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What is the ratio of a to b to c ?

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What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:05
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What is the ratio of a to b to c ?

(1) \frac{c}{a} =\frac{5}{2} and b=4a

(2) ac = 40 and b=16
[Reveal] Spoiler: OA

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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:25
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(1) Sufficient. we know that c=5x and a=2x for some number x. Therefore b=8x and the ratio a:b:c=2x:8x:5x=2:8:5.

(2) Insufficient. We don't know the ratio a:c.
For example a=4, c=10 or it could be in reverse order a=10, c=4.

The answer is A.
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:32
What is the ratio of a to b to c ?

a/b/c=\frac{a}{bc}

(1) \frac{c}{a} =\frac{5}{2} and b=4a
c=5/2a, b=4a
Substitute \frac{a}{4a*2.5a}=\frac{1}{10a}. It depends on a.
Not sufficient

(2) ac = 40 and b=16
a and c can have many values (2,20 or 4,10) so the ratio in not determined.
Not sufficient

1+2)b=16 a=4 c=10. We know all the numbers, we know also a/b/c
Sufficient

A is not the answer take c=50 a=20 and b= 80 ratio 1/200
take a=8 c=20 b=32 ratio 1/80. I disagree with the OA...
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:37
Zarrolou wrote:
What is the ratio of a to b to c ?

a/b/c=\frac{a}{bc}



I don't think that this is correct ?! all we know is the ratio of a to b is a/b .. that's why you must assign a new variable x and refer to it as smyarga did

It is the correct way to do this
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:48
Rock750 wrote:
I don't think that this is correct ?! all we know is the ratio of a to b is a/b .. that's why you must assign a new variable x and refer to it as smyarga did

It is the correct way to do this


My point is : the ratio a to b is a division a/b
The ration a to b is 2 means a/b=2
the ration (a to b) to c is also a division (I don't see why it sould not be). That's why I do not agree, maybe I have to review ratios
Look at the real numbers , all respect A.
a=20 , b= 80 c=50 a:b:c=1/200
a=8 , b=32 c=20 a:b:c=1/80
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:54
Zarrolou wrote:
Rock750 wrote:
I don't think that this is correct ?! all we know is the ratio of a to b is a/b .. that's why you must assign a new variable x and refer to it as smyarga did

It is the correct way to do this


My point is : the ratio a to b is a division a/b
The ration a to b is 2 means a/b=2
the ration (a to b) to c is also a division (I don't see why it sould not be). That's why I do not agree, maybe I have to review ratios
Look at the real numbers , all respect A.
a=20 , b= 80 c=50 a:b:c=1/200
a=8 , b=32 c=20 a:b:c=1/80


in your opinion, why should ratio of a to b to c equal to (a to b) to c and not equal to a to (b to c) ? we don't know if the operation "to" is transitive ..
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Last edited by Rock750 on 28 Apr 2013, 12:28, edited 1 time in total.
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 11:59
Rock750 wrote:
in your opinion, why should ratio of a to b to c equal to (a to b) to c and not equal to a to (b to c) ? we don't know if the operation "to" is commutative ..


I think we are going nowhere here but in my opinion we should not forget the meaning of ratio: it's a division
5:5:5=0.2 is equal to (5:5):5=0,2 and not to 5:(5:5)=5
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 12:00
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Zarrolou, in your examples the ratio is the same.
a=20 , b= 80 c=50 a:b:c=20:80:50=2:8:5
a=8 , b=32 c=20 a:b:c=8:32:20=2:8:5
You shouldn't mix them. Here is relative ratio of three numbers together. The ratio of three numbers a:b:c=n:m:k just means that there some integer x such that a=nx, b=mx, c=kx.
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 12:02
Zarrolou wrote:
Rock750 wrote:
in your opinion, why should ratio of a to b to c equal to (a to b) to c and not equal to a to (b to c) ? we don't know if the operation "to" is commutative ..


I think we are going nowhere here but in my opinion we should not forget the meaning of ratio: it's a division
5:5:5=0.2 is equal to (5:5):5=0,2 and not to 5:(5:5)=5


the ratio of 5:5:5 is equal to 1:1:1
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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 12:03
Oh yeah guys I see your point

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Re: What is the ratio of a to b to c ? [#permalink] New post 28 Apr 2013, 12:04
For example, the ratio of the number of women to the number of men to the number of children is 1 to 2 to 3.
Your ratio as 1 to 80 is meaningless, because you compare 3 objects and as result have comparison of two objects.
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 05:03
Hi

I am also having difficulty understanding this one, expressing all in terms of a
we get b=4a
c=\frac{5}{2} a
then a:b:c

\frac{\frac {a}{4a}}{\frac{5a}{2}}

so if a =1 \to a:b:c = \frac{\frac {1}{4}}{\frac{5}{2}}

if a = 2 \to a:b:c = \frac{\frac {2}{8}}{5}

if a=3 \to a:b:c = \frac{\frac {3}{12}}{\frac{15}{2}

if a=4 \to a:b:c = \frac{\frac {4}{16}}{{10}}

so a:b:c differs each time, does it not?


B alone is also insuff.

1+2

a:b:c 4:16:10

similar sum

what-is-the-ratio-of-x-y-z-56282.html#p400899

Can anybody clear this confusion? Thanks.
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 05:49
Rock750 wrote:
What is the ratio of a to b to c ?

(1) \frac{c}{a} =\frac{5}{2} and b=4a

(2) ac = 40 and b=16


st.1

a/c = 2/5, a/b = 1/4. to put these 3 in ratio we've to equate the common thing i.e. a.
it'll become a/c=2/5 and a/b=2/8. << I equated a in both the expression by multiplying and dividing 2
a:b:c=2:8:5.
simple.

st.2

It will clearly wont give the desired rsult.
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 06:39
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stne wrote:
Hi

I am also having difficulty understanding this one, expressing all in terms of a
we get b=4a
c=\frac{5}{2} a
then a:b:c

\frac{\frac {a}{4a}}{\frac{5a}{2}}

so if a =1 \to a:b:c = \frac{\frac {1}{4}}{\frac{5}{2}}

if a = 2 \to a:b:c = \frac{\frac {2}{8}}{5}

if a=3 \to a:b:c = \frac{\frac {3}{12}}{\frac{15}{2}

if a=4 \to a:b:c = \frac{\frac {4}{16}}{{10}}

so a:b:c differs each time, does it not?


B alone is also insuff.

1+2

a:b:c 4:16:10

similar sum

what-is-the-ratio-of-x-y-z-56282.html#p400899

Can anybody clear this confusion? Thanks.


What is the ratio of a to b to c ?

(1) \frac{c}{a} =\frac{5}{2} and b=4a. From b=4a, we have the \frac{a}{b}=\frac{1}{4}=\frac{2}{8}. So, we have that a:c=2:5 and a:b=2:8 --> a:b:c=2:8:5. Sufficient.

(2) ac = 40 and b=16. Not sufficient.

Answer: A.

As for your solution: the ratio is the same! You have a:b:c = a:4a:(2.5a) --> a reduces: a:b:c = 1:4:(2.5), which is the same ratio as a:b:c = 2:8:5 = 4:16:10 = 8:32:20 ...

Hope it's clear.
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 08:19
First of all thank you members,
But little confusion still remains, as long as I see this sum in isolation everything is clear but when I try and relate this sum to this sum given below I get a bit confused.

What is the ratio of x:y:z? (what-is-the-ratio-of-x-y-z-56282.html#p400899 )

(1) xy = 14
(2) yz = 21

now in this sum individually obviously both are insufficient
but when we take them together, expressing everything in terms of y

x = \frac {14}{y}
z =\frac {21}{y}

so \,x:y:z = \frac {\frac{\frac{14}{y}}{y}}{\frac{21}{y}}

so \, for \, Y=1 \to x:y:z =\frac {\frac{14}{1}}{21}

and
for \, Y=7 \to x:y:z =\frac {\frac{2}{7}}{3}


now using the same logic as above where we said "a reduces" and said that the ratio remains the same, can't we say in this sum too that ( x:y:z ) "Y reduces" hence 1+2 is sufficient to answer the ratio of x:y:z ?

if in a:b:c sum we say " as a:b:c = 2:8:5 = 4:16:10 = 8:32:20 ... " meaning in all the cases ratio is same

why in x:y:z = 14:1:21 and x:y:z = 2:7:3 , we are taking ratio to be different? shouldn't the ratio be same whether x:y:z= 14:1:21 or x:y:z = 2:7:3

we can say Y can be reduced and that 14:1:21 is the same as 2:7:3 answer should be C in this x:y:z sum, still here the answer is E. ?? Am I messing up something? Thanks.

Edit: my bad didn't see that 14:1:21 cannot be reduced to 2:7:3 so of course the answer is E
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Last edited by stne on 02 Sep 2013, 23:03, edited 1 time in total.
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 08:29
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stne wrote:
First of all thank you members,
But little confusion still remains, as long as I see this sum in isolation everything is clear but when I try and relate this sum to this sum given below I get a bit confused.

What is the ratio of x:y:z? (what-is-the-ratio-of-x-y-z-56282.html#p400899 )

(1) xy = 14
(2) yz = 21

now in this sum individually obviously both are insufficient
but when we take them together, expressing everything in terms of y

x = \frac {14}{y}
z =\frac {21}{y}

so \,x:y:z = \frac {\frac{\frac{14}{y}}{y}}{\frac{21}{y}}

so \, for \, Y=1 \to x:y:z =\frac {\frac{14}{1}}{21}

and
for \, Y=7 \to x:y:z =\frac {\frac{2}{7}}{3}


now using the same logic as above where we said "a reduces" and said that the ratio remains the same, can't we say in this sum too that ( x:y:z ) "Y reduces" hence 1+2 is sufficient to answer the ratio of x:y:z ?

if in a:b:c sum we say " as a:b:c = 2:8:5 = 4:16:10 = 8:32:20 ... " meaning in all the cases ratio is same

why in x:y:z = 14:1:21 and x:y:z = 2:7:3 , we are taking ratio to be different? shouldn't the ratio be same whether x:y:z= 14:1:21 or x:y:z = 2:7:3

we say say Y can be reduced and that 14:1:21 is the same as 2:7:3 answer should be C in this x:y:z sum, still here the answer is E. ?? Am I messing up something? Thanks.


Hey,

if you havent checked my post .. check it .. just above bunuel's post >> what-is-the-ratio-of-a-to-b-to-c-151800.html#p1240411

we have to equate the common term in two ratios to get a three no. ratio ... like a/b=1/2 and a/c=3/2 . here common term is a.

After equating a .. the expressions will be a/b=3/6 and a/c=3/2 .. here I made a equal in both the expressions. hence a:b:c=3:6:2.

In the question you posted there is no way you can do this. Hence answer is E. got it ?
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Re: What is the ratio of a to b to c ? [#permalink] New post 27 Jun 2013, 08:51
@stunn3r

Thank you, equating the common variable seems to make sense , doing it my way where I was expressing everything in terms of a single variable , does not seem to work here. Thank you
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