A certain alloy contains Lead, copper and tin. How many poun : GMAT Data Sufficiency (DS)
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# A certain alloy contains Lead, copper and tin. How many poun

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A certain alloy contains Lead, copper and tin. How many poun [#permalink]

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11 Feb 2011, 08:14
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A certain alloy contains Lead, copper and tin. How many pounds of tin are in 56 pounds of the alloy?

(1) By weight the alloy is 3/7 lead and 5/14 copper.
(2) By weight the alloy is 6 parts lead and 5 parts copper.
[Reveal] Spoiler: OA

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Re: DS word problem [#permalink]

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11 Feb 2011, 08:27
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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]

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11 Feb 2011, 08:36
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.

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Re: DS word problem [#permalink]

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12 Feb 2011, 08:07
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).

Regards,
Subhash
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Re: DS word problem [#permalink]

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12 Feb 2011, 08:18
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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]

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12 Feb 2011, 08:22
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]

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13 Feb 2011, 22:01
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.
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A certain alloy contains Lead, copper and tin. How many poun [#permalink]

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08 Jan 2016, 06:55
Weight of Tin - T lbs
Weight of Copper - C lbs
Weight of Lead - L lbs

T+C+L=56 lbs of the alloy

Q: Value Type

I: By weight the alloy is $$\frac{3}{7}$$ lead and $$\frac{5}{14}$$ copper

$$L=\frac{3}{7}*56$$
$$C=\frac{5}{14}*56$$
$$T=56-(\frac{3}{7}*56+\frac{5}{14}*56).$$
$$T=56-(\frac{11}{14}*56)$$
$$T=12 lbs$$

Statement I is Sufficient.

II: By weight the alloy is 6 parts lead and 5 parts copper

Let Alloy have y parts of tin.

x be the multiplier

$$6x+5x+yx=56$$
$$11x+yx=56$$
$$(11+y)x=56$$

$$y,x$$ can have multiples values namely $$(45, 1), (17, 2) and (3, 4).$$

Statement II is Not sufficient.

Thus A is the answer.
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A certain alloy contains Lead, copper and tin. How many poun   [#permalink] 08 Jan 2016, 06:55
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