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# The weight of every type A widget is the same, the weight of

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Manager
Joined: 10 Dec 2007
Posts: 53

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The weight of every type A widget is the same, the weight of [#permalink]

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01 Jan 2008, 18:27
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

12/23
21/40
2/3
77/96
10/7

[spoiler]OA is D[/spoiler]

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CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4661 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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02 Jan 2008, 02:35
D

8A=3B; 5B=7C; r=(A+B)/(B+C) ?

B=8/3A; C=5/7B=5/7*8/3A

r=(A+B)/(B+C)=(A+8/3A)/(8/3A+5/7*8/3A)=(11/3)/(8/3*12/7)=77/96
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Kudos [?]: 4661 [0], given: 360

VP
Joined: 22 Nov 2007
Posts: 1078

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02 Jan 2008, 06:02
AVakil wrote:
The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

12/23
21/40
2/3
77/96
10/7

[spoiler]OA is D[/spoiler]

8A=3B AND 5B=3C...let's put everything in terms of B, so ... A=3/8*B and C=5/7*B, thus (A+B)/B+C=11/8*7/12=77/96. OA must be D

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Director
Joined: 14 Oct 2007
Posts: 751

Kudos [?]: 235 [0], given: 8

Location: Oxford
Schools: Oxford'10

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02 Jan 2008, 08:42
A real easy way that doesn't involve any division:

A:B = 3:8
B:C = 7:5

therefore
A:B:C
= 3*7:7*8:5*8
=21:56:40

therefore (A+B)/(B+C) = (21+56)/(56+40) = 77/96

Kudos [?]: 235 [0], given: 8

SVP
Joined: 28 Dec 2005
Posts: 1543

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02 Jan 2008, 09:20
i ended up with 77/96 as well, so answer choice D

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Manager
Joined: 10 Dec 2007
Posts: 53

Kudos [?]: 21 [0], given: 0

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02 Jan 2008, 11:04
A real easy way that doesn't involve any division:

A:B = 3:8
B:C = 7:5

therefore
A:B:C
= 3*7:7*8:5*8
=21:56:40

therefore (A+B)/(B+C) = (21+56)/(56+40) = 77/96

Holy moly. By far the best method I have seen. Does this work for all problems involving ratios?

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Director
Joined: 12 Jul 2007
Posts: 857

Kudos [?]: 334 [0], given: 0

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02 Jan 2008, 11:37
A real easy way that doesn't involve any division:

A:B = 3:8
B:C = 7:5

therefore
A:B:C
= 3*7:7*8:5*8
=21:56:40

therefore (A+B)/(B+C) = (21+56)/(56+40) = 77/96

very, very slick. I love this approach

Kudos [?]: 334 [0], given: 0

Re: Ratios   [#permalink] 02 Jan 2008, 11:37
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# The weight of every type A widget is the same, the weight of

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