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Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms of oil are added to the 8 kilograms mixture A, how many kilograms of mixture A must be added to make a 70% material B in the new mixture? A) 6 B) 7 C) 8 D) 9 E) 10
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Originally posted by sva2300 on 06 Aug 2013, 08:55.
Last edited by Bunuel on 05 Mar 2018, 10:03, edited 2 times in total.
Edited the question, renamed the topic.




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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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18 Oct 2015, 23:53
sva2300 wrote: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms of oil are added to the 8 kilograms mixture A, how many kilograms of mixture A must be added to make a 70% material B in the new mixture?
A)6 B)7 C)8 D)9 E)10
Can someone please explain the solution to this problem... Use weighted averages concept discussed here: http://www.veritasprep.com/blog/2011/03 ... averages/http://www.veritasprep.com/blog/2011/04 ... mixtures/Old mix has 20% oil. You add 2 kgs of 100% oil. You get new mix with 30% oil (70% material B). w1/w2 = (100  30)/(30  20) = 70/10 = 7/1 For every 7 parts of mix A, you have added 1 part of oil. Since you have added 2 kgs oil, mix A must be 7*2 = 14 kgs. You already had 8 kgs of mix A, so extra mix A added must be 14  8 = 6 kgs. Answer (A)
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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06 Aug 2013, 09:05
To the 8 kg of solution A (which has \(0.2*8\) kg of oil), 2 kg of oil are added, so the resulting percentage of oil is: \(\frac{0.2*8+2}{8+2}=\frac{3.6}{10}\), the resulting solution is 36% oil (3.6 kg) and 64% B (6.4 B). The \(x\) kg of mixture A that must be added to the new mixture, to obtain a 70% of B, is: \(\frac{6.4 + 0.8x}{10+x}=0.7\), \(6.4+0.8x=7+0.7x\) so \(x=6\) AHope everything is clear
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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02 Oct 2014, 00:23
Oil ................ Material ................. Total 1.6 .................... 6.4 ........................... 8 (Initial Mix) After adding 2 units oil 3.6 ....................... 6.4 ........................ 10 Say "x" units of Initial mix is added to the new 3.6 ...................... \(6.4 + \frac{6.4x}{8}\) ................. 10+x Given that material has to be 70% in the new mix; setting up the equation accordingly \(\frac{70}{100} (10+x) = 6.4 + \frac{6.4x}{8}\) x = 6 Answer = A
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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21 May 2016, 10:29
sva2300 wrote: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms of oil are added to the 8 kilograms mixture A, how many kilograms of mixture A must be added to make a 70% material B in the new mixture?
A)6 B)7 C)8 D)9 E)10
Can someone please explain the solution to this problem... Can be done with weighted avg method. if 2 kg more of oil is added to 20% oil of 8 kg mix,then a 10kg of mix will have 3.6kg (or 36%) of oil . A1=20(initial mix.) Avg. is what we need..ie 70% of material B means 30% of oil A2=36(36% of 10 kg mix) w1/w2=(A2Avg)/(AvgA1) (3630)/(3020) 6/10 means for every 10 kg of new mix. we need 6 kg of mix A Ans A



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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Updated on: 06 Nov 2016, 19:23
let a=kilograms of mixture a to be added to new mixture total oil in new mixture will be 1.6+2+.2a=3.6+.2a kilograms total weight of new mixture will be 8+2+a=10+a kilograms (3.6+.2a)/(10+a)=.3 a=6 kilograms
Originally posted by gracie on 21 May 2016, 11:22.
Last edited by gracie on 06 Nov 2016, 19:23, edited 1 time in total.



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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27 Oct 2016, 18:31
0.20(8+x)+2=(10+x)(0.30) 3.6+0.20x=3+0.30x 0.6=0.10x x=6



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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05 Nov 2016, 19:23
hi there, i'm trying to solve this problem via the alligation method, but I'm stuck. Help is much appreciated!
A = 8070= 10 B = 0  70 = 70 70:10
solution=7???
not sure why I'm getting this wrong.



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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22 Nov 2017, 12:28
sva2300 wrote: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms of oil are added to the 8 kilograms mixture A, how many kilograms of mixture A must be added to make a 70% material B in the new mixture?
A)6 B)7 C)8 D)9 E)10 Since mixture A is 8 kilograms by weight originally, it has 0.2(8) = 1.6 kilograms of oil and 0.8(8) = 6.4 kilograms of material B. After 2 more kilograms of oil are added, there will be 3.6 kilograms of oil but there will still be 6.4 kilograms of material B. Notice now we will have 10 kilograms of mixture A. If we let x = the number of kilograms of mixture A to be added, we are adding 0.8x kilograms of material B, thus we have: (6.4 + 0.8x)/(10 + x) = 0.7 6.4 + 0.8x = 7 + 0.7x 0.1x = 0.6 x = 6 Answer: A
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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05 Mar 2018, 09:59
Let say X Kg to be added, then, ATQ, 8 + 2 + x = (1.6 +2 +0.20X) + 0.70( 8+2+X) >10 + X = 10.6 + 0.90X > 0.10X = 0.60 > X = 6 Kg to be added



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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05 Mar 2018, 11:19
VeritasPrepKarishma : How can this question be solved by the allegation method?



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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms
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05 Mar 2018, 22:54
ishpreetanand wrote: VeritasPrepKarishma : How can this question be solved by the allegation method? I have solved it by allegation here: https://gmatclub.com/forum/mixtureais ... l#p1588660The formula depicts the allegation method.
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