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If a+b=ab, then a*b must be equal to:
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Updated on: 04 Jul 2012, 01:35
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If a+b=ab, then a*b must be equal to: A. 1 B. 1 C. 0 D. 2 E. 2
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Originally posted by SergeNew on 03 Jul 2012, 21:27.
Last edited by Bunuel on 04 Jul 2012, 01:35, edited 1 time in total.
Edited the question.




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Re: If a+b=ab, then a*b must be equal to:
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04 Jul 2012, 01:37




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Re: Absolute number.
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03 Jul 2012, 21:40
This I solved by trial and error...with a logic that what RHS has (a  b) and LHS has addition of the same terms. So for this to be true one has to be be zero hence answer will be zero.
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Re: If a+b=ab, then a*b must be equal to:
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04 Jul 2012, 02:47
Clearly if \(a\) or \(b\) equal zero, \(ab = 0\)
so, let \(b\neq{0}\) Distance of \(a\) from \(b\), equals the distance of \(a\) from \(b\) Draw this on a number line, \(a\) must equal zero
same logic holds for \(a\neq{0}\)
So either \(a\) or \(b\) = 0, \(ab = 0\)
or just solve using our normal absolute value method, two cases:
\((a+b) = (ab)\) \(b = 0\)
\((a+b) = (ab)\) \(ab = ab\) \(a = 0\)
so \(ab = 0\)



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Re: If a+b=ab, then a*b must be equal to:
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07 Sep 2012, 13:39
Bunuel,
Can you always just square absolute values like you did on this problem? I am wondering if squaring is a valid operation on absolute values.



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Re: If a+b=ab, then a*b must be equal to:
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07 Sep 2012, 14:48
SergeNew wrote: If a+b=ab, then a*b must be equal to:
A. 1 B. 1 C. 0 D. 2 E. 2 If \(b=0\), the equality obviously holds. \(a+b=ab\) means the distance between \(a\) and \(b\) is the same as the distance between \(a\) and \(b\). For \(b\neq0,\) it means that \(a\) is the average of \(b\) and \(b\) (or the midpoint between \(b\) and \(b\)), so necessarily \(a=0.\) Altogether, the product \(ab\) must be \(0.\) Answer C
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Re: If a+b=ab, then a*b must be equal to:
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08 Sep 2012, 02:54
honggil wrote: Bunuel,
Can you always just square absolute values like you did on this problem? I am wondering if squaring is a valid operation on absolute values. Check this post: eliminationofradialsconfused138409.htmlAs for inequalities: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are nonnegative (the same for taking an even root of both sides of an inequality).For example: \(2<4\) > we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) > we can square both sides and write: \(x^2<y^2\); But if either of side is negative then raising to even power doesn't always work. For example: \(1>2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are nonnegative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: \(2<1\) > we can raise both sides to third power and write: \(2^3=8<1=1^3\) or \(5<1\) > \(5^2=125<1=1^3\); \(x<y\) > we can raise both sides to third power and write: \(x^3<y^3\). Hope it helps.
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Re: If a+b=ab, then a*b must be equal to:
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08 Sep 2012, 03:26
honggil wrote: Bunuel,
Can you always just square absolute values like you did on this problem? I am wondering if squaring is a valid operation on absolute values. Absolute value is always nonnegative, so if you have an equality between two absolute values, either they are both 0 or they are equal to the same positive number. Squared, they still remain equal.
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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 00:59
SergeNew wrote: If a+b=ab, then a*b must be equal to:
A. 1 B. 1 C. 0 D. 2 E. 2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = ab a+b = (ab) (a+b) = ab (a+b) = (ab) Simplifying each equation gets: b = b a = a a = a b = b From there, I concluded that a*b must be zero since the only way a = a is if a = 0 and the only way b = b is if b = 0. Did I do this problem incorrectly?



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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 06:56
HImba88 wrote: SergeNew wrote: If a+b=ab, then a*b must be equal to:
A. 1 B. 1 C. 0 D. 2 E. 2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = ab a+b = (ab) (a+b) = ab (a+b) = (ab) Simplifying each equation gets: b = b a = a a = a b = b From there, I concluded that a*b must be zero since the only way a = a is if a = 0 and the only way b = b is if b = 0. Did I do this problem incorrectly? It is absolutely correct. Good job!
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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 14:14
EvaJager wrote: HImba88 wrote: SergeNew wrote: If a+b=ab, then a*b must be equal to:
A. 1 B. 1 C. 0 D. 2 E. 2 I did this problem a little differently. Since both sides have absolute values, I used positives and negatives to get the following four possibilities: a+b = ab a+b = (ab) (a+b) = ab (a+b) = (ab) Simplifying each equation gets: b = b a = a a = a b = b From there, I concluded that a*b must be zero since the only way a = a is if a = 0 and the only way b = b is if b = 0. Did I do this problem incorrectly? It is absolutely correct. Good job! Thanks Eva. I didn't even think initially to square both sides. That way seems much more efficient than the way I approached the problem. Guess my brain is wired differently



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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 15:33
I found the right answer. But not sure if the procedure is right a+b = ab
a + b = a b
b = b this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to ab. so b=0 and a*b = 0. Please let me know if this is a right approach.



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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 16:03
ravipprasad wrote: I found the right answer. But not sure if the procedure is right a+b = ab
a + b = a b
b = b this is possible only with 0. So it is possible in only 2 cases. 1 and 0. But if it is 1, a+b neq to ab. so b=0 and a*b = 0. Please let me know if this is a right approach. Not sure if you can split the absolute value that way. For example: a = 5 b = 3 a+b gets you 2 while a + b gets you 8. That approach does get you the correct answer though so I may be incorrect



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Re: If a+b=ab, then a*b must be equal to:
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09 Sep 2012, 21:30
Then definitely my approach is wrong. But looking at the answer choices we can find that it should be 0.
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Re: If a+b=ab, then a*b must be equal to:
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05 Dec 2012, 02:04
Solution 1: Distance perspective ab = a+b ==> The distance of a and b is equal to the distance of a and b. <=======(b)=======0=======(b)======> Only 0 is the value that has a distance equal to b and b. Solution 2: ab = a+b (square both) a^2 2ab + b^2 = a^2 + 2ab + b^2 4ab = 0 ab = 0 Answer: 0
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Re: If a+b=ab, then a*b must be equal to:
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07 Dec 2012, 23:59
Can we plug in nos here? when we plug in random nos we realize that this can be equal only when a*b=0
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Re: If a+b=ab, then a*b must be equal to:
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04 Jul 2013, 01:44



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Re: If a+b=ab, then a*b must be equal to:
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04 Jul 2013, 03:28
SergeNew wrote: If a+b=ab, then a*b must be equal to:
A. 1 B. 1 C. 0 D. 2 E. 2 Think the equation without "a". We know that b and +b are equal in magnitude. We are adding "a" to each. That still doesn't change the equality of the magnitude. That is possible only when 0 is added or "a" is 0. We can say the reverse also and say that to a , we add b and b and the equality still holds. In this case b is 0. Either a or b is 0. So a*b must be equal to 0.
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Re: If a+b=ab, then a*b must be equal to:
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09 Jul 2013, 17:02
If a+b=ab, then a*b must be equal to:
a+b=ab (a+b)*(a+b) = (ab)*(ab) a^2+2ab+b^2 = a^22ab+b^2 4ab=0 In other words, a or b must = 0, therefore, the product of a*b is 0 regardless of what a or b are...one of them is 0.
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Re: If a+b=ab, then a*b must be equal to:
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21 Dec 2014, 15:38
I am very noob, please tell me whether this method is wrong or not
a+b = ab
Think positive values so, a+b = ab 2b=0 meaning b is 0
or
Think negative, then , a+b =  (ab) a+b= a+b 2a=o a=o
so a*b =0 either way




Re: If a+b=ab, then a*b must be equal to: &nbs
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