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which one must be true?

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by pemdas » Thu Jul 21, 2011 2:02 am
a=(a+b)/(a+b), a=1
considering that a and b are different positive integers a can be either greater or less than b.
a<b

e
finance wrote:If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?

I. a = 1
II. b = 1
III. a < b

- I only
- II Only
- III only
-I&II
-I&III
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by amit2k9 » Thu Jul 21, 2011 2:03 am
by solving we have,

(a+b)(a-1) = 0
hence either a=-b or a=-1.
now a and b are both positive integers. thus a=-b not possible.

hence a=1 has to be the solution.

A it is.
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by amit2k9 » Thu Jul 21, 2011 2:08 am
pemdas wrote:a=(a+b)/(a+b), a=1
considering that a and b are different positive integers a can be either greater or less than b.
a<b

e
finance wrote:If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?

I. a = 1
II. b = 1
III. a < b

- I only
- II Only
- III only
-I&II
-I&III

substitute for a=b=1 and you get 2=2 both sides. thus a<b need not be true.
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by pemdas » Thu Jul 21, 2011 2:31 am
think twice :)
finance wrote:If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?
amit2k9 wrote:
pemdas wrote:a=(a+b)/(a+b), a=1
considering that a and b are different positive integers a can be either greater or less than b.
a<b

e
finance wrote:If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?

I. a = 1
II. b = 1
III. a < b

- I only
- II Only
- III only
-I&II
-I&III

substitute for a=b=1 and you get 2=2 both sides. thus a<b need not be true.
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by GMATGuruNY » Thu Jul 21, 2011 2:43 am
finance wrote:If a and b are different positive integers and a + b = a(a + b), then which of the following must be true?

I. a = 1
II. b = 1
III. a < b

- I only
- II Only
- III only
-I&II
-I&III
Since a and b are positive, we can divide each side by (a+b):
(a+b)/(a+b) = a(a+b)/(a+b)
1=a.

Thus, I must be true.
Eliminate answers that don't include I.
Eliminate B and C.

Since a=1 and b must be a different positive integer, b>a.
Thus, III must be true.
Eliminate answers that don't include III.
Eliminate A and D.

The correct answer is E.
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by finance » Thu Jul 21, 2011 2:43 am
Yes the OA is E. I think I should read more carefully:((
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by winniethepooh » Thu Jul 21, 2011 10:00 am
Hey Mitch, thanks for your comment on the thread.
Can you please reply to my query:
Had the question mentioned that a & b are integers(nothing about their sign, then would A be true?
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by GMATGuruNY » Thu Jul 21, 2011 11:05 am
winniethepooh wrote:Hey Mitch, thanks for your comment on the thread.
Can you please reply to my query:
Had the question mentioned that a & b are integers(nothing about their sign, then would A be true?
If a and b don't have to be positive, then none of answers has to be true.
Solving algebraically:

a+b = a(a+b)
0 = a(a+b) - (a+b)
0 = (a+b)(a-1)
Thus, a = -b or a=1.

Thus, any combination of values that satisfies a = -b will work.
For example, plugging a=2 and b=-2 into a+b = a(a+b), we get:
(2 + -2) = 2(2 + -2)
0 = 0.

Since the combination a=2 and b=-2 works, none of the three options (a=1, b=1, a<b) must be true.
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