Alloy

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Tue Oct 19, 2010 9:13 am

Q: Stilium and Trilium are two metal alloys . Stilium consists of N,P,K and Trilium consists of N,P . An alloy of Stilium and Trilium is prepared in which the ratio of N,P,K is 26% ,68%, and 6% . The ratio of N,P,K in Stilium is 20% , 70% ,and 10% respectively. What is ratio of N , P in Trilium ?

a) 27% & 63 %

b) 33% & 67%

c) 35% & 65%

d) 70% & 30%

e) 75% & 25%

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Tue Oct 19, 2010 9:33 am

Aman verma wrote:Q: Stilium and Trilium are two metal alloys . Stilium consists of N,P,K and Trilium consists of N,P . An alloy of Stilium and Trilium is prepared in which the ratio of N,P,K is 26% ,68%, and 6% . The ratio of N,P,K in Stilium is 20% , 70% ,and 10% respectively. What is ratio of N , P in Trilium ?

a) 27% & 63 %

b) 33% & 67%

c) 35% & 65%

d) 70% & 30%

e) 75% & 25%

Hi,

We can't answer the question unless we know the ratio of Stilium to Trilium in the alloy - nowhere does it say that we use equal amounts of each.

What's the source?

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Tue Oct 19, 2010 9:46 am

We don't need the ratio of Stilium and Trilium in the aloy as K is not a common component of both of them. We can proceed as follows...

The ratio of N,P,K in Stilium is 20% , 70% ,and 10% respectively.
The ratio of N,P,K in the new alloy is 26% , 68% ,and 6% respectively.

Say, the ratio of N,P in Trilium is A% and B% respectively, thus A + B = 100.
Say, the ratio of Stilium and Trilium in the new aloy is x% and y% respectively, thus x + y = 100.

Therefore, for K, (0.1)x/100 = 0.06 => x = 60 => y = 40.

Therefore, for N, [(0.2)x + (A/100)y]/100 = 0.26 => 12 + (0.4)A = 26 => A = (26-12)/0.4 = 35.
Therefore, B = 100 - A = 65

The correct answer is C.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Wed Oct 20, 2010 5:03 am

Alternative Approach:

Imagine all numbers are (say) in 'grams' (g).

(1)

in 100g of Stilium, we have

N = 20 g
P = 70 g
K = 10 g

and multiplying "it all" by 6/10 (the reason is to "force" K to be 6 g, the motive will be understood shortly) we get:

in 60 g of Stilium, we have

N = 12 g
P = 42 g
K = 6 g

(2)

in 100 g of Trilium we have:

N = x g
P = (100-x) g

and multiplying "it all" by 4/10 (to "force" Trilium to contribute with 40 g, the motive will be seen shortly) we get:

in 40 g of Trilium, we have

N = (2/5 x) g
P = (2/5)(100-x) g

(3) In 100g of the alloy we have

N = 26 g
P = 68 g
K = 6 g

Please note that "forcingÂ´s" were done to guarantee I am allowed to conclude that, for instance (for N):

12+ (2/5) x = 26 therefore x = 35 (and 100-35 = 65).

P.S.: please note that the "key" was the fact that K came from Stilium only. This was also fundamental in RahulÂ´s (very nice) solution, by the way!

Regards,
Fabio.

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pzazz12: Master | Next Rank: 500 Posts; Posts: 155; Joined: Wed Sep 22, 2010 3:54 am; Thanked: 6 times

by pzazz12 » Fri Oct 22, 2010 4:04 am

fskilnik wrote:Alternative Approach:

Imagine all numbers are (say) in 'grams' (g).

(1)

in 100g of Stilium, we have

N = 20 g
P = 70 g
K = 10 g

and multiplying "it all" by 6/10 (the reason is to "force" K to be 6 g, the motive will be understood shortly) we get:

in 60 g of Stilium, we have

N = 12 g
P = 42 g
K = 6 g

(2)

in 100 g of Trilium we have:

N = x g
P = (100-x) g

and multiplying "it all" by 4/10 (to "force" Trilium to contribute with 40 g, the motive will be seen shortly) we get:

in 40 g of Trilium, we have

N = (2/5 x) g
P = (2/5)(100-x) g

(3) In 100g of the alloy we have

N = 26 g
P = 68 g
K = 6 g

Please note that "forcingÂ´s" were done to guarantee I am allowed to conclude that, for instance (for N):

12+ (2/5) x = 26 therefore x = 35 (and 100-35 = 65).

P.S.: please note that the "key" was the fact that K came from Stilium only. This was also fundamental in RahulÂ´s (very nice) solution, by the way!

Regards,
Fabio.

thank u.. but can u explain me little bit clear..........

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Fri Oct 22, 2010 4:25 am

pzazz12 wrote:thank u.. but can u explain me little bit clear..........

Well, I donÂ´t know which part(s) you found confusing/difficult, pzazz12, therefore I will say somethings about my solution and, if not enough, I will wait for your specific questions!

The beauty of the problem is the fact that you cannot deal with the proportions without care, because you must understand (as Rahul said at the begining of his post) that the alloy is MADE of two substances only, and therefore if you (say) take some grams of N of S to combine with some grams of N of T, you have to "respect" the number of grams taken from S and T in the other components of each substance, to "preserve" the proportions in which these components are presented in their corresponding substances. (I know, too much "wording" in all these, but this "philosophy" is what motivates my approach...)

Please note that IÂ´ve decided to force K to be 6g in S because I knew that T had 0g of this component, so it would be great to have 6g of K (6+0) at the end, where I really have 6g of K in the allow *PER 100g of alloy*! This last words are in "bold" because to have 6g of K (in S) at first, I had to use JUST 40g of S, but at the end I must have 6g of K in the alloy, therefore I understand that I have to "fill" with T what it takes to come out to 100g of alloy, and that means 60g of T, for sure, and thatÂ´s the second forcing.

The "premium" someone gets from doing this problem this way is not only the triviallity of the (only one) equation that solves the problem, but also because it is very easy to understand that it (the equation) must be satisfied. In other words, you are rewarded because you created a proper "structure" for the problem to be dealt with.

The problem is beautiful and I guess (no humble talk here) my solution attacks it beautifully...I hope you understand (and like) it, too.

Regards,
Fabio.

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Mon Nov 15, 2010 8:43 am

OA[spoiler]C)[/spoiler]. Thanks to all !!

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Nov 15, 2010 1:04 pm

fskilnik wrote:Alternative Approach:

Imagine all numbers are (say) in 'grams' (g).

(1)

in 100g of Stilium, we have

N = 20 g
P = 70 g
K = 10 g

and multiplying "it all" by 6/10 (the reason is to "force" K to be 6 g, the motive will be understood shortly) we get:

in 60 g of Stilium, we have

N = 12 g
P = 42 g
K = 6 g

(3) In 100g of the alloy we have

N = 26 g
P = 68 g
K = 6 g

That means we need 14 gm of N and 26 gm of P from 40 Gm of Trillium

14 gm of 40 gm is 35 % and 26 gm of P is 65 %.
I don't think much calculation is required after that...

fskilnik wrote:

in 100 g of Trilium we have:

N = x g
P = (100-x) g

and multiplying "it all" by 4/10 (to "force" Trilium to contribute with 40 g, the motive will be seen shortly) we get:

in 40 g of Trilium, we have

N = (2/5 x) g
P = (2/5)(100-x) g

Please note that "forcingÂ´s" were done to guarantee I am allowed to conclude that, for instance (for N):

12+ (2/5) x = 26 therefore x = 35 (and 100-35 = 65).

P.S.: please note that the "key" was the fact that K came from Stilium only. This was also fundamental in RahulÂ´s (very nice) solution, by the way!

Regards,
Fabio.

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by GMATGuruNY » Mon Nov 15, 2010 3:10 pm

Aman verma wrote:Q: Stilium and Trilium are two metal alloys . Stilium consists of N,P,K and Trilium consists of N,P . An alloy of Stilium and Trilium is prepared in which the ratio of N,P,K is 26% ,68%, and 6% . The ratio of N,P,K in Stilium is 20% , 70% ,and 10% respectively. What is ratio of N , P in Trilium ?

a) 27% & 63 %

b) 33% & 67%

c) 35% & 65%

d) 70% & 30%

e) 75% & 25%

I would plug in the answer choices -- which represent the ratio of N to P in Trilium -- and then use alligation.

Answer choice C: 35%N and 65%P in Trilium to be combined with 20%N and 70%P in Stilium

To combine 35%N with 20%N to achieve 26%N (the percentage of N in the alloy):
Proportion needed of 35%N = the difference of the other 2 percentages = 26-20 = 6.
Proportion needed of 20%N = the difference of the other 2 percentages = 35-26 = 9.
Thus, ratio of Trilium to Stilium = 6:9 = 2:3.

To combine 65%P with 70%P to achieve 68%P (the percentage of P in the alloy):
Proportion needed of 65%P = the difference of the other 2 percentages = 70-68 = 2.
Proportion needed of 70% = the difference of the other 2 percentages = 68-65 = 3.
Thus, the ratio of Trilium to Stilium = 2:3.

Success! The ratio is 2:3 in each case. This means that the percentages given in answer choice C will allow us to combine Trilium and Stilium in a 2:3 ratio so that the resulting alloy is 26%N and 68%P.

The correct answer is C.

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Nov 16, 2010 2:53 am

goyalsau wrote: That means we need 14 gm of N and 26 gm of P from 40 Gm of Trillium

14 gm of 40 gm is 35 % and 26 gm of P is 65 %.
I don't think much calculation is required after that...

You are absolutely right, goyalsau. Congrats (and thank you) for your careful analysis of my solution.

Regards,
Fabio.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Tue Nov 16, 2010 8:25 am

Rahul@gurome wrote:We don't need the ratio of Stilium and Trilium in the aloy as K is not a common component of both of them. We can proceed as follows...

Therefore, for K, (0.1)x/100 = 0.06 => x = 60 => y = 40.

The correct answer is C.

@Rahul Ji,

Plz explain the equation for K

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Tue Nov 16, 2010 9:32 am

gmatmachoman wrote:
Rahul@gurome wrote:We don't need the ratio of Stilium and Trilium in the alloy as K is not a common component of both of them. We can proceed as follows...

Therefore, for K, (0.1)x/100 = 0.06 => x = 60 => y = 40.

@Rahul Ji,

Plz explain the equation for K

We have assumed that the ratio of Stilium and Trilium in the new aloy is x% and y% respectively, thus x + y = 100. Now, as Trillium doesn't contain K, the amount of K in the new alloy is sole contribution of Stilium. We know there is 10% K in Stilium and 6% K in the new alloy.

As the new alloy contains x% Stillium and all of its K comes from Stillium only, amount of K in the new alloy = x% of 10% = (x/100)*(10/100) = (0.1)*x/100

But we know, the new alloy contains 6% K. Thus, amount of K in the new alloy = 6% = 0.06

Equating the above two we get the equation, (0.1)*x/100 = 0.06

Hope it is clear now.

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