Two varieties of steel, A and B, have a ratio of iron to chromium as 5

In 6 parts of alloy A, 5 parts are iron, and 1 part is chromium.

In 9 parts of alloy B, 7 parts are iron and 2 parts are chromium.

First, to compare the two alloys, get the same number of parts in total - we can use 18. So we have:

In 18 parts of alloy A, 15 parts are iron and 3 are chromium.
In 18 parts of alloy B, 14 parts are iron and 4 are chromium.

So combining 3 of A with 2 of B, we'd be combining 45 and 28 parts of iron, or 73 parts of iron, with 9 and 8 of chromium, or 17 of chromium, so 73 to 17 is the answer.

Finding the weighted average of one of the two :
Iron:
In solution A iron is 5/6
In solution B iron is 7/9

The ratio for Solution C is 3:2 for A and B respectively. This means that 60% of A was used and 40% of B

To find the weight of iron in the new solution: we do the weighted average:

\(\frac{5}{6}*\frac{3}{5}+\frac{7}{9}*\frac{2}{5}\)

We get \(\frac{73}{90}\)

73 is the weight of Iron and 90 is total
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Such questions test your knowledge on weighted average concept . Initially this technique may seem daunting but once you get familiar , you will find this technique very useful on GMAT.

Ok , so for this question we can focus on either Iron or chromium to get to the solution . i will show both ways .

in A , Iron = 5/6
in B , Iron= 7/9
resultant mixture has 3 portions of A for every 2 portions of B , so weighted average of Iron will lie closer to A than to B .
Specifically , w.Avg (Iron) in mixture will be 2 portion away from A and 3 portion away from B as shown in picture.

iron in 3A+2B=3(5/6)+2(7/9)=73/18
chromium in 3A+2B=3(1/6)+2(2/9)=17/18
iron:chromium ratio in C=(73/18)/(17/18)=73/17

Combine the 2 alloys in the given ratio.
We get (3(5/6)+2(7/9))/(3(1/6)+2(2/9))=>(5/2+14/9)/(1/2+4/9)=>73/17 choice D

reasoning:
we have a ratio I:C of group A and a different ratio of I:C for group B. We are given the ratio at which A and B are used to form a third group C (these are the weights). Then we are asked about the ratio of I:C of this latest group C.

what % of I do we have in A? And, how will be reflected this distribution (%) of A in group C?

(5/6)*3

what % of I do we have in I? And, how will be reflected this distribution (%) of B in group C?

(7/9)*2

Then, we apply the same reasoning for Chromium, and we end up having the following equation:


For I: (5/6)*3+(7/9)*2
For Chromium: (1/6)*3+(2/9)*2

Total Weighted Avg. ratio to get the ratio of I/C in group C:

(5/6)*3+(7/9)*2 73
------------------- = ----
(1/6)*3+(2/9)*2 17

Intuitive Representation
variables ---> Iron -- Chromium -- Total

Truth 1 -----> 5x -- x -- 6x
Truth 2 ------> 7y -- 2y -- 9y

Truth 3 ------> 3*5x --- 3*x -- - 18x
Truth 3 ------> 2*7y --- 2*2y --- 18y

what connections can we make?
6x=9y => x=3y/2

Iron = 3*5x+ 2*7y = 36.5y, chromium= 3*x+2*2y=8.5y
=> Iron:chromium = 73:17
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Using Alligation to knock out this one :

Lets take the Iron part :

[5][/6] [7][/9]

x
3 2

This simplifies to : [7-9x][/6x-5] = [9][/4]
which boils down to x = [73][/90]

But remember we started with part to the whole so the above is the expression for Iron to whole ( in C)
Therefore the desired ratio will be : [73][/90-73]= [73][/17], which is D

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