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Two alloys A and B are composed of two basic elements. The [#permalink]
30 Jun 2012, 20:15

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65% (hard)

Question Stats:

59% (03:06) correct
41% (02:07) wrong based on 185 sessions

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

Re: Two alloys A and B [#permalink]
01 Jul 2012, 00:33

farukqmul wrote:

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

(A) 1 : 1 (B) 2 : 3 (C) 5 : 2 (D) 4 : 3 (E) 7 : 9

Hi,

Proportion of 1st element in A = 5/8 Proportion of 1st element in B = 1/3 let, the proportion of 1st element in mixture = x using allegations:

Attachment:

all.jpg [ 7.83 KiB | Viewed 7092 times ]

\(\frac {x-1/3}{5/8-x}= \frac 43\) on solving, x=0.5 thus, proportion of two elements in the mixture is 1:1

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
15 Aug 2012, 04:36

3

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Expert's post

2

This post was BOOKMARKED

farukqmul wrote:

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

(A) 1 : 1 (B) 2 : 3 (C) 5 : 2 (D) 4 : 3 (E) 7 : 9

Responding to a pm:

When you need to find the average, it is better to use the formula in its original form: Avg = (C1*w1 + C2*w2)/(w1 + w2) Avg = [(5/8)*4 + (1/3)*3]/(4+3) = 1/2 Ratio of the 2 elements in the mixture = 1:1

Obviously the formula will work in the other form too though it is best to use that when you need to find the ratio of the weights. Using the formula in the other form: w1/w2 = (A2 - Aavg)/(Aavg - A1) element 1 in A = 5/8 element 1 in B = 1/3 element 1 in mixture = x

4/3 = (1/3 - x) / (x - 5/8) 4/3(x - 5/8) = 1/3 - x (7/3)x = 7/6 x = 1/2 (You made a calculation error)

Ratio of elements in the mixture = 1:1 _________________

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
15 Aug 2012, 05:00

4

This post received KUDOS

farukqmul wrote:

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

(A) 1 : 1 (B) 2 : 3 (C) 5 : 2 (D) 4 : 3 (E) 7 : 9

Mixture A has a total of 5 + 3 = 8 parts. If in the final mixture this represents 4 parts, then the total number of parts in mixture B should be (8/4)*3 = 6. So, we should take of mixture B a quantity with 2 and 4 parts, respectively.

This will give us in the final mixture (5 + 2) : (3 + 4), which means 7:7, or 1:1.

Answer A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
29 Oct 2012, 01:48

Alloy A: Since F:S = 5:3, and 5+3=8, F/total = 5/8.

Alloy B: Since F:S = 1:2, and 1+2 = 3, F/total = 1/3.

Mixture T: A:B = 4:3. If we mix 8 units of A with 6 units of B, we get: Amount of F in alloy A = (5/8)8 = 5 units Amount of F in alloy B = (1/3)6 = 2 units. (Total F)/(Mixture T) = (5+2)/(8+6) = 7/14 = 1/2. Since 1/2 of the mixture is composed of F, F:S = 1:1.

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
16 Dec 2012, 03:01

let the two elements be 1 and 2,element 1 in alloy A =5/8 , element 1 in alloy B=1/3, element 1 in mixture=x, therefore 4/3=(1/3-x)/(5/8-x)= ½ so the answer is (A). _________________

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
16 Dec 2012, 03:07

Expert's post

Another approach to this problem is by weighted average.

In such problems , it becomes easier if we stick to one metal only. So the compostion of first metal in A is 5/8 and and in B is 1/3. So when we combine A and B, the compostion will be between 5/8 and 1/3.

Now in the diagram attached: The "distance" between the compositions of first metal in A and B is 7/24. The composition of the final mixture is given as 4:3 i.e. A's 4 parts and B's 3 parts. But when we interpret in the distance format, the sequence flips i.e. the distance between final mixture and A will be 3 units and that of between final mixture and B will be 4 units.

Now to calculate the compostion of the first metal in X, add the 4 units to 1/3. i.e. \(4/7 *\)\(7/24\) \(+1/3\) OR subtract 3 units from 5/8 i.e. \(5/8 -\) \(4/7\)\(* 7/24\). This will come out as 50%. Therefore the other metal will also constitute 50%. Hence the ratio is 1:1. +1A

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
10 Apr 2013, 20:19

VeritasPrepKarishma wrote:

farukqmul wrote:

Two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. What is the ratio of the composition of the two basic elements in alloy X ?

(A) 1 : 1 (B) 2 : 3 (C) 5 : 2 (D) 4 : 3 (E) 7 : 9

Responding to a pm:

When you need to find the average, it is better to use the formula in its original form: Avg = (C1*w1 + C2*w2)/(w1 + w2) Avg = [(5/8)*4 + (1/3)*3]/(4+3) = 1/2 Ratio of the 2 elements in the mixture = 1:1

Obviously the formula will work in the other form too though it is best to use that when you need to find the ratio of the weights. Using the formula in the other form: w1/w2 = (A2 - Aavg)/(Aavg - A1) element 1 in A = 5/8 element 1 in B = 1/3 element 1 in mixture = x

4/3 = (1/3 - x) / (x - 5/8) 4/3(x - 5/8) = 1/3 - x (7/3)x = 7/6 x = 1/2 (You made a calculation error)

Ratio of elements in the mixture = 1:1

Dear karishma

thanks a lot. that's a great solution but one point is subtle to me. would you please explain which part of the question indicates that we should go with weighted average formula? thanks again in advance. Regards

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
11 Apr 2013, 06:15

Expert's post

marzan2011 wrote:

Dear karishma

thanks a lot. that's a great solution but one point is subtle to me. would you please explain which part of the question indicates that we should go with weighted average formula? thanks again in advance. Regards

Posted from my mobile device

It's a matter of practice. You will never be given 'find the weighted average'. When you have two things (classes, groups, mixtures, solutions) and they talk about the total, average or something, that should give you a hint that weighted average might work here. _________________

Re: Two alloys A and B are composed of two basic elements. The [#permalink]
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