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# A sum of money was distributed among Lyle, Bob, and Chloe. F

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A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  14 Mar 2019, 09:39
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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 [Reveal] Spoiler: OA _________________ Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests. Manager Joined: 04 Feb 2019 Posts: 204 Followers: 4 Kudos [?]: 147 [3] , given: 0 Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink] 20 Mar 2019, 15:40 3 This post received KUDOS Expert's post We don't care at all about what Lyle received. We only want to know how much Bob received, and we can work backwards from the$32 that Chloe received.

Let's say that x dollars remained after Lyle got his share. Bob, then, received $4, leaving us with x - 4. He received one-third of this, leaving two-thirds for Chloe. So Chloe received $$\frac{2}{3}(x - 4)$$. This is equal to$32, so:

$$\frac{2}{3}(x - 4) = 32$$

Multiply both sides by $$\frac{3}{2}$$:

$$x-4 = 48$$

$$x = 52$$

So there were 52 dollars left after Lyle got his share. Bob received $4, and then one-third of the remaining$48. One-third of 48 is 16, so Bob received 4 + 16 = 20 dollars in all.
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  20 Mar 2019, 16:07
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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 Notice that we need not consider Ann's portion in the solution. We can just let K = the money REMAINING after Ann has received her portion and go from there. Our equation will use the fact that, once we remove Bob's portion, we have$32 for Chloe.
So, we get K - Bob's $= 32 Bob received 4 dollars plus one-third of what remained Once Bob receives$4, the amount remaining is K-4 dollars. So, Bob gets a 1/3 of that as well.
1/3 of K-4 is (K-4)/3
So ALTOGETHER, Bob receives 4 + (K-4)/3

So, our equation becomes: K - [4 + (K-4)/3 ] = 32
Simplify to get: K - 4 - (K-4)/3 = 32
Multiply both sides by 3 to get: 3K - 12 - K + 4 = 96
Simplify: 2K - 8 = 96
Solve: K = 52

Plug this K-value into K - Bob's $= 32 to get: 52 - Bob's$ = 32
So, Bob's $= 20 Answer: B RELATED VIDEO FROM MY COURSE _________________ Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails GRE Instructor Joined: 10 Apr 2015 Posts: 2600 Followers: 96 Kudos [?]: 2800 [0], given: 41 Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink] 20 Mar 2019, 16:08 Expert's post Carcass wrote: A sum of money was distributed among Lyle, Bob, and Chloe. First, Lyle received$ 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

Cheers,
Brent
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  07 Aug 2019, 13:59
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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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daum_equation_1565215671335.png [ 165.37 KiB | Viewed 2269 times ]

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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  08 Aug 2019, 06:23
Expert's post

T - (0.5T + 2) - 4 simplifies to be T - 0.5T - 2 - 4

Cheers,
Brent
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  08 Aug 2019, 09:37
Thank you very much Brent.
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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  08 Aug 2019, 09:47
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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daum_equation_1565287061919.png [ 180.97 KiB | Viewed 2206 times ]

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Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]  16 Nov 2019, 10:46
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.
Re: A sum of money was distributed among Lyle, Bob, and Chloe. F   [#permalink] 16 Nov 2019, 10:46
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