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

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30 Jun 2012, 21:15

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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 are composed of two basic elements. The [#permalink]

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01 Jul 2012, 01:33

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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 23987 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

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]

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15 Aug 2012, 06:00

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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]

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29 Oct 2012, 02:48

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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]

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16 Dec 2012, 04:01

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]

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16 Dec 2012, 04:07

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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]

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10 Apr 2013, 21:19

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

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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13 Oct 2014, 06:40

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

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04 Aug 2015, 18:29

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

Hi,

Thanks alot for the great solution, just one quick question, if we are getting the x (which is the amount of A in mixture) = 1/2, then how are we getting to ratio as 1:1? Is it because 1/2 stands for A/A+B so 1/1+1 and hence A and B are in ration 1:1

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

Hi,

Thanks alot for the great solution, just one quick question, if we are getting the x (which is the amount of A in mixture) = 1/2, then how are we getting to ratio as 1:1? Is it because 1/2 stands for A/A+B so 1/1+1 and hence A and B are in ration 1:1

Thanks in advance!

Yes, element 1 in mixture is element 1 in total. So element 1 is 1 part out of 2 total parts. The other part must be the other basic element.
_________________

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

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09 Oct 2016, 10:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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

Responding to a pm:

Quote:

I don't completely understand your solution. Why did you do Avg = [(5/8)*4 + (1/3)*3]/(4+3) rather than Avg = [(5/8)*4 + (1/3)*4]/(4+3)? The 1/3 also refers to the first part of the ratio doesn't it?

The formula is

\(Cavg = \frac{(C1*w1 + C2*w2)}{(w1 + w2)}\)

The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 (first element is 5/8 of Total) and 1 : 2 (first element is 1/3 of Total)

"A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3" - This implies that w1:w2 = 4:3

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

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05 May 2017, 08:30

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

Edit: Source -> NOVA.

Alloy A 5:3 or \(\frac{5}{8}\) of Element 1

Alloy B 1:2 or \(\frac{1}{3}\) of Element 1

Alloy X 4:3 or \(\frac{4}{7}\) of Alloy A and \(\frac{3}{7}\) of Alloy B

Amount of Element 1 in both Alloys (A + B) \(\frac{4}{7}\) of the mixture will be \(\frac{5}{8}\) Element 1 + the remaining \(\frac{3}{7}\) of the mixture will be \(\frac{1}{3}\) of Element 1

Math \(\frac{4}{7}\) X \(\frac{5}{8}\) + \(\frac{3}{7}\) X \(\frac{1}{3}\) = \(\frac{28}{56}\) is Element 1 = a 1:1 ratio

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

Responding to a pm:

Quote:

I am trying to understand the approach to the solution you posted to the Alloy-composition question above (I know it is quite a while ago..). I understand how you get to the equation [5/8*4+1/3*3]/3+4 since 4 parts of the new mix need to contain 5/8 of element 1 from alloy A (correct? ) and the solution 1/2 but I don't understand what the 1/2 actually represent and how you the get to a ratio of 1:1. Could you please help me and explain the reasoning?

I unfortunately also do not understand how you get to the equation 4/3 = (1/3 - x) / (x - 5/8) (ie. Why do you subtract x from 1/3 to represent alloy a? )

We have a standard weighted average formula: Avg = (C1*w1 + C2*w2)/(w1 + w2) We can rearrange the terms in this formula to give us: w1/w2 = (A2 - Aavg)/(Aavg - A1)

Both are the same formula. A is the quantity that you need to average. w1 and w2 are the weights in which this quantity is mixed.

We are given the following: The ratios of the compositions of the two basic elements in the two alloys are 5 : 3 and 1 : 2, respectively. So we can say that the concentration of the first element is 5/(5+3) = 5/8 of alloy A and the concentration of the first element in alloy B is 1/(1+2) = 1/3 of alloy B. We need to find the average concentration Aavg.

A new alloy X is formed by mixing the two alloys A and B in the ratio 4 : 3. The two allows are mixed in the ratio 4:3. So w1/w2 = 4/3

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

Edit: Source -> NOVA.

1. Let the quantity of A and B be 4x and 3x

2. (Quantity of Alloy A *Proportion of element 1 + Quantity of Alloy B * proportion of element 1) / ( Quantity of Alloy A *Proportion of element 2 + Quantity of Alloy B * proportion of element 2) = Quantity of element 1 in mixture / Quantity of element 2 in mixture