Math questions from any Manhattan Prep GMAT Computer Adaptive Test.
rockrock
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IF 5a = 3b = 25 - PEDMAS question

by rockrock Mon Jun 14, 2010 3:16 pm

Silly, silly question here from the MGMAT CAT exam....

5a = 3b = 25

so (5a)(3b) = (25)(25)

Next the solution states that , 15b = 625.

This seems counterintuitive to PEMDAS strategy. I thought you couldn't separate unknown variables/terms if they are in parenthesis???
preetchilled
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Re: IF 5a = 3b = 25 - PEDMAS question

by preetchilled Thu Jun 17, 2010 1:11 am

rockrock Wrote:Silly, silly question here from the MGMAT CAT exam....

5a = 3b = 25

so (5a)(3b) = (25)(25)

Next the solution states that , 15b = 625.

This seems counterintuitive to PEMDAS strategy. I thought you couldn't separate unknown variables/terms if they are in parenthesis???


Hi Rock...5a=3b=25,

u could also solve it by substituting 3b with 5a to get 5a*5a=25*25
25a^2=625, a = 5...i dont know if that is where were you looking to go.

or alternatively

u could also solve it by substituting 5a with 3b to get 3b*3b=25*25, 9b^2=625
rockrock
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Re: IF 5a = 3b = 25 - PEDMAS question

by rockrock Thu Jun 17, 2010 4:28 pm

thanks, i just have to get used to spotting when equations can be combined to shorten the time to finish the problem! takes practice!
mschwrtz
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Re: IF 5a = 3b = 25 - PEDMAS question

by mschwrtz Sun Jun 27, 2010 2:15 am

Hey rockrock, next time please post the entire question. This is very hard to follow otherwise.

In any event, when you distribute one expression in parentheses over another expression in parentheses, you need to multiply each term in the first expression by every term in the second, that's why FOILing two binomial expressions involves four pieces of multiplication, because 2*2=4. (5a)(3b) is just the product of two terms; there's no addition or subtraction inside either set of parentheses. What we have here, then, is 5*a*3*b. When you multiply a bunch of factors together, so can reorder them and regroup them in whatever way is easiest to understand. Conventionally, we put "like" terms together, constants with constants, and variables with variables: (5a)(3b)=(5*a)*(3*b)=5*a*3*b=5*3*a*b=(5*3)*(a*b)=15ab.