S4-24 If x and y are integers between 10 and 99, inclusive, is (x-y) / 9 an integer?
(1) x and y have the same two digits, but in reverse order.
(2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.
Is (x-y)/9 an integer?
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D is ans here.
Both the statements are basically telling the same thing !
From 1 : they have same two digits but in reverse order.
let's take 12 which is greater than 10.Reverse of it's 21 and diff. is 9 and is divisile by 9 obviously.if you try with any other combination i.e. say 13 am 31 here the diff is 18 and so on ... so the series contains all multiples of 9.sufficient.
Stem 2: just says the same thing in a different ways.i.e. 13 and 31 proves that.So sufficient.
Both the statements are basically telling the same thing !
From 1 : they have same two digits but in reverse order.
let's take 12 which is greater than 10.Reverse of it's 21 and diff. is 9 and is divisile by 9 obviously.if you try with any other combination i.e. say 13 am 31 here the diff is 18 and so on ... so the series contains all multiples of 9.sufficient.
Stem 2: just says the same thing in a different ways.i.e. 13 and 31 proves that.So sufficient.
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There is a problem in your reasoning for (B). X can be 42 and Y can be 79amitansu wrote:D is ans here.
Both the statements are basically telling the same thing !
From 1 : they have same two digits but in reverse order.
let's take 12 which is greater than 10.Reverse of it's 21 and diff. is 9 and is divisile by 9 obviously.if you try with any other combination i.e. say 13 am 31 here the diff is 18 and so on ... so the series contains all multiples of 9.sufficient.
Stem 2: just says the same thing in a different ways.i.e. 13 and 31 proves that.So sufficient.
This can be solved with eq
(A) X = 10a+b
Y = 10b+a
X-Y = 9a-9b = 9 (a-b) so sufficient
(B) X = 10 (a+2) + a = 11a+20
Y = 10 (b-2) + b = 11b - 20
X-Y = 11(x-y) + 40
when (x-y)=2 then it is div when x-y = 1 it is not so insufficient
Ans should be A
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