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# Fraction Problem

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Hi,

It would be great if someone could explain the process entailed in solving this problem:

"Two glasses, C and D, have exactly the same size and shape. Glass C is empty, and glass D has some water in it. Half the water in D is then poured in to C. This process is repeated two more times. Each time, half the remaining water in D is poured into C. After three pourings, C is half full. What fraction of a glassfull is now left in D?"

Thanks!

Jedithious  May 17, 2014

### Best Answer

#1
+1792
+19

"Two glasses, C and D, have exactly the same size and shape. Glass C is empty, and glass D has some water in it. Half the water in D is then poured in to C. This process is repeated two more times. Each time, half the remaining water in D is poured into C. After three pourings, C is half full. What fraction of a glassfull is now left in D?"

Suppose glass D starts out with d amount of water in it.

after 1 pouring glass C has d/2, and glass D has d/2 water left in whatever units

after the 2nd pouring C has (d/2+d/4) and D has d/4

after the 3rd pouring C has (d/2+d/4+d/8) and D has d/8

We are told that glass C now is half full.

So 7/8d=1/2 a glass

d = 4/7 of a glass

and there is d/8 left in glass D at this point so

glass D now has 1/8(4/7) = 4/56 = 1/14 of a glassful

Rom  May 17, 2014
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### 11+0 Answers

#1
+1792
+19
Best Answer

"Two glasses, C and D, have exactly the same size and shape. Glass C is empty, and glass D has some water in it. Half the water in D is then poured in to C. This process is repeated two more times. Each time, half the remaining water in D is poured into C. After three pourings, C is half full. What fraction of a glassfull is now left in D?"

Suppose glass D starts out with d amount of water in it.

after 1 pouring glass C has d/2, and glass D has d/2 water left in whatever units

after the 2nd pouring C has (d/2+d/4) and D has d/4

after the 3rd pouring C has (d/2+d/4+d/8) and D has d/8

We are told that glass C now is half full.

So 7/8d=1/2 a glass

d = 4/7 of a glass

and there is d/8 left in glass D at this point so

glass D now has 1/8(4/7) = 4/56 = 1/14 of a glassful

Rom  May 17, 2014
#2
+3

Glass                  "C"                       "D"                Water Left

Initial                 0                         100

1st                     50                        50                    1/2

2nd              50+25=75                  25                     1/4

3rd              75+12.5=87.5            12.5                   1/8

4th               87.5+6.25=93.75       6.25                  1/16

Guest May 17, 2014
#3
+81027
+16

Well, Jedithious....this is kind of a tough one, huh??

Let's call the amount of water contained in D originally "x"....since we don't know how much it contains.

After the first pour, D loses(1/2)x and C gets (1/2)x

x -(1/2)x = (1/2)x

So D now contains (1/2)x

And, on the next pour, D loses (1/2)(1/2) x  = (1/4)x and C gets (1/4)x

So D now contains x - (1/2)x - (1/4)x = (1/4)x

And C contains  (1/2)x + (1/4)x = (3/4)x

On the last pour D loses(1/2)(1/4)x = (1/8)x  and C gets (1/8)x

So....let's look at what C contains, now.

It has (1/2)x + (1/4)x + (1/ 8)x = (7/ 8)x

But.....(7/8)x = (1/2)C = (1/2)D ....since both glasses are the same size and shape.

Therefore

(7/8)x = (1/2)D  and since we want to find out what x was, let's multiply both sides by (8/7)

So we have

x = (8/7)(1/2)D = (8/14) = (4/7)D

So D was (4/7) full when we started

So, since each pour "halves" what comes before, we have (4/7)(1/2)(1/2)(1/2) = 4/56 = 1/14 full...and that's the fraction of the glassful remaining in D.

Whew!!...some workout, huh?  Let me know if you spot any mistakes......but I think that's it......a pretty challenging problem!!

CPhill  May 17, 2014
#4
+91465
+8

Hi Jedithious,

It has been ages since I saw you on the forum.

I know that you have always had fantastic manners so I am expecting you to reappear with a thank you.  But, did you know that if you give the answers that you like a thumbs up you will be giving the answerer 5 points.  Everyone likes to be thank and also to be given points.

It is great to see you on the forum again!!!

Melody   CPhill's glasses look good on me, don't you think?

PS I have labelled your question as a puzzle one. And I think I will reference it in with the puzzles.

Melody  May 17, 2014
#5
+81027
+8

You sould feel honored, Jedithious........not many questions on here ever qualify for puzzle status !!!

CPhill  May 17, 2014
#6
+4145
+3
zegroes  May 17, 2014
#7
+91465
+8

I have referenced it at the end of this thread

http://web2.0calc.com/questions/puzzles_1

This is in with the sticky topics.

thanks Jedithious.

Melody  May 18, 2014
#8
+154
+8

Rom and CPhill:

Yes, evidently quite a workout! Thank you both for taking the time to do it and explain it so methodically. I feel like my comprehension of the answers you gave to this particular problem has increased my confidence in converting word problems to workable algebra formulas more than perhaps any other word problem I've recently encountered. Much appreciated!

Melody:

Thanks for the tip; done, and will be sure to do in the future. And its great to see you too! Haha, nice glasses, yes. How's the weather in Sydney this time of year?

Jedithious  May 18, 2014
#9
+91465
+3

Thanks Jedithious,

I gave you a thumbs up too but mine is only worth 3 points!

It is still sunny enough to need my sunnies.  Just as well Chris collects them and always has a few to spare!

I keep losing mine but I know where to go for more.  Besides, I am waiting on an accessory.  I was told I could have it ages ago - now i have to hassle apparently.

(This is not aimed at you jedithious, it is a private joke in a public areana)

Melody  May 19, 2014
#10
+81027
+3

Thanks for the kind note, Jedithious.......we never like to give "watered" down answers, here......

If Melody were to do a thorough search, she'd find a pair of "glasses" in this problem!!

CPhill  May 19, 2014
#11
+91465
+3

Oh! That's where they went - I've been looking for those!    Thanks Chris!!!!

Melody  May 19, 2014

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