moneyman wrote:If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random , what is the probability that their product will be of the form x^2-(by)^2, where b is an integer??

1/2

1/3

1/4

1/5

1/6

When a common formula applies, it's always good to start by writing down the formula.

Here, we know that:

probability = (# of desired outcomes)/(total # of possibilities)

We're choosing 2 items out of 4. So, the total # of possibilities is:

4C2 = 4!/2!2! = 24/4 = 6.

Therefore, the basic denominator is 6. Eliminate 1/4 and 1/5 (the answer could be 1/2 or 1/3 after cancelling).

Next, we need to calculate the # of products that can be written as the question demands.

If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:

(a + b) and (a - b).

The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.

Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.

Its given in the problem that b is an integer..I tried using 1 for b and got the expression right but I was stuck using values like 2,3,4..etc to see if there is any other expression..What is my mistake here??

Stuart Kovinsky wrote:

moneyman wrote:If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random , what is the probability that their product will be of the form x^2-(by)^2, where b is an integer??

1/2

1/3

1/4

1/5

1/6

When a common formula applies, it's always good to start by writing down the formula.

Here, we know that:

probability = (# of desired outcomes)/(total # of possibilities)

We're choosing 2 items out of 4. So, the total # of possibilities is:

4C2 = 4!/2!2! = 24/4 = 6.

Therefore, the basic denominator is 6. Eliminate 1/4 and 1/5 (the answer could be 1/2 or 1/3 after cancelling).

Next, we need to calculate the # of products that can be written as the question demands.

If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:

(a + b) and (a - b).

The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.

Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.

I'm also having trouble with this problem. 4 choose 2 gives the number of unordered 2-groups, right? (this may be incorrect- please let me know if I'm misremembering) If so, the answer should be (c) because both {(a+b),(a-b)} and {(a-b),(a+b)} are viable subsets...

...nevermind, i got it. been staring at gmat problems for too many hours =)

Stuart Kovinsky wrote:
Next, we need to calculate the # of products that can be written as the question demands.

If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:

(a + b) and (a - b).

The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.

Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.

Stuart:

One question: Why did you assume coeficient=1 (a+b) and (a-b)? The stems tells x^2 - by^2, and that b is an integer, different from 1, as in the case of x, so ... why not to choose (x+5y) (x-5y)??? .. I know th answer is the same, but, that will not always be the case.

Thank you,
Silvia.

ssuarezo wrote:

Stuart Kovinsky wrote:
Next, we need to calculate the # of products that can be written as the question demands.

If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:

(a + b) and (a - b).

The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.

Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.

Stuart:

One question: Why did you assume coeficient=1 (a+b) and (a-b)? The stems tells x^2 - by^2, and that b is an integer, different from 1, as in the case of x, so ... why not to choose (x+5y) (x-5y)??? .. I know th answer is the same, but, that will not always be the case.

Thank you,
Silvia.

The correct answer will be in the form of (x+y)(x-y); x and y can represent anything.

So, if there had been the expressions (56 + 7z) and (56 - 7z), those would also have given us a desired result.

Hi Stuart,

even though this discussion is 2 years old, I guess it might still be useful posting alternative solutions, as there might probably be others like me that are still reading the stuff..

So in this case, after one figured out that there is actually only 1 combination fulfilling the requirements, namely (x+y)*(x-y), one can also calculate:

1/2 * 1/3 = 1/6, because in the first draw we have a probability of 1/2 of catching either (x+y) or (x-y), and in the second draw a prob of 1/3 for catching the missing term.

We could also come up with it by calculating:
(1/4 * 1/3) * 2 = 1/6, this is the same logic as above, only that we consider the probabilities of catching either (x+y) or (x-y) in the first draw separately.

This is essentially the same as calculating 1/(4C2)

Hi @Stuart,

We have these expressions: x+y,x+5y,x-y and 5x-y

Why can't we choose either (x+y) and (x-y) OR (x+5y) and (x-y)? Why is the second option wrong to consider? Thank you!

(x+by)(x-by)

We are looking for a structure similar to x+by or x-by
Eliminate 5x-y, x+5y.
We are left with x+y, x-y.The probability of choosing x+ y or x-y from the four expressions is
2/4..Without replacement we are left with one, and there are three expressions left; So the probability of choosing the next one is 1/3...
When we multiply the two we get 2/4*1/3 = 1/6