Yogita25
[quote="GMATinsight"If \(10 - ax = bx\) then what is value of \(|a-b|\)?
1) \(a\) and \(b\) are prime numbers
2) \(x\) is an integer
\(10 - ax = bx\)
i.e. \(ax + bx = 10\)
i.e. \(x(a + b) = 10\)
Question : \(|a-b|\) = ?Statement 1: \(a\) and \(b\) are prime numbersSince, \(x(a + b) = 10\)
Case1: so a=3, b=7 and x=1 and \(|a-b| = 4\)
Case2: so a=2, b=3 and x=2 and \(|a-b| = 1\)
NOT SUFFICIENTStatement 2: \(x\) is an integerNo information about a and b so
NOT SUFFICIENTCombining the two statementsSince, \(x(a + b) = 10\) and \(x\) is an Integer
Case1: so a=3, b=7 and x=1 and \(|a-b| = 4\)
Case2: so a=2, b=3 and x=2 and \(|a-b| = 1\) so
NOT SUFFICIENTAnswer: Option E[/quote]
I tried solving it this way,
10-ax =bx..........10 = (a+b)x
option 1 and 2 are individually insufficient. but when combined...
(a + b) and x can be either 5 or 2 (considering X is an integer)
if x =5 then (a+b) =2 which means a and b should be 1..........not prime
hence (a+b) = 5 ....so a and b should be 2 and 3.......|a-b| =1
my answer is C
please help me find out the mistake in my solution.[/quote]
Factors of 10are 1, 2, 5, 10
So (a+b) can be 10 when x=1
For a+b= 10, a can be 3 and b can be 7 (both prime)
So absolute value of a+b can be 4 as well
Also it's nowhere given that x is an integers so we would have taken that into account at secondary step if this question had given soneheadway but we could prove without it that the information is NOT sufficient so solve question.