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Two alloys A and B are composed of two basic elements. The

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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
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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
all.jpg [ 7.83 KiB | Viewed 15241 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

Answer (A)

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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


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
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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.
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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.
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Ratio of Elements
A - 5:3 / 5x:3x
B - 1:2 / 1y:2y

Ratio of Alloys
A+B - 4:3
ie. A:B = 4:3
(5x+3x):(1y+2y) = 4:3
8x:3y = 4:3
x:y = 1:2 => y = 2x

Ratio of elements in (A+B)
1st element in A and B / 2nd Element in A and B
(5x + y)/(3x + 2y) but y = 2x
ie. 7x/7x = 1:1
Ans A
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New post 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).
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Re: Two alloys A and B are composed of two basic elements. The [#permalink]

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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
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solution.png [ 13.08 KiB | Viewed 13672 times ]


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

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New post 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

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New post 11 Apr 2013, 06:15
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

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

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New post 04 Aug 2015, 17:29
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

Thanks in advance!
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New post 04 Aug 2015, 20:39
neeraj609 wrote:
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

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

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New post 29 Sep 2015, 12:37
let E=element 1
4(5/8)+3(1/3)=7E
E=1/2
ratio between elements=1:1
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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


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

\(Cavg = \frac{(5/8)*4 + (1/3)*3}{4 + 3}\)
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Two alloys A and B are composed of two basic elements. The   [#permalink] 12 Oct 2016, 03:10
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