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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
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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 [#permalink]
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=(A2-Avg)/(Avg-A1)
(36-30)/(30-20)
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 [#permalink]
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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 [#permalink]
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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 [#permalink]
hi there, i'm trying to solve this problem via the alligation method, but I'm stuck. Help is much appreciated!

A = 80-70= 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 [#permalink]
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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 [#permalink]
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 [#permalink]
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 [#permalink]
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ishpreetanand wrote:
VeritasPrepKarishma : How can this question be solved by the allegation method?


I have solved it by allegation here: https://gmatclub.com/forum/mixture-a-is ... l#p1588660

The formula depicts the allegation method.
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
Hi, can somebody explain me why you ad 0.8 x and not simly x?
I wrote:

6,4 + x = (10+x)*0,7

Thanks for your help! :-)
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
Initial volume of material B

0.8X8 = 6.4

Let y be volume of mixture A added.

6.4 + 0.8y/10 + y = 70/100

64 + 8y = 70 + 7y

y = 6.

Posted from my mobile device
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
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Given : Mixture A is 20% oil and 80% material B by weight.
Asked: 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?

Mixture A:
Oil = 20%
Material B = 80%

8 kg mixture A
oil = 1.6 kg
Material B = 6.4 kg

After adding 2 kg oil:
Oil = 3.6 kg
Material B = 6.4 kg

Let x kg of mixture A be added.
Oil = 3.6 + .2x
Material B = 6.4 + .8x
Total = 10 + x

(6.4 + .8x)/(10+x) = 70%
6.4 + .8x = 7 + .7x
.1x = .6
x = 6 kg

IMO A
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
PaolaGRI wrote:
Hi, can somebody explain me why you ad 0.8 x and not simly x?
I wrote:

6,4 + x = (10+x)*0,7

Thanks for your help! :-)




It is because any volume of mixture A will contain 20% of oil and 80% of material B.

Posted from my mobile device
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
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Re: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms [#permalink]
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