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11 Jun 2012, 08:20
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (02:01) correct
33%
(01:42)
wrong
based on 581
sessions
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If x and y are positive integers and 1 + x + y +xy = 21, what is the value of x?
(1) y > 3
(2) y = 6
(1) y > 3
(2) y = 6
11 Jun 2012, 08:40
Kudos
Bookmarks
Good question. +1.
If x and y are positive integers and 1 + x + y +xy = 21, what is the value of x?
\(1 + x + y +xy = 21\) --> \((x+1)(y+1)=21\) --> either \(x+1=3\) and \(y+1=7\) OR \(x+1=7\) and \(y+1=3\). Notice that \(x+1=1\) and \(y+1=21\) OR \(x+1=21\) and \(y+1=1\) is not possible since in this case either \(x\) or \(y\) equals zero and we are told that \(x\) and \(y\) are positive integers.
Now, from \(x+1=3\) and \(y+1=7\) --> \(x=2\) and \(y=6\) AND from \(x+1=7\) and \(y+1=3\) --> \(x=6\) and \(y=2\).
(1) y>3 --> \(y=6\), so \(x=2\). Sufficient.
(2) y=6 --> \(x=2\). Sufficient.
Answer: D.
Hope it's clear.
If x and y are positive integers and 1 + x + y +xy = 21, what is the value of x?
\(1 + x + y +xy = 21\) --> \((x+1)(y+1)=21\) --> either \(x+1=3\) and \(y+1=7\) OR \(x+1=7\) and \(y+1=3\). Notice that \(x+1=1\) and \(y+1=21\) OR \(x+1=21\) and \(y+1=1\) is not possible since in this case either \(x\) or \(y\) equals zero and we are told that \(x\) and \(y\) are positive integers.
Now, from \(x+1=3\) and \(y+1=7\) --> \(x=2\) and \(y=6\) AND from \(x+1=7\) and \(y+1=3\) --> \(x=6\) and \(y=2\).
(1) y>3 --> \(y=6\), so \(x=2\). Sufficient.
(2) y=6 --> \(x=2\). Sufficient.
Answer: D.
Hope it's clear.
General Discussion
11 Jun 2012, 08:29
Kudos
Bookmarks
Hi,
We have, 1 + x + y +xy = 21 (where x & y are natural numbers/positive integers)
or x(1+y) = 20 -y
or x = (20-y)/(1+y)
Using (1)
y > 3, i.e., y = 4, 5, 6...
when y=4, x = (20-4)/5 = 16/5 (not a natural number)
when y=5, x = (20-5)/6 = 15/6 (not a natural number)
when y=6, x = (20-6)/7 = 14/7 = 2
when y=7, x = (20-7)/8 = 13/8 (not a natural number), and for remaining values of y, there would be no positive integral value of x.
so, (x,y) = (2,6). Sufficient
Using (2)
y = 6, => x = (20-6)/(1+6) = 2
so, (x,y) = (2,6). Sufficient.
Thus, Answer is (D)
Regards,
We have, 1 + x + y +xy = 21 (where x & y are natural numbers/positive integers)
or x(1+y) = 20 -y
or x = (20-y)/(1+y)
Using (1)
y > 3, i.e., y = 4, 5, 6...
when y=4, x = (20-4)/5 = 16/5 (not a natural number)
when y=5, x = (20-5)/6 = 15/6 (not a natural number)
when y=6, x = (20-6)/7 = 14/7 = 2
when y=7, x = (20-7)/8 = 13/8 (not a natural number), and for remaining values of y, there would be no positive integral value of x.
so, (x,y) = (2,6). Sufficient
Using (2)
y = 6, => x = (20-6)/(1+6) = 2
so, (x,y) = (2,6). Sufficient.
Thus, Answer is (D)
Regards,
