Tough word problem - dividing $ - Page 2 - The Beat The GMAT Forum

Tough word problem - dividing $

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[email protected]: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Jan 03, 2011 1:53 am

by [email protected] » Mon Jan 17, 2011 11:39 pm

Hi,
Answer is A. $20.
My approach is: A receives 4 +x/2. where let x be remaining amount.
after A receives the amount, remaining amount would be x/2.
Bob will receive now 4+1/3*(x/2)=4+x/6.
after deducting bob's amount leftover is given to C.
hence $32= x-{x/2+x/6}=x/3
thus x=$96
Therefore Bob receives=$4+$96/6=4+16=$20.

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DarylB: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Mar 05, 2011 6:44 am

by DarylB » Sun Mar 13, 2011 12:57 pm

kamu wrote:Bob took $4 + 1/3 of leftover..

that implies 2/3rds is left for Chloe..

if 2/3 of leftover = $32

1/3 of leftover = $16

$16+$4 = $20.

That's the way I did it, too. Well done, kamu, that's the quickest way.

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svsan_81: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Thu Mar 10, 2011 2:50 am; GMAT Score:540

by svsan_81 » Wed Mar 23, 2011 10:11 pm

4+(x-4)/2 = A

we have (X-4) /2 remaining from this 4 is subtracted so we have 4+1/3((x-12)/2)= B

now A+B+32 =X

ie x=108

so b= 20
--------------------------------

good question

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krishnakumar.ks: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Mon Apr 04, 2011 6:22 am; Thanked: 4 times

by krishnakumar.ks » Sun Apr 17, 2011 11:26 pm

Brent@GMATPrepNow wrote:Nice work, Logitech
My solution is below. Note that we donï¿½t need to consider Annï¿½s portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bobï¿½s portion, we have $32 for Chloe.
So, we get K ï¿½ Bobï¿½s $ = 32
The equation is K-4 ï¿½ (K-4)/3 = 32
Solve for K (K=52) and then determine Bobï¿½s portion ($20).
The answer is, indeed, A

Great approach. It really helps. Thanks Brent

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v_eshwar: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Mon Jun 22, 2009 11:49 am; Location: Columbus, Ohio, USA

by v_eshwar » Mon Apr 18, 2011 7:31 am

Thank you

Brent@GMATPrepNow wrote:Nice work, Logitech
My solution is below. Note that we donï¿½t need to consider Annï¿½s portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bobï¿½s portion, we have $32 for Chloe.
So, we get K ï¿½ Bobï¿½s $ = 32
The equation is K-4 ï¿½ (K-4)/3 = 32
Solve for K (K=52) and then determine Bobï¿½s portion ($20).
The answer is, indeed, A

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dv2020: Master | Next Rank: 500 Posts; Posts: 112; Joined: Sun Nov 08, 2009 9:39 pm; Location: Delhi; Thanked: 2 times

by dv2020 » Tue Apr 26, 2011 11:06 pm

That's brilliant indeed
I am solving algebraically somehow unable to solve it.....

kamu wrote:Bob took $4 + 1/3 of leftover..

that implies 2/3rds is left for Chloe..

if 2/3 of leftover = $32

1/3 of leftover = $16

$16+$4 = $20.

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sugmomo: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Wed Jan 06, 2010 3:36 am

by sugmomo » Wed Apr 27, 2011 1:22 am

Hi everyone..

I din't quite get this one..

kamu wrote:
Bob took $4 + 1/3 of leftover..

that implies 2/3rds is left for Chloe.. ( Shudn't it be 2/3rd - 4 is left for Chole ?? )

if 2/3 of leftover = $32

1/3 of leftover = $16

$16+$4 = $20.

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Warlock007: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Sat Jan 08, 2011 7:34 amFollowed by: 1 member; Thanked: 2 times

by Warlock007 » Mon May 02, 2011 10:39 am

Brent@GMATPrepNow wrote:Nice work, Logitech
My solution is below. Note that we donï¿½t need to consider Annï¿½s portion in the solution. We can just let K be the money remaining after Ann has received her portion and go from there.
Our equation will use the fact that, once we remove Bobï¿½s portion, we have $32 for Chloe.
So, we get K ï¿½ Bobï¿½s $ = 32
The equation is K-4 ï¿½ (K-4)/3 = 32
Solve for K (K=52) and then determine Bobï¿½s portion ($20).
The answer is, indeed, A

Thats cool explaination
couldn't even think this can be this much simple.......

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Keerthibme: Newbie | Next Rank: 10 Posts; Posts: 1; Joined: Sun Apr 03, 2011 11:28 pm; Thanked: 1 times

by Keerthibme » Mon May 09, 2011 7:58 pm

Please let me know where I went wrong
let total amount = x
Ann's share = 4+(x-4)/2
amount remaining now = x-ann's share = x-4-(x-4)/2= x/2-2=(x-4)/2
Bob's share = 4+1/3((x-4)/2)=4+(x-4)/6
chole's share = 32
Sum of Ann's, Bob's and chole's share is total amount
32+4+(x-4)/2+4+(x-4)/6 = x

This gives me x (total amount) as 112 and hence bob's share as 22.

Its quite long but still I am not getting how x-12/2 is obtained in the previous posts and same is asked by few other participants.

Regards,
Keerthi

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smackmartine: Legendary Member; Posts: 516; Joined: Fri Jul 31, 2009 3:22 pmFollowed by: 13 members; Thanked: 112 times

by smackmartine » Mon May 09, 2011 11:03 pm

I also did some kinda back solving :

As Bob had $4+1/3 of Remaining--------> Chole was left with 2/3 of Remaining = 32

=> Remaining =(3/2)*32 = 48
Now, Bob's share = 4+ 1/3 of 48 = 4+16=20 (A)

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worldpeace93: Junior | Next Rank: 30 Posts; Posts: 21; Joined: Sun Jun 05, 2011 10:39 pmFollowed by: 1 member; Thanked: 2 times

by worldpeace93 » Tue Jun 28, 2011 10:54 pm

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Warlock007: Senior | Next Rank: 100 Posts; Posts: 41; Joined: Sat Jan 08, 2011 7:34 amFollowed by: 1 member; Thanked: 2 times

by Warlock007 » Wed Jun 29, 2011 12:22 am

Brent@GMATPrepNow wrote:A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money did Bob receive?
A) 20
B) 22
C) 24
D) 26
E) 52

Here is my simplest magical solution
Bob receives $4 plus one-third of what remains.
Finally, Chloe receives the remaining $32

since bob receives $4 + 1/3 (what remains =suppose it is 'R')
then 32 should be equal to the 2/3rd of the sum 'R'
now R => 2/3(R)=32 so R = 48 and 1/3rd of R is =16
so BOB receives 4+16=20 $

A is the answer
hope it helps

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nubu: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Mon Jan 11, 2010 11:09 pm; Thanked: 2 times

by nubu » Wed Jun 29, 2011 1:13 am

kamu wrote:Bob took $4 + 1/3 of leftover..

that implies 2/3rds is left for Chloe..

if 2/3 of leftover = $32

1/3 of leftover = $16

$16+$4 = $20.

Excellent approach!
Thank you

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed Jun 29, 2011 6:27 am

Brent@GMATPrepNow wrote:A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives $4 plus one-half of what remains. Next, Bob receives $4 plus one-third of what remains. Finally, Chloe receives the remaining $32. How much money did Bob receive?
A) 20
B) 22
C) 24
D) 26
E) 52

I received a message asking me to explain how Logitech arrived at the following amount for Bob: 4 + 1/3 [(X-12)/2]

To do this, let's first simplify the amount that Logitech says Ann received.

If we let x = the original sum of money, then Ann receives 4 + (x-4)/2
To simplify this, we can rewrite this as 8/2 + (x-4)/2 which equals (8+x-4)/2 = (x+4)/2

So, Ann received (x+4)/2

Bob receives $4 plus 1/3 of what remains.
Well, what remains?
We began with x dollars and gave (x+4)/2 to Ann, which leaves x - (x+4)/2
Also, we just gave Bob $4, which leaves x - (x+4)/2 - 4
Bob gets 1/3 of x - (x+4)/2 - 4
To simplify x -(x+4)/2 - 4, we can rewrite it as 2x/2 - (x+4)/2 - 8/2
This equals (2x-x-4-8)/2 = (x-12)/2

So, Bob gets 4 plus 1/3 of (x-12)/2
In other words, Bob gets 4 + 1/3[(x-12)/2]

Of course, we've now seen a faster solution. But Logitech's method works too.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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winniethepooh: Master | Next Rank: 500 Posts; Posts: 370; Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members; Location: Arlington, MA.; Thanked: 27 times

by winniethepooh » Wed Jun 29, 2011 6:46 am

Hey Brent,
Thanks for the reply, you've been really helpful!
Does this mean that I am wrong if I say:
Bob's share = 4 + x-{[x+4/2]-4}* 1/3 ??
I am talking about the placement of brackets.

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