Problem solving

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Problem solving

by Baten80 » Tue Feb 08, 2011 11:39 am
A has $18 more than what B & C together would have had if both B and C had 1/4 of what A has. How much does A have?

(A) 20

(B) 48

(C) 32

(D) 36

(E) 40
Source: — Problem Solving |

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by Night reader » Tue Feb 08, 2011 12:47 pm
A=B+C+18; B+C=A/4
B+C=A-18; A/4=A-18 --> A=4A-72, A=72/3, A=24
OR
B=C=A/4
B+C=A/2; A-18=A/2 --> A=36

iom D


Baten80 wrote:A has $18 more than what B & C together would have had if both B and C had 1/4 of what A has. How much does A have?

(A) 20

(B) 48

(C) 32

(D) 36

(E) 40

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by Brian@VeritasPrep » Tue Feb 08, 2011 3:57 pm
Nice work, NightReader - and I'm with you that the question as written is a little unclear as to whether B and C each would have 1/4 of A or if B and C together have 1/4 of A, so I like that you plotted out both ways.

That aside, this one brings up a pretty good point about slowing down and setting up word problems step-by-step. It looks like a lot of action at first, but really what it's saying is:

A has ---> A =

18 more than ---> 18 +

what B & C would have if each had 1/4 of A ---> 1/4 A + 1/4 A

How much does A have ---> Solve for A

So if we put that back together, as NightReader says, it's:

A = 18 + 1/4 A + 1/4 A

A = 18 + 1/2 A

1/2 A = 18

A = 36

I think the tendency on these is to try to whip up an equation too quickly - if you more cautiously analyze what each phrase tells you mathematically, you set yourself up for quicker math.
Brian Galvin
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Chief Academic Officer
Veritas Prep

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