(1001^2 - 999^2)/(101^2 - 99^2) = [(1000+1)^2-(1000-1)^2]/[(100+1)^2-(100-1)^2] =
=> (1000^2+2000+1-1000^2+2000-1)/(100^2+200+1-100^2+200-1) =
=> 4000/400 = 10
Shortcut for long multiplication problem?
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AleksandrM
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You can also use an approximation on this problem. Since the final result is not an input for the next step in the problem, you need not obsess over precision.
You can rewrite the problem as the following:
1000(1.001^2 - .999^2)/100(1.01^2 - .99^2)
So from this you see that the only numbers worth looking at are:
1000/100 = 10
Hope this helps.
You can rewrite the problem as the following:
1000(1.001^2 - .999^2)/100(1.01^2 - .99^2)
So from this you see that the only numbers worth looking at are:
1000/100 = 10
Hope this helps.
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lunarpower
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whoa whoa already people.jplats wrote:I came across this question on a GMATPrep practice exam. Does anyone know of a shortcut beyond actually multiplying it out?
(1001^2 - 999^2)/(101^2 - 99^2) = ?
[^2 = squared and the answer is 10]
this is an absolutely classic, canonical presentation of the difference of squares pattern. it's not obscured, hidden, or even rearranged in any way; both the top and the bottom of the fraction are plain differences of squares, in all their glory.
so, ALL of you should be able to come up with the following solution. if you can't, then review the difference of squares formula, x^2 - y^2 = (x+y)(x-y), until you can.
no excuses.
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top of fraction = (1001 - 999)(1001 + 999) = (2)(2000)
bottom of fraction = (101 - 99)(101 + 99) = (2)(200)
fraction = 2000/200
= 10
Ron has been teaching various standardized tests for 20 years.
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lunarpower
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no, you used the expansion of a perfect square binomial. while that method is certainly viable, it has the following disadvantages vis-a-vis the solution i posted:xilef wrote:didn't I use the same method?
* it's not obvious
- meaning that you have to come up with the idea to express 99 as (100-1) and so on.
* it requires a great deal more work
- instead of factoring two differences of squares, both of which are already in canonical form, you have to discover, and then factor, four perfect square trinomials.
with that said, though, two comments:
(1) if you have a method that's going to work, don't hesitate; use it.
- this is super important. xilef's method is not the shortest method available - but it definitely solves the problem, and does so in under two minutes. therefore, if you come up with a method that WILL work, then just use it. NEVER just stare at a problem.
(2) drill the common factoring forms into your head until you can recognize them IMMEDIATELY.
- as stated above
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
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On peut poser des questions à Ron en français
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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Pueden hacerle preguntas a Ron en castellano
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On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
