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11 Nov 2017, 17:11
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
62% (02:20) correct
38%
(02:38)
wrong
based on 263
sessions
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Not Attempted Yet
xy and yx are a pair of two digit positive integers with reversed digits. For how many such pairs does yx - xy = y^2 - x^2?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
11 Nov 2017, 20:19
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gracie
\(yx-xy=y^2-x^2\)
\(=>10y+x-10x-y=(y-x)(y+x) \\
\\
=> 9y-9x-(y-x)(y+x)=0\)
or \((y-x)[9-(y+x)]=0 =>\) either \(y=x\) or \(y+x=9\)
Now it is given that \(xy\) & \(yx\) are a pair of two digit integers, hence \(y\) is not equal to \(x\) because in that case there will be only a single no and not a pair
so \(y+x=9\) the pairs possible are
(1,8), (2,7), (3,6), (4,5) = \(4\).........[here (4,5) & (5,4) represent the same pair]
Option C
General Discussion
11 Nov 2017, 19:53
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gracie
Hi..
=\(yx-xy=y^2-x^2...........10y+x-10x-y+(x-y)(x+y)=0.......9(x-y)+(x-y)(x+y)=0..\\
(y-x)(9-(x+y)=0\)
Either x=y or x+y=9.... But pair means x and y should be different otherwise pair would consist of just one number example 22
So numbers with sum of digits as 9 are
18 and 81
27 and 72
36 and 63
45 and 54
Since pair of numbers is asked 18 and 81 will be same as 81 and 18.
C
