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

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

Answer (A)

Regards,

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.

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.

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).

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

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.

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.

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

let e=first element's proportion of alloy X
4(5/8)+3(1/3)=7e
e=1/2 of Alloy X
ratio between two elements=1:1
A

Responding to a pm:


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

\(Cavg = \frac{(5/8)*4 + (1/3)*3}{4 + 3}\)

Tremendous Question.
Here is what i did on this one -->

The ratio of two alloys is 4/3

Let A be 4x and B be 3x.

For A --> 5k+3k=4x => k=x/2

Element 1-> 2.5x
Element 2->1.5x

For B --> p+2p=3x => p=x

Element 1->x
Element 2->2x

Hence in the combination -->
Element 1 ->3.5x
Element 2->3.5x
Ratio => 1:1

SMASH THAT A.

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

Responding to a pm:


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

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

3. 4x *(5/8) + 3x*(1/3) / 4x*(3/8) + 3x*(2/3) = 1/1

Hi

I am getting the below answer (23/18). Can someone explain me where I am making the mistake.

I understood where did I make the mistake