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May 15th, 2013, 10:42 AM   #2
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Joined: Jul 2012
From: DFW Area

Posts: 635
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Math Focus: Electrical Engineering Applications
Re: Confusing...

Hi [color=#00BF00]Denis[/color],

Here is what I get but it is lunch time and I did not get to check it:

$B_{me} \$ is the amount mommy bear eats.

$B_{be} \$ is the amount baby bear eats.

$B_{de} \$ is the amount daddy bear eats.

Quote:
 Now, Baby Bear originally had 30 spoonfuls in his bowl, Mummy Bear had 60 and Daddy Bear had 90.
Quote:
 Mummy Bear had eaten half the amount that Daddy Bear would have had left in his bowl if only he had eaten from his bowl and no-one else.
$B_{me}=\frac{90-B_{de}}{2} \ \rightarrow \ 2B_{me}=90-B_{de} \$ [1]

Quote:
 Baby Bear had eaten half the amount that Mummy Bear would have had left in her bowl if only Daddy Bear had eaten from her bowl and had eaten one third of his actual total consumption.
$B_{be}=\frac{60-\frac{B_{de}}{3}}{2} \ \rightarrow \ 2B_{be}=60-\frac{B_{de}}{3} \$ [2]

Quote:
 Daddy Bear had eaten the amount that Baby Bear would have had left if only Mummy Bear had eaten from Baby Bear's bowl and had eaten one third of her actual total consumption.
$B_{de}=30-\frac{B_{me}}{3} \$ [3]

Combining [1] and [3]:

$2B_{me}=90-\left(30-\frac{B_{me}}{3}\right)=60+\frac{B_{me}}{3}$

So:

$\fbox{B_{me}=36}$

Plugging this value for $\ B_{me} \$ into [3] and solving for the unknown gives:

$\fbox{B_{de}=18}$

Finally, $\ B_{be} \$ may be solved by plugging this value for $\ B_{de} \$ into [2] and solving which gives:

$\fbox{B_{be}=27}$

 May 15th, 2013, 04:11 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,143 Thanks: 1003 Re: Confusing... Gotcha! Thanks, JKS. I was assuming this meant each only ate from the other's bowls, not from their own: "... but the fact was, they'd eaten each others'! "

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### only daddy bear

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