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David has d books, which is 3 times as many as Jeff and 1/2

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David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 13 Dec 2012, 09:56
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David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?


(A) \(\frac{5}{6}*d\)

(B) \(\frac{7}{3}*d\)

(C) \(\frac{10}{3}*d\)

(D) \(\frac{7}{2}*d\)

(E) \(\frac{9}{2}*d\)
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 13 Dec 2012, 09:58
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Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d


David has d books;
Jeff has d/3 books;
Paula has 2d books;

Total = d+d/3+2d=10d/3.

Answer: C.
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 10 Jan 2014, 02:11
Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d



We are given: [(d/3) = J] and [(d/2) = P], and we also have Davids own books = d


Just add all three together and express them in common denominator of 3: (3/3)*d + (d/3) + (6/3)*d = (10/3)*d
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 11 Jun 2014, 05:58
1
It's also possible to solve with smart numbers
David has 120
Jeff 120/3 = 40
Paula 120*2 = 240
Total 400 Books --> 400/120 = 10/3
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 12 Sep 2014, 06:30
Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d




David = d ; jeff=f ; paula = p


d=3j ; d=p(1/2)

d+j+p = d+ d/3 + 2d = 10d /3
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 30 Sep 2014, 22:11
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David ................ Jeff ................... Paula

d ........................ \(\frac{d}{3}\) ....................... 2d

\(Total = 3d + \frac{d}{3} = \frac{10d}{3}\)

Answer = C
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 25 May 2016, 22:21
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Attached is a visual that should help.
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Screen Shot 2016-05-25 at 9.42.32 PM.png [ 110.22 KiB | Viewed 10213 times ]


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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 15 Jun 2016, 12:26
Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d


Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.

number of books David has = d

number of books Jeff has = j

number of books Paula has = p

We are given that David has 3 times as many books as Jeff. We can now express this in an equation.

d = 3j

d/3 = j

We are also given that David has ½ as many books as Paula. We can also express this in an equation.

d = (1/2)p

2d = p

Notice that we immediately solved for j in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:

d + d/3 + 2d

Getting a common denominator of 3, we have:

3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d

The answer is C.
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 15 Jul 2017, 07:59
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Hello Moderators,

Do you think the answers choices should be made math friendly? It will be helpful when solving.
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 15 Jul 2017, 09:50
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 25 Dec 2017, 11:37
JeffTargetTestPrep wrote:
Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d


Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.

number of books David has = d

number of books Jeff has = j

number of books Paula has = p

We are given that David has 3 times as many books as Jeff. We can now express this in an equation.

d = 3j

d/3 = j

We are also given that David has ½ as many books as Paula. We can also express this in an equation.

d = (1/2)p

2d = p

Notice that we immediately solved for j in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:

d + d/3 + 2d

Getting a common denominator of 3, we have:

3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d

The answer is C.


Hi. Can you please explain what does "j in terms of d and p in terms of d " mean ? I cant somehow digest the meaning of " in terms of " :) Thanks!
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 25 Dec 2017, 11:42
dave13 wrote:
JeffTargetTestPrep wrote:
Walkabout wrote:
David has d books, which is 3 times as many as Jeff and 1/2 as many as Paula. How many books do the three of them have altogether, in terms of d?

(A) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 9/2*d


Although we could plug in a real value for d, the problem can be just as easily solved by setting up equations. However, let’s start by defining some variables. Since we are given that David has d books, we can use variable d to represent how many books David has.

number of books David has = d

number of books Jeff has = j

number of books Paula has = p

We are given that David has 3 times as many books as Jeff. We can now express this in an equation.

d = 3j

d/3 = j

We are also given that David has ½ as many books as Paula. We can also express this in an equation.

d = (1/2)p

2d = p

Notice that we immediately solved for j in terms of d and p in terms of d. Getting j and p in terms of d is useful when setting up our final expression. We need to determine, in terms of d, the sum of the number of books for David, Jeff, and Paula. Thus, we have:

d + d/3 + 2d

Getting a common denominator of 3, we have:

3d/3 + d/3 + 6d/3 = 10d/3 = 10/3*d

The answer is C.


Hi. Can you please explain what does "j in terms of d and p in terms of d " mean ? I cant somehow digest the meaning of " in terms of " :) Thanks!


Express x in terms of y means to write x = some equation with y. For example, in x = 12y^2 - 3, x is expressed in terms of y.
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Re: David has d books, which is 3 times as many as Jeff and 1/2  [#permalink]

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New post 31 Jan 2018, 21:15
Hi All,

To start, there's a typo in your transcription - it's supposed to read "...and 1/2 as many as Paula...."

We can answer this question by TESTing VALUES or by doing algebra...

IF...
Jeff = 2 books
David = 6 books = D books
Paula = 12 books

The total number of books = 2+6+12 = 20

So we're looking for an answer that equals 20 when D=6. There's only one answer that matches...

Final Answer:

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Re: David has d books, which is 3 times as many as Jeff and 1/2 &nbs [#permalink] 31 Jan 2018, 21:15
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