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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.
Re: DS word problem [#permalink]
11 Feb 2011, 08:36
Expert's post
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.
Re: DS word problem [#permalink]
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 _________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
Re: DS word problem [#permalink]
12 Feb 2011, 08:18
2
This post received KUDOS
Expert's post
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. _________________
Re: DS word problem [#permalink]
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. _________________
Re: A certain alloy contains Lead, copper and tin. How many poun [#permalink]
24 Sep 2013, 17:43
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Re: A certain alloy contains Lead, copper and tin. How many poun [#permalink]
28 Nov 2014, 22:30
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