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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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Joined: 27 Feb 2015
Posts: 44
Concentration: General Management, Economics
GMAT 1: 630 Q42 V34
WE: Engineering (Transportation)
12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in  [#permalink]

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21 Oct 2016, 14:30
2
6
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Difficulty:

85% (hard)

Question Stats:

48% (03:08) correct 52% (02:15) wrong based on 66 sessions

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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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Joined: 02 Aug 2009
Posts: 7437
12kgs of alloy1 has P1% of iron in it. 6 kgs of alloy2 has P2% iron in  [#permalink]

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24 Oct 2016, 04:30
deepak268 wrote:
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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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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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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17 Feb 2019, 00:48
deepak268 wrote:
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] 17 Feb 2019, 00:48
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