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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # A sum of money was distributed among Lyle, Bob, and Chloe. F  Question banks Downloads My Bookmarks Reviews Important topics
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Founder  Joined: 18 Apr 2015
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A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Expert's post 00:00

Question Stats: 40% (04:15) correct 59% (02:48) wrong based on 37 sessions
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 [?]: 209  , given: 0 Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink] 4 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. GRE Instructor Joined: 10 Apr 2015
Posts: 3535
Followers: 133

Kudos [?]: 4009  , given: 65

Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
1
KUDOS
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 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 $= 20 Answer: B RELATED VIDEO FROM MY COURSE _________________ Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Day emails GRE Instructor Joined: 10 Apr 2015 Posts: 3535 Followers: 133 Kudos [?]: 4009 , given: 65 Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink] Expert's post Carcass wrote: A sum of money was distributed among Lyle, Bob, and Chloe. First, Lyle received$ 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
_________________

Brent Hanneson – Creator of greenlighttestprep.com  Manager  Joined: 19 Nov 2018
Posts: 102
Followers: 0

Kudos [?]: 100  , given: 53

Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
1
KUDOS
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)
Attachments daum_equation_1565215671335.png [ 165.37 KiB | Viewed 4466 times ]

GRE Instructor Joined: 10 Apr 2015
Posts: 3535
Followers: 133

Kudos [?]: 4009 , given: 65

Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Expert's post T - (0.5T + 2) - 4 simplifies to be T - 0.5T - 2 - 4

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Manager  Joined: 19 Nov 2018
Posts: 102
Followers: 0

Kudos [?]: 100 , given: 53

Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Thank you very much Brent.
Manager  Joined: 19 Nov 2018
Posts: 102
Followers: 0

Kudos [?]: 100 , given: 53

Re: A sum of money was distributed among Lyle, Bob, and Chloe. F [#permalink]
Here is the corrected version if anyone's interested. (Of course, as I said before, it's too long. Use the other method)
Attachments daum_equation_1565287061919.png [ 180.97 KiB | Viewed 4399 times ]

Director  Joined: 22 Jun 2019
Posts: 521
Followers: 4

Kudos [?]: 99 , given: 161

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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