Two containers of equal capacity are full of a mixture of oil and wat

Thanks, but I have a doubt, why you took the LCM ?

Easier for calculation here.

4/11 and 7/18 ..denominators cancel 11*18 giving integer values for Oil and Water quantity.

O1 = (4/11)(O1+W1)

W1 = (7/11)(O1+W1)

O2 = (7/18)(02+W2)

W2 = (11/18)(O2+W2)


O/W = (01+W1)/02+W2) = ((4/11)+(7/11))/((7/18)+(11/18)) [ EQUAL CAPACITY i.e O1+W1 = O2+W2 , so they
cancel out in numerator and denominator]

=149/247

Answer is A.

Can this question be done through allegation? thanks

when i read: "Two containers of equal capacity"...
then:
first container 4:7 so capacity 4+7 = 11
second container 7:11 so capacity 7+11 = 18

to create an equal capacity that would be easier to work with:
11*18 = 198 in each container

looking at the results i need the capacity of the two containers, so:
198*2 = 396

answer a fits perfect:
a. 149 : 247 -- 149+247 = 396

This is not an Official Guide question and cannot be. It has 4 options instead of 5. Tag removed.

I solved this question in a slightly different but much easier way.

I plugged in a value for the amount of each mixture added, which happens to be equal since the same amount is added.
To choose a good number I converted the Oil:Water ratio to a Oil:Solution ratio, by merely adding the numerator to the denominator, since the solution = Oil + Water.

Therefore, the two Oil:Solution ratios are = 4:11, and 7:18 respectively.
Then I picked a number which is divisible both by 11 and 18 = 11 x 18 = 198. (This gives us an easy answer for the Oil in each mixture)

Therefore
The amount of Oil in Mixture 1 = 4 x 198 / 11 = 4 x 18 = 72.
The amount of Oil in Mixture 2 = 7 x 198 / 18 = 7 x 11 = 77.
Total of Oil in the Final Mixture = 77 + 72 = 149

Total Volume of Final Mixture = 198 + 198 = 396.
Total Water in the Final Mixture = 396 - 149 = 247

Therefore, Answer = Total water / Total oil = 149/247
Answer = A

jamifahad

Thanks for the explanation. If I am reading correctly then aren't the highlighted steps incorrect

Please help, have same question

Let's work with the concentration of oil.
C1 = 4/11 th of total is oil
C2 = 7/18 th of total is oil

We know that
\(Cavg = \frac{(C1w1 + C2w2)}{(w1 + w2)}\)

\(Cavg = \frac{(4/11)*1 + (7/18)*1}{1 + 1}=\frac{149}{11*18*2}\)

Answer must be (A).

Check out these posts and videos for discussion on weighted averages and mixtures:

Weighted Avg and Mixtures Basics:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/
and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

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