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An alchemist discovered that the formula to turn ordinary me [#permalink]
 30 Dec 2012, 10:20
Question Stats:
66% (02:02) correct
33% (01:47) wrong based on 30 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
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]
30 Dec 2012, 12:15
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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
PS:Please consider kudos if it was helpful
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]
09 Jan 2013, 07:36
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]
09 Jan 2013, 08:07
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]
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, Answer : D
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Re: An alchemist discovered that the formula to turn ordinary me [#permalink]
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,
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]
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,
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]
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,
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]
11 Jan 2013, 05:31
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
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11 Jan 2013, 05:31
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