-
Target Test Prep 20% Off Flash Sale is on! Code: FLASH20
Redeem
Know the GMAT Code: Quadratics
Youve studied the math, but how well do you know how to decipher unusual math clues that the test writers hide in GMAT problems? To get a really high score on this test, youve got to Know the Code in order to make your way through the toughest problems.
If this is your first introduction to the Know the Code idea, follow the link in that last paragraph to learn more. Then, come back here and test your skills on this problem from the free GMATPrep exams.
Here you go:
*If two of the four expressions [pmath]x + y[/pmath], [pmath]x + 5y[/pmath], [pmath]x-y[/pmath], and [pmath]5x-y[/pmath] are chosen at random, what is the probability that their product will be of the form of [pmath]x^2-(by)^2[/pmath], where [pmath]b[/pmath] is an integer?(A) [pmath]1/2[/pmath]
(B) [pmath]1/3[/pmath]
(C) [pmath]1/4[/pmath]
(D) [pmath]1/5[/pmath]
(E) [pmath]1/6[/pmath]
Got an answer? Lets see what this problem is all about.
Glance at the problem. PS. Lots of text. Variables and some squares algebra? Time to Read and Jot.
All right, lets think about what weve got here.
Product means Ive got to multiply something together. Oh, I see: I multiple any two of the terms on the left, and thats going to give me something in a quadratic form (except not with an = 0 on the other side). And Im looking to match the form specified in the question.
Heres where you need to know something about quadratics that you may never have realized before.
For two of the common quadratic identities, you get 3 terms, like this:
[pmath](a+b)(a+b)=a^2+2ab+b^2[/pmath]
[pmath](a-b)(a-b)=a^2-2ab+b^2[/pmath]
But there is that one form where you get only two terms:
[pmath](a+b)(a-b)=a^2-b^2[/pmath]
How come that third one has only two terms? Where did the middle term go?
This only works when the Outer and Inner steps of FOIL cancel each other out. That third quadratic really plays out like this:
[pmath](a+b)(a-b)=a^2+ab-ab-b^2=a^2-b^2[/pmath]
What is it that allows those two middle terms to cancel out?
First, one has to be positive and one has to be negative. The positive and negative signs come from the two starting terms: one of those has to be addition and the other one has to be subtraction.
After that, one more thing has to happen. The coefficients have to match up so that the two middle terms exactly cancel out to 0.
Check out what happens in these examples:
[pmath](a + 3b)(a-b)[/pmath]
[pmath](a + b)(3a-b)[/pmath]
[pmath](a + 3b)(3a-b)[/pmath]
[pmath](3a + 3b)(a-b)[/pmath]
Go aheadtry those on your own scrap paper before you keep reading.
Ready? Heres what happens in each case:
[pmath](a+3b)(a-b)=a^2+3ab-ab-3b^2=a^2+2ab-3b^2[/pmath]
[pmath](a+b)(3a-b)=3a^2+3ab-ab-b^2=3a^2+2ab-b^2[/pmath]
[pmath](a+3b)(3a-b)=3a^2+9ab-ab-3b^2=3a^2+8ab-3b^2[/pmath]
[pmath](3a+3b)(a-b)=3a^2+3ab-3ab-b^2=3a^2-3b^2[/pmath]
Only in the fourth case did the two middle terms exactly cancel out. How come?
It turns out that you have to have the same coefficients for[pmath]a[/pmath]and [pmath]b[/pmath]in the same set of parentheses. In the fourth example, each variable has a coefficient of 3 in the first set of parentheses and each variable has a coefficient of 1 in the second set of parentheses. If theyre not the same within each set of parentheses, then the Outer and Inner steps of FOIL wont exactly cancel out.
Have you ever noticed that before? Well, you know now! :D Add that to your Know the Code flash cards.
Okay, knowing what you know now, lets actually do this problem.
To calculate probability, you have to find the number of desired outcomes divided by the total number of possible outcomes. Start with the second piece. How many ways are there to multiply together two of these four terms?
You can calculate this using combinatorics principles or you can just count it out. (I hate combinatorics. Im going to count it out!)
To save some effort, I labeled each statement with a number and then wrote out the combinations that way. There are 6 possible combinations.
Now, which of those combinations are going to get the two middle terms to cancel out completely?
First, I need to make sure that Ive got one addition term and one subtraction term.
The first and last cases dont qualify. Cross them off.
Next, note that the problem asks for the answer to be in the form [pmath]x^2-(by)^2[/pmath]. The [pmath]x^2[/pmath] term doesnt have a number or variable in front of it; in other words, the [pmath]x^2[/pmath]term has to have a coefficient of 1. Cross off any cases that wont have a coefficient of 1 for[pmath]x^2[/pmath]. The term I labeled (4), [pmath]5x-y[/pmath], wont work, since none of the other terms starts out [pmath](1/5)x[/pmath], so cross off any cases that use term 4.
Now, make sure that each set of parentheses has the same coefficients for the variables inside of those parentheses. The second term, [pmath]x + 5y[/pmath], doesnt qualify.
The only case that works is 1 & 3:
[pmath](x+y)(x-y)=x^2-y^2[/pmath]
There is 1 desired case out of a total of 6 possible cases, so the probability is 1/6.
The correct answer is (E).
Now you could just start trying to multiply out all of the different cases and stop whenever you realize a particular case is going to fail the test. Thats going to take a while though; you can shortcut that process by using the rules that will have to apply in order to reach the desired outcome.
Even if you dont know all of the rules, you can save some time (or at least cross off some answers) by knowing some of them. The most common one is the fact that the two signs will have to be opposite. The coefficient stuff is a bit more complicated, but if youre going for an exceptionally high score on this test, now you can Know the Code.
In order for the common quadratic identity[pmath](a+b)(a-b)=a^2-b^2[/pmath] to have no middle term, the coefficients within each set of parentheses have to be the same. For example: [pmath](3a + 3b)(a-b)[/pmath] works, as does [pmath](2a + 2b)(5a-5b)[/pmath].
Key Takeaways for Knowing the Code:
(1) When youre studying, dont just stop when you see the textbook answer to a problem. Push yourself to find the patterns or to articulate the underlying principles behind what youre seeing.
(2) Then turn that knowledge into Know the Code flash cards:
* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
Recent Articles
Archive
- April 2024
- March 2024
- February 2024
- January 2024
- December 2023
- November 2023
- October 2023
- September 2023
- July 2023
- June 2023
- May 2023
- April 2023
- March 2023
- February 2023
- January 2023
- December 2022
- November 2022
- October 2022
- September 2022
- August 2022
- July 2022
- June 2022
- May 2022
- April 2022
- March 2022
- February 2022
- January 2022
- December 2021
- November 2021
- October 2021
- September 2021
- August 2021
- July 2021
- June 2021
- May 2021
- April 2021
- March 2021
- February 2021
- January 2021
- December 2020
- November 2020
- October 2020
- September 2020
- August 2020
- July 2020
- June 2020
- May 2020
- April 2020
- March 2020
- February 2020
- January 2020
- December 2019
- November 2019
- October 2019
- September 2019
- August 2019
- July 2019
- June 2019
- May 2019
- April 2019
- March 2019
- February 2019
- January 2019
- December 2018
- November 2018
- October 2018
- September 2018
- August 2018
- July 2018
- June 2018
- May 2018
- April 2018
- March 2018
- February 2018
- January 2018
- December 2017
- November 2017
- October 2017
- September 2017
- August 2017
- July 2017
- June 2017
- May 2017
- April 2017
- March 2017
- February 2017
- January 2017
- December 2016
- November 2016
- October 2016
- September 2016
- August 2016
- July 2016
- June 2016
- May 2016
- April 2016
- March 2016
- February 2016
- January 2016
- December 2015
- November 2015
- October 2015
- September 2015
- August 2015
- July 2015
- June 2015
- May 2015
- April 2015
- March 2015
- February 2015
- January 2015
- December 2014
- November 2014
- October 2014
- September 2014
- August 2014
- July 2014
- June 2014
- May 2014
- April 2014
- March 2014
- February 2014
- January 2014
- December 2013
- November 2013
- October 2013
- September 2013
- August 2013
- July 2013
- June 2013
- May 2013
- April 2013
- March 2013
- February 2013
- January 2013
- December 2012
- November 2012
- October 2012
- September 2012
- August 2012
- July 2012
- June 2012
- May 2012
- April 2012
- March 2012
- February 2012
- January 2012
- December 2011
- November 2011
- October 2011
- September 2011
- August 2011
- July 2011
- June 2011
- May 2011
- April 2011
- March 2011
- February 2011
- January 2011
- December 2010
- November 2010
- October 2010
- September 2010
- August 2010
- July 2010
- June 2010
- May 2010
- April 2010
- March 2010
- February 2010
- January 2010
- December 2009
- November 2009
- October 2009
- September 2009
- August 2009