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Is (x-y)/9 an integer?

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Is (x-y)/9 an integer?

by mehravikas » Sun Jun 01, 2008 1:05 am
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.
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by amitansu » Sun Jun 01, 2008 8:38 am
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.
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by netigen » Sun Jun 01, 2008 11:59 am
amitansu 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.
There is a problem in your reasoning for (B). X can be 42 and Y can be 79

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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by chidcguy » Sun Jun 01, 2008 12:37 pm
X-Y = 11(a-b) + 40

a-b=1 51/9 No
a-b=2 62/9 No
a-b=0 40/9 No
a-b=-1 29/9 No
a-b=-2 18/9 Yes
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by mehravikas » Mon Jun 02, 2008 2:05 am
Official answer is 'A'

Thanks,
Vikas
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