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26 Apr 2006, 03:40
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(10x/x+y) + (20y/x+y) = k, where x, y and k are integers. If x<y. k=?
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26 Apr 2006, 04:02
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(10x/x+y) + (20y/x+y) = k
=> (10x + 20y/x+y) = k
=> 10(x + 2y/x+y) = k
as it is given x, y & k are integers we should select x & y such that it divides 10,
factors of 10 are 2 and 5
so we can take x=1 & y=1 or x=1 or y=4
as it is given x<y
x=1 and y=4
by substituting these values we get k=18
=> (10x + 20y/x+y) = k
=> 10(x + 2y/x+y) = k
as it is given x, y & k are integers we should select x & y such that it divides 10,
factors of 10 are 2 and 5
so we can take x=1 & y=1 or x=1 or y=4
as it is given x<y
x=1 and y=4
by substituting these values we get k=18
26 Apr 2006, 04:07
Kudos
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(10x+20y)/(x+y) = k
10(x+2y)/(x+y) = k
Since K is an integer, x+y must be a 10 to cancel out the 10 in the numerator.
x+y = 10 and we have x<y, so possible sets:
x,y
1,9 --> K = 19
2,8 --> K = 18
3,7 --> K = 17
4,6 --> K = 16
Only 18 is in the list of answers given.
10(x+2y)/(x+y) = k
Since K is an integer, x+y must be a 10 to cancel out the 10 in the numerator.
x+y = 10 and we have x<y, so possible sets:
x,y
1,9 --> K = 19
2,8 --> K = 18
3,7 --> K = 17
4,6 --> K = 16
Only 18 is in the list of answers given.
