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x,y are positive intergers, 80=22*x+y what is the value of

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x,y are positive intergers, 80=22*x+y what is the value of [#permalink] New post 11 Dec 2005, 06:38
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x,y are positive intergers, 80=22*x+y what is the value of y?

1) x=3
2) y<22
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 [#permalink] New post 11 Dec 2005, 09:43
I think the answer is D

1. statement 1 gives the value of x hence we can calculate the value of y. Therefore suff

2. y< 22. Both x and y are positive integers. If we pick the numbers for x we have the following options

if x=1, 22x=22. Therefore y =80-22=58
if x=2, 22x=44. Therefore y =80-44=36
if x=3, 22x=66. Therefore y =80-66=14
if x=4, 22x=88. Therefore x cannot take the value of 4 since y is a positive integer.

Given that y<22, we are left with just one option i.e. y=14
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Re: DS - Value of y [#permalink] New post 11 Dec 2005, 18:51
gamjatang wrote:
x,y are positive intergers, 80=22*x+y what is the value of y?

1) x=3
2) y<22


Answer is D

(1) 80 = 22*3 + y ==> y = 14. Sufficient
(2) y = 80 - 22*x ==> x must be less than 4
x = 1, y = 58
x = 2, y = 36
x = 3, y = 14
Re: DS - Value of y   [#permalink] 11 Dec 2005, 18:51
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