Fairly Simple One

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by Ian Stewart » Sat Jul 16, 2011 7:52 am
knight247 wrote:What are the values of x and y?
(1) xy=6 and x+y=5
(2) x^3y^2=72

The OA is B but i don't agree...I believe it should be C. Lets hear opinions
Where is the question from? It's not clear what the question even means. If I can work out that x and y are 2 and 3, in some order, is that sufficient? Or do I need to know which of my values is equal to x, and which is equal to y? Since the question is open to two different legitimate interpretations, there's no way to decide on the correct answer here.

In any case, if two numbers add to 5 and multiply to 6, the numbers must be 2 and 3. We encounter this situation whenever we factor a quadratic; when you see z^2 + 5z + 6, you look for two numbers with a sum of 5 and a product of 6. If there were multiple answers to this question, factoring simply wouldn't work, so there can only be one pair of numbers with the correct sum and product here. So from Statement 1, x and y need to be 2 and 3, in some order, but we don't know which value is x and which is y. From the wording of the question, it's impossible to say if this should be considered sufficient.

Statement 2 alone most certainly is not sufficient, however. We could have x = 2 and y = 3, or x = 2 and y = -3, or x = cube root of 2 and y = 6, among many other possibilities. But combining this with Statement 1, we know x = 2 and y = 3.

So the answer is either A or C, depending on how you interpret the question.
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by knight247 » Sat Jul 16, 2011 8:46 am
Hey Ian,
This problem is from the study material of a GMAT coaching institute in Mumbai. The same one we discussed the last time. And the order of x and y is important. So ur answer is going to have to say clearly which one is x and which is y. So i'm assuming from ur response that it is a C.

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by naveen451 » Fri Aug 05, 2011 4:49 am
imo b

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by inittowinit » Fri Aug 05, 2011 8:26 am
I initially thought C as well. Given that in 1: You know that 2 and 3 are the values but you cannot specifically assign them to X or Y. In 2: Y can be 3/-3 (even exponent). I was shocked to see the answer was B.

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by ColumbiaVC » Fri Aug 05, 2011 4:59 pm
I solved this problem as follows:
1) xy=6 and x+y=5
substitute x=(5-y)in xy=6
=> (5-y)y=6
=> y^2-5y+6=0
Solving this equation for y=3 or y=2
if y=3 then x=2 and if y=2 then x=3. So, x can be 2 or 3 and y can be 3 or 2. Insufficient

2) (x^3)*(y^2)=72 72 can be written as (2^3)*(3^2)
=> (x^3)*(y^2)=(2^3)*(3^2)

Thus can say x=2 and y=3 Sufficient

B

Correct me if I did wrong taking any assumptions into consideration.

Thanks in advance!