# Thread: this one bowled me......

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## this one bowled me......

The metallurgy department has found a new alloy having only metals S1,S2 and S3.The alloy is formed by mixing three alloys A,B,C(containing only metals S1 and S2) in the ratio 1:1:4 with S3. The new alloy has 50% S3 and 50% of the remaining is S2.If A contains 20% S1 and B contains 40% S2 then what is the ratio of S1 to S2 in C ?

A)11:9

B) 4:6

C)7:33

D)9:31
E)None of the above

SPOILER:

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## Re: this one bowled me......

In new mixture S1 = S2 as weight.
We should match the wieght of S1 and S2 mixing A,B,C in 1:1:4 ratio
this table explan you what could be the mixture C 11:9

New
S1 S2 S1 S2
A 1 20 80 20 80
B 1 60 40 60 40
C 4 55 45 220 180
300 300

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## Re: this one bowled me......

let's assume we have 48 grams of the alloy
50%, or 24 grams, is s3
50% of the remainder, or 12 grams, is s2
so there must be 12 grams of s1

total: 48g
s1: 12g
s2: 12g
s3: 24g

the three alloys A, B, and C are formed by combining s1 and s2, so we have 12 + 12, or 24 g of that
A : B: C = 1 : 1 : 4
1x + 1x + 4x = 24
x = 4

A, of which there is 4g, contains 20% s1
s1 : s2 = 20% : 80% = 1 : 4
x + 4x = 4
x = 4/5
s1: 4/5 * 1 = 4/5g
s2: 4/5 * 4 = 16/5g

B, of which there is 4g, contains 40% s2
s1 : s2 = 60% : 40% = 3 : 2
3x + 2x = 4
x = 4/5
s1: 4/5 * 3 = 12/5g
s2: 4/5 * 2 = 8/5g

since alloys A + B + C contain 12g of s1 and 12g of s2:
s1 in C: 12 - 4/5 - 12/5 = 60/5 - 4/5 - 12/5 = 44/5
s2 in C: 12 - 16/5 - 8/5 = 60/5 - 16/5 - 8/5 = 36/5

so s1 : s2 in C is 44/5 : 36/5 = 44/36 = 11/9

It took a bunch of thinking and a lot of work to get to this point. Anyone have a quicker way of reaching the answer?

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## Re: this one bowled me......

please explain the last part,how do A+B+C contains 12 of s1 and 12 of s2

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## Re: this one bowled me......

Originally Posted by shanmukh
please explain the last part,how do A+B+C contains 12 of s1 and 12 of s2
We assume that we start with 48g of the alloy. We're told that 50%, or 24g, of the alloy is s3. Then we're told that 50% of the remainder, or 12g, is s2. Since we only have three components, s1 is left and 48 - 24 - 12 = 12g of s1. The alloys A, B, and C contain only s1 and s2, and there is no s1 or s2 outside of those three alloys. Therefore, A + B + C has 12g of s1, 12g of s2, and a total of 24g.

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## Re: this one bowled me......

got it guys.good xplaination gmatgnat

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## Re: this one bowled me......

Gmat gnat,

prasanth showed a simpler way to do this problem .here is the elaboration ,pls check the attachment:

Alloy (A+B+C) + S3 has A,B and C in ratio1:1:4

Lets assume there is 1200 grams of Alloy ABCS3

50% is ABC = 600 gms (100, 100 and 400 gms of A, B and C)

Based on the % table,

25% is S1 = 25/100 x 600 =300 gms

25% is S2 = 25/100 x 600 =300 gms

thanks
salila.

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## Re: this one bowled me......

Thanks Salila. Prashanth, initially I didn't understand you explanation. I like the table idea - it lays the numbers out nicely.

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## Re: this one bowled me......

Let x - S1 in C.
Since the share of S1 and S2 in the resulting alloy is the same we have:
1 * 0.2 + 1 * 0.6 + 4x = 1 * 0.8 + 1 * 0.4 + 4 * (1 - x), or 8x = 4.4, or x = 0.55,
hence x / (1 - x) = 0.55 / 0.45 = 11:9.

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