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# A small beaker is 1/3 filled with salt water. Another beaker, which ha

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A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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Updated on: 20 Jun 2011, 22:28
5
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Difficulty:

5% (low)

Question Stats:

88% (02:00) correct 13% (02:17) wrong based on 100 sessions

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A small beaker is 1/3 filled with salt water. Another beaker, which has five times the capacity of the small beaker, is 1/4 filled with fresh water. After dumping all of the salt water from the small beaker into the large beaker, to what fraction of its capacity will the large beaker be filled?

A. 7/12
B. 7/24
C. 17/24
D. 17/60
E. 19/60

Originally posted by siddhans on 19 Jun 2011, 15:50.
Last edited by siddhans on 20 Jun 2011, 22:28, edited 1 time in total.
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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19 Jun 2011, 15:58
Let T(S) be the total capacity of small beaker
Let T(L) be the total capacity of large beaker

T(L)/T(S) = 5 ------------(1)

Small beaker filled = 1/3 T(S)

Large Beaker filled = 1/4 T(L)

After dumping small beaker into large beaker =>
1/3 T(S) + 1/4 T(L) = [ (1/3) (1/5) T(L) ] + 1/4 T(L) = 19/60 T(L)

Hence E
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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20 Jun 2011, 06:36
1
1/3*1/5+1/4
=1/15+1/4
=15+4/60
=19/60
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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23 Sep 2011, 12:29
Small = 1/3
Big = 1/4

S = 5B

small = 1/3 =5B => 1/15

1/15 + 1/ 4 = 19/60 ...
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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23 Sep 2011, 19:25
Its kind of implied that small beaker has only salt water which is 1/3 of its total capacity and
large beaker has only fresh water which is 1/4 of its total capacity.

Let capacity of the small beaker be C .
=> capacity of the large beaker = 5C (given)

=> salt water in the small beaker = C/3
fresh water in the large beaker = (5C)/4

After dumping all of the salt water from the small beaker into the large beaker,
the fraction to which large beaker will be filled = ((5C)/4 + C/3 ) / (5C)
= 19/60

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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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17 Oct 2011, 17:47
shrive555 wrote:
Small = 1/3
Big = 1/4

S = 5B

small = 1/3 =5B => 1/15

1/15 + 1/ 4 = 19/60 ...

There is another question from 'Veritas prep" I Tried in the same way but got it wrong. Question is
Jar X =1/6 full of water. Jar Y , which has half the capacity of jar X is 1/2 full . if the content of jar X is poured into jar Y . Jar Y will be filled to what fraction of its capacity.

X= 1/6
Y= 1/2
Also Y=X/2 => 1/12 ( since X=1/6)

1/2 +1/12 = 7/12

OA =
5/6

whats the difference between two questions ?
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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17 Oct 2011, 20:51
1
shrive555 wrote:
shrive555 wrote:
Small = 1/3
Big = 1/4

S = 5B

small = 1/3 =5B => 1/15

1/15 + 1/ 4 = 19/60 ...

There is another question from 'Veritas prep" I Tried in the same way but got it wrong. Question is
Jar X =1/6 full of water. Jar Y , which has half the capacity of jar X is 1/2 full . if the content of jar X is poured into jar Y . Jar Y will be filled to what fraction of its capacity.

X= 1/6
Y= 1/2
Also Y=X/2 => 1/12 ( since X=1/6)

1/2 +1/12 = 7/12

OA =
5/6

whats the difference between two questions ?

The difference between two Qs is that in the first one we are moving water from small jar to big jar. In your question water is moved from big jar to small jar.

JAR X is double the size of JAR Y so if we are moving water from JAR X to Y we need to multiply by 2.

Soln:

Jar X = 1/6 filled
To get the volume in terms of Y , we need to multiply 1/6 with 2 since the JAR X is double the size of JAR Y.

so in terms of Y, X's volume = (1/6)*2 = 2/6 =1/3

Now to know the amount of water in JAR Y = 1/2+1/3 = 5/6
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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25 Dec 2017, 06:52
niks18 amanvermagmat Bunuel

couple of queries in solution by shrive555

Quote:
Small = 1/3
Big = 1/4

S = 5B

small = 1/3 =5B => 1/15

1/15 + 1/ 4 = 19/60 ..

Should it not be B = 5 * S

Also for solution provided by Spidy001
Quote:
fresh water in the large beaker = (5C)/4

Should it not be 5 * (C/3) * 1/4 ie 5C/12 ??

We are given that Another beaker (ie bigger one), which has five times the capacity of the small beaker, is 1/4 filled with fresh water.
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A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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25 Dec 2017, 07:06
niks18 amanvermagmat Bunuel

couple of queries in solution by shrive555

Quote:
Small = 1/3
Big = 1/4

S = 5B

small = 1/3 =5B => 1/15

1/15 + 1/ 4 = 19/60 ..

Should it not be B = 5 * S

Also for solution provided by Spidy001
Quote:
fresh water in the large beaker = (5C)/4

Should it not be 5 * (C/3) * 1/4 ie 5C/12 ??

We are given that Another beaker (ie bigger one), which has five times the capacity of the small beaker, is 1/4 filled with fresh water.

Just a small suggestion, for these kind of fraction related questions it's Easiest to solve through SMART NUMBERS rather than algebra.

so here we have fractions $$\frac{1}{3}$$ & $$\frac{1}{4}$$, so let's assume capacity of small beaker be $$12$$ (LCM of 3 & 4)

So capacity of Large beaker $$= 12*5=60$$

Small is filled till $$12*\frac{1}{3}=4$$ units and large is filled till $$60*\frac{1}{4}=15$$ units.

So when contents of small beaker is poured in large beaker, it's total content will be $$= 4+15=19$$ units.

Hence fraction filled $$= \frac{19}{60}$$

Now coming to your question - I do not see any problem with Spidy001's solution. Can you explain why you think it should be C/3 and not C?

with shrive555's solution, I think the notation is wrong but somehow he/she has got the right answer
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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25 Dec 2017, 07:18
niks18

Quote:
Now coming to your question - I do not see any problem with Spidy001's solution. Can you explain why you think it should be C/3 and not C?

Large beaker is 5 times capacity of small beaker(given)

Since capacity of small beaker is C, and it is C/3 filled so Capacity of large beaker is 5C/3. Since large beaker is only 1/4 filled,
I got (5C/3) * 1/4 ie 5C/12.
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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25 Dec 2017, 07:27
1
niks18

Quote:
Now coming to your question - I do not see any problem with Spidy001's solution. Can you explain why you think it should be C/3 and not C?

Large beaker is 5 times capacity of small beaker(given)

Since capacity of small beaker is C, and it is C/3 filled so Capacity of large beaker is 5C/3. Since large beaker is only 1/4 filled,
I got (5C/3) * 1/4 ie 5C/12.

Hi,

the highlighted portion is not correct. Large beaker's capacity is 5 times that of small's. So as you mentioned that small's capacity is C so large's has to be 5C.

The question does not say that Large beaker's capacity if 5 times the filled capacity or contents of small beaker's. In that case because small is filled till C/3 so large's capacity would have been 5C/3.

I guess you are confused with the language of the question, kindly re-read the question slowly
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A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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25 Dec 2017, 15:39
siddhans wrote:
A small beaker is 1/3 filled with salt water. Another beaker, which has five times the capacity of the small beaker, is 1/4 filled with fresh water. After dumping all of the salt water from the small beaker into the large beaker, to what fraction of its capacity will the large beaker be filled?

A. 7/12
B. 7/24
C. 17/24
D. 17/60
E. 19/60

small beaker will fill (1/3)/5=1/15 of large beaker
1/15+1/4=19/60
E
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Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha  [#permalink]

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13 Oct 2018, 11:59
took number that was factor to 3.
Let small beaker be x = 24. 1/3x = 8. Big beaker =5*24=120. 1/4(120) = 30.

so small beaker amount poured into large--> 30+8 = 38 over total capacity --> 38/120 --19/60
Re: A small beaker is 1/3 filled with salt water. Another beaker, which ha   [#permalink] 13 Oct 2018, 11:59
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