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Word Problem - seems difficult

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muhtasimhassan: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Tue Jul 19, 2011 3:46 am

by muhtasimhassan » Sat Nov 08, 2014 10:25 pm

A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives a $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 got 20, but I am confused about the methodology. Can someone help me out (in details) here?

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Source: Beat The GMAT — Problem Solving | Sat Nov 08, 2014 10:25 pm

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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 » Sat Nov 08, 2014 10:57 pm

muhtasimhassan wrote:A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives a $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

Note: we don't need to consider Ann's portion in the solution.

Let K = the money remaining AFTER Ann has received her portion and then go from there.
We're told that, once we remove Bob's portion, we have $32 for Chloe.
So, we get K - Bob's $ = 32
Bob receives $4 plus one-third of what remains
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: A

Cheers,
Brent

Solve for K (K=52) and then determine Bob's portion ($20).
The answer is, indeed, A

Brent Hanneson - Creator of GMATPrepNow.com

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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Sat Nov 08, 2014 11:02 pm

Hi muhtasimhassan,

What is the source of this question? I ask because it does NOT follow any established GMAT patterns. Of the three people mentioned in the prompt, Ann (and the data that applies to Ann) is absolutely irrelevant to the question that is asked (meaning that it can be completely ignored). GMAT questions do NOT include "filler", so this question is NOT representative of what you'll see on Test Day, and you should be suspicious of whatever source this came from.

To answer the question, there are a couple of different approaches that you can use. You might find the algebra to be fairly straight-forward:

We're told that Bob receives $4 plus 1/3 of whatever money remains (re: exists). We're told that after Bob receives his money, Chloe receives the remaining $32.

Working backwards from this information, the 1/3 of the money that Bob received was based on a total that included Chloe's $32. Since Bob got 1/3 of what remained, Chloe got 2/3 of what remained:

(2/3)(X) = $32

X = $48

So, Bob received 1/3 of $48 + $4

Bob = $16 + $4 = $20

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Nov 09, 2014 4:26 am

muhtasimhassan wrote:A sum of money is to be divided among Ann, Bob and Chloe. First, Ann receives a $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

Let R = what remains after Ann takes her share and Bob takes $4.

Since, Bob receives (1/3)R, Chloe receives (2/3)R.
Since Chloe receives 32:
(2/3)R = 32
R = 48.

Since Bob receives 1/3 of what remains plus $4:
B = (1/3)*48 + 4 = 20.

The correct answer is A.

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by Brent@GMATPrepNow » Sun Nov 09, 2014 11:15 am

Hey Rich,

I created this question, based on a math contest question I quite enjoyed. The question isn't part of our GMAT prep course. That said, I don't agree with your assertion that it isn't a GMAT-style question.

Sure, we need not use the information about Ann to solve the question, but we COULD use it. In fact, if you check the original post (https://www.beatthegmat.com/tough-word-p ... 27218.html), you'll see that several people used the information about Ann to answer the question.

The fact that the question can be solved without using the "filler" information about Ann, doesn't mean it's not a GMAT-style question.

Consider question #172 from the OG13 (and the OG2015)

For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150

Here, we're given a formula for finding the sum of the first n positive integers. Do we need that formula to answer the question? No. If you check this thread https://www.beatthegmat.com/og13-q172-ps ... tml#624461, you'll find 3 different solutions, and ONLY 1 solution used the given formula.

So, the fact that we need not use the formula to answer the question does not make it less GMAT-like.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Mon Nov 10, 2014 9:20 am

Could someone please see what is flawed in this solution:
(Many thanks)

S = A + B + C
Ann: A = 4 + (S-4)/2] = 2 + S/2
Remainder1 = S - (2 +_ S/2) = S/2 - 2
Bob: B = 4 + (S/2 - 2 - 4)/3 = 2 + S/6
Remainder2 = S - (2 + S/6) = 5S/6 - 2
Chloe: C = 32

So 5S/6 - 2 = 32
5S/6 = 34
S/6 = 204/5 = 40.80
So B = 2 + 40.80 = 42.80

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Mon Nov 10, 2014 10:30 am

[email protected] wrote:GMAT questions do NOT include "filler", so this question is NOT representative of what you'll see on Test Day, and you should be suspicious of whatever source this came from.

Not true at all - plenty of questions contain superfluous information. I had a few inference questions on test day that didn't require most of the prompt, and heck, red herrings are the essence of data sufficiency, which is all about testing a student's ability to determine which statements are necessary, which are sufficient, and which are needless.

IMO, students are better off being suspicious of posters who have not been designated as experts by BTG.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Mon Nov 10, 2014 10:40 am

Mathsbuddy wrote:Could someone please see what is flawed in this solution:
(Many thanks)

S = A + B + C
Ann: A = 4 + (S-4)/2] = 2 + S/2
Remainder1 = S - (2 +_ S/2) = S/2 - 2
Bob: B = 4 + (S/2 - 2 - 4)/3 = 2 + S/6
Remainder2 = S - (2 + S/6) = 5S/6 - 2
Chloe: C = 32

So 5S/6 - 2 = 32
5S/6 = 34
S/6 = 204/5 = 40.80
So B = 2 + 40.80 = 42.80

Mathsbuddy, here's how I'd do it:

Total = x

Ann = 4 + (x - 4)/2

Remainder 1 = x - (4 + (x-4)/2) = x - 4 - (x-4)/2 = (x - 4)/2

Bob = 4 + ((x-4)/2 - 4)/3

At this point, we know that Chloe gets $32, so we can say that Ann + Bob + Chloe = x, or

4 + (x-4)/2 + 4 + ((x-4)/2 - 4)/3 + 32 = x

which gives x = 108. From there, find Ann's portion, since it's easier to compute ($56). Since Ann gets $56 and Chloe gets $32, Bob must get what's left ($20).

I like this solution better, since it doesn't require students to notice that Ann's portion is irrelevant. (That's not an easy deduction with time constraints, or even without them!)

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by Brent@GMATPrepNow » Mon Nov 10, 2014 11:04 am

Matt@VeritasPrep wrote:
[email protected] wrote:GMAT questions do NOT include "filler", so this question is NOT representative of what you'll see on Test Day, and you should be suspicious of whatever source this came from.
Not true at all - plenty of questions contain superfluous information. I had a few inference questions on test day that didn't require most of the prompt, and heck, red herrings are the essence of data sufficiency, which is all about testing a student's ability to determine which statements are necessary, which are sufficient, and which are needless.

IMO, students are better off being suspicious of posters who have not been designated as experts by BTG.

I'll add that the sole objective of many/most Integrated Reasoning questions is to wade through an A LOT of information (most of which is fluff/filler) and extract the key piece(s) of information needed to answer the question.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Mon Nov 10, 2014 11:15 am

Brent@GMATPrepNow wrote:I'll add that the sole objective of many/most Integrated Reasoning questions is to wade through an A LOT of information (most of which is fluff/filler) and extract the key piece(s) of information needed to answer the question.

Absolutely! Taking it that way, RC is also full of unhelpful fluff.

Nice question, Brent: I like that it has a plodding solution (mine) and a poetic one (yours), as any good tricky standardized test question should.

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Mathsbuddy: Master | Next Rank: 500 Posts; Posts: 447; Joined: Fri Nov 08, 2013 7:25 amFollowed by: 1 member; Thanked: 25 times

by Mathsbuddy » Tue Nov 11, 2014 5:11 am

Matt@VeritasPrep wrote:
Mathsbuddy wrote:Could someone please see what is flawed in this solution:
(Many thanks)

S = A + B + C
Ann: A = 4 + (S-4)/2] = 2 + S/2
Remainder1 = S - (2 +_ S/2) = S/2 - 2
Bob: B = 4 + (S/2 - 2 - 4)/3 = 2 + S/6
Remainder2 = S - (2 + S/6) = 5S/6 - 2
Chloe: C = 32

So 5S/6 - 2 = 32
5S/6 = 34
S/6 = 204/5 = 40.80
So B = 2 + 40.80 = 42.80
Mathsbuddy, here's how I'd do it:

Total = x

Ann = 4 + (x - 4)/2

Remainder 1 = x - (4 + (x-4)/2) = x - 4 - (x-4)/2 = (x - 4)/2

Bob = 4 + ((x-4)/2 - 4)/3

At this point, we know that Chloe gets $32, so we can say that Ann + Bob + Chloe = x, or

4 + (x-4)/2 + 4 + ((x-4)/2 - 4)/3 + 32 = x

which gives x = 108. From there, find Ann's portion, since it's easier to compute ($56). Since Ann gets $56 and Chloe gets $32, Bob must get what's left ($20).

I like this solution better, since it doesn't require students to notice that Ann's portion is irrelevant. (That's not an easy deduction with time constraints, or even without them!)

Thanks Matt,

I like what you have said; I fully understand and agree.
However, Remainder 2 must also equal Chloe's total of $32.
If I am not mistaken then
5S/6 - 2 = 32 which gives
b = 42.80
If this is not incorrect, then the question would be flawed for allowing 2 different answers.
Can you see my error, or am I right?
Thanks.

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