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A certain alloy contains Lead , copper and tin. how many

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A certain alloy contains Lead , copper and tin. how many [#permalink] New post 11 Feb 2011, 09:14
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A certain alloy contains Lead , copper and tin. how many pounds of tin are in 56 pounds of the alloy


S1 By weight the alloy is 3/7 lead and 5/14 copper


S2 By weight the alloy is 6 parts lead and 5 parts copper
[Reveal] Spoiler: OA

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Re: DS word problem [#permalink] New post 11 Feb 2011, 09:27
Weight of Tin - T lbs
Weight of Copper - C lbs
Weight of Lead - L lbs

T+C+L=56 lbs of the alloy

Q: T=?

S1: By weight the alloy is 3/7 lead and 5/14 copper

L=3/7*56
C=5/14*56
T=56-(3/7*56+5/14*56).
Sufficient.

S2: By weight the alloy is 6 parts lead and 5 parts copper
Alloy is y parts T
x be the multiplier
6x+5x+yx=56
11x+yx=56
y,x can have many values.
Not sufficient.

Ans: "A"
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Re: DS word problem [#permalink] New post 11 Feb 2011, 09:36
rxs0005 wrote:
A certain alloy contains Lead , copper and tin. how many pounds of tin are in 56 pounds of the alloy


S1 By weight the alloy is 3/7 lead and 5/14 copper


S2 By weight the alloy is 6 parts lead and 5 parts copper


A certain alloy contains Lead , copper and tin. How many pounds of tin are in 56 pounds of the alloy?

(1) By weight the alloy is 3/7 lead and 5/14 copper --> by weight tin will be 1-(3/7+5/14)=3/14 of the alloy, thus there are 56*3/14=12 pounds of tin in the alloy. Sufficient.

(2) By weight the alloy is 6 parts lead and 5 parts copper. Clearly insufficient: for example tin could be 9 parts (45% of 56 pounds) or 89 part (89% of 56 pounds). Not sufficient.

Answer: A.
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Re: DS word problem [#permalink] New post 12 Feb 2011, 09:07
Hi Bunuel

Please explain this, especially the underlined portion :

By weight the alloy is 6 parts lead and 5 parts copper. Clearly insufficient: for example tin could be 9 parts (45% of 56 pounds) or 89 part (89% of 56 pounds).

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Re: DS word problem [#permalink] New post 12 Feb 2011, 09:18
subhashghosh wrote:
Hi Bunuel

Please explain this, especially the underlined portion :

By weight the alloy is 6 parts lead and 5 parts copper. Clearly insufficient: for example tin could be 9 parts (45% of 56 pounds) or 89 part (89% of 56 pounds).

Regards,
Subhash


"By weight the alloy is 6 parts lead and 5 parts copper" means that the ratio lead/copper=6/5, as we don't know how many parts of tin are there then we can not find weight of each. Consider examples: if lead/copper/tin=6/5/9 --> total 20 parts --> tin=9/20=45% of 56 pounds or if lead/copper/tin=6/5/89 --> total 100 parts --> tin=89/100=89% of 56 pounds.
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Re: DS word problem [#permalink] New post 12 Feb 2011, 09:22
subhashghosh wrote:
Hi Bunuel

Please explain this, especially the underlined portion :

By weight the alloy is 6 parts lead and 5 parts copper. Clearly insufficient: for example tin could be 9 parts (45% of 56 pounds) or 89 part (89% of 56 pounds).

Regards,
Subhash


If Tin is 9 Parts; then the alloy will be 9 Part TIN+ 6 Parts Lead+ 5 Parts copper = 20
Tin's presence in the alloy will be: 9/20*100 = 45% (45% of the 56 lbs alloy is TIN = 56*45/100 = 25.2 lbs will be TIN)

If Tin is 89 Parts; then the alloy will be 89 Part TIN+ 6 Parts Lead+ 5 Parts copper = 100
Tin's presence in the alloy will be: 89/100*100 = 89% (89% of the 56 lbs alloy is TIN = 56*89/100 = 49.84lbs will be TIN)

Likewise; there will be infinitely many possibilities for TIN's weight in the alloy and exact weight of Tin can't be determined.
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Re: DS word problem [#permalink] New post 13 Feb 2011, 23:01
parts usually expresses a ratio hence, we do not know how many parts make a pound, we cannot deduct how much tin is present in the alloy per pound.
Re: DS word problem   [#permalink] 13 Feb 2011, 23:01
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