f2001290 wrote:2. Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?
(A) 2/3
(B) 7/3
(C) 5/2
(D) 8/3
(E) 3

Solution:

If "X" be the cups of milk poured in to 2-cup bottle
then "4-X" be the cups of milk poured in to 4-cup bottle

X/2 = (4-X)/4
X = 4/3
4-X = 4-(4/3) = 8/3.

What is wrong in my approach??

Isn't 8/3 the correct answer ?

You can either equate fractions (as you have done) or consider
fractions and have the total sum up to 4. Either way you should
get the same answer.

For the second part, 2x + 4x = 4 where x is the fraction that each
bottle is filled to. You can solve for x = 4/6 = 2/3 and hence get
8/3 as the total number of cups in the 4 cup bottle.

Last time you tried this problem, you didn't solve it completely.
Your solution can be seen below:

2) Let x be the fraction that each cup needs to be filled to
2x + 4x = 4
x = 2/3

I was assuming 2/3 to be the solution after looking at your approach.

f2001290 wrote:Last time you tried this problem, you didn't solve it completely.
Your solution can be seen below:

2) Let x be the fraction that each cup needs to be filled to
2x + 4x = 4
x = 2/3

I was assuming 2/3 to be the solution after looking at your approach.

Oops. My bad. Thanks for pointing this out. Yeah, I missed to post
the 4*2/3 part. Didn't realize 2/3 was one of the answers

2. Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?
(A) 2/3
(B) 7/3
(C) 5/2
(D) 8/3
(E) 3

Then capacity of the 4-cup bottle = 2 times the capacity of the 2-cup bottle...Hence capacity is 2x and x respectively....

So if y cups of milk is poured into the 4-cup bottle....the fraction of capacity becomes y/2x....Amount of milk poured in the 2-cup bottle = 4-y...

Now y/2x = 4-y/x
8-2y = y
3y = 8
Hence y = 8/3