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Updated on: 30 Jan 2022, 09:04
Originally posted by ArathiKrishnan on 30 Jan 2022, 06:01.
Last edited by Bunuel on 30 Jan 2022, 09:04, edited 1 time in total.
Last edited by Bunuel on 30 Jan 2022, 09:04, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Kudos
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E
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Difficulty:
95%
(hard)
Question Stats:
37% (02:00) correct
63%
(02:13)
wrong
based on 60
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?
(1) x = 2n, where n is an integer
(2) x > y
(1) x = 2n, where n is an integer
(2) x > y
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
Asked: If x and y are two distinct integers such that xy + x + y + 1 = 20, what is the value of x?
xy + x + y + 1 = (x+1)(y+1) = 20 = 2^2*5
(x,y) = {(3,4),(4,3),(1,9),(9,1),(0,19),(19,0),(-5,-6),(-6,-5),(-3,-11),(-11,-3),(-2,-21),(-21,-2)}
1) x =2n, where N is an integer
If x is even, then the possible solutions are when (x,y) = {(4,3),(0,19),(-6,-5),(-2,-21)}
NOT SUFFICIENT
2) x>y
(x,y) = {(4,3),(9,1),(19,0),(-5,-6),(-3,-11),(-2,-21)}
NOT SUFFICIENT
(1) + (2)
1) x =2n, where N is an integer
2) x>y
(x,y) = {(4,3),(-2,-21)}
NOT SUFFICIENT
IMO E
xy + x + y + 1 = (x+1)(y+1) = 20 = 2^2*5
(x,y) = {(3,4),(4,3),(1,9),(9,1),(0,19),(19,0),(-5,-6),(-6,-5),(-3,-11),(-11,-3),(-2,-21),(-21,-2)}
1) x =2n, where N is an integer
If x is even, then the possible solutions are when (x,y) = {(4,3),(0,19),(-6,-5),(-2,-21)}
NOT SUFFICIENT
2) x>y
(x,y) = {(4,3),(9,1),(19,0),(-5,-6),(-3,-11),(-2,-21)}
NOT SUFFICIENT
(1) + (2)
1) x =2n, where N is an integer
2) x>y
(x,y) = {(4,3),(-2,-21)}
NOT SUFFICIENT
IMO E
30 Jan 2022, 06:50
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ArathiKrishnan
xy + x + y + 1 = 20
(x+1)(y+1) = 20
So (x+1) and (y+1) are factors of 20
From 1: x = 2n, ie even; so (x+1) is odd
(x+1) is an odd factor of 20
=> x + 1 = 1 or 5 => x = 0 or 4 - not unique
From 2: x > y:
x+1 = 20 and y+1 = 1 ie x = 19 and y = 0
OR
x+1 = 10 and y+1 = 2 ie x = 9 and y = 1; etc.
Thus, x is not unique
Thus, either statement is not sufficient
Combining:
x + 1 = 1, y + 1 = 20 => x = 0 and y = 19 - doesn't satisfy that x > y; so rejected
OR
x + 1 = 5, y + 1 = 4 => x = 4 and y = 3 - satisfies x > y; so accepted
Thus, the solution is unique
Answer C
