Mixture A, which is 20% oil, contains 20 gallons of water

Question

Mixture A, which is 20% oil, contains 20 gallons of water and mixture B is 80% oil. If the combination of mixture A and B has a 4.5 : 3 ratio of oil to water, how many gallons of water does mixture B have?

A. 5
B. 20
C. 30
D. 40
E. 50

Source:gmatPrepster

I am not sure what the OA is.

A. 10
B. 20
C. 30
D. 40
E. 50

Source:gmatPrepster

OPTIONS ARE WRONG.

SOURH7WK · Answer

Combined mixture has oil: water = 4.5:3 ie. Oil = (4.5/7.5)*100% = 60%
Now by Alligation Rule:

A  . .........  B 
20% . . . 80%
 
. . 60% . .

(80-60) . (60-20)
=20 : =40
Qty A: Qty B = 1:2 ( A & B are mixed in 1: 2 ration)
So if A contains 20gallons water ie total Qty of A= 25 gallons (Since water is 80%)
So as per above ration Qty B mixed will be 50 gallons.
Qty of water in mixture B = 20% of 50 = 10 gallons.

Where i am doing mistake???

Indianblondienycgal · Answer

SOURH7WK,
I got the answer using Water (3/7.5)*100% = 40% and then use the alligator rule ,You should get the answer.

Bunuel · Answer

Answer D would be correct if we were asked about gallons of oil (not water) in mixture B.

Since 80% of water in mixture A amounts for 20 gallons, then mixture A is 25 gallons: 5 gallons of oil (20%) and 20 gallons of water (80%).

We are also told that mixture B is is 80% oil and mixture of A and B has the ratio of oil to water 4.5:3=3:2 so  \frac{oil}{water}=\frac{5+0.8B}{20+0.2B}=\frac{3}{2}  -->  B=50  --> so the amount of water in mixture B is  0.2B=10  gallons.

There is no correct answer listed.

Indianblondienycgal · Answer

I getting D.

Mixture A is 20% oil and contains 20 gallons water...hence 80 % water would 5 gallons.And Mixture A would be 20 + 5 =25 gallons.

Method 1:

Combined mixture has oil: water = 4.5:3 ie. water = (3/7.5)*100% = 40%
Now by Alligation Rule for WATER
A . ......... B
80% . . . 20%

. . 40% . .

20%.......40%

Since 40% (Wt avg. is closer to B ,the proportion of B:A in the mixture will be 2:1)

So if A contains 20 gallons water ,B should contain twice the water that will be 40 gallons.

Method 2:

Weighted Average Method:
Let the weight of Mixture A = A gallons.
Let the weight of Mixture A = B gallons.
Weighted average of water is 40% in the Mixture (A+B)

40 =80(A) + 20(B)/(A+B)
 40 A + 40 B = 80 A + 20 B
 40 B - 20 B = 80 A - 40 A
 20 B = 40 A
 B/A = 2/1

If A = 20 gallons of water,B = 20 * 2= 40 gallons of awter.

We are not interested in the proportion of Oil: water in the mixtures but rather what proportion of water should be mixture A and B to get 40 % water.

SOURH7WK · Answer

That marked line is wrong. The ratio of mixture A:B = 1:2 means, 25 Gallons of mixture A is mixed with 50 gallons of mixture B to get a resultant mixture where oil is 60% or water is 40%. No matter whether u used the water or oil %age in alligation rule it will give u the ration of A:B = 1:2

So in 50 gallons of mixture B: water is 20% i.e. 10 gallons. The Question clearly asks  how many gallons of water does mixture B have?

Hope it is clear to u??

Indianblondienycgal · Answer

Nope.I am looking at 1:2 Mixture Of WATER in A:B.

we are not comparing total gallons here.We are looking at the proportion of Water.

when you use Alligation rule,Use one component of the mixture like compare oil to oil ,Copper to copper etc.

D is the official answer om Gmatprepster.

Bunuel · Answer

Mixture A is 25 gallons:  5 gallons of oil (20%) and 20 gallons of water (80%) .
If there is 40 gallons of water in mixture B, then mixture B must be 200 gallons:  160 gallons of oil (80%) and 40 gallons of water (20%).

Ratio of oil to water, in combined mixture would be  \frac{oil}{water}=\frac{5+160}{20+40}=\frac{11}{4}  not 4.5:3 as given in the stem.

OA is wrong.

stne · Answer

I have tried to do this sum according to the method shown in this blog by karishma , example 2

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... -mixtures/

taking water only ( as suggested in the blog , to focus on one material only )
first mixture water is 80%, second mixture water is 20% and in the final mixture water is 40 % ( 3/7.5 * 100 )

so 80( water% in first mix ) .......40( water % in final mix.) ........ 20 ( water % in second mix)

quantity ( whole of A not only water ) of A : quantity of B ( not only water )= ( 20-40): (40 - 80) = -20 / -40 = 20/40 = 1/2 ( scale method allegation as shown in the blog )

so qty. A : qty. B = 1/2

But given qty. A = 25 ( as in A 80% = 20 so 100% = 25 )

so 1/2 = 25/x
x=50

so quantity of B = 50 and 20% of B is water , so 20% of 50 = 10 ( the required answer)

Now focusing on oil and trying to solve
taking oil only ( as suggested in the blog , to focus on one material at one time only )
first mixture oil is 20%, second mixture oil is 80% and in the final mixture oil is 60 % ( 4.5/7.5 * 100 )

so 20( oil % in first mix ) .......60( oil% in final mix.) ........ 80 ( oil% in second mix)

quantity A ( whole of A not only oil ) : qty. B( again whole B not only oil ) = 20/ 40 = 1/2 ( scale method as shown in the blog )

so qty A: qty B = 1/2
But given qty A = 25 ( as in A 80% = 20 so 100% = 25 )

so Qty A: Qty B = 1/2 = 25/x

hence qty B = 50 again 20% of 50 = 10 ( Required answer )

Indianblondienycgal · Answer

Thanks for all your inputs.I agree that the official answer should be 10 .In fact thats what I got when I solved this question first ,I did get "10" but when I found that "40" was he official answer
i thought i was doing something wrong so changed my approach.
I am gonna call Gmatprepster to resolve this mistake on their website.

ankur1901 · Answer

while you call gmatprepster to resolve the mistake, pls can you also update the and choices here too

kinjiGC · Answer

@Mods. Please remove this wrong question. I just wasted 10 mins on this question. :cry:

Please remove the tag "Difficulty: 700-Level".

TGC · Answer

Should be 10.

In A

Water = 20

0.8A = 20 => A=25

A: 5 - O and 20 - W

B: 80% O and 20% W

Oil: Water = 45/30

5 + 0.8B / 20 + 0.2 B = 45/30

B=50

So water is 0.2 * B = 10

Maxirosario2012 · Answer

Mixture A, which is 20% oil, contains 20 gallons of water and mixture B is 80% oil. If the combination of mixture A and B has a 4.5 : 3 ratio of oil to water, how many gallons of water does mixture B have?

A. 5
B. 20
C. 30
D. 40
E. 50

Source:gmatPrepster

I am not sure what the OA is.

A. 5
B. 20
C. 30
D. 40
E. 50

My answer: 10
As we have 20 gallons of water (W) in mixture A, which represent 80%, then:
20 gallons ...................... 80%
Total gallons = T .............100% 
Then:
 T= 100*\frac{0,2}{0,8}=\frac{20}{8}*10 = 25 --> then total oil + water = 25 --> Oil = 25-20=5
As a result:
Mix A = 20 water (W) + 5 Oil (O)
Mix B = 80% (O) + 20% (W).

Then:
 Combination A + B (AVG) = \frac{Oil}{Water} = \frac{4,5}{3} = \frac{9}{6} = \frac{3}{2} = \frac{5}{20} (Mix A) + \frac{4}{1} (Mix B)

Then:  \frac{(AVG - Mix A)}{(Mix B - AVG)} = (\frac{3}{2} - \frac{5}{20}) / (\frac{4}{1} - \frac{3}{2}) = \frac{25}{50}

Then Mix B = 50, and 20% water ox Mix B = 10

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Mixture A, which is 20% oil, contains 20 gallons of water

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