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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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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) 5/6*d
(B) 7/3*d
(C) 10/3*d
(D) 7/2*d
(E) 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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13 Dec 2012, 09:58
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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) 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.

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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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10 Jan 2014, 02:11
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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11 Jun 2014, 05:58
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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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12 Sep 2014, 06:30
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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30 Sep 2014, 22:11
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David ................ Jeff ................... Paula

d ........................ $$\frac{d}{3}$$ ....................... 2d

$$Total = 3d + \frac{d}{3} = \frac{10d}{3}$$

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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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04 Oct 2015, 01:47
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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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25 May 2016, 22:21
Attached is a visual that should help.
Attachments

Screen Shot 2016-05-25 at 9.42.32 PM.png [ 110.22 KiB | Viewed 888 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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15 Jun 2016, 12:26
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

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Jeffrey Miller
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Re: David has d books, which is 3 times as many as Jeff and 1/2   [#permalink] 15 Jun 2016, 12:26
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