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

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

67% (02:51) correct
33% (01:35) wrong based on 56 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?

Re: An alchemist discovered that the formula to turn ordinary me [#permalink]
30 Dec 2012, 11:15

1

This post received KUDOS

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

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

GMAT - Practice, Patience, Persistence Kudos if u like

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

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

If you found my post useful and/or interesting - you are welcome to give kudos!

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

KUDOS is the good manner to help the entire community.

"If you don't change your life, your life will change you"

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

If you found my post useful and/or interesting - you are welcome to give kudos!

Re: An alchemist discovered that the formula to turn ordinary me [#permalink]
11 Jan 2013, 04:31

Expert's post

1

This post was BOOKMARKED

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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