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Sub 505 (Easy)|   Word Problems|                                       
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General Discussion
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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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Walkabout
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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David ................ Jeff ................... Paula

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

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

Answer = C
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Attached is a visual that should help.
Attachments

Screen Shot 2016-05-25 at 9.42.32 PM.png
Screen Shot 2016-05-25 at 9.42.32 PM.png [ 110.22 KiB | Viewed 53756 times ]

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Walkabout
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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Hello Moderators,

Do you think the answers choices should be made math friendly? It will be helpful when solving.
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susheelh
Hello Moderators,

Do you think the answers choices should be made math friendly? It will be helpful when solving.
_________________
Edited. Thank you.
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Walkabout
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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dave13
JeffTargetTestPrep
Walkabout
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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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:
GMAT assassins aren't born, they're made,
Rich
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dave13
JeffTargetTestPrep
Walkabout
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!
dave13
''In terms of d'' means--> You must keep the alphabet ''d'' in the answer options. You're NOT allowed to put the value of ''d'' in the answer option.
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Quote:
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) \(\frac{9}{2}×d\)
Total books has to be more than \(d\) books. But, choice A carries less than \(d\) books because of proper fraction. So, we can cross choice A.
--AA
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given that
david = d books
jeff = d/3 books
paula = 2d

total books = d+(d/3)+2d
LCM --> 10d/3
Answer C
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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David = d books
Jeff = d/3 books
Paula = 2d

total books = d+(d/3)+2d = 10d/3 (C)

Posted from my mobile device
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­No sweat. Just express everything in terms of d:

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Walkabout
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) \(\frac{9}{2}*d\)





Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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