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12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in

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deepak268
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12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   21 Oct 2016, 14:30
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12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in it ( P1 not equal to P2). X kgs of each alloy is cut and then fused with the remaining part of the other alloy. If two resulting alloys have equal percentage of iron, find X?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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D
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   24 Oct 2016, 04:30
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deepak268
12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in it ( P1 not equal to P2). X kgs of each alloy is cut and then fused with the remaining part of the other alloy. If two resulting alloys have equal percentage of iron, find X?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Consider Kudos if you like the question. :)

Let P1 be a and P2 be b for simplicity.....
Given that x kg is taken out and mixed with other...
So two different alloys will have iron content as...
1) ((12-x)*a+x*b)
% iron content \(\frac{(12-x)*a+x*b}{12}\)
2) ((6-x)*b+x*a)
%iron content=\(\frac{(6-x)*b+x*a}{6}\)

Now the above two must be equal..
SO
\(\frac{(12-x)*a+x*b}{12}=\frac{(6-x)*b+x*a}{6}\)
\(\frac{12*a-x*a+x*b}{12}=\frac{6*b-x*b+x*a}{6}\)
72a-6xa+6xb=72b-12xb+12xa
72(a-b)=18x(a-b).....
72=18x.....x=72/18=4..
D


Another way slightly easier is to substitute choices And find ratio Alloy1:Alloy2....
A) 1..
A1:A2...(12-1):1and 1:(6-1)..... ratios are not same..
Similarly for B and C..
D) 4..
(12-4):4 and 4:(6-4), which is 8:4 and 4:2...
Now both are in ratio 2:1..
So this is what we are looking for..

Ans D
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   17 Feb 2019, 00:48
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deepak268
12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in it ( P1 not equal to P2). X kgs of each alloy is cut and then fused with the remaining part of the other alloy. If two resulting alloys have equal percentage of iron, find X?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Consider Kudos if you like the question. :)

Let us consider the final percentage of iron be P3.

We can form the following two equations from the information given:

For Iron content in 1st Alloy A
(12 - X) * P1 + X * P2 = 12 * P3

For Iron content in 1st Alloy B
(6 - X) * P2 + X * P1 = 6 * P3
=> (6 - X) * 2 * P2 + X * 2 * P1 = 12 * P3

Now equating both the equations we get:
(12 - X) * P1 + X * P2 = (6 - X) * 2 * P2 + X * 2 * P1

12P1 - XP1 + XP2 = 12P2 - 2XP2 + 2XP1
=> 12P1 - 12P2 = 3XP1 - 3XP2
=> 4 * (P1 - P2) = X (P1 - P2)
=> X = 4

Hence D is the correct answer.
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   17 May 2020, 05:58
By taking a weighted average and solving the question, we get:

\(\frac{(12-X)P1+XP2}{12}\) = \(\frac{2*(12-X)P2 + XP1}{6}\)

On solving and transposing, we get:

(12-X)P1 - 2XP1 = (12-2X)P2 - XP2,
which implies that P1(12-3X)=P2(12-3X) - (i)

Now the question categorically says that \(P1\neq{P2}\). So what's the way forward?

Think about this - in what possible scenario will P1(12-3X)=P2(12-3X), given that \(P1\neq{P2}\)? When (12-X) = 0.

Solving the equation, we get that X=4, hence the answer will be (D).
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   23 May 2020, 08:53
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deepak268
12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in it ( P1 not equal to P2). X kgs of each alloy is cut and then fused with the remaining part of the other alloy. If two resulting alloys have equal percentage of iron, find X?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Consider Kudos if you like the question. :)


I think the fastest way to solve this is using alligation which takes <1 min to solve this.

Picture this :

case 1: P1% and P2% are mixed in the proportion of x and 6-x to give average A.

case 2: P1% and P2% are mixed in the proportion of 12-x and x to give average A.

Since, the concentrations are same and resulting average of mixture is same in both cases; then ratio of weights in case 1 = ratio of weights in case 2

essentially the question is nothing but x/(6-x) = 12-x/x ; solving for x

or, x^2 = 72-18x + x^2

or, 18x = 72

or, x= 4
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   10 Aug 2023, 06:10
nice question. works with percentages and words. needs some nuance to catch it.
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink] 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in   28 Sep 2024, 13:43
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Re: 12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in [#permalink]
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