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hard PS multi choice

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by jsl » Thu Sep 11, 2008 9:04 am
If abc = b^3 which of the following must be true?

I ac = b^2
II b = 0
III ac = 1

A. none
B. I
C. II
D. I and III
E. II and III

My question is.... why can't I cancel out the original equation so I am left with ac = b^2 ? I thought one can always simplify if the variable are the same.

Thanks,
Jon
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Source: — Problem Solving |

by mim3 » Thu Sep 11, 2008 9:56 am
If b=0 that won't work. Example:

abc=b^3

a=2
b=0
c=3

abc=0, b^3=0
but
ac=6, b^2=0

so that eliminates I.

if:

a=2
b=4
c=8

abc=64, b^3=64

which eliminates II and III also

So, I'd go with A. OA?
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by gautamberry » Thu Sep 11, 2008 10:07 am
u cannot simplify it like that coz if u wanna cancel b from both sides u gotta divide both sides by b......but if b=0 then it is not defined.......and not defined cannot be equal to not defined.......thats y u cannot cancel out b from both sides
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by superrrom » Thu Sep 11, 2008 10:14 am
mim3 wrote:If b=0 that won't work. Example:

abc=b^3

a=2
b=0
c=3

abc=0, b^3=0
but
ac=6, b^2=0

so that eliminates I.

if:

a=2
b=4
c=8

abc=64, b^3=64

which eliminates II and III also

So, I'd go with A. OA?

it is stated that abc=b^3, so choosing b=0 is obviously irrelevant.
I think the answer is D.
1. ac=b^2
2. if ac =1, then b could be only 1.
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by mim3 » Thu Sep 11, 2008 11:04 am
I'm not with you, Superrrom.

I don't think b=0 is irrelevant because it disqualifies the first statement. The question asks which of the following MUST be true. So we're trying to find values that disprove each statement.

abc=b^3

I. ac = b^2
If a=2, b=0, c=3
acb=b^3, 0=0
ac=6, b^3=0
Not valid

II. b=0
If a=2, b=4, c=8
abc=b^3, 64=64
b=4
Not valid

III. ac=1
same example above, ac=16 and abc=b^3
not valid

So, I'd go with A. None

Let me know if I did something wrong/misunderstood something (very possible)

Can we get an OA?
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by jsl » Fri Sep 12, 2008 12:55 am
I still don't understand why you can't simplify the following equation. Am I mistaken in the belief that you can simplify all equations where there is the same variable on both sides. I understand that by setting b to 1 and 0 yields different results.

When simplifying equations, should I always assume that variables can equal 0 as well as another integer/number?

abc = b^3
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by mim3 » Fri Sep 12, 2008 8:07 am
That would work if it were 3b instead of b^3. As it is, they're different numbers. In this case because abc=b^3, you can't take away a b from one side and make it b^2.

Does that make sense?

What's the OA?
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by Sunny22uk » Sun Sep 14, 2008 1:27 pm
jsl wrote:I still don't understand why you can't simplify the following equation. Am I mistaken in the belief that you can simplify all equations where there is the same variable on both sides. I understand that by setting b to 1 and 0 yields different results.

When simplifying equations, should I always assume that variables can equal 0 as well as another integer/number?

abc = b^3
abc = b^3 is a disguised quadratic equation. :wink:
You cannot discover new oceans unless you have the courage to loose sight of the shore.
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by jsl » Mon Sep 15, 2008 1:42 am
OA is A....

I still don't understand why the equation can't cancel out to ac=b^2

Whenever you cancel and reduce an equation, is it necessary to know whether one of the variables equals zero?
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by Ian Stewart » Mon Sep 15, 2008 4:07 am
jsl wrote:OA is A....

I still don't understand why the equation can't cancel out to ac=b^2

Whenever you cancel and reduce an equation, is it necessary to know whether one of the variables equals zero?
Think of a simpler equation:

x^2 = x

Can we cancel (divide) x on both sides to get x = 1? Well, we can, as long as we know x is not equal to zero: you can never divide by zero. If you don't know that x is not zero, you can still divide by x, as long as you realize you're losing one of the solutions when we do this- x could also be zero, as is clear by plugging in x = 0. The equation has two solutions, x = 1 or x = 0. In general, if

x * (other stuff) = x * (other stuff)

then either x = 0, or (other stuff) = (other stuff).

So if b^3 = abc, we have b*(b^2) = b*(ac), and either b = 0, or b^2 = ac.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by jsl » Tue Sep 16, 2008 4:08 am
Ian Stewart wrote:Think of a simpler equation:

x^2 = x

Can we cancel (divide) x on both sides to get x = 1? Well, we can, as long as we know x is not equal to zero: you can never divide by zero. If you don't know that x is not zero, you can still divide by x, as long as you realize you're losing one of the solutions when we do this- x could also be zero, as is clear by plugging in x = 0. The equation has two solutions, x = 1 or x = 0. In general, if

x * (other stuff) = x * (other stuff)

then either x = 0, or (other stuff) = (other stuff).

So if b^3 = abc, we have b*(b^2) = b*(ac), and either b = 0, or b^2 = ac.
Hey Ian,

As always you have a way in explaining things clearly! Thanks so much! The light appeared for me when I saw that the equation is actually a quadratic. I then set this equal to zero and could very clearly see that there are multiple values for this equation.

Thanks for your input! I really appreciate it!
Jon
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by kris610 » Tue Sep 16, 2008 8:41 am
jsl wrote:OA is A....

I still don't understand why the equation can't cancel out to ac=b^2

Whenever you cancel and reduce an equation, is it necessary to know whether one of the variables equals zero?
On the GMAT, if you are given an equation like this, and if you are dividing both sides by a variable, you've to make sure that the variable is not equal to 0 -- GMAT would say that explicitly if the variable is not 0. Dividing by zero is not defined.

Your claim that choosing 0 is not relevant is not valid.

if b = 0, abc = 0 and so is b^3 satisfying the given equation -- abc = b^3.
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by lunarpower » Fri Sep 19, 2008 3:16 am
one thing i feel compelled to add:
there is really no such thing as "cancel".
in other words, there are exactly 6 operations of which you should be aware on the gmat: add, subtract, multiply, divide, raise to a power, take a root.
that is all.

if you have any other, more exotic operations, such as "cancel" or "cross-multiply", YOU NEED TO UNDERSTAND WHICH OPERATION YOU'RE REALLY DOING when you do those.
examples:
if you "cancel" the b in abc = b^3, to give ac = b^2, then you're actually dividing both sides by b.
if you "cancel" the x's in x + y = x - y to give y = -y, then you're actually subtracting x from both sides.
if you "cross multiply" a/b = c/d to give ad = bc, then you're actually multiplying both sides by (bd).

you MUST understand what operation you're really doing in every case; otherwise, you won't understand the restrictions on those operations.
for instance, if you think of the first operation as "cancelling", then you won't understand why it's not ok. on the other hand, if you understand the fact that you're dividing by b on both sides, it should be abundantly clear why you need to make sure that b is non-zero before performing this operation.

similarly, if you're given a/b > c/d, you may want to "cross multiply" to give ad > bc. however, if you understand that "cross multiplication" is really multiplication by the product of both denominators (bd), then you'll realize that you can't do this, either, unless you know whether bd is positive or negative.

basically, what it boils down to is "always know what you're doing when you're doing what you're doing".
Ron has been teaching various standardized tests for 20 years.

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by cramya » Fri Sep 19, 2008 2:32 pm
Ron,
Thanks!

Are we going with none i.e choice A) because I) and II) toegther is not listed as a choice.

abc-b^3=0;
b(ab-b^2) = 0(I can understand the dividing by b part we cant do since b cane be 0 but what about factoring it)

b=0 or ac = b^2

Is this wrong?? Please help me understand
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by earth@work » Fri Sep 19, 2008 4:28 pm
was thinking same as cramya... why cant we factorize it. pls help understand.
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