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units and tens digits

by sairamGmat » Mon Jul 19, 2010 4:44 pm
. 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.

OA is A. But not sure why it is not D

(1) from 1, if K1 and K2 are two digits, then x = 10k1+k2 and y = 10K2+K1. (x-y) = 9(k1-k2) and (x-y)/9 is an integer. If k1<k2, then it is a negative integer and if k1>k2, then it is positive integer. - SUFFICIENT

(2) from 2, if K1 is tens' digit and K2 is units digit

K1 = 2+K2;
K2 = K1-2;

Now as x = 10K1+K2 and y = 10K2+K1; x-y = [10k1+(K1-2)] - [10 (k1-2) + k1] = [11k1-2] - [11k1 -20] = -2 + 20 = 18.

(x-y)/9 is 2 which is an integer. So, why not it is D?
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by Rahul@gurome » Mon Jul 19, 2010 5:02 pm
sairamGmat wrote:. 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.

OA is A. But not sure why it is not D

(1) from 1, if K1 and K2 are two digits, then x = 10k1+k2 and y = 10K2+K1. (x-y) = 9(k1-k2) and (x-y)/9 is an integer. If k1<k2, then it is a negative integer and if k1>k2, then it is positive integer. - SUFFICIENT

(2) from 2, if K1 is tens' digit and K2 is units digit

K1 = 2+K2;
K2 = K1-2;

Now as x = 10K1+K2 and y = 10K2+K1; x-y = [10k1+(K1-2)] - [10 (k1-2) + k1] = [11k1-2] - [11k1 -20] = -2 + 20 = 18.

(x-y)/9 is 2 which is an integer. So, why not it is D?
(1) Your method is correct, so (1) is SUFFICIENT.
(2) Let the ten's digit of x is 'a' and tens digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20 = 11(a - b), which will be a multiple of 11, so (2) is NOT SUFFICIENT.

The correct answer is (A).
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by sumanr84 » Tue Jul 20, 2010 2:03 am
Rahul@gurome wrote: (2) Let the ten's digit of x is 'a' and tens digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20 = 11(a - b), which will be a multiple of 11, so (2) is NOT SUFFICIENT.

The correct answer is (A).
Correcting a typo,
Let the unit digit of x is 'a' and unit digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20
x - y = 11(a-b) +40

Let try getting a YES and a NO.
1. a = 4, b =3
11 * 1 + 40 = 51 not divisible by 9 - NO

2. a = 8 , b = 1
11 * 7 + 40 = 117 divisible by 9 - YES

So, B is not sufficient
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by Rahul@gurome » Tue Jul 20, 2010 3:54 am
sumanr84 wrote: Correcting a typo,
Let the unit digit of x is 'a' and unit digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20
x - y = 11(a-b) +40

Let try getting a YES and a NO.
1. a = 4, b =3
11 * 1 + 40 = 51 not divisible by 9 - NO

2. a = 8 , b = 1
11 * 7 + 40 = 117 divisible by 9 - YES

So, B is not sufficient
Thanks for the same!
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by ankurmit » Tue Jul 20, 2010 8:10 pm
1) Your method is correct, so (1) is SUFFICIENT.
(2) Let the ten's digit of x is 'a' and tens digit of y is 'b'.
Then x = 10(a+2) + a and y = 10(b-2) + b
x - y = 11a +20 - 11b + 20 = 11(a - b), which will be a multiple of 11, so (2) is NOT SUFFICIENT.


Mr Rahul quistion stem is asking if x-y is a multiple of 9.

And from 2nd x-y is a multiple of 11.In case of data sufficiency we can find from 2nd that x-y is not multiple of 9.Hence sufficient
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by ccassel » Mon Apr 25, 2011 9:47 am
I would use plug-in method to solve this question.

(1)
x=21 y=12
(21-12)/9=9/9=1 integer
x=31 y=13
(31-13)/9=18/2=2 integer
Therefore, notice the pattern for any possible variables x & y and declare sufficient.

(2)
x = 31,42,53,64,75,86,97
y = 13,24,35,46,57,68,79
(31-24)/9=7/9 not an integer
(31-13)/9=18/9=2 integer

Therefore, you have one integer and one non-integer declaring 2 insufficient.

Therefore, answer is A.
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