., 99 so basically we are told that m is divisible by 11. Though still insufficient, as no info about z.
(2) m - z is divisible by 11. Clearly insufficient.
(1)+(2) {multiple of 11}-z={multiple of 11} --> z={multiple of 11}. Sufficient.
Answer: C.
Below might help to understand this concept better.
If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.
If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3;
OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5;
OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.
Hope it's clear.
