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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Carcass wrote:
A sum of money was distributed among Lyle, Bob, and Chloe. First, Lyle received $ 4 plus one-half of what remained. Next, Bob received $ 4 plus one-third of what remained. Finally, Chloe received the remaining $ 32. How many dollars did Bob receive?

A. 10

B. 20

C. 26

D. 40

E. 52


Another approach:

This time, let K = the money REMAINING after Ann has received her portion AND after Bob has taken $4.
At this point, Bob receives 1/3 of K, and Chloe gets the rest.
This means that Chloe receives 2/3 of K

Since Chloe receives $32, we can say that: (2/3)K = 32
Multiply both sides by 3/2 to get: K = 48

Since Bob receives 1/3 of K plus $4, we can see that the amount Bob gets = (1/3)(48) + 4 = $20

Answer: B

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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
1
I looked at Magoosh's and Brent's solution, which makes sense and is relatively quick. When I tried it I went about it a long way. Specifically, I thought that:

Lyle + Bob + Chloe = Total

I could write Lyle and Bob in terms of Total with the given info and write in 32 for Chloe and then solve for Total. With that I could find Bob.

That was the idea. You can see in the attached calculations I got the wrong answer. I don't know why. Yes, in the future I will do the problem the quick way, as shown by others, but if someone could tell me where I went wrong in my long calculations I would appreciate it (or maybe their is a conceptual flaw in my thinking about it)
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
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T - (0.5T + 2) - 4 simplifies to be T - 0.5T - 2 - 4

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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Thank you very much Brent.
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
1
Here is the corrected version if anyone's interested. (Of course, as I said before, it's too long. Use the other method)
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Carcass wrote:
A sum of money was distributed among Lyle, Bob, and Chloe. First, Lyle received $ 4 plus one-half of what remained. Next, Bob received $ 4 plus one-third of what remained. Finally, Chloe received the remaining $ 32. How many dollars did Bob receive?

A. 10

B. 20

C. 26

D. 40

E. 52



Lets, we (L, B, and C) have total 100$.

Now L received 4 + \(\frac{1}{2}\) of 96 ; (96, because after receiving 4 from 100, remaining is 96);

After that B received 4 + \(\frac{1}{3}\) of 48 ; (48, because after receiving \(\frac{1}{2 }\)of 96 we have remaining 48 only)

Now, we have remain only 32 which is C received. (32, because \(\frac{48}{3}\) = 16; 48 - 16 = 32 remaining lastly )

So, from B, ( 4 + \(\frac{1}{3}\) 48 ), we get, 4 + 16 = 20.
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A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Quick and easy method:
Since Bob received $4 plus 1/3 of the remainder, the total between Bob and Chloe can be represented as $32=2/3(T-4).
Simplify.
96=2(T-4), 48=T-4, T=52
Then, subtract Chloe's sum from the total between Bob and Chloe.
52-32=$20
B
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
1
Lets say initial amount is x.
Lyle gets an amount of 4+(x-4)/2.
Bob gets an amount of 4+(x-4)/(2x3).
Chloe gets an amount of 2/3 * (x-4)/2 = (x-4)/3 which is in turn = 32 (given in question). This implies (x-4)/6 = 16 -- (Eqn 1)
Therefore, without even solving for x, we can get Bob's amount = 4+(x-4)/6 = 4+16 (Using Eqn 1) = 20.
Hence, 20 is the answer.
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
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