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gmat prep equations + probability

Expert replies
Source: — Problem Solving |

by zacharyz » Tue Jun 03, 2008 10:45 am
Viper - first, can you please actually retype the questions. Makes it easier for people to find these, as well as respond when looking in this form. Pictures don't always show up.

For this problem:

The manual approach ----------------

Look at the combinations:
(x-y)(x+y) = x^2 - y^2
(x-y)(x+5y) = x^2 +4xy - 5y^2
(x-y)(5x-y) = 5x^2 -4xy + y^2
(x+y)(x+5y) = x^2 + 6xy + 5y^2
(x+y)(5x-y) = 5x^2 +4xy - y^2
(5x-y)(x+5y) = 5x^2 +24xy - 5y^2

Now you can see that x^2 - (by)^2 is only 1 of the 6 cases.

The theoretical approach ---------------

You have to recognize that the only to end up with that form is multiplying something of the form
(ax-by)(ax+by)
where the the constants a and b stick with x or y, respectively.
The only combination is the (x-y) and (x+y). If you really wanted nothing covering x, then a has to be 1 but b has to be constant as the multiplier of y across the two equations.

Therefore, the chance of choosing one of those to begin with is:
1/2 (2 options out of the four)
Then, to choose its match, the probability is:
1/3 (only 3 equations left, with one of them being the right one)

Multiply this through and you get
1/2 * 1/3 =

1/6 probability of getting that form.


Hope this helps
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by vinviper1 » Fri Jun 06, 2008 9:15 am
Will do! Thanks for the information.
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