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PS Question: Division of money

Problem Solving — algebra and arithmetic
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by baalok88 » Thu Mar 29, 2018 10:12 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 could not resist but put a 'X' and then put the equations . But even after lengthy calculations, I could not reach the answer. Let me know the trick. I even tried putting the answer choices. Still I got stuck. Can you help?
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by Jay@ManhattanReview » Fri Mar 30, 2018 12:11 am
baalok88 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

I could not resist but put a 'X' and then put the equations . But even after lengthy calculations, I could not reach the answer. Let me know the trick. I even tried putting the answer choices. Still I got stuck. Can you help?
Say the sum of money = $x

Ann got 4 + 1/2*(x – 4) = 4 + x/2 – 2 = x/2 + 2

Bob got 4 + 1/3*[x –4 – (x/2 + 2 )] = x/6 + 2

Thus, Chloe got x – [(x/2 + 2) – (x/6 + 2)] = x – x/2 – 2 – x/6 –2 = x/3 – 4

=> 32 = x/3 – 4

=> x/3 = 36

=> x = $108

Bob got x/6 + 2 = 108/6 +2 = $20

The correct answer: A

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Fri Mar 30, 2018 5:39 am
baalok88 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
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by Brent@GMATPrepNow » Fri Mar 30, 2018 5:43 am
baalok88 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

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: A

Cheers,
Brent
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by Jake@ThePrincetonReview » Fri Mar 30, 2018 12:16 pm
I'd definitely be using the answers here. In general, when I'm testing out the answers, I start with the middle choice, answer C. What if Bob got $24? That's $4 plus 1/3 of what remains, so 1/3 of what remains would have to be $20 and therefore what was remaining at that point would be $60. Now, we know Chloe gets what's left and Bob just got $20 out of $60, so that would leave Chloe with $40, which is too much. C is not the answer. Notice that Ann is completely irrelevant.

Since C gave us a number for Chloe that was too much, go lower. Try Answer B, what if Bob got $22? That's $4 plus 1/3 of what remains, so 1/3 of what remains would have to be $18 and therefore what was remaining at that point would be $54. Now, we know Chloe gets what's left and Bob just got $18 out of $54, so that would leave Chloe with $36, which is too much. B is not the answer.

But we're getting closer so I'd feel pretty comfortable at this point that A is the answer. But to prove it, walk A through the problem. What if Bob got $20? That's $4 plus 1/3 of what remains, so 1/3 of what remains would have to be $16 and therefore what was remaining at that point would be $48. Now, we know Chloe gets what's left and Bob just got $16 out of $48, so that would leave Chloe with $32. Bingo.

Sure, I enjoy algebra as much as the next guy, but when there are integers in the answer choices, I usually try Plugging in the Answers first.

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by baalok88 » Sat Mar 31, 2018 4:16 am
Thank you very much for the explanation given! Is it a 700 level question or I just made a mistake of getting into many equations?

[quote="Brent@GMATPrepNow"][quote="baalok88"]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
[/quote]


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: A

Cheers,
Brent[/quote]
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by Brent@GMATPrepNow » Sat Mar 31, 2018 5:16 am
baalok88 wrote:Thank you very much for the explanation given! Is it a 700 level question or I just made a mistake of getting into many equations?
This, I believe, is a 700+ level question.

Cheers,
Brent
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by Jeff@TargetTestPrep » Thu Apr 05, 2018 4:16 pm
baalok88 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
We can let the total amount of money = n.

Thus, Ann receives:

4 + (1/2)(n - 4)

4 + (1/2)n - 2

(1/2)n + 2

Bob receives:

4 + (1/3)[n - 4 - Ann]

4 + (1/3)[n - 4 - ((1/2)n + 2)]

4 + (1/3)[n - 4 - (1/2)n - 2]

4 + (1/3)[(1/2)n - 6]

4 + (1/6)n - 2

(1/6)n + 2

Since we are given that Chloe receives the remaining $32, we can create the following equation, which combines Ann’s money, Bob’s money, and Chloe’s money:

(1/2)n + 2 + (1/6)n + 2 + 32 = n

Multiplying both sides by 6, we have:

3n + 12 + n + 12 + 192 = 6n

4n + 216 = 6n

2n = 216

n = 108

Thus, Bob receives:

(1/6)108) + 2 = 18 + 2 = $20

Alternate Solution:

Let the money Bob receives be x. Then, x - 4 represents “one third of what remains” and thus, there were 3(x - 4) dollars remaining after Bob took his $4. Since Chloe takes what’s left after Ann and Bob take their money, the money Chloe takes will be “two thirds of what remains;” in other words:

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

x - 4 = 16

x = 20

Answer: A

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