8 years from now, the bottle of wine labeled 'Aged' will be

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8 years from now, the bottle of wine labeled 'Aged' will be [#permalink]

05 Mar 2010, 22:21

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8 years from now, the bottle of wine labeled 'Aged' will be 7 times older than the bottle of wine labe1ed 'Table.' I year ago, the bottle of wine labeled 'Table' was one-fourth as old as the bottle of wine labeled 'Vintage.' If the'Aged' bottle was 20 times older than the 'Vintage' bottle 2 years ago, then how old is each bottle now?

this question from manhattan word prep
in solution the equations are given as:
a+8 = 7( t+8)
t-1 = ( v-1)/4
& a-2 =20( v-2)

i am just curious to know if first and third equations are translated as given in problem.
what i perceived is it should be
a+8 = (a+8) + 7( t+8)
t-1 = ( v-1)/4
& a-2 =(v-2)+20( v-2)
as question stem says 7 times older than.. and 20times older than..

please, any one can clarify this.
Thanks

Last edited by Bunuel on 14 May 2013, 02:09, edited 1 time in total.

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Re: age problem [#permalink]

08 Mar 2010, 11:40

Quote:

i am just curious to know if first and third equations are translated as given in problem.
what i perceived is it should be
a+8 = (a+8) + 7( t+8)

That can't logically make any sense unless t=-8. It has to be translated as you had it above. Here’s what I got.

Wines 2 yrs ago 1 yr ago Now 8 yrs future
aged a-2 a-1 a a+8
table t-2 t-1 t t+8
vintage v-2 v-1 v v+8

What the question tells us
(a-2)=20(v-2)
4(t-1)=v-1
(a+8)=7(t+8)

a=20v-40+2
a=20v-38
so 20v-38+8=7t+56
20v-30=7t+56
20v=7t+86
v=7t/20 + 4+ 3/10
so

4(t-1) = 7t/20 + 3.3
4t-4 = 7t/20 + 3.3
4t-7.3 = 7t/20
4t-7t/20=7.3
80t/20 - 7t/20=7.3
73t/20=7.3
73t=146
t=2 so
4(2-1)=v-1
v=3
so
(a+8)=7(2+8)
a+8=70
a=62
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Re: age problem [#permalink]

08 Mar 2010, 13:20

sandeep25398 wrote:

Let the current age of each wine be A - Aged, V - Vintage & T - Table.

You get three eq from the wordings:
A+8 = 7(t+8)
A-2 = 20(V-2)
T-1 = 0.25(V-1)

Solving them you get T = 2, V = 5, A = 62
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Re: age problem [#permalink]

09 Mar 2010, 01:50

vannbj wrote:

That can't logically make any sense unless t=-8. It has to be translated as you had it above. Here’s what I got.

thank you for the reply.
my doubt is whether the wording of question stem is correct. first equation is equivalent to saying "x is 7 times as old as y", not as mentioned in question stem. similarly 3rd equation..

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Re: age problem [#permalink]

09 Mar 2010, 09:10

8 years from now, the bottle of wine labeled 'Aged' will be 7 times older than the bottle of wine labe1ed 'Table.'
==> when you say 8 years from now, we should add 8 to the current value. So it will be Aged+8 and Table+8.
At this time first one is 7 times older than second. Times is always a multiplication.
So Equation is Aged+8 = 7 (Table+8)

If the'Aged' bottle was 20 times older than the 'Vintage' bottle 2 years ago
==> With the same logic, 2 years ago means subtract 2 from those values. So it will be Aged-2 and Vintage-2.
20 times older again will be a multiplication.
So Equation is Aged-2 = 20 (Vintage-2)

Hope it helps to clarify your question.

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Re: age problem [#permalink]

11 Mar 2010, 07:45

sandeep25398 wrote:

vannbj wrote:

That can't logically make any sense unless t=-8. It has to be translated as you had it above. Here’s what I got.

No the first equation is not equivalent to saying "x is 7 times as old as y",
take for example, x is 10 yrs old now and y is 118 yrs old, so right now x:y is 10:118 which is 5:59
but after 8 years x will be 18 and y will 126 and now the ratio x:y is 18:126 and if u simplify it is 1:7
Age always increase linearly with years so you cannot say that the multiplicative ratio holds good as the years pass by.
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Re: age problem [#permalink]

14 May 2013, 01:58

This question took me 4+ minutes!!! Do we expect this kind of problems? Is there a faster way ?
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Re: age problem [#permalink] 14 May 2013, 01:58

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8 years from now, the bottle of wine labeled 'Aged' will be

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8 years from now, the bottle of wine labeled 'Aged' will be

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