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If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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12 Dec 2014, 07:17
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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. DS00502.01
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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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12 Dec 2014, 10:03
Statement 1: x and y have the same two digits, but in reverse order. Let the digits be ab and ba xy = 10a+b  10b a = 9a  9b = 9*(ab)
therefore, (xy)/9 = ab = Integer. Sufficient
Statement 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.
Let say the numbers are 42 and 24, then (xy)/9 = (42  24)/9 = 18/9 = 3 (integer). Therefore, (x  y)/9 an integer Let say the numbers are 42 and 46, (xy)/9 = (42  46)/9 = 4/9 (not an integer). Therefore , (x  y)/9 is not an integer Hence insufficient
Therefore, A) should be the answer



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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12 Dec 2014, 17:21
Alt Rephrase  Is (xy) a multiple of 9?
Statement 1:
x = 10a + b y = 10b  a
xy = 9a  9b x  y = 9 (ab) As such, xy is a multiple of 9.
S1 is sufficient
Statement 2: I utilized the test case approach, and I looked to prove insufficiency.
Test 1: x is 53, y is 13 (xy) in this case (40) is not a multiple of 9
Test 2: x is 42, y is 24 (xy) in this case (18) is a multiple of 9
As such, only Statement 1 is sufficient.



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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19 Dec 2014, 12:51
Solution: Whether (x  y) is a multiple of 9 Statement 1: x and y have the same two digits, but in reverse order Say x= ab i.e. x=10a + b y=ba i.e. y = 10b + a (xy) = 9(ab)  multiple of 9  Sufficient Statement 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 No relationship is defined between x and y  Insufficient Answer A
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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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19 Dec 2014, 12:54
Bunuel wrote: Tough and Tricky questions: Number Properties. 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. Kudos for a correct solution.The correct answer is A.
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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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08 Dec 2016, 02:44
kinghyts wrote: Statement 1: x and y have the same two digits, but in reverse order. Let the digits be ab and ba xy = 10a+b  10b a = 9a  9b = 9*(ab)
therefore, (xy)/9 = ab = Integer. Sufficient
Statement 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.
Let say the numbers are 42 and 24, then (xy)/9 = (42  24)/9 = 18/9 = 3 (integer). Therefore, (x  y)/9 an integer Let say the numbers are 42 and 46, (xy)/9 = (42  46)/9 = 4/9 (not an integer). Therefore , (x  y)/9 is not an integer Hence insufficient
Therefore, A) should be the answer how did you get 10a+b?



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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03 Jan 2017, 06:52
nishaadnak wrote: kinghyts wrote: Statement 1: x and y have the same two digits, but in reverse order. Let the digits be ab and ba xy = 10a+b  10b a = 9a  9b = 9*(ab)
therefore, (xy)/9 = ab = Integer. Sufficient
Statement 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.
Let say the numbers are 42 and 24, then (xy)/9 = (42  24)/9 = 18/9 = 3 (integer). Therefore, (x  y)/9 an integer Let say the numbers are 42 and 46, (xy)/9 = (42  46)/9 = 4/9 (not an integer). Therefore , (x  y)/9 is not an integer Hence insufficient
Therefore, A) should be the answer how did you get 10a+b? Hi nishaadnak, Any two digit number can be written in the form of 10a+b . for example 45 = 10*4 + 5, a =4 and b =5. Similarly three digit number can be written as 100a+ 10b + c. 451 = 100*4 + 10*5 + 1; a=4, b =5 , and c=1 Thanks.



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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03 Jan 2017, 17:42
Hello Team, I am a bit confused.... but why we are not considering a=b as a possibility in 9(ab) assumption ( As it have never been mentioned x=!y) in stmt 01. If we consider that x could be equal to y, then stmt 1 together with stmt 2 will take us to the conclusion. Regards
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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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03 Jan 2017, 19:01
ushapdatta wrote: Hello Team,
I am a bit confused.... but why we are not considering a=b as a possibility in 9(ab) assumption ( As it have never been mentioned x=!y) in stmt 01.
If we consider that x could be equal to y, then stmt 1 together with stmt 2 will take us to the conclusion.
Regards Hi ushapdatta, You are right, it's not mentioned that x != y, and we don't have to assume this either. If x = y, then also statement 1 alone is sufficient, you don't need statement 2 to get the unique answer. x = y => xy = 0 => (xy)/9 = 0/9 = 0 ( please note that 0 is an integer). Hence statement 1 alone is sufficinet. If you consider x = y, that will be a special case of the abovediscussed solution. Thanks.



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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07 Mar 2017, 10:37
kinghyts wrote: Statement 1: x and y have the same two digits, but in reverse order. Let the digits be ab and ba xy = 10a+b  10b a = 9a  9b = 9*(ab)
therefore, (xy)/9 = ab = Integer. Sufficient
Statement 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.
Let say the numbers are 42 and 24, then (xy)/9 = (42  24)/9 = 18/9 = 3 (integer). Therefore, (x  y)/9 an integer Let say the numbers are 42 and 46, (xy)/9 = (42  46)/9 = 4/9 (not an integer). Therefore , (x  y)/9 is not an integer Hence insufficient
Therefore, A) should be the answer I understand how it is 10a+b, but how is it 10b  a? Shouldn't that too be 10b + a? Thank you!



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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21 Mar 2017, 04:29
menontion wrote: kinghyts wrote: Statement 1: x and y have the same two digits, but in reverse order. Let the digits be ab and ba xy = 10a+b  10b a = 9a  9b = 9*(ab)
therefore, (xy)/9 = ab = Integer. Sufficient
Statement 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.
Let say the numbers are 42 and 24, then (xy)/9 = (42  24)/9 = 18/9 = 3 (integer). Therefore, (x  y)/9 an integer Let say the numbers are 42 and 46, (xy)/9 = (42  46)/9 = 4/9 (not an integer). Therefore , (x  y)/9 is not an integer Hence insufficient
Therefore, A) should be the answer I understand how it is 10a+b, but how is it 10b  a? Shouldn't that too be 10b + a? Thank you! Hi menontion, Please note that y is \(10b + a\) (as you have mentioned). \(x  y = 10a +b  (10b +a) = 10a + b  10b  a = 9a  9b = 9(ab)\) Hope this helps.



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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26 Feb 2019, 23:06
Bunuel VeritasKarishma chetan2uFor Statement(2): (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. If we approach analytically, Lets assume units digit of X is x...SO tens digit of X is x+2...X=10(x+2)+x=11x+20...(1) Lets assume units digit of Y is y...SO tens digit of Y is y2... Y=10(y2)+y =11y20...(2) From (1) and (2)...XY=11(xy)+40... so (XY)/9= [11(xy)+40]/9.... 40/9 gives a remainder of 4.....So if 11(xy)/9 gives a remainder of 5 or 4.....only then will the entire expression be divisible by 9... (xy) will range from 0 to 9 (+/)....so 11(xy) can have values 0,11,22,33,..99.....which will never give a remainder of 5 when divided by 9.... So this entire expression will never be divisible by 9.... Plz explain where did I go wrong.. Thanks



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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27 Feb 2019, 01:24
Debashis Roy wrote: Bunuel VeritasKarishma chetan2uFor Statement(2): (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. If we approach analytically, Lets assume units digit of X is x...SO tens digit of X is x+2...X=10(x+2)+x=11x+20...(1) Lets assume units digit of Y is y...SO tens digit of Y is y2... Y=10(y2)+y =11y20...(2) From (1) and (2)...XY=11(xy)+40... so (XY)/9= [11(xy)+40]/9.... 40/9 gives a remainder of 4.....So if 11(xy)/9 gives a remainder of 5 or 4.....only then will the entire expression be divisible by 9... (xy) will range from 0 to 9 (+/)....so 11(xy) can have values 0,11,22,33,..99.....which will never give a remainder of 5 when divided by 9.... So this entire expression will never be divisible by 9.... Plz explain where did I go wrong.. Thanks Considering x and y to be single digits as you defined them, x and (x+2) need to be single digits so the value of x can vary from 0 to 7. y and (y2) need to be single digits such that (y2) is not 0 so the value of y can vary from 3 to 9 Value of (x y) can go from 9, 8, ... 3, 2, 1, 0, 1, 2, 3, 4 When (xy) = 2, 11*(x  y) = 22 which gives remainder 5. In other words, when say x = 1 and y = 3, the two twodigit numbers X and Y are 31 and 13. (31  13)/9 = 18/9 = 2 is an integer.
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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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27 Feb 2019, 02:58
VeritasKarishmaThanks!....Gross mistake on my part to wrongly assess the range of x and y... Thanks again...you have been very helpful.. Cheers!



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Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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01 Mar 2019, 10:19
Bunuel wrote: Tough and Tricky questions: Number Properties. 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. Kudos for a correct solution.#1 x & y have same 2 digit ; reverse order x=ab and y = ba 10a+b=10b+a 9a=9b 9(ab)=0 yes sufficeint #2 x=42 , y = 40 4240/9 ; no x=42,y=24 yes in sufficient IMO A




Re: If x and y are integers between 10 and 99, inclusive, is (x  y)/9 an
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