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If (x+y)^2 = (x+y^2)^2, what is the value of y?
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07 Nov 2019, 03:15
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If \((x+y)^2 = (x+y^2)^2\), what is the value of y? (1) \(x = y^2\) (2) \(xy^2 = 0\) Are You Up For the Challenge: 700 Level Questions
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If (x+y)^2 = (x+y^2)^2, what is the value of y?
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Updated on: 08 Nov 2019, 09:43
Bunuel wrote: If \((x+y)^2 = (x+y^2)^2\), what is the value of y? (1) \(x = y^2\) (2) \(xy^2 = 0\) Are You Up For the Challenge: 700 Level Questionsgiven \((x+y)^2 = (x+y^2)^2\) we get y=y^2 y=1 or 0 #1 x=y^2 insufficient as y could be 1 or 0 #2 xy^2=0 either of x or y is 0 insufficient from 1 &2 y has to be 0 IMO C
Originally posted by Archit3110 on 07 Nov 2019, 12:51.
Last edited by Archit3110 on 08 Nov 2019, 09:43, edited 1 time in total.



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Re: If (x+y)^2 = (x+y^2)^2, what is the value of y?
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07 Nov 2019, 13:32
Bunuel wrote: If \((x+y)^2 = (x+y^2)^2\), what is the value of y?
(1) \(x = y^2\)
(2) \(xy^2 = 0\)
Lots of algebra, so let's simplify the given info in the question before we dive in. \((x+y)^2 = (x+y^2)^2\) This can mean two different things, because negative and positive numbers become the same when they're squared. Since we don't know whether the expressions were positive or negative in the first place, we have to think about both possibilities. Here's the first possibility: \(x+y = x+y^2\) \(y = y^2\) y = 0 or 1 When is this true? Only when x + y and x + y^2 have the same sign. In other words, only when y and y^2 have the same sign, which happens when y is nonnegative. Here's the second possibility, which happens when y is negative: \((x+y) = x + y^2\) \(x  y = x + y^2\) \(2x = y^2 + y\) In this case, the value of y would depend on the value of x. So, here's where we stand as far as finding the value of y. If y is nonnegative, then y will be either 0 or 1. However, if y is negative, then we don't know much about the value of y, but we do have an equation for it. Statement 1: \(x = y^2\) This is definitely not sufficient. It doesn't tell us whether y is positive or negative, for starters. It also doesn't let us tell the difference between the y = 0 and y = 1 cases. (Note that x = y = 0 and x = y = 1 both fit this statement and the given info, so the question has at least two different answers.) Statement 2: \(xy^2 = 0\) Okay, that's interesting! If the product of two values is 0, then at least one of those values needs to be 0. So, either x = 0 (and y could be anything), or y^2 = 0 (and x could be anything.) We don't know which. Again, try some easy cases: x = y = 0 works and fits all the info we have. x = 0, y = 1 also works and fits all the info we have. Two different answers, so it's insufficient. Statements 1 and 2 together: We know two things: x = y^2, and xy^2 = 0. When will this happen? Well, for the second statement, either x has to be 0, or y^2 has to be 0. However, they're equal, so they BOTH have to be zero! So, x = y = 0. The value of y can be found, so the statements are sufficient together and the answer is C. You can get to the same place with algebra only, but it involves working with some 4th and 3rd powers, which isn't very fun...
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If (x+y)^2 = (x+y^2)^2, what is the value of y?
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07 Nov 2019, 14:19
Thank you Ccooley ,Archit3110 answer above got me wondering where I actually went wrong .Used almost Two pages of my book canceling and solving ccooley wrote: Bunuel wrote: If \((x+y)^2 = (x+y^2)^2\), what is the value of y?
(1) \(x = y^2\)
(2) \(xy^2 = 0\)
Lots of algebra, so let's simplify the given info in the question before we dive in. \((x+y)^2 = (x+y^2)^2\) This can mean two different things, because negative and positive numbers become the same when they're squared. Since we don't know whether the expressions were positive or negative in the first place, we have to think about both possibilities. Here's the first possibility: \(x+y = x+y^2\) \(y = y^2\) y = 0 or 1 When is this true? Only when x + y and x + y^2 have the same sign. In other words, only when y and y^2 have the same sign, which happens when y is nonnegative. Here's the second possibility, which happens when y is negative: \((x+y) = x + y^2\) \(x  y = x + y^2\) \(2x = y^2 + y\) In this case, the value of y would depend on the value of x. So, here's where we stand as far as finding the value of y. If y is nonnegative, then y will be either 0 or 1. However, if y is negative, then we don't know much about the value of y, but we do have an equation for it. Statement 1: \(x = y^2\) This is definitely not sufficient. It doesn't tell us whether y is positive or negative, for starters. It also doesn't let us tell the difference between the y = 0 and y = 1 cases. (Note that x = y = 0 and x = y = 1 both fit this statement and the given info, so the question has at least two different answers.) Statement 2: \(xy^2 = 0\) Okay, that's interesting! If the product of two values is 0, then at least one of those values needs to be 0. So, either x = 0 (and y could be anything), or y^2 = 0 (and x could be anything.) We don't know which. Again, try some easy cases: x = y = 0 works and fits all the info we have. x = 0, y = 1 also works and fits all the info we have. Two different answers, so it's insufficient. Statements 1 and 2 together: We know two things: x = y^2, and xy^2 = 0. When will this happen? Well, for the second statement, either x has to be 0, or y^2 has to be 0. However, they're equal, so they BOTH have to be zero! So, x = y = 0. The value of y can be found, so the statements are sufficient together and the answer is C. You can get to the same place with algebra only, but it involves working with some 4th and 3rd powers, which isn't very fun... Posted from my mobile device



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Re: If (x+y)^2 = (x+y^2)^2, what is the value of y?
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08 Nov 2019, 09:44
Staphykyeah i did mistake initially .. . edited now thanks ccooleyStaphyk wrote: Thank you Ccooley ,Archit3110 answer above got me wondering where I actually went wrong .Used almost Two pages of my book canceling and solving ccooley wrote: Bunuel wrote: If \((x+y)^2 = (x+y^2)^2\), what is the value of y?
(1) \(x = y^2\)
(2) \(xy^2 = 0\)
Lots of algebra, so let's simplify the given info in the question before we dive in. \((x+y)^2 = (x+y^2)^2\) This can mean two different things, because negative and positive numbers become the same when they're squared. Since we don't know whether the expressions were positive or negative in the first place, we have to think about both possibilities. Here's the first possibility: \(x+y = x+y^2\) \(y = y^2\) y = 0 or 1 When is this true? Only when x + y and x + y^2 have the same sign. In other words, only when y and y^2 have the same sign, which happens when y is nonnegative. Here's the second possibility, which happens when y is negative: \((x+y) = x + y^2\) \(x  y = x + y^2\) \(2x = y^2 + y\) In this case, the value of y would depend on the value of x. So, here's where we stand as far as finding the value of y. If y is nonnegative, then y will be either 0 or 1. However, if y is negative, then we don't know much about the value of y, but we do have an equation for it. Statement 1: \(x = y^2\) This is definitely not sufficient. It doesn't tell us whether y is positive or negative, for starters. It also doesn't let us tell the difference between the y = 0 and y = 1 cases. (Note that x = y = 0 and x = y = 1 both fit this statement and the given info, so the question has at least two different answers.) Statement 2: \(xy^2 = 0\) Okay, that's interesting! If the product of two values is 0, then at least one of those values needs to be 0. So, either x = 0 (and y could be anything), or y^2 = 0 (and x could be anything.) We don't know which. Again, try some easy cases: x = y = 0 works and fits all the info we have. x = 0, y = 1 also works and fits all the info we have. Two different answers, so it's insufficient. Statements 1 and 2 together: We know two things: x = y^2, and xy^2 = 0. When will this happen? Well, for the second statement, either x has to be 0, or y^2 has to be 0. However, they're equal, so they BOTH have to be zero! So, x = y = 0. The value of y can be found, so the statements are sufficient together and the answer is C. You can get to the same place with algebra only, but it involves working with some 4th and 3rd powers, which isn't very fun... Posted from my mobile device




Re: If (x+y)^2 = (x+y^2)^2, what is the value of y?
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08 Nov 2019, 09:44






