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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
Please, could someone help me understand why solving like this is wrong?

1/a+1/b+1/c = (a+b+c)/(abc)

(1) a+b+c = 1
(2) abc=1

(1) and (2) gives us = (a+b+c)/(abc)=1/1
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
milanezr wrote:
Please, could someone help me understand why solving like this is wrong?

1/a+1/b+1/c = (a+b+c)/(abc)

(1) a+b+c = 1
(2) abc=1

(1) and (2) gives us = (a+b+c)/(abc)=1/1


1/a + 1/b + 1/c = (bc + ac + ab) / (abc)

and NOT the way you have solved.
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
Given:
1/a + 1/b + 1/c

Combine to a single fraction, multiply by abc/abc to every single term:
1/a + 1/b + 1/c = (bc + ac+ ab) / abc

There is no way to derive bc / ac / ab from any of the below options:
(1) a + b + c = 1
(2) abc = 1

Answer: E, Neither
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
AM*HM = GM^2

If we apply this formula, the answer should be C. Since option A is 3*AM of a,b,c and option B is GM^3 of numbers a,b,c.
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
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satyadev123 wrote:
AM*HM = GM^2

If we apply this formula, the answer should be C. Since option A is 3*AM of a,b,c and option B is GM^3 of numbers a,b,c.


Unfortunately you can't use anything like that here, because any relationships you might have learned about geometric means are only true for sets of positive numbers, and here, two of the unknowns must be negative.

The answer is genuinely E to this question. If you imagine first that a=5, then you have these equations:

b + c = -4
bc = 1/5

and the solutions for b and c are the same as the solutions to the quadratic x^2 + 4x + (1/5). You can immediately see this will have two negative solutions (the discriminant is positive), and if you solve for them, or just think about what they would need to approximately equal, they are roughly -0.05 and -3.95. In this case, if we add 1/a + 1/b + 1/c, we get something close to 1/5 - 1/4 - 1/(0.05) which is close in value to -20.

Instead if you imagine a = 101, then we find b and c by solving x^2 + 100x + (1/101). The solutions here are roughly -100 and -1/10,000. Now if we add the three required reciprocals, the sum is very close to -10,000.

As I said in my post above from several years ago ( :) ), the question is really beyond the algebraic scope of the GMAT, so test takers shouldn't be concerned about it.
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
Expert Reply
IanStewart wrote:
satyadev123 wrote:
AM*HM = GM^2

If we apply this formula, the answer should be C. Since option A is 3*AM of a,b,c and option B is GM^3 of numbers a,b,c.


Unfortunately you can't use anything like that here, because any relationships you might have learned about geometric means are only true for sets of positive numbers, and here, two of the unknowns must be negative.

The answer is genuinely E to this question. If you imagine first that a=5, then you have these equations:

b + c = -4
bc = 1/5

and the solutions for b and c are the same as the solutions to the quadratic x^2 + 4x + (1/5). You can immediately see this will have two negative solutions (the discriminant is positive), and if you solve for them, or just think about what they would need to approximately equal, they are roughly -0.05 and -3.95. In this case, if we add 1/a + 1/b + 1/c, we get something close to 1/5 - 1/4 - 1/(0.05) which is close in value to -20.

Instead if you imagine a = 101, then we find b and c by solving x^2 + 100x + (1/101). The solutions here are roughly -100 and -1/10,000. Now if we add the three required reciprocals, the sum is very close to -10,000.

As I said in my post above from several years ago ( :) ), the question is really beyond the algebraic scope of the GMAT, so test takers shouldn't be concerned about it.



But, it does not say that a, b, c are integers. We can always have a = b = c = 1/3 (from statement 1)
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
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sujoykrdatta wrote:
But, it does not say that a, b, c are integers. We can always have a = b = c = 1/3 (from statement 1)


I do not understand why you have quoted me in your reply -- I obviously haven't assumed a, b and c are integers when my solutions are not integers. Using both Statements, there are obviously no integer solutions for a, b and c. Perhaps you meant to quote a different post?
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
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BeavisMan wrote:
What is the numerical value of 1/a + 1/b + 1/c ?

(1) a + b + c = 1
(2) abc = 1

See notes below

This question comes from the Kaplan GMAT Advanced 2009 - 2010 edition, question #3 on page 350. Kaplan's explanation is not very thorough - it says that there are 3 unknowns and 2 equations, thus multiple solutions are possible. It took me 20 minutes of trial and error to understand the answer. Can anyone offer a methodical explanation?

Also, does the GMAT ever provide a set of equations to which there is no solution? What would the answer be if that were the case, (E)?



We have:

a + b + c = 1 ... (i)
abc = 1 ... (ii)

Simple logic: There are 3 unknowns and 2 equations => We cannot solve for a, b and c => We cannot determine the value of (1/a + 1/b + 1/c)

Counter logic: Often, we don't really need to determine the values of a, b, c and still can determine the value of an expression. For example, had the question been to determine the value of (1/ab + 1/bc + 1/ac), we can easily get the answer by combining the statements.

Hence, we need to do some analysis:

Assuming a, b, c are positive and using AM ≥ GM for (i):

(a + b + c)/3 ≥ (a * b * c)^(1/3)

=> (a + b + c)/3 ≥ (a * b * c)^(1/3)

=> (a * b * c)^(1/3) ≤ 1/3

=> abc ≤ 1/27

Clearly, this contradicts with the second statement: abc = 1

This leads to 2 possibilities:

Two of the terms of a, b, c are negative (note - we cannot have only one negative)
OR
a, b, c are complex numbers (beyond the scope of GMAT)

Considering the first case: We can at best simplify as follows:

a = 1 - b - c

Thus, abc = 1 implies: (1 - b - c) * bc = 1

=> (b^2) * c + b * (c^2 - c) + 1 = 0

Which is a quadratic in 'b' and CAN have multiple real solutions => there are many such values of a, b, c possible.

A couple of such values I obtained:

Say a = 2 => b + c = 1 - a = -1 and bc = 1/a = 1/2
=> 1/b + 1/c = (b + c)/bc = -2
=> 1/a + 1/b + 1/c = 1/2 - 2 = -3/2

Say a = 3 => b + c = 1 - a = -2 and bc = 1/a = 1/3
=> 1/b + 1/c = (b + c)/bc = -6
=> 1/a + 1/b + 1/c = 1/3 - 6 = -17/3

Thus, the value of 1/a + 1/b + 1/c is NOT unique - Not sufficient

Answer E

Note: This is NOT going to be asked in the GMAT - don't lose your sleep over it :)
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
[quote="BeavisMan"]What is the numerical value of 1/a + 1/b + 1/c ?

(1) a + b + c = 1
(2) abc = 1

bc+ac+ab/abc

Is this an illogical way to solve the equation stated in the question?
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
Expert Reply
IanStewart wrote:
sujoykrdatta wrote:
But, it does not say that a, b, c are integers. We can always have a = b = c = 1/3 (from statement 1)


I do not understand why you have quoted me in your reply -- I obviously haven't assumed a, b and c are integers when my solutions are not integers. Using both Statements, there are obviously no integer solutions for a, b and c. Perhaps you meant to quote a different post?



Oh, yes ... I think I tagged a different post than the one I wanted to :)
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
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Re: What is the numerical value of 1/a + 1/b + 1/c ? [#permalink]
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