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b – [b - (b – a) – {b – (b + (b – a)) + 3 a}] =

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b – [b - (b – a) – {b – (b + (b – a)) + 3 a}] =

by sanju09 » Mon Oct 18, 2010 5:49 am
b - [b - (b - a) - {b - (b + (b - a)) + 3 a}] =
(A) 3 a
(B) 3 a + b
(C) 5 a
(D) 5 a - b
(E) 5 a - 3 b

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by shovan85 » Mon Oct 18, 2010 6:11 am
b - [b - (b - a) - {b - (b + (b - a)) + 3 a} ]

Solving the bold Part: b - (b + (b - a)) + 3 a = b - (2b - a) +3a = - b + 4a

Substitute in main eqn b - [b - (b - a) - {- b + 4a} ]

Solving the underlined part: [b - (b - a) - {- b + 4a} ] = [b - b + a + b -4a] = b - 3a

Substitute in main eqn b - [ b - 3a] = 3a

IMO A
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by mshameen » Mon Oct 18, 2010 6:17 am
The answer is 3a. You just have to resolve the parenthesis -
b-[b-(b-a)-{b-(b+(b-a))+3a}]
=b-[b-b+a -{b-(b+b-a)+3a}]
=b-[a-{b-(2b-a)+3a}]
=b-[a-{b-2b+a+3a}]
=b-[a-{-b+4a}]
=b-[a+b-4a]
=b-[b-3a]
=b-b+3a
=3a
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by Brian@VeritasPrep » Mon Oct 18, 2010 3:53 pm
Nice work, everyone. If I can add just one thing, it's this:

On problems like this with multiple sets of parentheses, perhaps the easiest mistake to make is to not distribute the negative sign as multiplication across parentheses!

To illustrate, let me first get rid of the innermost parentheses here:

(b + (b-a)) = 2b - a

So we have inside the { } brackets:

b - (2b -a) + 3a

It's incredibly easy here to only multiply the - before that parentheses to the 2b term, and to forget to multiply the - * -a.

If you made a mistake like this when attempting this problem you know what I'm talking about, so one thing I recommend is to put a 1 coefficient in front of the parentheses to remind yourself that it's multiplication when you have a - sign before parentheses:

b + -1(2b - a) + 3a

b -1(2b) +-1(-a) + 3a
b - 2b +a + 3a
-b + 4a


It may even sound too basic, but when you're working quickly that distribution mistake is an easy one to make, so if you think you're in danger of doing so you may want to get in the habit of using the 1 coefficient to remind yourself that you have no choice but to multiply that -1 across all terms!
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