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22 Apr 2015, 03:35
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:31) correct
50%
(02:16)
wrong
based on 417
sessions
History
Date
Time
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Not Attempted Yet
What is the sum of the digits of two-digit positive integer M?
(1) If the digits of M are reversed, the resulting integer is 27 greater than M.
(2) The difference between the squares of the digits of M is 39.
(1) If the digits of M are reversed, the resulting integer is 27 greater than M.
(2) The difference between the squares of the digits of M is 39.
22 Apr 2015, 03:53
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Bunuel
1) let's our number be \(ab\) (a = tens and b = units), than we can write this number as \(a*10 + b*1\)
and as we know that difference \(ba-ab = 27\) we can write this statement in such way:
\((b*10+a)-(a*10+b) = 27\)
\(9b-9a=27\)
\(b-a=3\)
and we see that there is more than one combination: \(4-1\); \(5-2\) etc
Insufficient
2) Picking numbers which squares more than 39: 49, 64, 81 the only possible difference of squares that gives us result \(39\) is \(64-25=39\)
Sufficient
And answer is B
General Discussion
22 Apr 2015, 05:07
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What is the sum of the digits of two-digit positive integer M?
(1) If the digits of M are reversed, the resulting integer is 27 greater than M.
(2) The difference between the squares of the digits of M is 39.
Took me 2 minutes 50 seconds, but what the heck.
1) [x] [y] + 27 = [y] [x] where x and y represent digits of a two number digits
I thought x was 0 and y was 3, 3+27 = 30, y = 3 and x = 0. But after working on the second statement, I realised 5 and 8 was also a possibility.
I.
2) I just wrote down the squares of all integers from 1 - 9
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
The only possibility is 64 - 25 = 39. S.
B.
(1) If the digits of M are reversed, the resulting integer is 27 greater than M.
(2) The difference between the squares of the digits of M is 39.
Took me 2 minutes 50 seconds, but what the heck.
1) [x] [y] + 27 = [y] [x] where x and y represent digits of a two number digits
I thought x was 0 and y was 3, 3+27 = 30, y = 3 and x = 0. But after working on the second statement, I realised 5 and 8 was also a possibility.
I.
2) I just wrote down the squares of all integers from 1 - 9
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49
8^2 = 64
9^2 = 81
The only possibility is 64 - 25 = 39. S.
B.
