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# An alchemist discovered that the formula to turn ordinary me

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Manager
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An alchemist discovered that the formula to turn ordinary me [#permalink]

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30 Dec 2012, 10:20
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Question Stats:

64% (01:49) correct 36% (01:58) wrong based on 149 sessions

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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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Manager
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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30 Dec 2012, 12:15
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best approach would be to plug in answer choices.

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

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Manager
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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09 Jan 2013, 07:36
mansoorfarooqui wrote:
best approach would be to plug in answer choices.

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

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]

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09 Jan 2013, 08:07
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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.
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Senior Manager
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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09 Jan 2013, 08:39
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,

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Manager
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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10 Jan 2013, 11:38
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,

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]

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10 Jan 2013, 15:17
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,

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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Kudos [?]: 398 [0], given: 85

Manager
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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11 Jan 2013, 03:14
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,

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]

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11 Jan 2013, 05:31
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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,

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]

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22 Jun 2015, 05:36
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]

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28 Dec 2016, 05:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: An alchemist discovered that the formula to turn ordinary me   [#permalink] 28 Dec 2016, 05:19
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