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An alchemist discovered that the formula to turn ordinary me

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An alchemist discovered that the formula to turn ordinary me [#permalink] New post 30 Dec 2012, 09:20
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C
D
E

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67% (02:30) correct 33% (01:43) wrong based on 49 sessions
An alchemist discovered that the formula to turn ordinary metal into gold is G = \frac{3}{2}M + 15, where G is the number of gold bars and M is the number of metal bars. If a metal bar weighs twice as much as a gold bar, how many metal bars will yield an equal weight of gold?

A. 10
B. 15
C. 22.5
D. 30
E. 67.5
[Reveal] Spoiler: OA
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 30 Dec 2012, 11:15
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best approach would be to plug in answer choices.

start with C:
M=22.5
2M=45
G = 22.5*3/2+15 as it would have a fractional part.. skip it
choose a number which wont result in fractional part so A and D remains

Take D:
M=30
2M=60 which should be our target value

put M=30 we get G=60
so D is the answer



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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 09 Jan 2013, 06:36
mansoorfarooqui wrote:
best approach would be to plug in answer choices.

start with C:
M=22.5
2M=45
G = 22.5*3/2+15 as it would have a fractional part.. skip it
choose a number which wont result in fractional part so A and D remains

Take D:
M=30
2M=60 which should be our target value

put M=30 we get G=60
so D is the answer



PS:Please consider kudos if it was helpful


Hey Mansoor,
"a metal bar weighs twice as much as a gold bar"
this doesnt mean
M=2G ?
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 09 Jan 2013, 07:07
Expert's post
danzig wrote:
An alchemist discovered that the formula to turn ordinary metal into gold is G = \frac{3}{2}M + 15, where G is the number of gold bars and M is the number of metal bars. If a metal bar weighs twice as much as a gold bar, how many metal bars will yield an equal weight of gold?

A. 10
B. 15
C. 22.5
D. 30
E. 67.5


A time saving method:
Since we know that Integer+Integer=Integer, and number of bars can't be in fractional form, therefore 3M/2 must be an integer or M must be a multiple of 2.
The only options that we have is A and D.
On putting 10 and 30 respectively, we find that 30 fits the conditions.
Hence 30 is the answer.
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 09 Jan 2013, 07:39
Since a metal bar weighs twice as much as a gold bar, 2M = G

Thus, by the equation given :
2M = (3/2 *M) + 15
M/2 = 15

M = 30

Hence,

Answer : D
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 10 Jan 2013, 10:38
Rock750 wrote:
Since a metal bar weighs twice as much as a gold bar, 2M = G

Thus, by the equation given :
2M = (3/2 *M) + 15
M/2 = 15

M = 30

Hence,

Answer : D


Well here i faced confusion, i thought if 1 bar of mettal is twice havier than 1 bar of gold, for example 1 bar of M=4 kg then 1 bar of G=2 kg, that means 4=2*2 --> M=2G but not 2M=G, So according to my initial calculation i stuck then i decided to plug in and found D plausable.

But could anyone suggest whether my initial thinking was wrong?
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 10 Jan 2013, 14:17
ziko wrote:
Rock750 wrote:
Since a metal bar weighs twice as much as a gold bar, 2M = G

Thus, by the equation given :
2M = (3/2 *M) + 15
M/2 = 15

M = 30

Hence,

Answer : D


Well here i faced confusion, i thought if 1 bar of mettal is twice havier than 1 bar of gold, for example 1 bar of M=4 kg then 1 bar of G=2 kg, that means 4=2*2 --> M=2G but not 2M=G, So according to my initial calculation i stuck then i decided to plug in and found D plausable.

But could anyone suggest whether my initial thinking was wrong?

Did u get D using M=2G ?
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 11 Jan 2013, 02:14
Rock750 wrote:
ziko wrote:
Rock750 wrote:
Since a metal bar weighs twice as much as a gold bar, 2M = G

Thus, by the equation given :
2M = (3/2 *M) + 15
M/2 = 15

M = 30

Hence,

Answer : D


Well here i faced confusion, i thought if 1 bar of mettal is twice havier than 1 bar of gold, for example 1 bar of M=4 kg then 1 bar of G=2 kg, that means 4=2*2 --> M=2G but not 2M=G, So according to my initial calculation i stuck then i decided to plug in and found D plausable.

But could anyone suggest whether my initial thinking was wrong?

Did u get D using M=2G ?


No i did not, that s why i raised this question, to clarify where i did wrong? or where exactly i am loosing the track.
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink] New post 11 Jan 2013, 04:31
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ziko wrote:
Rock750 wrote:
Since a metal bar weighs twice as much as a gold bar, 2M = G

Thus, by the equation given :
2M = (3/2 *M) + 15
M/2 = 15

M = 30

Hence,

Answer : D


Well here i faced confusion, i thought if 1 bar of mettal is twice havier than 1 bar of gold, for example 1 bar of M=4 kg then 1 bar of G=2 kg, that means 4=2*2 --> M=2G but not 2M=G, So according to my initial calculation i stuck then i decided to plug in and found D plausable.

But could anyone suggest whether my initial thinking was wrong?


Notice that M is the number of metal bars (not their weight)
And G is the number of gold bars.

If weight of a metal bar is twice the weight of a gold bar, and if you want to equate their weights, you will need twice the number of gold bars to make their weights equal.

That is how you get 2M = G (Number of gold bars should be twice the number of metal bars)
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Re: An alchemist discovered that the formula to turn ordinary me   [#permalink] 11 Jan 2013, 04:31
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