BTG quistion-tough one!!

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BTG quistion-tough one!!

by ankurmit » Thu Apr 14, 2011 3:28 am
A sum of money was among Lyle, Bob and Chloe. First, Lyle received 4 dollars plus one-half of what remained. Next, Bob received 4 dollars plus one-third of what remained. Finally, Chloe received the remaining $32. How many dollars did Bob receive?


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by force5 » Thu Apr 14, 2011 3:38 am
Lyle = 4+(x-4)/2
Bob= 4+(x-12)/6 because remaining was x-4/2 and after removing $4 he received 1/3 or the amount.

hence money left is 2/3 or x-12/2 = (x-12)/3
now this is equal to $32

hence x= 108
Bob's share can be calculated by putting the value of x
hence bob's share is = $20

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by Anurag@Gurome » Thu Apr 14, 2011 4:16 am
ankurmit wrote:A sum of money was among Lyle, Bob and Chloe. First, Lyle received 4 dollars plus one-half of what remained. Next, Bob received 4 dollars plus one-third of what remained. Finally, Chloe received the remaining $32. How many dollars did Bob receive?

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Let the Sum of money among Lyle, Bob and Chloe = $S
Lyle received = 4 + (1/2) × (S - 4) = 2 + S/2
Remaining = S - (2 + S/2) = S/2 - 2
Bob received = 4 + (1/3) × (S/2 - 6) = 2 + S/6
Remaining = S/2 - 2 - 2 - S/6 = S/3 - 4 = 32 implies S = 36 × 3 = $108

Therefore, Bob received = 2 + 108/6 = 2 + 18 = [spoiler]$20[/spoiler]

The correct answer is A.
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by GMATGuruNY » Thu Apr 14, 2011 4:19 am
ankurmit wrote:A sum of money was among Lyle, Bob and Chloe. First, Lyle received 4 dollars plus one-half of what remained. Next, Bob received 4 dollars plus one-third of what remained. Finally, Chloe received the remaining $32. How many dollars did Bob receive?


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Use the only known value -- that Chloe received 32 -- to determine Bob's amount.

Let C = 32, B = Bob, R = remainder after Bob took 4.
Since Bob received (1/3)R, Chloe received (2/3)R.
Thus, 32 = (2/3)R, and R = 48.
Thus, B = (1/3)R + 4 = (1/3)*48 + 4 = 20.

The correct answer is A.
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by SarahLiz » Thu Apr 14, 2011 1:14 pm
I worked backwards. You know that $32 is 2/3 of what was in the pot when Bob took money. That means 1/3 of the pot was $16. So Bob took $4 + $16 = $20.

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by ankurmit » Mon Apr 18, 2011 8:08 am
Thanks everyone!

Correct ans is indeed A
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by tomada » Mon Apr 18, 2011 9:34 am
Hi Anurag,

For the statement Bob received = 4 + (1/3) × (S/2 - 6) = 2 + S/6,
how did you determine that the 1/3 is being multiplied by S/2 - 6 rather than s/2 - 2, when s/2 - 2 is what remains after Lyle gets his money?


Anurag@Gurome wrote:
ankurmit wrote:A sum of money was among Lyle, Bob and Chloe. First, Lyle received 4 dollars plus one-half of what remained. Next, Bob received 4 dollars plus one-third of what remained. Finally, Chloe received the remaining $32. How many dollars did Bob receive?

20
26
35
40
45
Let the Sum of money among Lyle, Bob and Chloe = $S
Lyle received = 4 + (1/2) × (S - 4) = 2 + S/2
Remaining = S - (2 + S/2) = S/2 - 2
Bob received = 4 + (1/3) × (S/2 - 6) = 2 + S/6
Remaining = S/2 - 2 - 2 - S/6 = S/3 - 4 = 32 implies S = 36 × 3 = $108

Therefore, Bob received = 2 + 108/6 = 2 + 18 = [spoiler]$20[/spoiler]

The correct answer is A.
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by amisahoo » Mon Apr 18, 2011 10:45 am
The answer lies in thinking simple .

Lyle's share = x
Bob's share = y
Chloe's share = 32

Total = x + y + 32

Bob's share => y = 4 + ( remaining after Lyle took his share - 4 ) / 3
=> y = 4 + (32+y-4)/3
Solving this we get
y = 20 -> Bob's share

Amit