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What is the value of b + c ? 12 Edition Number 83 DS

Expert replies
by bookjive » Sat Feb 11, 2012 10:10 am
What is the value of b + c ?
(1) ab + cd + ac + bd = 6
(2) a + d = 4

(1) Th e equation ab + cd + ac + bd = 6 can be
simplified to isolate b + c by regrouping and
then factoring as follows:
(ab + bd ) + (ac + cd ) = 6
b(a + d ) + c(a + d ) = 6
(a + d) (b + c ) = 6

I understand until the last step. How do you factor to get back to the answer (a+d) (b+c) = 6
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Source: — Data Sufficiency |

by Mike@Magoosh » Sat Feb 11, 2012 12:55 pm
Hi, there. I'm happy to help with this. :)

What you ask is a fantastic question. This is really one of the subtle points of math, ultimately what separates mathematical thinkers from the rest of the population. It's about the ability to extend patterns through abstraction.

Here, the law in question is something called the Distributive Law

ab + ac = a(b = c)

Incidentally, technically it's called "distributing" when you go from right to left, and "factoring out" when you go from left to right. Those two are the yin/yang of the Distributive law.

When mathematicians write something in variable form like that, the variables can stand for numbers, or for other single variables, or for function or expressions or just about any other mathematical object. It might be more helpful to write that same equation as

(thing #1)*(thing #2) + (thing #1)*(thing #3) = (thing #1)*[(thing #2)+(thing #3)]

That's considerably more cumbersome, but for folks not as used to mathematical abstraction, I think it illuminates what a mathematician sees when he looks at "ab + ac = a(b = c)"

So, now to your question:

First you regrouped, a very good move:

(ab + bd) + (ac + cd) = 6

Then, you factored out d from the first parentheses, and factored out c from the second parentheses, another excellent move.

b(a + d) + c(a + d) = 6

Now, here's where the mathematical abstraction comes in. We have to look at (a + d) and see simply (thing #1). Therefore, what we have is:

b(thing #1) + c(thing #1) = 6

and of course, we can factor out (thing #1)

(b + c)(thing #1) = 6

(b + c)(a + d) = 6


The Distributive Law (and in fact, all laws of arithmetic and algebra) apply not only to individual numbers & individual variables, but also equally well to all algebraic expressions, no matter how complex. The leap from using individual variables to using algebraic expression in any particular mathematical pattern is one of the major thresholds of advanced mathematical thinking.

If you were embarking on getting a B.S. in mathematics, there would be nothing but this at more and more abstract levels. Fortunately, if your mathematical ambitions are simply commensurate with the GMAT, then this problem here is about as tough as it's going to get.

Does all this make sense? Here's another challenging algebra question, for practice.

https://gmat.magoosh.com/questions/981

The question at that link, once you submit your answer, should be followed by a complete video solution.

If you have any questions on what I've said, please do not hesitate to ask.

Mike :)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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by pappueshwar » Wed Feb 29, 2012 7:52 am
so is the final answer C ?
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by Mike@Magoosh » Wed Feb 29, 2012 9:56 am
pappueshwar wrote:so is the final answer C ?
Yes, combined the statements are sufficient. The answer to the DS question is C.

Mike :)
Magoosh GMAT Instructor
https://gmat.magoosh.com/
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